Maths new edition

["12.8 Chapter 12 \u2009\u2009(ii)\t BC + AC > AB A (iii)\t AB + AC > BC \u2009\u2009\u2009\u2009\u20098.\tThe difference of any two sides of a triangle is less than the third side. \t In \u0394ABC (see Fig. 12.21): BC \u2009\u2009\u2009\u2009(i)\t (BC \u2013 AB) < AC \u2009\u2009(ii)\t (AC \u2013 BC) < AB Figure 12.21 (iii)\t (AC \u2013 AB) < BC \u2009\u2009\u2009\u2009\u20099.\t In triangle ABC (see Fig. 12.21), if \u2220B > \u2220C, then the side opposite to \u2220B is longer than the side opposite to \u2220C, i.e., AC > AB. 10.\tIn triangle ABC (see Fig. 12.21), if AC > BC, then the angle opposite to side AC is greater than the angle opposite to side BC, i.e., \u2220B > \u2220A. Congruence of Triangles Two geometrical figures are congruent if they have the same shape and the same size. The three angles of a triangle determine its shape and its three sides determine its size. If the three angles and the three sides of a triangle are respectively equal to the corresponding angles and sides of another triangle, then the two triangles are congruent. However, it is not necessary that each of the six elements of one triangle have to be equal to the corresponding elements of the other triangle in order to conclude that the two triangles are congruent. Based on the study and experiments, the following results can be applied to establish the congruence of two triangles. Side-Side-Side (SSS) A P Congruence Property By the SSS congruence property, two B CQ R triangles are congruent if the three Figure 12.22 sides of one triangle are respectively equal to the corresponding three sides of the other triangle. ABC and PQR are two triangles (see Fig. 12.22) such that AB = PQ, BC = QR and CA = RP, then \u0394ABC is congruent to \u0394PQR. We write this as \u2206ABC \u2245 \u2206PQR. Side-Angle-Side (SAS) Congruence Property By the SAS congruence property, two triangles are congruent if the two sides and the included angle of one triangle are respectively equal to the corresponding two sides A P and the included angle of the other triangle. If ABC and PQR are two triangles B CQ R Figure 12.23 (see Fig. 12.23) such that AB = PQ, BC = QR\u2220ABC = \u2220PQR, then \u2206ABC \u2245 \u2206PQR.","Geometry 12.9 Angle-Side-Angle (ASA) Congruence Property By the ASA congruence property, two A P triangles are congruent if any two angles and a side of one triangle are respectively B CQ R equal to the corresponding parts of the Figure 12.24 R other triangle. In Fig. 12.24, \u2220BAC = \u2220QPR and \u2220ABC = \u2220PQR and AB = PQ. It follows that \u2206ABC \u2245 \u2206PQR. Right Angle-Hypotenuse-Side Congruency Property The RHS congruence property states that two A P right triangles are congruent, if the hypotenuse and one side of a triangle are respectively equal to the hypotenuse and the corresponding side of the other right triangle. In DABC and DPQR (see Fig. 12.25), AB B CQ = PQ, AC = PR and \u2220ABC = \u2220PQR = (90\u00b0), Figure 12.25 then \u2206ABC \u2245 \u2206PQR. Quadrilaterals S R A closed figure bounded by four line segments is called a quadrilateral. A quadrilateral PQRS has the following elements (see Fig. 12.26): 1.\t Four vertices P, Q, R and S. 2.\t Four sides PQ, QR, RS and SP. P Q 3.\t Four angles \u2220P, \u2220Q, \u2220R and \u2220S. Figure 12.26 B 4.\t Two diagonals PR and QS. 5.\t Four pairs of adjacent sides (PQ, QR), (QR, RS), (RS, SP ) and (SP, PQ). 6.\t Two pairs of opposite sides (PQ, RS) and (QR, PS). 7.\t Four pairs of adjacent angles (\u2220P, \u2220Q) (\u2220Q, \u2220R), (\u2220R, \u2220S) and (\u2220S, \u2220P). 8.\t Two pairs of opposite angles (\u2220P, \u2220R) and (\u2220Q, \u2220S). 9.\t The sum of the four angles of a quadrilateral is 360\u00b0. Different Types of Quadrilaterals DC Figure 12.27 Trapezium In a quadrilateral, if two opposite sides are parallel to each other, then it is called a trapezium. In Fig. 12.27, AB CD, hence ABCD is a trapezium. Parallelogram A In a quadrilateral if both the pairs of opposite sides are parallel, then it is called a parallelogram.","12.10 Chapter 12 In Fig. 12.28: D C 1.\t AB = CD and BC = AD. A B 2.\t AB CD and BC AD. Figure 12.28 Hence, ABCD is a parallelogram. \u2002Note\u2002\u2002In a parallelogram, diagonals need not be equal, but they bisect each other. Rectangle D C A B In a parallelogram, if each angle is a right angle (90\u00b0), then it is called a rectangle. In Fig. 12.29, \u2220A = \u2220B = \u2220C = \u2220D = 90\u00b0, AB = CD and BC = AD. Hence, ABCD is a rectangle. \u2002Note\u2002\u2002In a rectangle, the diagonals are equal, i.e., AC Figure 12.29 = BD. Rhombus C In a parallelogram, if all the sides are equal, then it is called a rhombus. In Fig. 12.30, AB = BC = CD = AD, hence ABCD is a rhombus. D O B \u2002Notes\u2002 1.\t In a rhombus, the diagonals need not be equal. A Figure 12.30 2.\t I\u0007n a rhombus, the diagonals bisect each other at right angles, i.e., AO = OC, BO = OD and AC \u22a5 DB. Square DC In a rhombus, if each angle is a right angle, then it is called a O square, or otherwise in a rectangle, if all the sides are equal, then it is called a square. A B In Fig. 12.31, AB = BC = CD = DA and \u2220A = \u2220B = \u2220C = Figure 12.31 \u2220D = 90\u00b0. Hence, ABCD is a square. \u2002Notes\u2002 1.\t In a square, the diagonals bisect each other at right angles. 2.\t In a square, the diagonals are equal. DC Isosceles Trapezium A B In a trapezium, if the non-parallel opposite sides are equal, Figure 12.32 then it is called an isosceles trapezium. In Fig. 12.32, AB CD and BC = AD. Hence, ABCD is an isosceles trapezium. \u21d2 \u2220A = \u2220B and \u2220C = \u2220D.","Geometry 12.11 Kite In a quadrilateral, if two pairs of adjacent sides are equal, then it is called a kite. In Fig. 12.33, AB = AD and BC = CD. Hence, ABCD is a kite. C DB A Figure 12.33 Geometrical Results on Areas 1.\t Parallelograms on the same base and between the same parallels are equal in area. DF CE AB Figure 12.34 \t \u0007\t In Fig. 12.34, parallelogram ABCD and parallelogram ABEF are on the same base AB and between the same parallels AB and CD. \t \u2234 Area of parallelogram ABCD = Area of parallelogram ABEF. \u2002Note\u2002\u2002 A parallelogram and a rectangle on the same base and between the same parallels are equal in area. 2.\t The area of a triangle is half the area of the parallelogram, if they lie on the same base and between the same parallels. DE C AB Figure 12.35 \t \u0007\t In Fig. 12.35, parallelogram ABCD and \u0394ABE are on the same base AB and between the same parallels AB and CD. \t \u2234 Area of \u2206ABE = 1 \u2009Area of parallelogram ABCD. 2","12.12 Chapter 12 3.\t Triangles on the same base and between the same parallels are equal in area. DC AB Figure 12.36 \t \u0007\t In Fig. 12.36, \u0394ABC and \u0394ABD are on the same base AB and between the same parallels AB and CD. \t \u2234 Area of \u0394ABC = Area of \u0394ABD. \u2002Note\u2002\u2002 Triangles with equal bases and between the same parallels are equal in area. CF A BD E Figure 12.37 \t \u0007\t In Fig. 12.37, in \u0394ABC and \u0394DEF, CD \t AB = DE and AE CF. \t \u2234 Area of \u0394ABC = Area of \u0394DEF. AE B F 4.\t Triangles with equal bases and with equal Figure 12.38 areas lie between the same parallels. \t \u0007\t In Fig. 12.38, if Area of \u0394ABC = Area of \u0394ABD, then AB CD. \u2002Note\u2002\u2002 In this case, altitudes CE and DF are equal. 5.\t A diagonal of a parallelogram divides the parallelogram into two triangles of equal area. \t \u0007\t In Fig. 12.39, diagonal AC divides parallelogram ABCD into two triangles, \u0394ABC and \u0394ACD. \t \t Here, area of \u0394ABC = area of \u0394ACD. \t \t Similarly, diagonal BD divides the parallelogram into two triangles, \u0394ABD and \u0394BDC. \t \t Hence, the area of \u0394ABD = area of \u0394BCD. DC AB Figure 12.39","Geometry 12.13 Some Theorems on Triangles Mid-point Theorem P In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side and also half of it. Given: In \u0394PQR, A and B are the mid-points of PQ and PR. A B C respectively. RTP: AB QR and AB = 1 QR 2 Q R Figure 12.40 Construction: Draw\u2009RC \u2009parallel to\u2009QA \u2009to meet produced AB at C (see Fig. 12.40). Proof: 1.\t In \u0394ABP and \u0394CBR, \t \u2220PBA = \u2220RBC (vertically opposite angles) \t \u2220PAB = \u2220RCB (alternate angles and CR PQ) \t PB = BR (B is mid-point of PR). \t By AAS congruence property, \t \u2206ABP \u2245 \u2206CBR \t \u2234 PA = CR and AB = BC (corresponding parts of congruent triangles) \u21d2 AQ = CR.\u2003\u2003(\u2235PA = AQ) \t In quadrilateral ACRQ, AQ = CR and AQ CR. \t \u2234 ACRQ is a parallelogram. \u2234 AC QR \u21d2 AB QR 2.\t AC = QR\t(opposite sides of parallelogram) \u21d2 QR = AB + BC \u21d2 QR = 2AB ( AB = BC) \u21d2 AB = 1 QR. 2 Basic Proportionality Theorem In a triangle, if a line is drawn parallel to one side of the A F triangle, then it divides the other two sides in the same E ratio. P D Given: In \u0394ABC, DE is drawn parallel to BC. RTP: AD = AE DB EC Construction: Draw EP \u22a5 AB and DF \u22a5 AC. Join B C DC and BE (see Fig. 12.41). Figure 12.41","12.14 Chapter 12 Proof: Area of triangle ADE 1 \u00d7 AD \u00d7 PE AD \t Area of triangle BDE 2 BD = 1 = 2 \u00d7 BD \u00d7 PE Area of triangle ADE 1 \u00d7 AE \u00d7 DF AE Area of triangle CDE 2 EC and = 1 = . 2 \u00d7 EC \u00d7 DF But, area of triangles BDE and CDE are equal. ( Two triangles lying on the same base and between the same parallel lines are equal in areas). \u2234 AD = AE . DB EC Similarly, it can be prov=ed that AADB AA=CE and BADB AC . CE Converse of Basic Proportionality Theorem A DE If a line divides two sides of a triangle in the same ratio, then that line is parallel to the third side. In Fig. 12.42, AD = AE \u21d2 DE BC. DB EC \u2002Note\u2002\u2002 The intercepts made by three or more parallel lines on B C any two transversals are proportional. Figure 12.42 ts A P l1 B Q C l2 R l3 Figure 12.43 In Fig. 12.43, the parallel lines l1, l2 and l3 make intercepts AB and BC on transversal t and intercepts PQ and QR on transversal s. Then, AB = PQ . BC QR","Geometry 12.15 Example 12.1 Divide line segment AB = 10 cm into six equal parts. Solution Step 1:\u2002 Draw AB = 10 cm. L Step 2:\u2002 Draw a ray AL such that AL does not coincide with AB. H G Step 3:\u2002 Mark six equal line segments AC, F CD, DE, EF, FG and GH on ray AL E with convenient length, using a compass. D C Step 4:\u2002 Join HB. PQ RS TB Step 5:\u2002 Draw lines parallel to HB through A the points C, D, E, F and G intersecting the line segment AB at P, Q, R, S and T. Figure 12.44 respectively. The line segments AP, PQ, QR, RS, ST and TB are the required six equal parts of the line segment AB (see Fig. 12.44). Similarity Two figures are said to be congruent, if they have the same shape and same size. But the figures of the same shape need not have the same size. The figures of the same shape, but not necessarily of the same size are called similar figures. Examples: AB C D 1.\t Any two line segments are similar (see Fig. 12.45). Figure 12.45 2.\t Any two squares are similar. Figure 12.46 3.\t Any two equilateral triangles are similar. 60\u00b0 60\u00b0 60\u00b0 60\u00b0 60\u00b0 60\u00b0 4.\t Any two circles are similar. Figure 12.47 \u2022\u2022 C1 C2 Figure 12.48","12.16 Chapter 12 Two polygons are said to be similar to each other, if: 1.\t Their corresponding angles are equal, and 2.\t The lengths of their corresponding sides are proportional. \u2002Note\u2002\u2002 \u2018~\u2019 is the symbol used for \u2018is similar to\u2019. If \u0394ABC is similar to \u0394PQR, we denote it as \u0394ABC ~ \u0394PQR. The relation \u2018is similar to\u2019 satisfies the following properties: 1.\t It is reflexive as every figure is similar to itself. 2.\t It is symmetric as, if A is similar to B, then B is also similar to A. 3.\t It is transitive as, if A is similar to B and B is similar to C, then A is similar to C. \t\u2234 The relation \u2018is similar to\u2019 is an equivalence relation. Criteria for Similarity of Triangles In two triangles, if either corresponding angles are equal or the ratio of corresponding sides are proportional, then the two triangles are similar to each other. D A B CE F Figure 12.49 In \u0394ABC and \u0394DEF (see Fig. 12.49), 1.\t If \u2220A = \u2220D, \u2220B = \u2220E and \u2220C = \u2220F, then \u0394ABC ~ \u0394DEF. This property is called AAA criterion. 2.\t If DA=EB BE=CF AC , then \u0394ABC ~ \u0394DEF. This property is called SSS criterion. DF 3.\t If AB = AC and \u2220A = \u2220D, then \u0394ABC ~ \u0394DEF. This property is called SAS criterion. DE DF Results on Areas of Similar Triangles 1.\t The ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides of the triangles. In Fig. 12.50, \u0394ABC ~ \u0394DEF. AD 70\u00b0 70\u00b0 50\u00b0 60\u00b0 50\u00b0 60\u00b0 B C E F Figure 12.50","Geometry 12.17 \u2206ABC \u223c \u2206DEF \u21d2 Area of \u2206ABC = AB 2 = BC 2 = CA2 . Area of \u2206DEF DE 2 EF 2 FD2 2.\t The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding altitudes. \t \t In Fig. 12.51, \u0394ABC ~ \u0394DEF and AX, DY are the altitudes. \t Then, Area of \u2206ABC = AX 2 . Area of \u2206DEF DY 2 D A BX CE Y F Figure 12.51 3.\t The ratio of the areas of two similar triangles is equal to the ratio of the squares on their corresponding medians. \t \t In Fig. 12.52, \u0394ABC ~ \u0394PQR and AD and PS are medians. \t Then, Area of \u2206ABC = AD2 . Area of \u2206PQR PS2 P A B D CQ S R Figure 12.52 4.\t The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding angle bisector segments. \t \u0007\t In Fig. 12.53, \u0394ABC ~ \u0394DEF and AP, DQ are bisectors of \u2220A and \u2220D respectively, then D \t Area of \u2206ABC = AP 2 . A Area of \u2206DEF DQ2 B P CE Q F Figure 12.53","12.18 Chapter 12 Pythagorean Theorem In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. A Given: In \u0394ABC, \u2220B = 90\u00b0. To prove: AC2 = AB2 + BC2 Construction: Draw BP perpendicular to AC (see Fig. 12.54). P Proof: In \u0394APB and \u0394ABC, \u2220APB = \u2220ABC (right angles) BC \u2220A = \u2220A (common) Figure 12.54 \u2234Triangle APB is similar to triangle ABC. \u21d2 AP = AB AB AC \u21d2 AB2 = (AP)(AC). Similarly, BC2 = (PC)(AC) \u2234 AB2 + BC2 = (AP)(AC) + (PC)(AC) AB2 + BC2 = (AC)(AP + PC) AB2 + BC2 = (AC)(AC) \u21d2 AC2 = AB2 + BC2. Hence, proved. Converse of Pythagorean Theorem X In a triangle, if the square of one side is equal to the sum of the squares of YZ the other two sides, then the angle opposite to the first side is a right angle. Figure 12.55 Given: In \u0394XYZ, XZ\u20092 = XY\u20092 + YZ\u20092 (see Fig. 12.55). To prove: \u2220Y = 90\u00b0 Construction: Construct \u0394MNO, such that XY = MN, YZ = NO and \u2220N = 90\u00b0 (see Fig. 12.56). Proof: In \u0394MNO, \u2220N = 90\u00b0. \u2234OM2 = MN2 + NO2 (Pythagorean theorem) \t \u21d2 OM2 = XY2 + YZ2 (construction) \b (1) \tBut, XZ2 = XY2 + YZ\u20092 (Given) \b (2) From Eqs. (1) and (2), we get: OM\u20092 = XZ\u20092 \u21d2 OM = XZ By SSS congruence criterion, we get: \u0394XYZ\u22c5 \u2245 \u0394MNO \u2234 \u2220Y = \u2220N = 90\u00b0. Hence, \u2220Y = 90\u00b0.","Geometry 12.19 Example 12.2 In the given figure (not to scale), ABC is an A E isosceles triangle in which AB = AC. AEDC is a F parallelogram. If \u2220CDF = 70\u00b0 and \u2220BFE = 100\u00b0, G D then find \u2220FBA. C (a) 30\u00b0\t (b) 40\u00b0\t (c) 50\u00b0\t (d) 80\u00b0 B Hints \u2009\u2009\u2009\u2009(i) I\u0007n a parallelogram, any pair of adjacent angles are supplementary. \u2009\u2009(ii) \u2220EFG = \u2220FGC = \u2220AGB and \u2220ACB ( )\t\t\u2003\u2003 = \u2220ABC = \u2220CDE. \u2235 AC ||DE (iii) Use the above data and find \u2220BAC, and then \u2220ABG. Example 12.3 E In the given figure, ABCD is a cyclic quadrilateral, \u2220DAB = 50\u00b0 and \u2220ABC = 80\u00b0. EG and FG are the angle bisectors of \u2220DEC D and \u2220BFC. Find \u2220FHG. G (a) 80\u00b0\t (b) 90\u00b0\t (c) 75\u00b0\t (d) 105\u00b0 Solution H 90\u00b0 \u2220DAB = 50\u00b0 \u21d2 \u2220DCE = 50\u00b0 (exterior angle of a cyclic quadrilateral) C \u2220ABC = 80\u00b0 \u21d2 \u2220EDC = 80\u00b0 (exterior angle of a cyclic quadrilateral) A B \\\\ In DDEC, \u2220DEC = 180\u00b0 \u2013 (50\u00b0 + 80\u00b0) = 50\u00b0. F EG is the bisector of \u2220DEC \u21d2 \u2220DEH = 50\u00b0 = 25\u00b0. 2 \\\\ \u2220DHE = 180\u00b0 \u2013 (\u2220HDE + \u2220DEH) = 180\u00b0 \u2013 (80\u00b0 + 25\u00b0) = 75\u00b0. \u2220FHG = \u2220DHE (vertically opposite angles) \u21d2 \u2220FHG = 75\u00b0. Polygons A closed plane figure bounded by three or more line segments is called a polygon. 1.\t Each line segment is called a side of the polygon. 2.\t The point at which any two adjacent sides intersect is called a vertex of the polygon. Different polygons, the number of their sides and their names are given in the following table. Number 34 5 6 78 9 10 of Sides Decagon Triangle Quadrilateral Pentagon Hexagon Septagon Octagon Nonagon Name of the polygon Corresponding figure","12.20 Chapter 12 Polygons can also be classified as 1.\t Convex polygons and 2.\t Concave polygons. Convex Polygon and Concave Polygon A polygon in which each interior angle is less than 180\u00b0 is called a convex polygon (Fig. 12.57). Otherwise it is called concave polygon (Fig. 12.58). Figure 12.57 \u2003\u2003\u2003\u2003\u2003 Figure 12.58 \u2002Note\u2002Unless otherwise mentioned, we refer to convex polygons simply as polygons. Some Important Results on Polygons 1.\t D\u0007 iagonals of a polygon: A line segment joining any two non-consecutive vertices of a polygon is called its diagonal. \u2002Note\u2002The number of diagonals of a polygon with n sides is given by n(n \u2212 3) . 2 Example 12.4 Find the number of diagonals of a 10-sided polygon. Solution 10(10 \u2212 3) Here, n = 10 2 \u2234 Number of diagonals = = 35. 2.\t The sum of the interior angles of an n-sided polygon is (2n \u2013 4) 90\u00b0 or (2n \u2013 4) right angles. Example 12.5 Find the sum of the interior angles of a polygon of 8 sides. Solution Here, n = 8 \u2234 Sum of interior angles = [(2)(8) \u2013 4]90\u00b0 = 1080\u00b0. 3.\t The sum of all the exterior angles of a polygon is 360\u00b0. 4.\t In a polygon, if all the sides are equal and all the angles are equal, then it is called a regular polygon. \t In case, all the sides are not equal, then the polygon is called an irregular polygon. 5.\t Each interior angle of regular polygon of n sides is \uf8eb (2n \u2212 4)90 \uf8f6 \u00b0. \uf8ed\uf8ec n \uf8f7\uf8f8","Geometry 12.21 6.\t Each exterior angle of a regular polygon of n sides is \uf8eb 360 \uf8f6 \u00b0. \uf8ec\uf8ed n \uf8f7\uf8f8 \u2002Note\u2002\u2002 In case of a regular polygon, all the interior angles are equal and all the exterior angles are also equal. 360 \uf8f6 x \uf8f7\uf8f8 7.\t The number of sides n of a regular polygon whose exterior angle x\u00b0 is = \uf8eb \u00b0. \uf8ed\uf8ec Construction of Triangles In the earlier classes, we have learnt construction of triangles, when 1.\t all the three sides are given. 2.\t two sides and an angle are given. 3.\t two angles and an included side are given. 4.\t a side and the hypotenuse of a right triangle are given. In the present section, we will discuss construction of triangles in some more cases. 1.\t To construct a triangle when the base and the sum of the other two sides and one base angle are given. Example 12.6 Construct a triangle ABC, in which base BC = 6 cm, AB + AC = 9 cm and \u2220ABC = 45\u00b0. Solution X D Step 1:\u2002 Draw BC = 6 cm. A Q Step 2:\u2002 Draw BX, such that \u2220CBX = 45\u00b0. P Step 3:\u2002 With B as the centre and the radius as AB + AC = 9 cm, draw an arc to meet BX at D. Join CD. BC Figure 12.59 Step 4:\u2002 Draw the perpendicular bisector of CD to intersect BD at A. Join CA. \u0394ABC is the required triangle. Proof: As AQ is the perpendicular bisector of CD, ACD is an isosceles triangle, hence AD = AC and AB + AC = AB + AD = BD = 9 cm. 2.\t Construction of a triangle when its perimeter and the base angles are given. Example 12.7 Construct \u0394ABC, whose perimeter is 10 cm and base angles are 60\u00b0 and 44\u00b0. Solution L A MF D E C Q B G P Figure 12.60","12.22 Chapter 12 Step 1:\u2002 Draw PQ = AB + BC + CA = 10 cm. Step 2:\u2002 Draw \u2220LPQ equal to \u2220B and \u2220MQP equal to \u2220C. Step 3:\u2002 Draw the angle bisectors of \u2220LPQ and \u2220MQP. Let these bisectors meet at A. Step 4:\u2002 Draw the perpendicular bisectors of AP and AQ to meet PQ at B and C respectively. Step 5:\u2002 Join AB and AC to form the required \u0394ABC. 3.\t To construct a triangle when its base, one base angle and the difference of the remaining sides are given. Example 12.8 X Construct a triangle PQR, in which QR = 4.5 cm, P \u2220Q\u00a0=\u00a044\u00b0 and PQ \u2013 PR = 2 cm. S Solution Step 1:\u2002 Draw a line segment QR = 4.5 cm. 44\u00b0 R Step 2:\u2002 Draw QX, such that \u2220RQX = 44\u00b0. Q Step 3:\u2002 With Q as the centre and radius = 2 cm, draw an arc to intersect QX at S. Figure 12.61 Step 4:\u2002 Join SR and draw the perpendicular bisector of SR to intersect QX at P. Step 5:\u2002 Join P and R. PQR is the required triangle. 4.\t Construction of a triangle, when the base, the angle at the vertex and the sum of the other two sides are given. Example 12.9 Construct \u0394PQR in which QR = 2.1 cm, \u2220P = 46\u00b0 and PQ + PR = 5.1 cm. Solution Step 1:\u2002 Construct \u2220QS = PQ + PR = 5.1 cm. X Step 2:\u2002 Construct \u2220QSR = 1 (46\u00b0) = 23\u00b0. R\u2032 R 2 QP Step 3:\u2002 With Q as the centre and the radius 2.1 Figure 12.62 S cm (QR = 2.1 cm), draw an arc to intersect SX at R and R\u2032. Step 4:\u2002Join QR (or QR\u2032) and draw the perpendicular bisector of SR (or SR\u2032) to intersect QS at P (or P\u2032). PQR (or P\u2032QR\u2032) is the required triangle.","Geometry 12.23 5.\t To construct a triangle when two sides and the median drawn on the third side are given. Example 12.10 Construct a triangle PQR, such that PQ = 3.8 cm, QR = 4.3 cm and the median from Q to PR is 3.5 cm. Solution Step 1:\u2002 Draw a line segment QR of length 4.3 cm. X Step 2:\u2002 Draw the perpendicular bisector of QR and mark the P mid-point of QR as S. Step 3:\u2002 Taking Q as the centre and radius equal to 3.5 cm, T draw an arc. Step 4:\u2002 Taking S as the centre and radius = 1.9 cm \uf8eb 1 PQ\uf8f7\uf8f6\uf8f8 , 3.5 cm 1.9 cm \uf8ed\uf8ec 2 SR draw another arc intersecting the already drawn arc (in Step\u00a03) Q at T. Step 5:\u2002 Join R and T and produce RT to X. Figure 12.63 Step 6:\u2002 With Q as the centre and radius = 3.8 cm, draw an arc to intersect RX at P. Join PQ. \u0394PQR is the required triangle. 6.\t Construction of a triangle when the base, the angle between the other two sides and the difference of the other two sides are given. Example 12.11 Construct \u0394PQR in which QR = 3.4 cm, \u2220P = 50\u00b0 and PQ \u2013 PR = 1.2 cm. Solution Y Step 1:\u2002 Draw the ray, QX. R Step 2:\u2002 Mark the point S on QX, such that QS = 4 cm 1.2 cm. 115\u00b0 Step 3:\u2002 Construct an angle equal to QS 180 \u2212 1 (180 \u2212 50) = 115\u00b0 Figure 12.64 2 at S, such that \u2220QSY = 115\u00b0. P X Step 4:\u2002 With Q as the centre and radius = 3.4 cm, draw an arc to meet SY at R. Step 5:\u2002 Join Q and R. Step 6:\u2002 Draw the perpendicular bisector of SR intersecting QX at P. Step 7:\u2002 Join PR. PQR is the required triangle. 7.\t Construction of a triangle when the base, sum of the remaining sides and the difference of the base angles are given.","12.24 Chapter 12 Example 12.12 Construct \u0394PQR in which QR = 3.2 cm, PQ + PR = 5.9 cm and \u2220R \u2013 \u2220Q = 60\u00b0. Solution Step 1:\u2002 Draw line segment QR = 3.2 cm. X Step 2:\u2002 Draw RX, such that S \u2220QRX = \uf8eb 90\u00b0 + \u2220R \u2212 \u2220Q \uf8f6 = 90\u00b0 + 30\u00b0 = 120\u00b0. \uf8ed\uf8ec 2 \uf8f7\uf8f8 P Step 3:\u2002 Taking Q as the centre and a radius equal to Y T 5.9 cm (PQ + PR), draw an arc intersecting RX at S. Q R Step 4:\u2002 Join Q and S. Figure 12.65 Step 5:\u2002 Draw RY, such that \u2220QRY = \uf8eb \u2220R \u2212 \u2220Q \uf8f6 = 30\u00b0, intersecting QS at T. \uf8ed\uf8ec 2 \uf8f7\uf8f8 Step 6:\u2002Draw the perpendicular bisector of RT intersecting QS at P. Step 7:\u2002 Join P and R. PQR is the required triangle. 8.\t Construction of a triangle when the base, difference of the remaining two sides and the difference of the base angles are given. Example 12.13 Construct a triangle PQR in which QR = 4.7 cm, PR \u2013 PQ = 2.3 cm and \u2220Q \u2013 \u2220R = 50\u00b0. Solution P Step 1:\u2002 Draw line segment QR = 4.7 cm. Step 2:\u2002 Draw\u2009QX, \u2009such that \u2220RQX = 1 (\u2220Q \u2212 \u2220R ) = 25\u00b0. X 2 \u2022 x Step 3:\u2002 With R as the centre and the radius = 2.3 cm, draw x A an arc intersecting QX at A. Q Step 4:\u2002 Join RA and produce RA. Step 5:\u2002 Draw the perpendicular bisector of QA to intersect R RA produced at P. Figure 12.66 Step 6:\u2002 Join PQ. PQR is the required triangle. 9.\t Construction of a triangle when two angles and the sum of the sides are given. Example 12.14 Construct \u0394PQR in which \u2220P = 96\u00b0, \u2220Q = 40\u00b0 and PQ + PR = 6 cm. Solution Step 1:\u2002 Draw a line segment RS = 6 cm. Step 2:\u2002 Draw RX, such that \u2220SRX = 44\u00b0.","Geometry 12.25 Step 3:\u2002 Draw SY , such that \u2220RSY = 1 \u2220P = 48\u00b0, intersecting RX at Q. Step 4:\u2002 Draw the 2 of QS to intersect RS at P. perpendicular bisector Step 5:\u2002 Join P and Q. PQR is the required triangle. Y X Q 40\u00b0 48\u00b0 44\u00b0 96\u00b0 48\u00b0 RP S Figure 12.67 10.\tConstruction of a triangle with two angles and the difference between two sides are given. Example 12.15 Construct a triangle PQR, such that \u2220P = 50\u00b0, \u2220Q = 30\u00b0and PQ \u2013 QR = 1.4 cm. Solution Step 1:\u2002 Draw PX. Step 2:\u2002 Locate the point S on PX, such that Y PS = 1.4 cm. Z Step 3:\u2002 Draw SY , such that \u2220PSY = 180 \u2212 1 2 R (180\u00b0 \u2212 \u2220Q ) = 180\u00b0 \u2212 1 (180\u00b0 \u2212 30\u00b0) = 105\u00b0. S 2 Figure 12.68 Step 4:\u2002 Draw PZ, such that \u2220SPZ = 50\u00b0, QX intersecting SY at R. Step 5:\u2002 Draw perpendicular bisector of RS P intersecting PX at Q. Step 6:\u2002 Join Q and R. PQR is the required triangle. 11.\tConstruct a triangle equal to the area of a given P R T convex quadrilateral. S \t Let PQRS be the given quadrilateral. Q Step 1:\u2002 Join QS. Figure 12.69 Step 2:\u2002 Through R, draw a line parallel to QS. Step 3:\u2002 Produce PQ to meet the parallel line drawn through R at T. Step 4:\u2002 Join ST. PST is the required triangle.","12.26 Chapter 12 \t Justification \t Ar(\u0394QSR) = Ar(\u0394QST) \u2192 Eq. 1. (\u2234 They lie on the same base SQ , and lie between the same parallels). \t Area of quadrilateral PQRS = Ar(\u0394PQS) + Ar(\u0394QSR). \t = Ar(\u0394PQS) + Ar(\u0394QST) (From Eq. 1) \t \u2234 Area of quadrilateral PQRS = Area of \u0394PST. 12.\tConstruct a triangle equal to the area of a D given pentagon. EC Step 1:\u2002 Construct a pentagon ABCDE. Step 2:\u2002 Join BD and AD. GA BF Step 3:\u2002 Draw lines parallel to BD and AD through C and E respectively to meet AB Figure 12.70 produced and BA produced at F and G. respectively. Step 4:\u2002 Join DF and DG to form \u0394DGF which is equal to the area of the pentagon ABCDE. Justification Area of \u0394BDC = Area of \u0394BDF Area of \u0394ADE = Area of \u0394ADG ( They lie on the same bases, BD and AD, and between the same parallels.) \tArea of pentagon ABCDE = Sum of the areas of \u0394AED, \u0394ABD, \u0394DBC \t = Sum of the areas of \u0394ADG, \u0394ABD and \u0394BDF \t = Area of \u0394GDF \u2234 The area of the pentagon ABCDE is equal to the area of \u0394GDF. Construction of Quadrilaterals 1.\t When four sides and one angle are given. Example 12.16 Construct a quadrilateral ABCD in which AB = 4.2 cm, \u2220A = 80\u00b0, BC = 2.4 cm, CD = 3.3 cm and AD = 2.4 cm. Solution DX 3.3 cm C 2.4 cm Step 1:\u2002 Draw a line segment AB = 4.2 cm. 2.4 cm Step 2:\u2002 Draw \u2220BAX = 80\u00b0. B Step 3:\u2002 Mark D on AX, such that AD = 2.4 cm. 80\u00b0 4.2 cm Step 4:\u2002 Taking D as the centre and 3.3 cm as the A radius, draw an arc, and taking B as the centre and 2.4 cm as radius, draw another arc to intersect the Figure 12.71 previous arc at C. Step 5:\u2002 Join CD and BC. ABCD is the required quadrilateral.","Geometry 12.27 2.\t When three consecutive sides and two included angles are given. Example 12.17 Construct a quadrilateral ABCD with AB = 4 cm, BC = 2.8 cm, CD = 4 cm, \u2220B = 75\u00b0 and \u2220C = 105\u00b0. Solution D 4 cm X Step 1:\u2002 Draw a line segment AB = 4 cm. Y C Step 2:\u2002 Draw BX , such that \u2220ABX = 75\u00b0. 105\u00b0 Step 3:\u2002 With B as the centre and a radius of 2.8 cm, draw an arc to cut BX at C. 2.8 cm Step 4:\u2002 Draw CY which makes an angle A 4 cm 75\u00b0 105\u00b0 with BC. Figure 12.72 B Step 5:\u2002 Mark D on CY , such that CD = 4 cm. Step 6:\u2002Join AD. ABCD is the required quadrilateral. 3.\t When four sides and one diagonal are given. Example 12.18 Construct a quadrilateral ABCD in which AB = 4.6 cm, BC = 2.6 cm, CD = 3.5 cm, AD = 2.6 cm and the diagonal AC = 4.9 cm. Solution D 3.5 cm C 2.6 cm 4.9 cm 2.6 cm Step 1:\u2002 Draw a line segment AB = 4.6 cm. B Step 2:\u2002 With A and B as the centres, draw two A 4.6 cm arcs of radii 4.9 cm and 2.6 cm respectively to Figure 12.73 intersect each other at C. Step 3:\u2002 With C and A as the centres, draw two arcs of radii 3.5 cm and 2.6 cm respectively to intersect at D. Step 4:\u2002Join BC, CD and AD to form quadrilateral ABCD. ABCD is the required quadrilateral. 4.\t To construct a parallelogram, when two consecutive sides and the included angle are given. Example 12.19 Construct a parallelogram ABCD, when AB = 4 cm, BC = 2.5 cm and \u2220B = 100\u00b0. Solution Step 1:\u2002 Draw line segment AB = 4 cm.","12.28 Chapter 12 Step 2:\u2002 Construct line BX, such that \u2220ABX = 100\u00b0. X Step 3:\u2002 Taking B as the centre and the radius = 2.5 cm, cut BX \u2009at the point C with an arc. D C Step 4:\u2002 Draw two arcs taking C and A as centres and A 4 cm and 2.5 cm as radii respectively to intersect at D. 2.5 cm Step 5:\u2002Join AD and CD. ABCD is the required 100\u00b0 parallelogram. B 4 cm Figure 12.74 5.\t Construction of a parallelogram when two adjacent sides and one diagonal are given. Example 12.20 Construct a parallelogram PQRS, when PQ = 3.7 S 3.7 cm R cm, QR = 2.3 cm and PR = 4.8 cm. 2.3 cm 4.8 cm 2.3 cm Step 1:\u2002 Draw a line segment PQ = 3.7 cm. Step 2:\u2002 Draw an arc with P as the centre and a radius of 4.8 cm. Step 3:\u2002 With Q as the centre and QR = 2.3 cm, P 3.7 cm Q draw another arc to intersect the previous arc of Step 2 at R and join QR. Figure 12.75 Step 4:\u2002 With R as the centre, draw an arc of radius 3.7 cm. Step 5:\u2002 With P as the centre, draw another arc of radius 2.3 cm to intersect the arc in Step 4 at S. Join RS and PS. PQRS is the required parallelogram. 6.\t Construction of a parallelogram when both the diagonals and the angle between them are given. Example 12.21 Construct a parallelogram PQRS with PR = 3 cm, QS = 4.2 cm and the angle between the diagonals equal to 75\u00b0. \u2022X Solution S Step 1:\u2002 Draw the diagonal PR = 3 cm. Step 2:\u2002 Bisect PR to mark the mid-point of PR as O. Step 3:\u2002 Construct an angle of 75\u00b0at O, such that \u2220POX = 75\u00b0. P 75\u00b0 R O Step 4:\u2002 Taking O as the centre and radius\u2009= 1 (QS ) = 1 \u00d7 4.2 = 2.1 2 2 Q cm, draw arcs on the angular line constructed in Step 3 to cut at Figure 12.76 Q and S. Step 5:\u2002Join PQ, QR, RS and SP to obtain the required parallelogram PQRS.","Geometry 12.29 7.\t Construction of a rectangle when two adjacent sides are given. Example 12.22 S X P R Construct a rectangle PQRS with PQ = 5.2 cm and QR = 2.6 cm. Q Step 1:\u2002 Draw PQ = 5.2 cm. Step 2:\u2002At Q, construct a right angle, such that Figure 12.77 \u2220PQX = 90\u00b0. Step 3:\u2002 Taking Q as the centre and 2.6 cm as radius, draw an arc to cut QX at R. Step 4:\u2002 With R and P as centres, draw two arcs with radii 5.2 cm and 2.6 cm respectively to cut each other at S. Join PS and RS. PQRS is the required rectangle. 8.\t Construction of a rectangle when a side and a diagonal are given. Example 12.23 Construct a rectangle PQRS with PQ = 5.3 cm and diagonal PR = 5.8 cm. Solution S 5.8 cm \u2022X P R Step 1:\u2002 Draw a line segment PQ = 5.3 cm. 5.3 cm Step 2:\u2002 At Q, construct \u2220PQX = 90\u00b0. Figure 12.78 Q Step 3:\u2002 Taking P as the centre and 5.8 cm as radius, draw an arc to cut QX at R. Step 4:\u2002 With R and Q as centres, 5.3 cm and 5.8 cm respectively as radii, draw two arcs to intersect each other at S. Step 5:\u2002Join RS and PS to form the required rectangle PQRS. 9.\t Construction of a rectangle when one diagonal and the angle between two diagonals are given. Example 12.24 Construct a rectangle PQRS, such that PR = 5.2 cm and the angle between the diagonals is 50\u00b0. Step 1:\u2002 Draw a line segment PR = 5.2 cm. Step 2:\u2002 Mark the midpoint of PR as O. Step 3:\u2002 Draw XY which makes an angle of 50\u00b0 with PR at the point O. Step 4:\u2002 With O as the centre and with radius equal to 1 (PR ) = 2.6 cm, cut OX and OY at S and Q respectively. 2 Step 5:\u2002 Join PQ, QR, RS and PS to form the required rectangle PQRS.","12.30 Chapter 12 X S 50\u00b0 R PO YQ Figure 12.79 10.\tConstruction of a square when one side is given. Example 12.25 S 3 cm X Construct a square of side 3 cm. 3 cm R 3 cm Solution Step 1:\u2002 Draw a line segment PQ = 3 cm. P 3 cm Q Step 2:\u2002 Construct \u2220PQX = 90\u00b0. Step 3:\u2002 Mark the point R on QX, such that QR = 3 cm. Figure 12.80 Step 4:\u2002 With R and P as centres and with radii of 3 cm each draw two arcs to intersect each other at S. Step 5:\u2002 Join PS and RS to form the required square PQRS. 11.\tConstruction of a square when a diagonal is given. Example 12.26 X Construct a square with its diagonal as 4 cm. S Solution P O R Step 1:\u2002 Draw a line segment PR = 4 cm. Q Step 2:\u2002 Draw perpendicular bisector XY of PR to bisect Y PR at O. Figure 12.81 Step 3:\u2002Mark the points Q and S on OY and OX, respectively, such that OQ = OS = 2 cm. Step 4:\u2002 Join PS, RS, PQ and QR to form the required square PQRS.","Geometry 12.31 12.\tConstruction of a rhombus when one side and one angle are given. Example 12.27 Construct a rhombus PQRS with PQ = 3.6 cm and \u2220P = 50\u00b0. Solution X R S 3.6 cm 3.6 cm Step 1:\u2002 Draw a line segment PQ = 3.6 cm. Step 2:\u2002 Construct \u2220QPX = 50\u00b0. 3.6 cm Step 3:\u2002 Taking P as the centre and a radius equal to 3.6 cm, draw an arc to cut PX at S, 50\u00b0 Q such that PS = 3.6 cm. P 3.6 cm Step 4:\u2002 From Q and S, draw two arcs with radii 3.6 cm each to meet each other at R. Figure 12.82 Step 5:\u2002 Join QR and SR to form the required rhombus PQRS. 13.\tConstruction of rhombus when one side and one diagonal are given. Example 12.28 S 3.2 cm R Construct a rhombus PQRS, such that PQ = 3.2 cm 4.2 cm 3.2 cm and PR = 4.2 cm. Q 3.2 cm Solution 3.2 cm Figure 12.83 Step 1:\u2002 Draw a line segment PQ = 3.2 cm. Step 2:\u2002 Taking P as the centre and radius equal to P 4.2 cm, draw an arc and taking Q as centre, radius as 3.2 cm draw another arc to cut the previous arc at R. Step 3:\u2002 With R and P as centres and the radii equal to 3.2 cm each, draw two arcs to meet at S. Step 4:\u2002 Join PS, RS and QR to form rhombus PQRS. 14.\tConstruction of a rhombus when both the diagonals are given. Example 12.29 S Construct a rhombus PQRS with diagonal PR = 3.4 cm and QS = 3.6 cm. Solution PO 90\u00b0 R Step 1:\u2002 Draw a line segment PR = 3.4 cm. Step 2:\u2002 Bisect PR and mark its mid-point as O. Step 3:\u2002 With O as the centre and radii 1.8 cm each, draw arcs on Q either sides of PR to cut perpendicular bisector of PR at Q and S. Step 4:\u2002 Join PS, PQ, QR and RS to form the required rhombus Figure 12.84 PQRS.","12.32 Chapter 12 Circles In the lower classes, you have learnt some properties of a circle, like: \u2022\u2022 A circle is a simple closed round figure. \u2022\u2022 Three non-collinear points are required to construct a unique circle. \u2022\u2022 The diameter of a circle is its longest chord. \u2022\u2022 Equal chords are equidistant from its centre. \u2022\u2022 The perpendicular from the centre of a circle to a chord in the circle bisects it. Now, you shall learn more about circles. Arc of a Circle P An arc is a part of a circle. In Fig. 12.85, the two parts of the circle between \u2022 A and B are arcs. When the two arcs are unequal, the shorter one is called the minor arc and the longer one is called the major arc. \u2002Note\u2002\u2002 In Fig. 12.85, AQB is the minor arc and APB is the major A\u2022 \u2022B arc. \u2022 Semi-circle Q The diameter of a circle divides it into two equal arcs. Each arc is called Figure 12.85 a semi-circle. D Segment of a Circle AB C A segment of a circle is the region between the arc of the circle and the chord joining the endpoints of the arc. Figure 12.86 In Fig. 12.86, AB is a chord of the circle. Chord AB divides the circle into two regions. The smaller region is the minor segment and the larger region is the major segment. The minor segment lies between the minor arc ACB and chord AB. The major segment lies between the major arc ADB and the chord AB. Congruence of Circles Circles having equal radii are congruent. Two arcs of a circle (or of congruent circles), which subtend equal angles at the centre (or at the corresponding centres) are equal in length. Two chords of the same or different circles are said to be congruent, if they have the same length. Arcs\u2013Chords C\u2022 \u2022D A\u2022 \u2022B If two arcs of a circle are congruent, then their corresponding chords are congruent. Conversely, if two chords of a circle are congruent, then their corresponding arcs are congruent (see Fig. 12.87). 1.\t If AB \u2245 CD, then chord AB \u2245 chord CD. 2.\t If chord AB \u2245 chord CD, then AB \u2245 CD. \u2002Note\u2002\u2002 This theorem holds good for two congruent circles Figure 12.87 also.","Geometry 12.33 If two circles C1 and C2 are congruent, and PQ is a chord in C1 and RS is a chord in C2, such that PQ \u2245 RS, then PQ \u2245 RS. Theorem 1 The angle subtended by an arc at the centre of a circle is double the C angle subtended by the same arc at any point on the remaining part of the circle. x\u00b0 O Given: In Fig. 12.88, \u2220AOB is subtended by the\u2009AB \u2009at the centre 2x\u00b0 O and \u2220ACB is subtended by the same arc AB at a point C on the remaining part of the circle. The centre of the circle O may lie in the interior, the exterior or on \u2220ACB. AB Case 1:\tCentre O lies in the interior of \u2220ACB. D Case 2:\tCentre O lies on \u2220ACB. Case 3:\tCentre O lies in the exterior of \u2220ACB. Figure 12.88 Construction: Join C and O and produce CO to D. Proof Case 1:\u2003Centre O lies in the interior of \u2220ACB. In triangle OCA, \t\u2220OCA = \u2220OAC (Angles opposite to equal sides OC, OA, the radii of the circle.) \t\u2220AOD = \u2220OCA + \u2220OAC (Exterior angle is equal to the sum of the interior opposite angles) \t = 2\u2220OCA. Similarly, \u2220DOB = 2\u2220OCB \u2234 \u2220AOB = 2\u2220OCA + 2\u2220OCB \u21d2 \u2220AOB = 2\u2220ACB. Hence, proved. Similarly, it can be proved in the other two cases mentioned above. Example 12.30 A In Fig. 12.89, O is the centre of the circle. Find the angles of 120\u00b0 \u0394ABC. O\u2022 Solution 110\u00b0 We know that the sum of angles at a point is 360\u00b0. BC \t \u2234 \u2220AOB + \u2220BOC + \u2220COA = 360\u00b0 Figure 12.89 \t \u2220AOB = 360\u00b0 \u2013 (\u2220BOC + \u2220COA) = 360\u00b0 \u2013 (110 + 120)\u00b0. \t \u2234 \u2220AOB = 130\u00b0. An angle subtended by an arc at the centre of the circle is double the angle subtended by the same arc at any point on the remaining circle. \u2234 \u2220C = 1 \u2220AOB = 1 (130\u00b0) = 65\u00b0, 2 2","12.34 Chapter 12 \u2220A = 1 \u2220BOC = 1 (110\u00b0) = 155\u00b0 and \u2220B = 1 \u2220AOC = 1 (120\u00b0) = 60\u00b0. 2 2 2 2 \u2234 The angles of the triangle ABC are \u2220A = 55\u00b0, \u2220B = 60\u00b0 and \u2220C = 65\u00b0. Example 12.31 \u2022P In Fig. 12.90, O is the centre of the circle. Find the value of x. Solution O \u2022 In the given figure, reflex \u2220AOC is the angle subtended by\u2009 APC. x \u2220ABC is the angle subtended by the\u2009APC at point B on the A 120\u00b0 C remaining part of the circle. \u2234 Reflex \u2220AOC = 2\u2220ABC = 2(120\u00b0) = 240\u00b0. B \u2234 x = 360\u00b0 \u2013 240\u00b0 = 120\u00b0. Figure 12.90 Theorem 2 Angles in the same segment of a circle are equal. D O Given: O is the centre of the circle. A, B, C and D are the \u2022 points on the circle as shown in Fig. 12.91. To prove: \u2220ACB = \u2220ADB. Proof:\u2002 \u2220ACB = 1 \u2220AOB (Angle subtended by an arc at C 2 the centre is double the angle subtended at any point on the remaining part of the circle.) Also, \u2220ADB = 1 \u2220AOB (Using the above rule). A B 2 \u2234 \u2220ACB = \u2220ADB. Figure 12.91 C Theorem 3 A O B An angle in a semi-circle is a right angle. \u2022 Given: AB is the diameter of the circle (see Fig. 12.92). To prove: \u2220ACB = 90\u00b0 D\u2022 Figure 12.92 Proof:\u2002 ADB is an arc, making an angle 180\u00b0 at the centre O. (AOB is a straight line). \u2234 \u2220ACB = 1 \u2220AOB = 1 (180\u00b0). 2 2 \u2234 \u2220ACB = 90\u00b0. Corollary\u2002 If an arc of a circle subtends a right angle at any point on the remaining part of the circle, it is a semi-circle.","Geometry 12.35 Cyclic Quadrilateral D \u2022C \u2022 If all the four vertices of a quadrilateral lie on one circle, then the quadrilateral is called a cyclic quadrilateral. O\u2022 \u2002Notes\u2002 A 1.\t Opposite angles in a cyclic quadrilateral are supplementary. B 2.\t I\u0007n a quadrilateral, if the opposite angles are supplementary, Figure 12.93 then the quadrilateral is a cyclic quadrilateral. \t In Fig. 12.93, ABCD is a cyclic quadrilateral. \u2220A + \u2220C = 180\u00b0 and \u2220B + \u2220D = 180\u00b0 D\u2022 x C 3.\t E\u0007 xterior angle of a cyclic quadrilateral is equal to the interior opposite angle. In Fig. 12.94, ABCD is a cyclic quadrilateral. AB is produced to E to form an exterior angle, \u2220CBE and it A x is equal to the interior angle at the opposite vertex, i.e., \u2220ADC. BE \u2234 \u2220CBE = \u2220ADC Figure 12.94 Example 12.32 C In Fig. 12.95, find the value of x. D Solution x\u00b0 In the given figure, \u2220CBE is an exterior angle which is equal to the opposite interior angle at the opposite vertex, \u2220ADC. B E \u2234 \u2220CBE = \u2220ADC \t (1) A 80\u00b0 \u2220CBE + \u2220EBY = 180\u00b0 (\u2235 linear pair) \u2234 \u2220CBE = 180\u00b0 \u2013 80\u00b0 = 100\u00b0 Y x\u00b0 = \u2220ADC = \u2220CBE = 100\u00b0. Figure 12.95 Example 12.33 A and B are the centres of the circles as shown in the given figure. The circles intersect at C and D. Find \u2220CED + \u2220CFD. C (a) 90\u00b0\t (b) 135\u00b0\t (c) 120\u00b0\t (d) 150\u00b0 \u2022 Hints E\u2022 A \u2022 \u2022 B \u2022F \u2009\u2009\u2009\u2009(i)\u2002Join AB, which is the radius of both the circles. D\u2022 \u2009\u2009(ii)\u2002 AC = AD = BC = BD = radius. (iii)\u2002\u0007\u0394ABC and \u0394ABD are equilateral triangles. \u2220CAD = \u2220BAD = 120\u00b0. (iv)\u2002 \u2220CED = \u2220CFD = 1 \u2220CAD = 1 \u2220CBD. 2 2 (\u0007Angle at the centre is equal to half the angle at the circumference).","12.36 Chapter 12 Example 12.34 In the given figure, AC is the diameter of the circle on which the point D E lies. A, B, C and D are concyclic. If \u2220ADC = 55\u00b0, find the sum of E \u2220DAE and \u2220DCE. (a) 35\u00b0\t (b) 55\u00b0\t (c) 45\u00b0\t (d) 65\u00b0 AC Solution B ABCD is a cyclic quadrilateral \u21d2 \u2220A + \u2220C = 180\u00b0 and \u2220B + \u2220D = 180\u00b0 (sum of the opposite angles of acyclic quadrilateral) \u2220D = 55\u00b0 (given) \u21d2 \u2220B = 180\u00b0 \u2013 55\u00b0 = 125\u00b0. In quadrilateral ABCE, \u2220A + \u2220B + \u2220C + \u2220E = 360\u00b0 \\\\ \u2220A + \u2220C + 125\u00b0 + 90\u00b0 = 360\u00b0 (\\\\ AC is the diameter) \\\\ \u2220A + \u2220C = 360\u00b0 \u2013 215\u00b0 = 145\u00b0. \u2220DAE + \u2220DCE = \u2220BAD + \u2220BCD \u2013 (\u2220BAE + \u2220BCE) = 180\u00b0 \u2013 (145\u00b0) = 35\u00b0. Constructions Related to Circles Construction 1: To construct a segment of a circle, on a given chord and containing a given angle. Example 12.35 Construct a segment of a circle with a chord of length 8.5 cm and containing an angle of 55\u00b0 (\u03b8). A Solution Step 1:\u2002 Draw a line segment BC of the given length, 8.5 cm. Y O X Step 2:\u2002 Draw BX and CY , such that \u2220CBX = BCY = B C \u2022 180 \u2212 2\u03b8 = 35\u00b0. 2 110\u00b0 35\u00b0 35\u00b0 Step 3:\u2002 Mark the intersection of BX and CY as O. 8.5 cm Step 4:\u2002 Taking O as the centre and OB or OC as radius, draw BAC. Figure 12.96 Step 5:\u2002 In \u0394BOC, \u2220BOC = 110\u00b0 \u21d2 \u2220BAC = \uf8eb 1 \uf8f6 \u2220BOC = \uf8eb 1 \uf8f6\uf8f8\uf8f7 (110\u00b0) = 55\u00b0. \uf8ed\uf8ec 2 \uf8f8\uf8f7 \uf8ed\uf8ec 2 \u2234 \u2220BAC = 55\u00b0. The region bounded by BAC and BC is the required segment.","Geometry 12.37 Construction 2: Construct an equilateral triangle inscribed in a circle A of radius 3.5 cm. 120\u00b0 \u2022 O C Step 1:\u2002 Draw a csircle of radius 3.5 cm and mark its centre as O. 120\u00b0 Step 2:\u2002Draw radii OA, OB and OC, such that \u2220AOB = \u2220BOC = 120\u00b0. B Step 3:\u2002 Join AB, BC and CA, which is the required equilateral \u0394ABC in the given circle. Figure 12.97 Construction 3: Construct an equilateral triangle circumscribing a circle of radius 3 cm. Step 1:\u2002 Draw a circle of radius 3 cm with centre O. A Q R O B PC Figure 12.98 Step 2:\u2002 Draw radii OP, OQ and OR, such that \u2220POR = \u2220ROQ = 120\u00b0. C Step 3:\u2002 At P, Q and R draw perpendiculars to OP, OQ and OR respectively to form \u0394ABC. \u0394ABC is the required circumscribing equilateral triangle. Construction 4: Construct a square inscribed in a circle of radius D O B 3 cm. Step 1:\u2002 Draw a circle of radius 3 cm and mark the centre as O. A Step 2:\u2002 Draw diameters AC and BD, such that AC \u22a5 BD. Figure 12.99 Step 3:\u2002Join AB, BC, CD and DA. Quadrilateral ABCD is the required square inscribed in the given circle. Construction 5: Construct a square circumscribing a circle of radius 2.5 cm. DR C SO Q A PB Figure 12.100","12.38 Chapter 12 Step 1:\u2002 Draw a circle of radius 2.5 cm. Step 2:\u2002 Draw two mutually perpendicular diameters PR and SQ. Step 3:\u2002 At P, Q, R and S, draw lines perpendicular to OP, OQ, OR and OS to form square ABCD as shown in Fig. 12.100. Construction 6: Inscribe a regular hexagon in a circle of radius 3 cm. D E C O\u2022 F B A Figure 12.101 Step 1:\u2002 Draw a circle of radius 3 cm taking the centre as O. Step 2:\u2002 Draw radius OA. With radius equal to OA and starting with A as the centre, mark points B, C, D, E and F one after the other. Step 3:\u2002 Join A, B, C, D, E and F. Polygon ABCDEF is the required hexagon. Construction 7: Construct a regular hexagon circumscribing a circle of radius 3.5 cm. Step 1:\u2002 Draw a circle of radius 3.5 cm and mark its centre as O. Step 2:\u2002 Draw radii OP, OQ, OR, OS, OT and OU , such that the angle between any two adjacent radii is 60\u00b0. Step 3:\u2002 Draw lines at P, Q, R, S, T and U perpendicular to OP, OQ, OR, OS, OT and OU respectively to form the required hexagon ABCDEF. D SC T R E \u2022 B U O Q F P A Figure 12.102","Geometry 12.39 Construction 8: Draw the circum-circle of a given triangle. Step 1:\u2002 Draw \u0394ABC with the given measurements. Step 2:\u2002 Draw perpendicular bisectors of AB and AC to intersect each other at S. Step 3:\u2002 Taking S as the centre and radius equal to AS or BS or CS, draw a circle. The circle passes through all the vertices A, B and C of the triangle. \u2234 The circle drawn is the required circum-circle. \u2002Note\u2002\u2002 The circum-centre is equidistant from the vertices of the triangle. C S\u2022 B A Figure 12.103 Construction 9: Construct the in-circle of a given triangle ABC. A I \u2022 B MC Figure 12.104 Step 1:\u2002 Draw a \u0394ABC with the given measurements. Step 2:\u2002 Draw bisectors of \u2220B and \u2220C to intersect at I. Step 3:\u2002 Draw perpendicular IM from I onto BC. Step 4:\u2002 Taking I as centre and IM as the radius, draw a circle. This circle touches all the sides of the triangle. This is the in-circle of the triangle. \u2002Note\u2002\u2002 The in-centre is equidistant from all the sides of the triangle.","12.40 Chapter 12 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t Can a triangle be formed by line segments of 1\t 7.\t If lines l1, l2 and l3 pass through a point P, then they lengths a, b and c, such that a > b \u2013 c? are called _________ lines. \t2.\t Can a triangle be formed by line segments of 1\t 8.\t In an isosceles right triangle ABC, \u2220B = lengths a, b and c, such that a = b \u2013 c? 90\u00b0 and BD \u22a5 AC. Then BD = ______. \t3.\t The areas of parallelograms on the same base and between the same parallel lines are _______. \uf8eb AB \/ BC \/ AC \uf8f6 . \uf8ed\uf8ec 2 2 2 \uf8f8\uf8f7 \t4.\t In a regular polygon, are all the exterior angles equal? 1\t 9.\t The sum of all the altitudes in a triangle is _______ \t5.\t Can the sum of the two angles of a triangle be less than the third angle? the sum of all the sides. (equal to\/less than\/greater \t6.\t If all the sides of a polygon are equal, then all its than) interior angles must be equal. Is the given state- ment true? 2\t 0.\t An angle is 2 times its supplementary angle. What 3 \t7.\t If a circle passes through four points, then the four is the angle? points are said to be _________. 2\t 1.\t Are all the diagonals of a regular polygon always \t8.\t Two circles cannot intersect in more than two concurrent? points. [True\/False] 2\t 2.\t Two lines AB and CD intersect at point O. \u2220AOD \t9.\t Two quadrilaterals of equal perimeters occupy : \u2220BOD = 3 : 1, then \u2220AOD = _________. equal areas. Is this statement always true? \t23.\t Are the diagonals of a regular polygon equal in 1\t 0.\t Can a polygon have the sum of all its interior length? angles equal to 810\u00b0? \t24.\t In a triangle ABC, if \u2220B > \u2220C, then AB > AC. Is 1\t 1.\t The exterior angle of a regular polygon is 60\u00b0. the given statement true? The number of sides of the polygon is ______. PRACTICE QUESTIONS \t25.\t Can the length of the median in a triangle to a side 1\t 2.\t A line l intersects a pair of parallel lines. The exte- be less than the corresponding altitude? rior angles on the same side of line l are in the ratio 5 : 4. The measure of the bigger angle of the two \t26.\t If d is the distance from a point P to the centre of is _________. the circle of radius r and d \u2013 r > 0, then the point P lies _________ the circle. (outside\/inside\/on) \t13.\t When all the sides of a quadrilateral are equal, then it is either a ________ or a _______. \t27.\t If the longest chord of a circle is of length 14 cm, then the circumference of the circle is ________ 1\t 4.\t In a quadrilateral, ABCD, \u2220DAB + \u2220BCD = cm. 180\u00b0, then the quadrilateral ABCD is ________. \t28.\t In a triangle ABC, D is a point on BC, such that \t15.\t If four lines intersect in a plane, at the maximum AD is the shortest distance from A to BC and how many triangles are formed? \u2220ADC = 70\u00b0. Is it possible? \t16.\t If all the angles of a quadrilateral are equal, then \t29.\t Can the exterior angle of a polygon be three times all its sides must be equal. Is the above statement its interior angle? true? 3\t 0.\t If AB and CD are two chords of a circle and AB > CD, then the chord which is nearer to the centre is _________.","Geometry 12.41 Short Answer Type Questions 3\t 1.\t In the following figure, PQ |\u2009| RS. If \u2220TRS = 105\u00b0, D \u2220PTR = 35\u00b0, \u2220QPT = a\u00b0, find the value of a. E S\u2022 Q \u2022 C P a\u00b0 46\u00b0 105\u00b0 x\u00b0 x\u00b0 R A x\u00b0 B 30\u00b0 \t37.\t In the following figure, CB is produced to the point A. Find \u2220BDC. T ED \t32.\t In triangle ABC, \u2220A < \u2220B, \u2220B > \u2220C and \u2220A = 2\u2220C, prove that AC > BC > AB. \t33.\t If two complementary angles are in the ratio 7 : 100\u00b0 60\u00b0 11, find the supplement of the bigger angle. AB C \t34.\t In the given quadrilateral ABCD, p\u00b0 + q\u00b0 = 100\u00b0, a\u00b0 = 140\u00b0 and r\u00b0 = 1 (a\u00b0 + q\u00b0). Find the angles p\u00b0, q\u00b0, r\u00b0 and s\u00b0. 2 \t38.\t If the sum of the interior angles of a polygon DC is 2340\u00b0, then find the number of sides of the s\u00b0 r \u00b0 polygon. p\u00b0 a\u00b0 b\u00b0 q\u00b0 3\t 9.\t In the following figure, if O is the centre of the PRACTICE QUESTIONS A B circle and AB is the diameter, then find \u2220ACB. C D \t35.\t In the following figure, O is the centre of the 80\u00b0 30\u00b0 circle. Find the value of x + y O A B E F xD 70\u00b0 C \t40.\t l O m x Ay B Pz y \t36.\t In the following figure, BE is the diameter of the \t\tIn the above figure, find the value of z, if x is two- third of y which is a complement of 45\u00b0. circle. Find the value of \u2018x\u2019 and \u2220DAB.","12.42 Chapter 12 \t41.\t Find the length of the sides of the given parallelo- DC gram, if the perimeter of the parallelogram is 24 O cm, the length of perpendicular OF = 3 cm, OB = 5 cm and OC = 18 cm. CF D AB O \t44.\t In the following figure, O is the centre of the BA circle. If x = 40\u00b0 and x : y = 4 : 3, find the value of z. C \t42.\t In the figure given below, OD = OC = OB. Find x O \u2220DCB. A zy C B D \t45.\t In the following figure, AD = 5.6 cm, AE = (x + 1) 100\u00b0 cm, AB = 8.4 cm and EC = (x \u2013 1) cm, find AC. 80\u00b0 Given that DE |\u2009| BC. O A B A \t43.\t If the sum of the lengths of two diagonals of the D E given square ABCD is 24 cm, then find the side of B C the square and the magnitude of AO + OB. PRACTICE QUESTIONS 30\u00b0 Essay Type Questions A FE \t46.\t ABCD and BFDC are cyclic quadrilaterals. CD is produced to E. If \u2220BCE = 45\u00b0, then find \u2220BFD. A F BOD BDC C \t48.\t Find \u2220DBA, \u2220DAB and \u2220AED in the following E figure, where ABCDE is a semi-circle. D C 20\u00b0 \t47.\t In the following triangle ABC, D, E and F are the E 30\u00b0 mid-points of sides BC, CA and AB respectively. A B Prove that AB \u2212 BC < AE < AB + BC . 2 2","Geometry 12.43 \t49.\t In the following figure, ABDG is a rectangle with \t50. \tIn the following figure, BD \u2009is produced to E. If AB = 10 cm and AG = 6 cm. Find the areas of AD is the angle bisector of \u2220A, then find \u2220BCD. parallelograms ABCF and ABEH. Also find the area of \u0394AEB. C A H GF E D C B 120\u00b0 DE AB CONCEPT APPLICATION Level 1 \t1.\t In the following figure, A \t4.\t In the following, CDEF F E \u0394ABC is right-angled C is a cyclic quadrilateral. at C, and M is the mid- point of hypotenuse AB. M CG and DH are the angle HG If AC = 32 cm and BC = 60 cm, then find the bisectors of \u2220C and \u2220D P length of CM. respectively. If \u2220E = 100\u00b0 and \u2220F = 110\u00b0, then find \t\t(a) 32 cm C D B \u2220CPD. \t\t(a) 105\u00b0\t\t (b) 80\u00b0 \t\t(b) 30 cm \t\t(c) 150\u00b0\t\t (d) 90\u00b0 \t\t(c) 17 cm \t5.\t In the following figure, ABC is an equilateral triangle. DE is parallel to BC and equal to half the \t\t(d) 34 cm length of BC. If AD + EC + CB = 24 cm, then what is the perimeter of triangle ADE\u2009? \t2.\t A cyclic polygon has n sides such that each of its PRACTICE QUESTIONS interior angle measures 144\u00b0. What is the measure A of the angle subtended by each of its side at the geometrical centre of the polygon? \t\t(a) 144\u00b0\t\t (b) 30\u00b0 DE \t\t(c) 36\u00b0\t\t (d) 54\u00b0 \t3.\t In the following, PQRS is a rhombus, SQ and PR BC are the diagonals of the rhombus intersecting at O. If angle OPQ = 35\u00b0, then find the value of angle \t\t(a) 12 cm\t ORS + angle OQP. \t\t(b) 16 cm SR \t\t(c) 18 cm \t\t(d) Cannot be determined O P Q \t6.\t In \u0394PQR, M and N are points on PQ and PR, respectively, such that MN ||QR. If PM = x, PR \t\t(a) 90\u00b0\t\t (b) 180\u00b0 = x + 9, PQ = x + 13 and PN = x \u2013 2, then find x. \t\t(c) 135\u00b0\t\t (d) 45\u00b0 \t\t(a) 10\t\t (b) 11 \t\t(c) 13\t\t (d) 15","12.44 Chapter 12 \t7.\t In the following figure (not to scale), the chords P AC and BD intersect at E and \u2220BAE = \u2220ECD + 20\u00b0. If \u2220CDE = 60\u00b0, find \u2220ABE. O AD MN \u2022 Q E \t\t(a) 145\u00b0\t\t (b) 162 1 \u00b0 2 \u22c5 BC \t\t(c) 158 1 \u00b0 \t\u22c5 \t (d) 180\u00b0 2 \t\t(a) 40\u00b0\t\t (b) 60\u00b0 1\t 2.\t In the following figure, \u0394PQR is right-angled at \t\t(c) 80\u00b0\t\t (d) None of these R and S is the mid-point of hypotenuse PQ. If RS \t8.\t In the following figure, DE and FG are equal = 25 cm and PR = 48 cm, then find QR. chords of the circle subtending \u2220DHE and \u2220FHG P at the point H on the circle. If \u2220DHE = 23 1 \u00b0, S then find \u2220FHG. 2 D QR EH \t\t(a) 7 cm\t\t (b) 25 cm F \t\t(c) 14 cm\t\t (d) Cannot be determined G \t13.\t In a cyclic quadrilateral PQRS, PS = PQ, RS = RQ and \u2220PSQ = 2\u2220QSR. Find \u2220QSR. 1 \t\t(a) 27 2 \u00b0 \u22c5 \t\t (b) 30\u00b0 \t\t(a) 20\u00b0\t\t (b) 30\u00b0 PRACTICE QUESTIONS \t\t(c) 23 1 \t\t (d) 60\u00b0 \t\t(c) 40\u00b0\t\t (d) 50\u00b0 2 \u00b0 \u22c5 \t9.\t The bisectors of two adjacent angles in a parallelo- 1\t 4.\t In the following figure, two isosceles right triangles, gram meet at a point P inside the parallelogram. DEF and HGI are on the same base DH and DH is parallel to FI. If DE = GH = 9 cm and DH = The angle made by these bisectors at a point is 20 cm, then the area of the quadrilateral FEGI is ______. ________. FI \t\t(a) 180\u00b0\t\t (b) 90\u00b0 \t\t(c) 45\u00b0\t\t (d) None of these \t10.\t If x\u00b0 is the measure of an angle which is equal to its complement and y\u00b0 is the measure of an angle which is equal to its supplement, then x\u00b0 \u22c5 is_____. D EG H y\u00b0 \t\t(a) 1\t\t (b) 3 \t\t(a) 99 cm2\t\t (b) 40.5 cm2 \t\t(c) 0.5\t\t (d) 2 \t \t (c) 81 cm2\t\t (d) 180 cm2 \t11.\t In the following figure, O is the centre of the \t15.\t A pole of height 14 m casts a 10 m long shadow on circle. If \u2220MPN = 55\u00b0, then find the value of the ground. At the same time, a tower casts a 70 m long shadow on the ground. Find the height of the \u2220MON + \u2220OMN + 1 \u2220MNO. tower. 2","Geometry 12.45 \t\t(a) 50 m\t\t (b) 78 m \t\t(a) 70\u00b0\t\t (b) 50\u00b0 \t\t(c) 90 m\t\t (d) 98 m \t\t(c) 110\u00b0\t\t (d) 35\u00b0 \t16.\t The angle subtended by a minor arc in its alternate \t21.\t In the following figure (not to scale), AB||CD. If segment is ________. \u2220BAE = 25\u00b0 and \u2220DCE = 30\u00b0, then find \u2220AEC. \t\t(a) acute\t\t (b) obtuse AB \t\t(c) 90\u00b0\t\t (d) reflex angle 1\t 7.\t The number of diagonals of a regular polygon is E 27. Then, each of the interior angles of the poly- gon is __________. CD \t\t(a) \uf8eb 500\uf8f6 \u00b0\t (b) 140\u00b0 \t\t(a) 30\u00b0\t\t (b) 45\u00b0 \uf8ec\uf8ed 3 \uf8f8\uf8f7 (d) 154\u00b0 \t\t(c) 50\u00b0\t\t (d) 55\u00b0 \t\t(c) 128\u00b0\t\t \t18.\t ABC is a triangle inscribed in a circle, AC being \t22.\t A tower of height 60 m casts a 40 m long shadow the diameter of the circle. The length of AC is as on the ground. At the same time, a needle of much more than the length of BC as the length of height 12 cm casts a x cm long shadow the ground. BC is more than the length of AB. Find AC : AB. Find x. \t\t(a) 5 : 3\t\t (b) 5 : 4 \t\t(a) 6\t\t (b) 8 \t\t(c) 6 : 5\t\t (d) 3 : 2 \t\t(c) 10\t\t (d) 14 \t19.\t MN is the arc of the circle with centre O. \t23.\t In the given figure, AC is the diameter. AB and AD are equal chords. If \u2220AED = 110\u00b0, then find \u2220BAD. If \u2220MOR = 100\u00b0 and \u2220NOR = 135\u00b0, then A 1 1 2 \u2220ORN + 4 \u2220ORM is ________. E R PRACTICE QUESTIONS BD C O \t\t(a) 40\u00b0\t\t (b) 55\u00b0 MN \t\t(c) 110\u00b0\t\t (d) 120\u00b0 \t\t(a) 22 1 \u00b0 \t\u22c5 \t (b) 40\u00b0 2\t 4.\t In the given rectangle ABCD, the sum of the 2 lengths of two diagonals is equal to 52 cm and E 1 \t\t(c) 125\u00b0\t\t (d) 21 4 \u00b0 \u22c5 is a point in AB, such that OE is perpendicular to \t20.\t In the following figure (not to scale), \u2220BCD = AB. Find the lengths of the sides of the rectangle, 40\u00b0, \u2220EDC = 35\u00b0, \u2220CBF = 30\u00b0 and \u2220DEG = if OE = 5 cm. 40\u00b0. Find \u2220BAE. DC A O BE AEB FG D \t\t(a) 24 cm, 10 cm\t (b) 12 cm, 10 cm C \t\t(c) 24 cm, 5 cm\t (d) 12 cm, 15 cm","12.46 Chapter 12 \t25.\t In the following figure (not to scale), AD bisects E \u2220BAC. If \u2220BAD = 45\u00b0 and \u0394ABC is inscribed in 85\u00b0 a circle, then which of the following is the longest? A AC D 40\u00b0 B \t\t(a) 45\u00b0\t\t (b) 125\u00b0 BD C \t\t(c) 55\u00b0\t\t (d) 110\u00b0 \t\t(a) AB\t\t (b) AD 2\t 8.\t In the given figure, DEF is a D triangle. If DF is the longest E \t \t (c) AC\t\t (d) BC side and EF is the shortest side, then which of the fol- 2\t 6.\t In the given figure, AB||DE and area of the lowing is true? parallelogram ABFD is 24 cm2. Find the areas of \u0394AFB, \u0394AGB and \u0394AEB. \t\t(a) \u2220E > \u2220D > \u2220F F D GF E \t\t(b) \u2220D < \u2220F < \u2220E \t \t (c) \u2220D < \u2220E < \u2220F \t\t(d) None of these AB 2\t 9.\t The ratio between the exterior angle and the inte- rior angle of a regular polygon is 1 : 3. Find the \t\t(a) 8 cm2 number of the sides of the polygon. \t\t(b) 12 cm2 \t \t (c) 10 cm2 \t\t(a) 12\t\t (b) 6 \t\t(d) 14 cm2 2\t 7.\tIn the given figure, AD and BE intersect at C, \t\t(c) 8\t\t (d) 10 PRACTICE QUESTIONS such that BC = CE, \u2220ABC = 40\u00b0 and \u2220DEC = 3\t 0.\t Find each interior and exterior angle of a regular 85\u00b0. Find \u2220BAC \u2212 \u2220CDE. polygon having 30 sides. \t\t(a) 144\u00b0, 36\u00b0\t (b) 156\u00b0, 24\u00b0 \t\t(c) 164\u00b0, 16\u00b0\t (d) 168\u00b0, 12\u00b0 Level 2 \t31.\t If BC : CD = 2 : 3, AE A A D : EC = 3 : 4 and BC : E B P AE = 2 : 3, then find the ratio of the area of C O \u0394ECD to the area of C \u0394AEB. B D \t\t(a) 2 : 1\u2003\u2003\u2003 (b) 2 : 3 \t\t(a) 10\u00b0 \t\t(c) 3 : 5\u2003\u2003\u2003 (d) 4 : 3 \t\t(b) 30\u00b0 \t\t(c) 40\u00b0 \t32.\t In the given figure (not to scale), O is the centre of \t\t(d) 50\u00b0 the circle. If PB = PC, \u2220PBO = 25\u00b0 and \u2220BOC = 130\u00b0, then find \u2220ABP + \u2220DCP.","Geometry 12.47 \t33.\t In a polygon, the greatest angle is 110\u00b0 and all the \t\t(a) 4 \t\t (b) 1 angles are distinct in integral measures (in degrees) 3 Find the maximum number of sides it can have. 5 2 \t\t(a) 4\t\t (b) 5 \t\t(c) 3 \t\t (d) 3 \t\t(c) 6\t\t (d) 7 \t37.\t In the given figure (not to scale), the points M, R, N, S and Q are concyclic. Find \u2220PQR + \u2220OPR 3\t 4.\t In the given figure, ABCD is a rectangle inscribed + \u2220NMS + \u2220OSN, if O is the centre of the circle. in a semi-circle. If the length and the breadth of the rectangle are in the ratio 2 : 1. What is the ratio M of the perimeter of the rectangle to the diameter of the semicircle? RQ DC O \u2022 B NS AO P \t\t(a) 3 : 2 \t\t (b) 2 : 3 \t\t(a) 90\u00b0\t\t (b) 180\u00b0 \t\t(c) 2 : 5 \t\t (d) 3 : 5 \t\t(c) 270\u00b0\t\t (d) Data inadequate \t35.\t In the given figure, AB is the diameter of the circle 3\t 8.\t In the given figure (not to scale), AC is the diameter with area p sq. units. Another circle is drawn with of the circle and \u2220ADB = 20\u00b0, then find \u2220BPC. C as centre, which is on the given circle and passing through A and B. Find the area of the shaded region. E D A AC C\u2022 B P PRACTICE QUESTIONS \t\t(a) 50\u00b0\t\t (b) 70\u00b0 B \t\t(c) 90\u00b0\t\t (d) 110\u00b0 \t\t(a) \u03c0 \u22c5 sq. units\t (b) 2\u03c0 s\u22c5 q. units 3\t 9.\t In the following figure, O is the centre of the circle 3 3 and CD = DE = EF = GF. If \u2220COD = 40\u00b0, then find reflex \u2220COG. \t\t(c) 1 sq. units\t (d) 1.2 sq. units \t36.\t In the following figure (not to scale), C1 and C2 O are two congruent circles with centres O1 and O2. CG respectively. Each circle passes through the centre of the other circle. If the circumference of each D F circle is 2 cm, the perimeter of the shaded region E is _______ cm. A \t\t(a) 200\u00b0\t\t (b) 90\u00b0 O1 O2 \t\t(c) 80\u00b0\t\t (d) 160\u00b0 C1 B C2 4\t 0.\t In the given figure (not to scale), AC is the median as well as altitude to BD. In \u0394ACE, AD is the median to CE. Which of the following is true?","12.48 Chapter 12 \t\t(a) AB + CD > AE A A \t\t(b) AB + BC = AE \t \t (c) AB + DE < AE E \t\t(d) None of the above \t41.\t In the given figure, (not B C DE to scale), rectangle ABCD and triangle ABE are BD C inscribed in the circle with centre O. If \u2220AEB = 40\u00b0, then find \u2220BOC. \t\t(a) 15\t\t (b) 18 E \t\t(c) 20\t\t (d) 24 40\u00b0 A D \t45. \tIn DABC, P is the mid-point of BC and Q is the mid-point of AP. Find the ratio of the area of O DABQ and the area of DABC. The following are the steps involved in solving the above problem. BC \t\t(a) 60\u00b0\t\t (b) 80\u00b0 \t\t(A) W\u0007 e know that a median of a triangle divides a triangle into two triangles of equal area. \t\t(c) 100\u00b0\t\t (d) 120\u00b0 \t\t(B) \u21d2 Ar(DABP) = 1 [Ar(DABC)] 2 4\t 2.\t In the given figure (not to D scale), O is the centre of A \t\t(C) Ar(DABQ) = 1 [Ar(DABP)] = 1 [ArDABC] the circle. BC and CD are O 2 4 equal chords. If \u2220OBC = C 55\u00b0, then find \u2220BAC. \t\t(D) \u21d2 Ar(DABQ) : Ar(DABC) = 1 : 4 B \t\t(a) 60\u00b0 \t\t(a) A\u0007 CBD\t\t (b) ADBC \t\t(b) 70\u00b0 \t \t (c) ABCD\t \t (d) ADCB PRACTICE QUESTIONS \t\t(c) 80\u00b0 4\t 6. \tABCD is a cyclic quadrilateral, ABC is a minor arc and O is the centre of the circle. If \u2220AOC = 160\u00b0, \t\t(d) 90\u00b0 then find \u2220ABC. 4\t 3.\t In the given figure (not to scale), O is the centre of \t\tThe following are the steps involved in solving the the circle C1 and AB is the diameter of the circle above problem. Arrange them in sequential order. C2. Quadrilateral PQRS is inscribed in the circle with centre O. Find \u2220QRS. \t\t(A) \u0007We have, \u2220ABC + \u2220ADC = 180\u00b0 P B \t\t(B) \u2220\u0007 ABC + 1 \u2220AOC = 180\u00b0 ( \u2220ADC = 1 C1 O 2 2 S R C2 \u2220AOC) Q \t\t(C) \u2220ABC = 180\u00b0 \u2013 80\u00b0 A \t\t(D) \u2220ABC + 160\u00b0 = 180\u00b0 D 2 O \t\t(a) 105\u00b0\t\t (b) 115\u00b0 \t\t(E) \\\\ \u2220ABC = 100\u00b0 A C \t\t(c) 135\u00b0\t\t (d) 145\u00b0 \t\t(a) ABDEC \t44.\t In the given figure (not to scale), E and D are the \t\t(b) ABDCE B mid-points of AB and BC respectively. Also, \u2220B = \t\t(c) BCDAE 90\u00b0, AD = 292 cm and CM = 208 cm . Find \t\t(d) BACDE AC.","Geometry 12.49 \t47. \tShow that each diagonal of a parallelogram divide SB R it into two congruent triangles. CD \t\tThe following are the steps involved in showing the above result. Arrange them in sequential order. P AQ \t\t(A) \u0007In DABC and DCDA, AB = DC and BC = AD \t\t(a) rhombus\t\t (b) trapezium ( opposite angles of parallelogram) AC = AC (common side). \t\t(c) square\t\t (d) rectangle \t\t(B) Let ABCD be a parallelogram. Join AC. 5\t 1. \tThe sides of a triangle are 2006 cm, 6002 cm and m cm, where m is a positive integer. Find the num- \t\t(C) By SSS congruence property, DABC \u2245 DCDA. ber of such possible triangles. \t\t(D) S\u0007 imilarly, BD divides the triangle into two \t\t(a) 1\t\t (b) 2006 congruent triangles. \t\t(a) BACD\t\t (b) BDAC \t\t(c) 3996\t\t (d) 4011 \t \t (c) BADC\t\t (d) BDCA \t52. \tIf a, b and c are the lengths of the sides of a right triangle ABC with c = 2a and b2 \u2013 3a2 = 0, then 4\t 8. \tShow that any angle in C \u2220ABC = ______. a semi-circle is a right A angle. \t\t(a) 60\u00b0\t\t (b) 30\u00b0 \t\tThe following are the D B \t\t(c) 45\u00b0\t\t (d) 90\u00b0 steps involved in showing the above result. Arrange \t53. \tIn DABC, AC = BC, S is the circum-centre and them in sequential order. \u2220ASB = 150\u00b0. Find \u2220CAB. 180\u00b0 A 2 \t\t(A) \u2234 \u2220ACB = = 90\u00b0 \t\t(B) \u0007The angle subtended by an arc at the centre is S double of the angle subtended by the same arc at any point on the remaining part of the circle. PRACTICE QUESTIONS \t\t(C) L\u0007 et AB be a diameter of a circle with centre D BC and C be any point on the circle. Join AC and BC. \t\t(a) 55 1 \u00b0 \t\u22c5 \t (b) 52 1 \u00b0 \u22c5 2 2 \t\t(D) \u2234 \u2220ADB = 2 \u00d7 \u2220ACB 1 1 2 2 \t\t180\u00b0 = 2\u2220ACB ( \u2220ADB = 180\u00b0) \t\t(c) 62 \u00b0 \t\u22c5 \t (d) 35 \u00b0 \u22c5 \t\t(a) DBAC\t\t (b) DBCA \t54. \tIn the given figure, P, Q, R and S are concyclic \t\t(c) CBAD\t\t (d) CBDA points, and O is the mid-point of the diameter QS. 4\t 9.\t A, B, C and D are concyclic. AC bisects BD. If AB \t\tIf \u2220QPR = 25\u00b0, then find \u2220SOR. = 9 cm, BC = 8 cm, and CD = 6 cm, then find the measure of AD. S \t\t(a) 7 cm\t\t (b) 10 cm O TR \t\t(c) 12 cm\t\t (d) 15 cm P Q \t50. \tIn the given figure, PQRS is a parallelogram. A \t\t(a) 130\u00b0\t\t \t\t(c) 75\u00b0\t\t (b) 120\u00b0 and B are the mid-points of PQ and SR respec- (d) 100\u00b0 tively. If PS = BR, then the quadrilateral ADBC is a ______.","12.50 Chapter 12 Level 3 \t55. \tIn \u0394ABC, \u2220B = 90\u00b0. P, Q and R are the mid- 5\t 9. \tIn the given figure, the angles \u2220ADE and \u2220ABC differ by 15\u00b0. Find \u2220CAE. points of AB, BC , and AC respectively. Then which of the following is true? DB \t\t(a) A, P, Q and R are concyclic points \t\t(b) B, P, R and Q are concyclic points O AE \t\t(c) C, Q, P and R are concyclic points C \t\t(d) All of these \t56. \tIf p, q and r are the lengths of the sides of a right \t\t(a) 10\u00b0\t\t (b) 7 1 \u00b0 \u22c5 2 triangle, PQR, and the hypotenuse r = 2pq, then \u2220QPR = _________. \t\t(c) 15\u00b0\t\t (d) 30\u00b0 \t\t(a) 50\u00b0\t\t (b) 45\u00b0 6\t 0. \tIn the given figure, ABCD is a cyclic quadrilat- \t \t (c) 60\u00b0\t\t (d) 30\u00b0 eral, \u2220ABC = 70\u00b0, FG bisects \u2220CFA, EG bisects \u2220DEB, \u2220DCE = 60\u00b0 and \u2220EGF = 90\u00b0. Find 5\t 7. \tIn a triangle PQR, PQ = QR. A and B are the \u2220HEC. mid-points of QR and PR respectively. A circle E passes through P, Q, A and B. Then which of the following is necessarily true? \t\t(a) \u0394PQR is equilateral D \t\t(b) \u0394PQR is right isosceles \t\t(c) PQ is a diameter HC 90\u00b0 \t\t(d) Both (a) and (c) G A 5\t 8. \tIn the figure given below (not to scale), D is a point F on the circle with centre A and C is a point on the B PRACTICE QUESTIONS circle with centre B. AD \u22a5 BD and BC \u22a5 CA. \t\t(a) 20\u00b0 Then which of the following is true? \t\t(b) 40\u00b0 \t\t(c) 25\u00b0 D \t\t(d) 45\u00b0 A \u2022 \u2022B 6\t 1. \tIn the given figure, A, D, B, E and C are concyclic. C If \u2220ACB = 60\u00b0 and \u2220AED = 50\u00b0, then find \u2220DEB. \t\t(a) BD = AC, when AD = BC BE \t\t(b) BD = AC, when AD||BC \t\t(a) 15\u00b0 \t\t(c) Both (a) and (b) ?50\u00b0 \t\t(d) BD = AC is always true \t\t(b) 10\u00b0 D \t \t (c) 20\u00b0 60\u00b0 C \t\t(d) 5\u00b0 A","Geometry 12.51 TEST YOUR CONCEPTS 1\t 6. \tNo Very Short Answer Type Questions \t17. \tconcurrent \t1. \tYes 1\t 8.\t AC \t2. \tNo 2 \t3. \tequal \t4. \tYes \t19. \tless than \t5. \tYes \t6. \tNo \t20. \t72\u00b0 \t7. \tconcyclic \t8. \tTrue \t21. \tNo \t9. \tNo 1\t 0. \tNo \t22. \t135\u00b0 1\t 1. \t6 1\t 2. \t100\u00b0 \t23. \tNeed not be 1\t 3. \tsquare, rhombus \t14. \tcyclic \t24. \tNo \t15. \tFour \t25. \tNo Shot Answer Type Questions \t26. \toutside \t31. \ta = 135 \t33. \t125\u00b0 \t27. \t44 \t34. \tp = 40, q = 60, r = 100, s = 160 \t35. \t250 \t28.\t No 3\t 6. \t\u2220DAB = 46\u00b0, x = 44\u00b0 3\t 7. \t40\u00b0 2\t 9. \tYes 3\t 8. \t15 \t39. \t60\u00b0 \t30. \tAB Essay Type Questions 4\t 0. \t75\u00b0 ANSWER KEYS 4\t 1. \tBC = 5 cm = AD and AB = 7 cm = CD \t46. \t\u2220BFD = 45\u00b0 4\t 2. \t90\u00b0 4\t 8. \t\u2220ABD = 40\u00b0 4\t 3. \tAO + OB = 12 cm, and side of the square \t\t\u2220DAB = 50\u00b0 \t\t\u2220AED = 140\u00b0 = 6 2 cm. \t44. \tz = 140\u00b0 4\t 5. \t6 cm \t49. \tArea of parallelogram ABCF = 60 cm2 \t\tArea of \u0394AEB = 30 cm2. \t50. \t\u2220BCD = 60\u00b0","12.52 Chapter 12 CONCEPT APPLICATION Level 1 \t1.\u2002(d)\t 2.\u2002 (c)\t 3.\u2002 (a) \t 4.\u2002 (a)\t 5.\u2002 (c)\t 6.\u2002 (c)\t 7.\u2002 (a)\t 8.\u2002 (c)\t 9.\u2002 (b) \t 10.\u2002 (c) \t11.\u2002 (b)\t 12.\u2002 (c)\t 13.\u2002 (b)\t 14.\u2002 (a)\t 15.\u2002 (d)\t 16.\u2002 (a)\t 17.\u2002 (b)\t 18.\u2002 (a)\t 19.\u2002 (d)\t 20.\u2002 (d) \t21.\u2002 (d)\t 22.\u2002 (b)\t 23.\u2002 (a)\t 24.\u2002 (a)\t 25.\u2002 (d)\t 26.\u2002 (b)\t 27.\u2002 (a)\t 28.\u2002 (b)\t 29.\u2002 (c)\t 30.\u2002 (d) Level 2 \t31.\u2002 (a)\t 32.\u2002 (b)\t 33.\u2002 (b)\t 34.\u2002 (a)\t 35.\u2002 (c)\t 36.\u2002 (a)\t 37.\u2002 (b)\t 38.\u2002 (d)\t 39.\u2002 (a)\t 40.\u2002 (a) \t41.\u2002 (c)\t 42.\u2002 (b)\t 43.\u2002 (c)\t 44.\u2002 (c)\t 45.\u2002 (c)\t 46.\u2002 (b)\t 47.\u2002 (a)\t 48.\u2002 (d)\t 49.\u2002 (c)\t 50.\u2002 (d) \t51.\u2002 (d)\t 52.\u2002 (a)\t 53.\u2002 (b)\t 54.\u2002 (a) Level 3 57.\u2002 (d)\t 58.\u2002 (c)\t 59.\u2002 (c)\t 60.\u2002 (c)\t 61.\u2002 (b) 55.\u2002 (b)\t 56.\u2002 (b)\t ANSWER KEYS","Geometry 12.53 CONCEPT APPLICATION Level 1 \t1.\t Find AB using Pythagoras\u2019s theorem and CM = 1 \t15.\t Use the concept of similar triangles. 2 AB. 1\t 6.\t Angle subtended by a minor arc is always less than 90\u00b0. \t2.\t Find the number of sides of the polygon, angle 1\t 7.\t Number of diagonals of a polygon with n sides subtended at the centre 360\u00b0 . = n(n \u2212 3) . n 2 = \t3.\t In a rhombus, diagonals bisect at 90\u00b0. \t18.\t AC2 = AB2 + BC2.\tConsider AB as x, BC as x + k and AC as x + 2k. \t4.\t In a cyclic quadrilateral, opposite angles are supplementary. \t19.\t \u0394MOR and \u0394NOR are isosceles triangles. \t20.\t (i)\tProduce AF and AG to meet CD. \t5.\t In \u0394ABC, AA=DB AA=CE DE . BC \t\t(ii)\tExterior angle of a triangle is equal to the sum \t6.\t (i) Use BPT. of its interior opposite angles. \t\t(ii) By BPT, PM = PN . \t\t(iii)\tSum of the angles in a triangle is always 180\u00b0. PQ PR \t21.\t Draw a parallel line to AB through E. \t7.\t \u2220CDE = \u2220BAE ( angles in the same segment). \t22.\t Use the concept of similar triangles. \t8.\t Equal chords subtend equal angles on the circum- 2\t 3.\t Join BD, ABDE is a cyclic quadrilateral. ference of the circle. \t9.\t The angle bisectors of two adjacent angles in a \t24.\t Find the length of the diagonal. parallelogram form a right angle at their point of intersection. \t25.\t \u2220BAC = 90\u00b0 and BC is a diameter. \t10.\t Recall complement and supplement angles. \t26.\t Areas of the parallelograms lying on the same base and between same parallel lines are equal. \t11.\t Angle subtended by an arc at the centre is double the angle subtended by it on the remaining part of 2\t 7.\t Use the angle sum property of a triangle. the circle. \t28.\t Apply the concept of inequality property of a triangle. 1\t 2.\t RS = 1 PQ. 2 ANSWER KEYS 1\t 3.\t In a cyclic quadrilateral, opposite angles are \t29.\t Sum of the exterior and interior angles of a poly- gon = 180\u00b0. supplementary. 1\t 4.\t Area of trapezium, FEGI = 1 (FI + EG ) \u00d7 FD. 3\t 0.\t Find one exterior angle. 2 Level 2 \t31.\t \u2002 (i) A\u0007 s BC : CD = 2 : 3, Area (\u0394ABC) : Area(\u0394ACD) \t32.\t \u2002 (i) As PB = PC, \u2220PBO = \u2220PCO. = 2 : 3. \t\t(ii) As \u2220BOC = 130\u00b0, \u2220BAC = \u2220BDC = 65\u00b0. 3\t 3.\t \u2002 (i) A\u0007 s the greatest interior angle is 110\u00b0, the least \t\t\t Let area of \u0394ABC = 2x and (\u0394ACD) = 3x. exterior angle is (180\u00b0 \u2013 110\u00b0). \t\t(ii) \u0007As AE : EC = 3 : 4, Area (\u0394AEB) = 3 (2x) and \t\t(ii) Sum of all the exterior angles = 360\u00b0. 7 \t\t(iii) All exterior angles are distinct. 4 Area (\u0394ECD) = 7 (3x).","12.54 Chapter 12 \t34.\t \u2002 (i) J\u0007 oin OC and find the relation between OB \t\t(ii) \u2220QPS = 1 (\u2220QOS). and BC. 2 \t\t(ii) OB = BC and OC :OB = 2 : 1. \t\t(iii) PQRS is a cyclic quadrilateral. \t35.\t \u2002 (i) Find the diameter of the smaller circle. 4\t 4.\t \u2002 (i) Use Pythagoras\u2019s theorem. \t\t(ii) Join AB, AC and BC. \t\t(ii) Let AE = BE = x and BD = CD = y. \t\t(iii) \u2220ACB = 90\u00b0 and AC = BC. \t\t(iii) U\u0007 se Pythagoras\u2019s theorem in the right triangles ABD and BEC and obtain the equations in 3\t 6.\t \u2002 (i) J\u0007 oin O1, A, O2 and B and find the angles of terms of x and y. rhombus O1AO2B. \t\t(iv) A\u0007 dd the above equations and obtain AC, i.e., \t\t(ii) Join O1A, O2A, O1B and O2B. 4x2 + 4y2. \t\t(iii) O1A = O1B = O1O2. 4\t 5.\t ABCD is the required sequential order. \t\t\u2009(iv) \u2220O1AO2 = 60\u00b0. 4\t 6.\t ABDCE is the required sequential order. 1 1 4\t 7.\t BACD is the required sequential order. 2 2 \t37.\t \u2002 (i) \u2220PQR = (\u2220POR); \u2220NMS = (\u2220NOS). 4\t 8.\t CBDA is the required sequential order. \t \t(ii) If \u2220OPR = x, then \u2220ROP = 180\u00b0 \u2013 2x. \t49.\t Let \u2018O\u2019 be the intersection point of AC and BD 3\t 8.\t \u2002 (i) \u0007Join DC and use properties of cyclic \t\tBy AA similarity, quadrilateral \t\t\u0394AOB ~ \u0394DOC \t\t(ii) Use \u2220ADC = 90\u00b0 and find \u2220BDC. AB = OB \t\tCD OC \t\t(iii) BDCP is cyclic. \t\tBy AA similarity, Hints and Explanation \t39.\t \u2002 (i) Join OD, OE and OF. \t\t\u0394AOD ~ \u0394BOC \t\t(ii) E\u0007qual chords subtend equal angles at the AD = OD centre. \t\tABCD = OC OB 4\t 0.\t \u2002 (i) \u0007In a triangle median is same as altitude drawn BC OC on same side, then the remaining two sides are equal. \t\t(AB)(BC) = (CD)(AD) \t\t(ii) \u0007As AC is median as well as altitude to BD, AB \t\tAD = 12 cm. = AD. 5\t 0.\t Since B is the mid-point of SR and PS = BR, PS \t\t(iii) In \u0394ADE, AD + DE > AE. = BS = PA = AB. \t\t\u2009(iv) As AD is median, CD = DE. \t\t\\\\\u0007 PABS is a rhombus, and similarly, AQRB is a \t41.\t \u2002 (i) Join OA and use properties of arcs. rhombus. \t\t(ii) \u2220AOB = 2\u2220AEB. \t\t\\\\\u2220BCA = \u2220ADB = 90\u00b0. \t\t(iii) \u0007AOC is the diameter, i.e., \u2220AOB + \u2220BOC = \t\tAs PB and QB are bisectors of \u2220P and \u2220Q, \u2220PBQ 180\u00b0. = 90\u00b0. 4\t 2.\t \u2002 (i) J\u0007 oin OC, and apply the concept of equal \t\t\\\\CADB is a rectangle. chords. 5\t 1.\t From the inequality of triangles, we have: \t\t(ii) \u2220OBC = \u2220OCB. \t\t6002 + 2006 > m and \t\t(iii) As BC = CD, \u2220BOC = \u2220COD. \t\t6002 \u2013 2006 < m \t\t\u2009(iv) \u2220BAC = 1 \u2220BOC. \t\t\u21d2 3996 < m < 8008 2 \t\tSince m is an integer, it can take (8008 \u2013 3996 \u2013 1) \t43.\t \u2002 (i) As AB is the diameter, \u2220AOB = 90\u00b0. or 4011 values.","Geometry 12.55 \t52.\t B \t53.\t Since, SA = SB = SC, S is the circum-centre a \t\t\u2220ASB = 150\u00b0 (given) \u21d2 \u2220ACB = 75\u00b0. c \t\tSince, AC BC, \u2220CAB \u2220CBA \uf8eb 180\u00b0 \u2212 75\u00b0 \uf8f6 \uf8ec\uf8ed 2 \uf8f7\uf8f8 = = = Cb A = 52 1 \u00b0. 2 \t\t\u0007Given that, c = 2a and b2 \u2013 3a2 = 0. \u21d2 b = 3a. 5\t 4.\tGiven, \u2220QPR = 25\u00b0. \t=\ta : b : c a=: 3a : 2a 1 : 3 : 2. \t\tSince, QS is the diameter, \u2220QPS = 90\u00b0. \t\t\u2234 \u2220A = 30\u00b0, \u2220B = 60\u00b0 and \u2220C = 90\u00b0. \t\t\u21d2 \u2220SPR = 90 \u2013 25\u00b0 = 65\u00b0 \t\tSo, \u2220ABC = 60\u00b0. \t\t\\\\ \u2220SOR = 2 \u00d7 65\u00b0 = 130\u00b0. Level 3 \t55.\t \u2002 (i) D\u0007 raw the figure and mark the points P, Q and \t59.\t D B R on sides AB, BC and AC respectively. O \t\t(ii) A\u0007 ngle B = 90\u00b0 and \u2220PRQ is formed by lines AE PR and RQ, which are parallel to lines BC and AB (mid-point theorem). C \t\t(iii) \u0007Since BC and AB are perpendicular, PR \t\tFrom the given figure, Hints and Explanation and RQ are also perpendicular, i.e., \u2220PRQ = 90\u00b0. \t\tAE > AC \t\t\u2009 (iv) Q\u0007 uadrilateral joining BPRQ is a cyclic \t\t\u21d2 \u2220ADE > \u2220ABC quadrilateral. \t\t\u21d2 (\u2220ADE \u2013 \u2220ABC) = 15\u00b0 \t56.\t \u2002 (i) Draw the figure and mark the points on it. \t\t\u21d2 \u2220AOE \u2212 \u2220AOC = 15\u00b0 \t\t(ii) Use Pythagoras\u2019s theorem. 2 2 \t\t(iii) We will get p2 + q2 = 2pq which gives p = q. \u21d2 \u2220AOE \u2212 \u2220AOC = 30\u00b0 \t57.\t\u2002 (i) Draw the figure and mark the points on it. \t\t\u21d2 \u2220COE = 30\u00b0. \t\t(ii) \u0007Join BQ, now BQ is the median to PR. Since \u2234 \u2220COE = 1 \u2220COE = 15\u00b0. PQ = QR, BQ is also the altitude to PR. 2 \t\t(iii) \u0007PBQ is a right triangle, right angled at B.\t 6\t 0.\t E Thus, PQ is the diameter. D F \t\t(iv) \u0007Joining PA, we get another right triangle PAQ H 60\u00b0 C which is right angled at A.\t PBQ and PAQ G 90\u00b0 are congruent angles. A 70\u00b0 \t\t\u2002 (v) P\u0007 R = QR, so DPQR is an equilateral triangle. B \t58.\t \u2002 (i) J\u0007 oin AB. Now using Pythagoras\u2019s theorem, \t\t\u2220DAB = \u2220DCE = 60\u00b0 and \u2220EDC = \u2220CBA AB2 = AD2 + DB2 and AB2 = AC2 + CB2. = 70\u00b0. (Exterior angle of a cyclic quadrilateral.) \t\t(ii) If BD = AC, then AD = BC. \t\tIn DABE, \u2220DEC = 180 \u2013 (60\u00b0 + 70\u00b0) = 50\u00b0. \t\t(iii) I\u0007 f AD||BD, then AC ||BD, so ABCD is a rectangle.","12.56 Chapter 12 \t\tSince, FG bisects \u2220F \t\tJoin CD and EB \t\t\u2220ACD = \u2220AED = 50\u00b0 (angles in the same seg- \t\t\u2234 \u2220HEC = \u2220DEC = 50\u00b0 = 25\u00b0. 2 2 ment and \u2220AED = 50\u00b0) \t\t\u2220ACB = 60\u00b0 (given) \t61.\t C E \t\t\u21d2 \u2220DCB = \u2220ACB \u2013\u2220ACD = 60\u00b0 \u2013 50\u00b0 = 10\u00b0. \t\t\u2220DEB = \u2220DCB = 10\u00b0. (angles in the same B AD segment) Hints and Explanation","1132CChhaapptteerr MKiennesmuartaictsion REMEMBER Before beginning this chapter, you should be able to: \u2022 Understand the concepts of plane figures \u2022 Calculate the areas of plane figures \u2022 Obtain volume of simple solid figures KEY IDEAS After completing this chapter, you should be able to: \u2022 Know about polygons and calculate the area of different types of polygons \u2022 Learn about circles, circumference of a circle, sector of a circle and area of a ring \u2022 Understand different solids and find their volumes \u2022 Find lateral surface areas, total surface areas and volumes of prism, cube, cuboid, cylinder, pyramid, cone, torus, sphere and polyhedrons (tetrahedron and octahedron) Figure 1.1"]