202110722-PERFORM-STUDENT-WORKBOOK-MATHEMATICS-G09-FY_Optimized

PRACTICE SHEET - 5 (PS-5) III. Long answer questions. 1. (i) The perimeter of a parallelogram is 30 cm. If the longer side measures 2 cm, then measure of the shorter side is equal to (ii) In a square ABCD, AB = (3x - 9) cm and BC = (8x + 11) cm. Then, what is the value of x? 2. In the given figure, the side AC of ∆ABC is produced to E such that CE = 1 AC. If D is the midpoint of BC 2 and ED produced meets AB at F and CP, DQ are drawn parallel to BA, then find FD. 87

Self-Evaluation Sheet Marks: 15 Time: 30 Mins 1. Four non parallel lines p, q, r and s intersect as 3. Check if the following statements are true or shown in the figure. false? i) A quadrilateral can have 3 obtuse angles. Find the sum of the angles ii) A quadrilateral can have 2 right angles and i) ∠a, ∠b, ∠c and ∠d ii) ∠1, ∠2, ∠3 and ∠4  two acute angles. (2 Marks) (1 Mark) 4. PQRS is a quadrilateral as shown, find the missing angles of the quadrilateral if QA and BS intersect at point R. (2 Marks) 2. P, Q are midpoints of AB and BC respectively. What is the nature of the quadrilateral ABPQ? What is the nature of quadrilateral ABQP?  (1 Mark) 88

Self-Evaluation Sheet Marks: 15 Time: 30 Mins 5. Show that diagonals of a square divide it into 4 6. ABCD is a rhombus whose diagonals intersect congruent triangles. (3 Marks) at O. If P, Q, R and S are midpoints of OA, OB, OC and OD, then show that PQRS is a rhombus.  (3 Marks) 89

Self-Evaluation Sheet Marks: 15 Time: 30 Mins 7. E,F,G and H are the midpoints of the sides of the rectangle ABCD as shown. Show that i) AE = PQ  ii) PQ = 2PH = 2QF. (3 Marks) 90

9. Areas of Parallelograms and Triangles Learning Outcome By the end of this lesson, a student will be able to: • Determine relationship between lengths and areas • Determine relationship between lengths and areas of triangles with same base and between same of parallelograms with same base and between parallels. same parallels. Concept Map Key Points In the figure below, XZWU is made of two non overlapping regions, UXYV and VYZW, hence • The part of a plane enclosed by a simple closed figure is called a planar region corresponding to ar=( XZWU) ar (UXYV ) + ar (VYZW ) that figure. The magnitude or measure of this planar region is called its area. • Area of a parallelogram is the product of any of its side and the corresponding altitude. • Two figures are called congruent if they have the same shape and same size. • Two figures are said to be on the same base and between same parallels, if they have a common • If two figures are congruent, then they must have base (side) and the vertices (or the vertex) opposite equal areas. If two figures have equal areas, then to the common base of each figure lie on a line they need not be congruent. parallel to the base. • Area of a figure is a number associated with the Example: part of the plane enclosed by the figure with two properties: o If A and B are two congruent figures, then ar ( A) = ar (B) o If a planar region formed by a figure P, is made up of two non-overlapping planar regions formed by figures C, D, then a=r (P) ar (C) + ar (D) Example: If the figure, ∆XYZ and ∆UVW are congruent figures, then ar ( XYZ) = ar (UVW ) 91

9. Areas of Parallelograms and Triangles D SC R D SC D CS R P Q R R S AB AB APB A CB i) ABCD, ABRS have iv) ABQ, ABR have same base and same base AB and are ii) ABCD, ABRS iii) ABCD, APRS arebetween same parallels between same ABS, ABR have same base and not parallels have same base AB donot have same base between same parallels ACP, ABS donot have same base and and not between same and are between same arenot between same parallels parallels parallels • Theorem 9.1 In the figure AF and DG are parallel lines and AB = EF Parallelograms on the same base and between the a=r ( ABD) a=r ( ABC) ar (EFG) same parallels are equal in area. The theorem can also be stated as, parallelograms on the same base or equal bases and between the same parallels are equal in area. The converse of the theorem, Parallelograms on the same base (or equal bases) and having equal areas lie between the same parallels. Example In the figure PU || AD and AB = CD , ABSQ, ABRP,CDUT are all parallelograms. a=r ( ABSQ) a=r ( ABRP) ar (CDUT ) • Theorem 9.2 Two triangles on the same base (or equal bases) and between the same parallels are equal in area. • Theorem 9.3 Two triangles having the same base (or equal bases) and equal areas lie between the same parallels. The theorem can also be stated as, two triangles with same base (or equal bases) and equal areas will have equal corresponding altitudes. Example 92

9. Areas of Parallelograms and Triangles Work plan CONCEPT COVERAGE COVERAGE DETAILS PRACTICE SHEET Pre-requisites • Area of Congruent figures PS – 1 • Figures on Same Base and Between Same parallels PS – 2 PS - 3 Areas of • Parallelograms on the same Base and between the same Self Evaluation Sheet Parallelograms and Parallels Triangles • Triangles on the same base and between same Parallels Worksheet for “Areas of Parellelograms and Triangles” Evaluation with Self ---- Check or Peer Check* 93

PRACTICE SHEET - 1 (PS-1) 1. In the rectangle ABCD as shown, ar (BCD) = 22 sq. 5. In the given figure ar (ABO) = 910 cm2, ar (BCO) = 525 cm2, ar(COD) cm2 and ar(AOD) = 2 × ar(BCO). units. Find the value of the following: Determine the area of the following regions: i) ar (ABD) (i) ar(ABCD) (ii) ar(ACD) ii) ar (ABCD) 2. In the given figure ar(ABC) = 30 sq units and ar(ABD) = 7sq units. Find ar(BDC). 3. Determine the areas of the following parallelogram ABCD. 4. AD, BE and CF are the altitudes of the triangles over the sides BC, CA and AB of DABC. If AB = 8 cm, BC = 7 cm and AC = 9 cm and BE=6 cm, determine the lengths of AD and CF. 94

PRACTICE SHEET - 2 (PS-2) 1. In each of the four given figures, determine the geometries that lie on same base and between same parallels. Mention, the common base and the parallels. 6. In the given figure, ABCD and ABED are two parallelograms of equal areas. Determine if point E,D,C are collinear. 2. Looking at the given figure, justify whether the statement Ar(∆DCP)=1/2Ar(ABCD), is true. Ar (∆DCP ) =12 Ar (ABCD) . 7. Prove that two triangles on the base and between the same parallels are equal in area. 3. Point E is the midpoint of side AD of the 8. A foot over bridge in a railway station is made parallelogram ABCD as shown in the figure. If F is any point on the side BC of the parallelogram, using 5 horizontal steel bars of 3.5 m lengths joined by 6 inclined bars as shown. The height of ar (DEF ) the bridge is 4 m. Prove the following: then find ar (ABCD) . i) ar (ABP) = ar (PQC) = ar (QDE) ii) ar (ACP) = ar(BCQP) iii) ar (ACQP) = ar (BDQP) 4 iv) aarr ((AADDQP)) == 5 aarr((AAEEQQPP) 4. If P, Q, R and S are the midpoints of the sides of parallelogram ABCD, then show that Ar(PQRS) = 1 2 × Ar(ABCD) 5. In the give figure, AB || CD, AD || CE and DF || BC and the perpendicular distance between them is 5 units. The lengths of the sides AB = 10 units and CD = 4 units. If ar (ABCD) = 35 sq units, determine i) ar (AFD) ii) ar (BEC) 95

PRACTICE SHEET - 2 (PS-2) 9. In parallelogram ABCD, P, Q are the midpoints of sides BC and CD as shown. Determine the following: i) ar (AQD) ii) ar (ABPQ) ar (ABCD) ar (ABCD) 10. In the given figure, ar (AED) = ar (BEC) and E is point of intersection of AC and BD. Prove that AB || CD. 11. In the given figure C is a point on side BD of the quadrilateral ABDE such that AE = BC. Prove that ar (ABD) = ar (ACDE) 96

PRACTICE SHEET - 3 (PS-3) I. Choose the correct option. 1. Parallelograms on the same base and between the same parallels are equal in: (A) perimeter (B) volume (C) weight (D) area 2. The area of a rhombus is 20 cm2. If one of its diagonals is 5 cm, the other diagonal is: (A) 10 cm (B) 5 cm (C) 8 cm (D) 7 cm 3. The area of a trapezium whose parallel sides are 9 cm and 16 cm and the distance between these sides is 8cm, is: (A) 60 cm2 (B) 100 cm2 (C) 56 cm2 (D) 120 cm2 4. Which of the following figures lies on the same base and between the same parallels? (A) (B) (C) (D) All of these 5. In figure, ABCD is a trapezium in which AB||DC. Find the length of DC. (A) 15 cm (B) 12 cm (C) 13 cm (D) 11 cm 6. If PS is median of the triangle PQR, then ar(∆PQS) : ar(∆QRP) is: (A) 1 :1 (B) 1 : 2 (C) 2 : 1 (D) Can’t be determined 7. Assertion: In the given figure, the point D divides the sides BC of ∆ABC in the ratio m : n. Then ar(∆ABD) = m ar(∆ADC) n Reason: Area of triangle = 1 × Base × Height 2 (A) If both assertion and reason are true and reason is the correct explanation of assertion. (B) If both assertion and reason are true but reason is not the correct explanation of the assertion. (C) If assertion is true but reason is false. (D) If assertion is false but reason is true. 97

PRACTICE SHEET - 3 (PS-3) 8. A canal, 30 m long has a depth of water of 80 cm at one end and 2.4 m at the other end. Find the area of the vertical cross-section of the pool along the length. (A) 54 cm2 (B) 42 cm2 (C) 36 cm2 (D) 48 cm2 9. ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. If area of ∆DFA is 3 cm2, then find the area of parallelogram ABCD. (A) 9 cm2 (B) 12 cm2 (C) 18 cm2 (D) 6 cm2 10. In ∆ABC, if L and M are points on AB and AC respectively such that LM || BC. Then match equal areas in both lists. List – I List - II (P) ar(∆LCM) (1) ar(∆ACL) (Q) ar(∆LBC) (2) ar(∆LBM) (R) ar(∆ABM) (3) ar(∆MOC) (S) ar(∆LOB) (4) ar(∆MBC) Code: P Q R S (A) 4 2 3 1 (B) 3 2 4 1 (C) 2 4 1 3 (D) 1 2 3 4 II. Short answer questions. 1. The area of trapezium PQRS is k sq.cm. Then value of k is: 20 2. ABCD is a trapezium in which AB||DC and DC = 40 cm and AB = 60 cm. If x and y are respectively, the midpoints of AD and BC, then find xy. 3. In the given figure, ABCD is a trapezium in which AB||DC. DC is produced to E such that CE = AB, prove that ar(∆ABD) = ar(∆BCE). 98

PRACTICE SHEET - 3 (PS-3) III. Long answer questions. 1. In parallelogram ABCD, AB = 10 cm. The altitudes corresponding to the sides AB and AD are respectively 7 cm and 8 cm, If AD is k cm. Then value of k – 0.75 is: 2. In ∆ABC, AD is the median and p is a point on AD such that AP : PD = 1 : 2, then find the area of ∆ABP. 99

Self-Evaluation Sheet Marks: 15 Time: 30 Mins 1. In the figure given, AD and BC are parallel lines. 3) ACDE is a trapezium with sides AC || DE. B is a point Determine on side AC such that BC = DE. Prove the following: i) Pairs of triangles having same base and lie i) ar (AGE) = ar (CGD) between same parallels ii) ar (BFD) = ar (AFE) iii) ar (EFG) = ar (DHC) – ar (BFGH) ii) Pairs of triangles having same base and donot lie between same parallels (4 Marks) iii) Pairs of triangles that lie between same parallels and not having same base. (3 Marks) 2) In the given figure, ABCD and BEFG are parallelograms between the parallel lines DF and AE. If ar (ABCD) = ar (BEFG) = ar (BCG) the determine the relation between the sides DC, CG, GF, AB and DE. (2 Marks) 100

Self-Evaluation Sheet Marks: 15 Time: 30 Mins 4) A farmer divides his rectangular plot OACE as shown in the figure such that B, D are the midpoints of the sides AC and CE respectively. Prove that the four triangular portions are equal in area. (3 Marks) 5) In the given figure, points D, E are points on the line CF such that EF = DC and O is the point of intersection of BE and AD. If ar (OAFE) = ar (ODCB) prove that AB || FC. (3 Marks) 101

10. Circles Learning Outcome By the end of this lesson, a student will be able to: • Determine the relation between the lengths of • Determine the relationship between the angles and chords and a line drawn from the centre of a circle chord lengths of circles and geometrical shapes to the chord. drawn using points on circles. • Determine the relation between the angles subtended by the arc of a circle. • Determine the angles of a cyclic quadrilateral. Concept Map Key Points • A line joining two points on the circle is called chord. A chord which passes through the center of • The collection of all the points in a plane, which are the circle is called diameter of the circle. Diameter at a fixed distance from a fixed point in the plane, is is the longest chord and all diameters have the called a circle. The fixed point is called the centre of same length which is equal to two times the radius. the circle and the fixed distance is called the radius. The line segment joining the center and any point is also called radius of the circle. • The piece of circle between two points is called an Arc. • A circle divides a plane on which it lies into three 102 parts: o Inside the circle, called interior of the circle o Outside the circle, called exterior of the circle o The circle. The circle and its interior make up the circular region. The length of the complete circle is called its circumference.

10. Circles CHAPTER: Quadratic Equations • When the two arcs are equal, then each is known as semicircle, then both segments and both sectors When two points or a chord divide a circle into become the same and each is known as semicircular two parts, the longer arc is called major arc and region. the shorter one is called minor arc. Major arc is represented using the three points, i.e., two end • PQ is a chord of a circle with center at O. points and an intermediate point. Minor arc is ∠PPORQQ is the angle subtended by the line segment represented using the two end points of the arc. PQ at the point R, on major arc. If the end points of the arc form the diameter, then ∠PPOOQQ is the angle subtended by the chord at both arcs are equal and each is called a semi circle. centre O. The region between a chord and either of its arcs ∠PPOSQQ is the angle subtended by the chord at S, on is called segment of circular region or simply a segment of the circle. The region between chord minor arc. and minor arc is called minor segment and the region between major arc and the chord is called • Theorem 10.1 major segment Equal chords of a circle subtend equal angles at the Example: P, Q are any two points on the circle centre. dividing it into two arcs. R is a point on the circle as Example: ⇒ AB|| FC shown. Theorem 10.2 If the angles subtended by the chords of a circle at • The region between arc and two radii joining the center to the end points of the arc is called a sector. the centre are equal, then the chords are equal. The region between the major arc and two radii is Example: ∠AOB=∠COD ⇒ AB = CD called major sector and is represented using three points (Two end points and one point in between). • Theorem 10.3 The region between minor arc and two radii joining The perpendicular from the center of a circle to a the end points is called minor sector. chord bisects the chord. Example: P, Q are any two points on the circle Example: BC is chord and AD is a line from center dividing it into two arcs. R is a point on the circle as AD ⏊ BC BD = DC shown. Let O be the center of the circle. Theorem 10.4 The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord. Example: BC is chord and AD is a line from center A. BC = BD ⇒ AD ⏊ BC • Theorem 10.5 There is one and only one circle passing through three given non-collinear points. Example: Let A, B, C are three non collinear points 103

10. Circles Congruent arcs (equal arcs) of a circle subtend equal angles at the centre. as shown in fig (i) , then only one circle can be drawn through them. Example: AB and CD are two equal arcs. If A, B, C are collinear points as shown in fig (ii), then • Theorem 10.8 a circle (of finite radius) cannot be drawn through The angle subtended by an arc at the centre is three points. But through any two points infinite number of circles can be drawn. double the angle subtended by it any point on the remaining part of the circle. • To locate the centre of a given arc or a given circle Example: ∠BOC = 2 x ∠BAC the following steps can be followed: • Theorem 10.9 Angles in the same segment of a circle are equal. o Step 1: Draw two chords at angle to one another Example: ∠APB =∠AQB for the given arc or circle. • Theorem 10.10 If a line segment joining two points subtends equal o Step 2: Draw perpendicular bisectors to the angles at two other points lying on the same side chords drawn. of the line containing the line segment, the four points lie on a circle. (i.e., they are concyclic.) o Step 3: The point of intersection perpendicular Example: ∠PRQ = ∠PSQ⇒P,Q,R,S are concyclic bisectors is the center of the arc or circle. Or ∠PQS = ∠PRS⇒P,Q,R,S are concyclic This procedure can also be used to draw a circle through three non collinear points. Draw line segments through pair of points. Draw perpendicular bisectors to these line segments and te point of intersection will be the center of the circle. With the center and distance to the point as radius draw a circle and it will pass through the given points. • A circle passing through the vertices of a polygon is called circumcircle. The center of circumcircle is circumcenter and its radius is called circumradius. • The length of the perpendicular from a point to a line is the distance of the line from the point. If a point lies on the line the distance of the line from the point is zero. • Theorem 10.6 Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centers). Example: AB= CD OP=OQ Theorem 10.7 • Theorem 10.11 Chords equidistant from the centre of a circle are The sum of either pair of opposite angles of a cyclic equal in length. quadrilateral is 180° . Example: OP= OQ ⇒AB=CD Example: If ABCD is a cyclic quadrilateral, then • If two chords of a circle are equal, then their ∠ABC + ∠AD=C 180° and ∠DAB + ∠DC=B 180° Theorem 10.12 corresponding arcs are congruent. If the sum of a pair of opposite angles of a If two arcs are congruent, then their corresponding quadrilateral is 180o, the quadrilateral is cyclic. chords are equal. 104

10. Circles CHAPTER: Quadratic Equations Example: ADC = 180o then ABCD is cyclic quadrilateral. If ABC + DCB = 180o then ABCD is cyclic quadrilateral. If DAB + Work plan CONCEPT COVERAGE COVERAGE DETAILS PRACTICE SHEET PS – 1 Pre-requisites • Circles and its related terms: A Review PS – 2 Pair of Linear Equations in • Angle Subtended by a Chord at a Point PS – 3 Two Variables • Perpendicular from the Center to a Chord PS – 4 • Circle through Three Points PS – 5 • Equal Chords and their Distances from the Centre PS – 6 • Angle Subtended by an Arc of a Circle Self Evaluation Sheet • Cyclic Quadrilaterals Worksheet for “Circles” Evaluation with Self ---- Check or Peer Check* 105

PRACTICE SHEET - 1 (PS-1) 1. A circle of 4 cm radius is drawn with point O as the center and P is a point on the circle. i) Whether the line OP is radius of circle? ii) What is the value of the diameter of circle? iii) What is the largest size of chord that can be drawn in the circle? 2. Three points P, Q , R line in a plane. A circle is drawn with center O and radius 4 cm such that OP = 5 cm, OQ = 4 cm and OR = 3 cm. Determine the location of the points with respect to the circle. 3. State whether true or false. Explain your answers. i) In a semicircle, the length of major arc is more than that of minor arc. ii) If major sector is same as minor sector, then the two radii forming the sector will be along a straight line. iii) For any point on the circle, a point inside the circle is always closer than the point outside the circle. 106

PRACTICE SHEET - 2 (PS-2) 1. A circle with center O has two equal chords PQ and RS. If ∠OQP = 52°, then find the all remaining angles of DOPQ and DORS. Are the two triangles congruent? 2. A circle is drawn through the vertices of a regular pentagon. Find the values of the angles of DAOB and DBOC. 3. Raju wanted to cut 3 pieces from a large pizza and wanted to share it with his friends. He takes a measure using his fingers and marks four points A, B, C and D which are equidistant on the circumference of the pizza and then cuts along OA, OB, OC and OD as shown. Are the three slices equal? 4. In the given figure, AC = BC and ∠OBC = 20°, determine the acute ∠AOB. 5. Four points are marked on the circle such that AD = BC as shown in the figure. Determine the angles of the quadrilateral ABCD. 6. In the given figure below, if O is the center of circle and also the point of intersection of the lines AC and BD, determine the relation between AB and CD. 107

PRACTICE SHEET - 3 (PS-3) 1. Find the intersection of the perpendicular bisectors of the chords of the circle. 2. Draw a sketch to show that a circle cannot always be drawn though four given points. 3. In the given figure, O is the center of the circle and A, B, C are three points on the circle. If BD = DC then prove that ∠AOB = ∠AOC. 4. Two circles of different diameters and centers at P and Q intersect at A, B as shown in the figure. If C is the midpoint of AB, then prove that P, C, Q are collinear points. 5. The length of the 3 sides of a triangle are 7 cm, 5 cm and 6 cm. Draw a circle passing through the three vertices of a triangle. 6. Rahul brought a cake for his birthday party. He wanted to place the candle exactly at its centre. Explain a method to locate the centre of the cake with illustration. 7. In the given figure OP = OQ , then prove that DABC is an isosceles triangle and PQ || BC. 8. In a circle of 6 cm diameter, if the distance from the center to a chord is 2 cm, determine the length of the chord. (Use ar ( ABCD) ) 108

PRACTICE SHEET - 4 (PS-4) 1. BD is a diameter of the circle with center O. AB, BC, CD and AD are the four chords of the circle as shown. Determine all the angles of DOAB, DOBC, DOCD and DOAD 2. CD is the diameter of the circle as shown in the figure. O is the point of intersection of the chords AC and BD. Determine the following angles. i) ∠DOA ii) ∠DOC iii) ∠ACB 3. Two chords AB and CD of a circle intersect perpendicularly at point O as shown in the figure such that AD = BC. Prove that AB = CD. 4. In the given figure BQ = BR. If PQRS is a straight line, then find the three angles of DAQR and DBQR. 5. Check whether the points A, B, C and D lie on a circle. 109

PRACTICE SHEET - 5 (PS-5) 1. Check if the angles given in the quadrilateral ABCD are correct if the points A, B, C and D lie on the circumference of a circle. 2. Determine the missing angles of the quadrilateral ABCD. 3. Two circles of equal radius intersect at points B, D as shown in the figure. A, C, E, F are points on the circles as shown. Determine the following angles i) ∠BAD ii) ∠BCD iii) ∠BFD iv) ∠BED 4. In the figure, ABCD is cyclic quadrilateral with center O, find the following: i) ∠DCB ii) angles in DODB 5. A pentagonal shape is to be carved out of a circular block of wood as shown. If AC is the line segment representing the diameter of the circular block, determine the corner angles of the pentagon. 110

PRACTICE SHEET - 6 (PS-6) I. Choose the correct option. 1. Equal chords of a circle subtend equal angles at (A) Circumference (B) Centre (C) Both (a) and (b) (D) None of these 2. There is one and only one circle passing through three given ______ points. (A) perpendicular to chord (B) parallel to the chord (C) tangent to the chord (D) equal to the chord 3. If the length of an arc is less than the length of the arc of the semicircle then it is called _______. (A) a minor segment (B) a major arc (C) a major segment (D) a minor arc 4. Which of the following statement is true for a regular polygon? (A) All vertices are concyclic (B) only four vertices are concyclic (C) All vertices are not concyclic (D) Cannot say anything about regular polygon 5. In the given figure, AB is a diameter with centre O and OP || BQ. If ∠ABQ = 40°, then the value of x is. (A) 50° (B) 70° (C) 80° (D) None of these 6. In the given figure, AEDF is a cyclic quadrilateral. The values of x and y respectively are (A) 79°, 37° (B) 89°, 47° (C) 89°, 37° (D) 79°, 37° 7. In the given figure, MNQS is a cyclic quadrilateral in which ∠QNR = 61° and x : y is 2 : 1. The values of x and y respectively are (A) 19 1 o , 38 1 o (B) 38 2 o ,19 1o (C) 1o , 33 2o (D) 18 1 o , 37 4 o 44 33 21 44 33 8. A crescent is formed of two circular arcs ACB, ADB of equal radius, centers E and F in the given figure. The perpendicular bisector of AB cuts the crescent at C and D, where CD = 12 cm, AB = 16 cm. The radius of arc ACB is (A) 18 cm (B) 16 cm (C) 10 cm (D) 12 cm 111

PRACTICE SHEET - 6 (PS-6) 9. When two chords of a circle bisect each other, then which of the following statement is true? (A) both chords are perpendicular to each other. (B) both chords are parallel to each other. (C) both chords are unequal. (D) both are diameters of the circle. 10. Assertion: In a isosceles triangle ABC with AB = AC, a circle is passing through B and C intersects the sides AB and AC at D and E respectively. Then DE||BC. Reason: Exterior angle of a cyclic quadrilateral is equal to interior opposite angle of that quadrilateral. (A) If both assertion and reason are true and reason is the correct explanation of assertion. (B) If both assertion and reason are true but reason is not the correct explanation of the assertion. (C) If assertion is true but reason is false. (D) If assertion is false but reason is true. II. Short answer questions. 1. In given figure, m ∠PQB where O is the centre of the circle. 2. Two chords AB and CD of a circle are parallel and a line l is the perpendicular bisector of AB. Show that l bisects CD. 3. In the given figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠BAC = 45°, find ∠BCD. III. Long answer questions. 1. (a) What is an arc of circle called if the ends of the arc are the ends of diameter? (b) What is longest chord of circle known as? (c) What is the sum of either pair of the opposite angles of a cyclic quadrilateral. (d) How many circles can pass through three non-collinear points? 2. AB and CD are two chords of a circle such that AB = 6 cm, CD = 12 cm and AB||CD. If the distance between AB and CD is 3 cm, find the radius of the circle. 112

Self-Evaluation Sheet Marks: 15 Time: 30 Mins 1. Determine the angle subtended by a chord at a 3. Prove that the line drawn perpendicular to a centre of a circle whose length is equal to the chord though the centre of a circle bisects the radius of the circle (1 Mark) angle subtended by the chord at the center. (2 Marks) 2. Answer the following. Give reasons. i) Can two circles of different radius can be drawn passing through three given points. ii) A, B, C are three collinear points each 2cm apart from one another on a line. What is the distance of point B from the line AC. 4. A farmer wanted to dig a well at the centre (2 Marks) of his circular field of 100 m diameter. Find the centre of the field. (4 Marks) 113

Self-Evaluation Sheet Marks: 15 Time: 30 Mins 5. AB and CD are diameters, O is the centre of the 6. In the quadrilateral ABCD as shown in the circle and P is any point as shown between A,C. figure AB || CD, determine the angles of the If ∠APC = x and ∠BOC = y then prove that quadrilateral. y = 2x – 180°. (3 Marks) (3 Marks) 114

11. Constructions Learning Outcome By the end of this chapter, a student will be able to: • Draw triangles using the given angles and • Draw perpendicular bisectors of given line relationship between the lengths of its sides. segments. • Construct 30°, 60°, 90°, 45°, etc angles using compass and straight edge. Concept Map Perpendicular Marking angles bisector using Protractor Constructions Basic Constructions Construction of Triangles Bisector of Angle Using Base, Base angle and Perpendicular bisector Sum of other two sides of line segment Using Base, base angle and Constructing given Difference of other two sides angle Using perimeter and Two base angles Key Points radius such that they intersect OA and OB at C, D respectively. (OC = OD) • A geometry box will have the following tools ■ Step 3: Draw arcs from points C and D such that ■ A graduated scale on one side of which the radius is more than half of CD. centimeters and millimeters are marked off B and on the other side inches and their parts are marked off. D ■ A pair of set-squares one with angles 90°, 60°, E 30° and other with angles 90°, 45° , 45°. ■ A pair of dividers (or a divider) with adjustments O CA ■ A pair of compasses (or a compass) with The two arcs intersect at point E. (CE = DE) provision of fitting a pencil at one end. ■ Step 4: Join the points O and E. OE is the angular ■ A protractor • Geometrical construction is the process of bisector of ∠AOB using only two instruments – an ungraduated ⇒ ∠AOE =∠EOB = ∠AOB ruler also called straight edge and a compass. 2 In constructions where measurements are also required, a graduated scale and protractor are • Procedure to construct the perpendicular used. bisector of a line segment • Procedure to construct the bisector of a given angle 115 ■ Step 1: ∠AOB is the angle whose angular bisector is to be drawn. The angle is formed by the lines OA and OB. ■ Step 2: Using the compass, draw arcs of any

11. Constructions P • A triangle is unique if ■ Two sides and the included angle are given. A OB ■ Three sides are given ■ Two angles and the included side are given Q ■ In right triangle, hypotenuse and one side are ■ Step 1: AB is the line segment for which the given. • Procedure to construct a triangle, given its base, a perpendicular bisector has to drawn. base angle and sum of other two sides ■ Step 2: From the points A, B draw arcs of radius In DABC, base AB and base angle ∠A are given The sum of two other sides i.e.,AC + BC = L is given. more than half of AB on both sides of the line P segment. Method 1: ■ Step 3: The arcs intersect at points P, Q. ■ Step 4: Join PQ. Q ■ Step 5: PQ is the perpendicular bisector of the line AB. C • Procedure to construct an angle of 60° at a point on a ray AB ■ Step 1: Draw a ray with A as the initial point. ■ Step 1: Draw line AB of given length. ■ Step 2: From point A draw an arc of some radius ■ Step 2: Draw a line AP so that ∠BAP = ∠A and intersecting the line AB at C and extend this arc. AP = L ■ Step 3: From point C draw an of radius equal to ■ Step 3: Join PB. the arc already drawn, such that the two arcs ■ Step 4: From B draw a line BQ such that intersect at point D. ■ Step 4: Join AD. AD is a line at angle of 60° to AB. ∠BPA = ∠PBQ The three sides of DABC are formed by equal radius ■ Step 5: Line BQ intersect PA at C. and thus the three angles are equal. ■ Step 6: DABC is the required triangle. Method 2: D A CB • 90°and 60° can be constructed using the methods ■ Step 1: Draw line AB of given length. discussed above. Any angle which is a fraction of ■ Step 2: Draw a line AP so that ∠BAP = ∠A and AP these angles can be constructed by using principle of angular bisectors. =L ■ Step 3: Join PB. Example: ■ Step 4: Draw the perpendicular bisector of PB i) 30° =60° (Angular bisector of 60°) 2 and extend it to intersect PA at C. ■ Step 5: Join BC DABC is the required triangle. ii) 7.5=° 6=0 3=0 15 8 4 8 (Bisecting the angle 60° to • Procedure to construct a triangle gives its base, get 30°, bisecting 30°to get 15°, bisecting 15° to get 7.5°) a base angle and the difference of the other two sides In DABC, AB is the base, ∠A is base angle. The difference of two other sides AC – BC or BC – AC iii) 120°= 90° + 60° (Adding 90° to bisected angle is given. 2 Case 1: AC – BC is given, i.e., AC – BC of 60°) 116

11. Constructions X DABC is the required triangle, and say ∠B and ∠C are C given along with the perimeter of the triangle, i.e., AB + BC + CA D Perpendicular A Bisector of DB ■ Step 1: Draw a line PQ such that PQ = AB + BC + B CA ■ Step 1: Draw the base AB of required length. ■ Step 2: Draw a line PX at point P such that ∠XPQ ■ Step 2: Draw a line AX such that ∠BAX = ∠A. = ∠B ■ Step 3: Mark D on the line AX such that AD = AC – ■ Step 3: Draw a line QY at point Q such that ∠YQP BC. = ∠C ■ Step 4: Join BD. ■ Step 5: Draw the perpendicular bisector of BD ■ Step 4: Draw the angular bisectors of ∠XPQ and ∠YQP and their point of intersection is point A. and extend it to meet the line AX at C. ■ Step 6: Join BC. DABC is the required triangle. ■ Step 5: Draw perpendicular bisector of AP and it Note: The method will give correct results when intersects PQ at B. ∠BDC < 90° ■ Step 6: Draw perpendicular bisector of AQ and it Case 2: BC – AC is given, i.e., AC < BC intersects PQ at C. ■ Step 1: Draw the base AB of required length. ■ Step 2: Draw a line X such that ∠BAX = ∠A. ■ Step 7: Join AB, AC. ■ Step 3: Mark D on the extension of AX and on the ■ Step 8: DABC is the required triangle. other side (opposite side) of AX such that AD = BC – AC ■ Step 4: Join BD. ■ Step 5: Draw the perpendicular bisector of BD and extend it to meet the line AX at C. Step 6: Join BC. DABC is the required triangle. X C A D Perpendicular Bisector of DB • Procedure to construct a triangle, given its perimeter and two base angles. X Perpendicular Y Perpendicular bisector of AP bisector of AQ Angular bisector B C Angular of XPQ bisector of YQP P Q 117

11. Constructions Work Plan CONCEPT COVERAGE DETAILS PRACTICE SHEET COVERAGE PS – 1 Pre-requisites • Use of geometrical tools • Perpendicular bisector, Marking angles PS – 2 PS – 3 Constructions • Basic Constructions PS – 4 • Some constructions of triangles Self Evaluation Sheet Worksheet for \"Constructions\" Evaluation with Self ---- Check or Peer Check* 118

PRACTICE SHEET - 1 (PS-1) 1. Draw a perpendicular bisector of a line whose length is 6 cm explain the steps. 2. Mark the given angles with respect to the line OA using a protractor. i) 70° ii) 30° iii) 90° iv) 150° v) 170° vi) 50° 3. Construct a triangle whose sides are 6 cm, 7 cm and 8 cm. 119

PRACTICE SHEET - 2 (PS-2) 1. Using a protractor draw two lines at an angle of 48° and then construct the angular bisector of the angle. Explain the procedure and justify the construction. 2. Using a protractor draw two lines at an angle of 172° and then construct the angular bisector of the angle. Explain the procedure. 3. Using a protractor draw two lines at an angle of 100° and divide it into 4 equal parts. Explain the procedure. 4. Draw a perpendicular bisector to a line measuring 7.6 cm and justify the construction. 5. Draw a perpendicular bisector for a line of 6 cm length which is at an angle of 40° (measure using protractor) to a horizontal line. Also draw the angular bisector of the 40° angle. 6. Draw a line at 60° to a given line and justify the construction. 7. Draw an angle at 120° to the given line and explain the procedure. 8. Construct an angle of 37.5° and explain the procedure. 9. Construct an angle of 150° and justify the construction. 10. Explain the construction of 52.5° in two different ways and prove the measure of the angle. 11. Explain the construction of a right angled isosceles and justify its construction. 120

PRACTICE SHEET - 3 (PS-3) 1. Draw a ∆ABCsuch that AB = 4 cm and ∠B = 60°. Measure the length of the remaining two sides if, i) AC + BC = 9cm ii) AC + BC = 6cm 2. Draw a triangle ABC, such that AC = 3 cm, ∠A = 30°and the perimeter of the triangle is 10 cm. 3. Draw a triangle ABC, such BC = 5cm, ∠C = 120° and the sum of the remaining two sides is 8 cm. 4. Construct a triangle ABC, such that AB = 7cm, ∠A = 45°and i) AC – BC = 3cm ii) BC – AC = 3 cm Give the measure of the remaining two sides of the triangle. 5. Construct a triangle PQR, such that QR = 5 cm, ∠Q = 90°and PR – PQ = 2cm. Give the measure of the remaining two sides of the triangle. 6. Construct a triangle PQR, such that QR = 5.5 cm, ∠Q = 75° and PR – PQ = 2.5 cm. Give the measure of the remaining two sides of the triangle. 7. Construct a triangle whose perimeter is 10 cm and two of its angles are 60° and 60°. Give the length of the sides of the triangle. 8. Construct a triangle whose perimeter is 9 cm and two of its angles are 60° and 90°. Determine the length of the sides of the triangle. 121

PRACTICE SHEET - 4 (PS-4) I. Choose the correct option. 1. A geometrical construction is the process of drawing a geometrical figure using only: (A) a divider and a compass (B) an ungraduated ruler and a compass (C) an ungraduated ruler and a divider (D) a protractor and a compass 2. Atleast ___________ parts of a triangle have to be given for constructing it. (A) one (B) three (C) two (D) four 3. In which of the following cases the triangle cannot be constructed? (A) If three sides are given (B) If two angles and the included side is given (C) If two sides and the angle not included between them are given (D) If two sides and the included angle is given 4. In constructing a ∆ABC, given BC = 8 cm and ∠B = 55° the difference of AB and AC should not be: (A) 7 cm (B) 7.5 cm (C) 5 cm (D) 8.5 cm 5. Which of the following set of angles can be the angles of a triangle? (A) 50°, 60°, 60° (B) 50°, 80°, 60° (C) 40°, 60°, 80° (D) 50°, 90°, 60° 6. Following are the steps of construction of a ∆ABC in which BC = 4 cm, ∠B = 75° and ∠C = 60°. Arrange them and select the right option. (i) Draw a ray CY making an angle of 60° with CB such that CY intersects BX at A. (ii) ABC is the required triangle formed. (iii) Draw a line segment BC of length 4 cm. (iv) Draw a BX ray making an angle of 75° with BC . II. Very short answer questions. 1. The construction of a triangle ABC, given that BC = 4 cm, ∠C = 60° is possible when difference of AB and AC is equal to 4.2 cm or 3.8 cm? 2. Is it possible to construct a ∆ABC in which BC = 5 cm, ∠B = 120° and ∠C = 60°? III. Short answer questions. 1. A triangle ABC is constructed whose perimeter is 12 cm. If the sides of a triangle are in the ratio 2 : 3 : 4, find the sides. 2. State whether the following is true or false with a reason. (a) An angle of 52.5° can be constructed with the help of a ruler and a compass (b) A triangle ABC can be constructed in which AB = 5 cm, ∠A = 45° and BC + AC = 5 cm. IV. Long answer questions. 1. Draw a perpendicular bisector of line segment AB of length 8.6 cm. 2. Construct a ∆ABC in which BC = 6 cm, ∠B = 30° and AB – AC = 2.4 3. Construct a ∆XYZ given YZ = 5 cm, ∠Y = 60° and XY + XZ = 9 cm. 122

Self-Evaluation Sheet Marks: 15 Time: 30 Mins 1. Explain the construction of an angle of 30°. 3. Construct a triangle ABC, such that AB = 8 cm, (3 Marks) ∠B = 60°and BC – AC = 1.5cm. Give the measure of the remaining two sides of the triangle. (4 Marks) 2. Draw a triangle PQR such that QR = 4 cm,∠Q 4. Draw a triangle such that its perimeter is = 45° and the perimeter of the triangle is 10 10 cm and two of its angles are 60° and 45°. cm. Determine the length of the sides of the Determine the lengths of the sides of the triangle. (4 Marks) triangle. (4 Marks) 123

12. Herons Formula Learning Outcome By the end of this chapter, a student will be able to: • Calculate the area of any given triangle using Heron’s Formula. • Calculate the area of the given geometrical shape by dividing it into triangles. Concept Map Base and Height of Heron’s Formula triangle Area of triangle Area of Quadrilateral Area of Geometrical shapes Key Points • The area of quadrilateral or any other shape made • Area of a triangle = 1 × base × height, base is of line segments can be computed by using Heron’s formula if the lengths of the line segments are 2 known. The given area is first split into triangles and the area of each triangle is computed using length of a side of triangle and height is the Heron’s formula and total area is equal to sum of perpendicular distance of the side from the vertex all individual areas. opposite to it. Example: i) In right angled triangle ABC, if BC is considered Example: as base, then AB will be height. If AB is Area of the shape shown in the figure (shaded considered as base, then BC will be the height region) is equal to the sum of the areas of the four of the triangle. triangular portions, A, B, C and D. ii)  In any triangle, say ∆PQR, PS is the height 124 of the triangle on the base PQ. It is drawn perpendicular to base PQ and passing through the opposite vertex of PQ, i.e., R. • The formula given by Heron about the area of a triangle is also known as Heron’s formula. It is stated as Area of a triangle = s(s − a)(s − b)(s − c) where a, b and c are the sides of the triangle and s is semi perimeter of the triangle. (a + b+c) s= 2

12. Herons Formula Work Plan CONCEPT COVERAGE COVERAGE DETAILS PRACTICE SHEET PS – 1 Pre-requisites • Altitude of a Triangle • Area of Triangle PS – 2 PS – 3 Heron’s Formula • Area of Triangle by Heron’s Formula Self Evaluation Sheet • Application of Heron’s formula in finding area of Quadrilateral Worksheet for “Herons Formula” Evaluation with Self Check ---- or Peer Check* 125

PRACTICE SHEET - 1 (PS-1) 1. If the sides of a triangle are 2 cm, 4 cm and 3 cm, find the perimeter of the triangle. 2. Find the area of the right angled triangles shown in the figure. 3. In the given figure, AC is the median of ∆ACD and AE ⟘ BE. Find the area of i) ∆ABC  ii) ∆ACD   iii) ∆ADE. iv) State the relation between the areas of ∆ABC and ∆ACD v) If x = y, state the relation between the area of ∆ABC, ∆ACD, ∆ADE. 4. From the given figure, find the following: i) Ratio of the length of the sides of ∆ OAC ii) Area of ∆ OAC iii) Perimeter of ∆ OAC 126

PRACTICE SHEET - 2 (PS-2) 7. During measurement of a triangle it was perimeter was 50 m and value of perimeter was 1. Find the area of the triangle. 30 m greater than first side, 38 m greater than second side and 32 m more than the third side. Find the sides and area of the triangle. 8. The wrapping paper covering a box was removed and the various lengths are measured. Find the area of the paper. 2. An equilateral triangle has a side of 6 cm. Find the area of the triangle. 3. A vacant land as shown in the figure is to be turned into garden after fencing it. The cost of fencing was Rs. 100 per m and the cost of gardening is Rs. 500 per sq m. Find the total cost involved. (Use 14 =3.74) 4. A tailor finds an old piece of cloth whose shape is shown in the figure. Find the area of the cloth and the total perimeter of the cloth. 5. King asks a farmer to choose between two fields which have same perimeter of 600 m. One field was in the shape of isosceles triangle with the unequal side being 100 m and the other field was in the shape of an equilateral triangle. Which field should the farmer select to get more area of land? 6. The ratio of the three sides of a triangle was found to be in the ratio of 8 : 6 : 10. What is the ratio of the area and perimeter of the triangle? 127

PRACTICE SHEET - 3 (PS-3) I. Choose the correct option. 1. Diagonals of rhombus are d1 and d2, then side of a rhombus is (A) 1 d12 d 2 (B) d12 + d 2 (C) d1d2 (D) 1 d12 + d 2 2 2 2 2 2 2. Area of triangle given by Heron’s formula is (A) 2s(s - a)(s - b)(s - c) (B) s (s - a)(s + b)(s - c) (C) s(s - a)(s - b)(s - c) (D) s(s - a)(s + b)(s - c) 3. Area of a quadrilateral whose sides and one diagonal are given, can be calculated by dividing the quadrilateral into _____triangles. (A) two (B) three (C) four (D) many 4. In the given figure, the area of the ∆ABC is (A) 15.37 cm2 (B) 12.28 cm2 (C) 111.32cm2 (D) 13.24 cm2 5. The percentage increase in the area of a triangle, if its each side quadrupled is equal to (A) 1000% (B) 1500% (C) 900% (D) 800% 6. An isosceles right triangle has an area 400cm2. What is the length of its hypotenuse? (A) 2 40 cm (B) 4 20 cm (C) 20 3 cm (D) 40 2 cm 7. The perimeter of a triangle is 540 m and its sides are in the ratio 25 : 17 : 12. Find its area. (A) 9200 m2 (B) 9000 m2 (C) 9500 m2 (D) 9100 m2 8. The adjacent sides of a parallelogram are 5 cm and 11 cm. The ratio of its altitudes is (A) 10 : 7 (B) 55 : 22 (C) 25 : 33 (D) 11 : 5 9. A rhombus shaped feeding area has seeds for 12 birds. If each side of the rhombus is 30 m and its diagonal is 48 m, then the area of the feeding place which each bird can have seeds. (A) 70 m2 (B) 74 m2 (C) 72 m2 (D) 864 m2 10. Assertion : The perimeter of a right triangle is 60cm and its hypotenuse is 26 cm. The other sides of the triangle are 10 cm and 24 cm. Reason : (Base)2 + (Perpendicular)2 = (Hypotenuse)2 (By Pythagoras theorem) (A) If both assertion and reason are true and reason is the correct explanation of assertion. (B) If both assertion and reason are true but reason is not the correct explanation of the assertion. (C) If assertion is true but reason is false. (D) If assertion is false but reason is true. 128

PRACTICE SHEET - 3 (PS-3) II. Short answer questions. 1. In the adjoining trapezium ABCD, AB = 8 cm, DC = BC = 13 cm and AB is parallel to DC. Find area of the trapezium. 2. Find the cost of levelling the grass field in the form of a triangle having its sides 75m, 55 m and 50 m at ` 7 per square metre. 3. A rectangular field is 22 m long and 11 m wide. How many blocks is of 10 cm length and 5 cm breadth are required for the field. If the rate of blocks is ` 1200 per thousand, then find the total cost. III. Long answer questions. 1. The area of a parallelogram ABCD in which AB = 8 cm, BC = 6 cm and diagonal AC =10 cm is k cm2. Find the value of k - 80 . 8 2. (a) The area of triangle, two sides of which are 6 cm and 9 cm and the perimeter is 24 cm is cm2. Find the value of k. (b) The area and base of a right-angled triangle are 32 sq.cm and 8 cm respectively. What is the length of the perpendicular? 129

Self-Evaluation Sheet Marks: 15 Time: 30 Mins 1. Calculate the area of the triangle with sides 10m, 5. Half the perimeter of a triangle is 12 cm and 5m, 5m. (1 Mark) the sides are in the ratio 5:3:4. Find the area of the triangle. (2 Marks) 2. Calculate the area of a triangle with sides 12m, 13m and 5m. (2 Marks) 3. Thread of length 20cm and three pins are 6. Find area of the quadrilateral ABCD as shown provided to two students Karishma and Kareena. in the figure. (4 Marks) The students are asked to press the pins on the board and wrap the thread around them to form a triangle and they did so as shown in the figure. Who has wrapped the thread around a larger area? (3 Marks) 4. The perimeter of an isosceles triangle is 50 m. The length of the unequal side is 10 m. Find area of the triangle. (3 Marks) 130

13. Surface Areas and Volumes Learning Outcome By the end of this lesson, a student will be able to: • Determine the surface area and volume of Right • Determine the surface area and volume of Cuboids. Circular Cone. • Determine the surface area and volume of Right • Determine the surface area and volume of Sphere Cylinders. and Hemisphere Concept Map Area and Perimeter Rectangle Circle Surface Area and Volume Cuboid Right Cylinder Right Circular Cone Sphere Hemisphere Key Points 1000 cm3 = 1 litre 1000 litres = 1 m3 • Face: A face is a flat surface of a 3D object. It is 2 • Mass of a solid object = Volume of the object × dimensional in nature. Density of the object. Edge: Two faces meet at a line segment called an • Volume of liquid flowing in a unit time is equal to edge. the velocity multiplied by the area of the flow. Vertex: Three edges meet at a point called vertex. • The outer surface of a cuboid is made up of six Face rectangles (rectangular regions, called faces of Edge cuboid). Length of the cuboid be l, breadth is b and height is h. Vertex l h b • Area of a circle of radius r is given is πr2 Total Surface area of a cuboid = 2(lb+ bh+ hl) Circumference of a circle of radius r is 2πr Total length of all edges of a cuboid 4(l + b+ h) • Area of the rectangle whose length is and breadth Volume of a cuboid with length (l), breadth (b) and is b is given by l x b or b height (h) is given by Perimeter of the rectangle whose length is and Volume = lbh = base area × height • In a cuboid if the length, breadth and height are all breadth is is given by 2(l +b) • If an object is solid, then the space occupied by such equal then it is called a cube. If a is the side of the cube, then an object is termed as the volume of the object. Surface area of cube = 6a2 The volume of substance that can fill the interior of a solid object is called the capacity of the container. The unit of measurement of volume and capacity is cubic unit. • Unit conversions 131

13. Surface Areas and Volumes a Surface area of the sphere =4πr2 where r is the aa radius of the sphere Volume of a cube = a3, where a is the length of its A sphere has only curved surface and no flat edge. surfaces. • In a Right Cylinder, with radius of the base r and Volume of the sphere = 4 π r3 height of cylinder h , 3 Curved surface area =2πrh Sphere Hemisphere Total surface area of a cylinder =2πr (r + h) Here π is approximately taken as 22 or 3.14 If a sphere cut into two equal halves, then each half Volume of a cylinder whose base7radius is r and is called hemisphere. Each of the hemispheres will have a flat circular base whose radius is equal to height is h is given by, the radius of the sphere. Volume of cylinder = πr2h Curved surface of the hemisphere =2πr2 rr Total surface area of hemisphere =3πr2 Volume of the hemisphere is half the volume of the sphere of same radius. h Volume of hemisphere = 2 π r3 3 r Not Right cylinder r Right cylinder • In a Right Circular Cone, the height of the cone, radius and slant height (length of slant edge) of the cone are denoted by h , r and l . l2 = r2 + h2 Curved surface area of the cone = πrl Total surface area of the cone = πr(l + r) Volume of a cone is 1 of the volume of the cylinder with 3 same base circle and height. Volume of the cone = 1 π r2h 3 Vertex or apex Slant edge Circular base Right Circular Cone Not Right Circular Cone • Sphere is a three dimensional solid figure made of up of all points in the space which lie at a constant distance called radius from a fixed point called the centre of the sphere. 132

13. Surface Areas and Volumes Work plan CONCEPT COVERAGE COVERAGE DETAILS PRACTICE SHEET PS – 1 Prerequisites  Area of circle, rectangle PS – 2  Surface area of Cuboid and Cylinder PS – 3 PS – 4 Surface Areas and  Surface Area of Cuboid and Cube Self Evaluation Sheet Volumes  Surface Area of Right Circular Cylinder  Surface Area of Right Circular Cone  Surface Area of a Sphere  Volume of Cuboid  Volume of Cylinder  Volume of Right circular Cone  Volume of Sphere and Hemisphere Worksheet for “Surface Areas and Volumes” Evaluation with Self Check or ---- Peer Check* 133

PRACTICE SHEET - 1 (PS-1) 1. If the radius of a circle is 100 cm, then find its area and circumference. What will be the area and circumference of the circle, semicircle and that of a quarter of a circle? (Use p = 3.14) rr Circle rr r Semicircle Quarter of circle 2. The diagonal of a square is 20 m. Find its area. If two such squares are joined along their edges, what is the perimeter and area of the geometry so formed? A BA B E D CD C F Given square Geometry formed by joining squares 3. Two items are to be wrapped in a special paper. One of them is a box of size 10 cm × 10 cm × 20 cm and the other is a cylinder of 10 cm diameter and 20 cm height. Determine which item requires more paper? Find the volume of the box and cylinder. 134

PRACTICE SHEET - 2 (PS-2) 1. A steel box of length 8 m, breadth 6 m and 8. A steel container of 40 cm diameter and 80 cm height 5 m is available in a paint shop. One of height was made by bending a sheet to form the smallest faces is to be removed from the box the curved surface and then welding the top before painting and then replaced with a special and bottom with circular plates. 5% of material door which does not require painting work. What is needed for overlaps determine the total area will be the cost of painting the outer surface of of the sheet needed for making the container. the box if the cost of painting is Rs. 150 per sq. m? (Neglect the thickness of the sheet) Use p = 3.14 2. The cost of gold plating silver boxes is Rs. 5 per 9. A large cylindrical water storage tank was made sq. cm. Ramesh and Suresh wanted to do gold using concrete. The thickness of the tank on all plating of the boxes available with them whose sides as 0.5 m and its outer diameter was 10 sizes are 7 cm × 8 cm × 6 cm and 5 cm × 9 cm m and its height was 5 m. The tank was to be × 7 cm respectively. Find out who will pay more sealed with cement on the entire outer surface money and how much more money does they at a cost Rs. 100 per sq. m and is to be sealed pay for gold plating? with special coating in the inside at a cost of Rs. 200 per sq. m. Find the total cost involved. Use 3. Saritha wanted to get her son’s room painted as p = 3.14 it was all scribbled. The size of the room was 5 m long, 4 m wide and 3 m tall. Saritha checked the 10. Ramesh and Mahesh are two dosa experts paints in a shop and it costed her Rs. 16 per sq. m. who roll dosas into a conical shape. The dosa A painter tells her that the cost of painting would made by Ramesh had a slant height of 15 cm be Rs. 2220 including the paint and Rs. 1110 if and a height of 10 cm while the dosa made she gets the paint of her choice. Determine the by Mahesh was of 20 cm diameter and 10 cm cost of painting mentioned by the painter per sq. height. If the cost of the two dosa is same, then m. What should Saritha do to save money? which doas can satisfy your hunger better? Use 5 = 2.2 and 2 =1.4 4. Students wanted to create boxes by applying gum to the edges of the boxes whose size is 10 11. Seema makes ice cream cones for a particular cm × 8 cm × 8 cm. The cost of glue would be Rs. brand. She can make 100 cones with 1 kg of flour 0.01 for every cm length. What is the total cost of and whose total area would be 0.66 m2. If the glue that is needed to make 500 boxes. diameter of each cone is 6 cm then find the height of the cone made by Seema. If the company fills 5. The length, breadth and height of a large storage the cones with ice cream and then packs them box are in the ratio of 10:6:2. If the total surface with special paper, what is the surface area of area of the box is 736 sq. units, determine the paper needed for one cone? (Use 10 = 3.1) size of the box. 12. A school wanted to plant trees on a nearby hill. 6. A long steel pipe is needed to carry special liquid The plan was to put 1 sapling for every 10 sq. m in an industry. The inner diameter of the pipe of the area of the hill. If the total height of the hill was 8 cm and the outer diameter of the pipe was 120 m and its slant height was 130 m then was 10 cm and its length was 0.5 m. If the cost finds the number saplings that were planted by of painting the inner surface is Rs. 0.8 per sq. cm the school. (Use p = p = 3.14. Consider the hill as and the cost of painting the outer surface was a cone) Rs. 0.6 per sq. cm, determine the total cost of painting the pipe. Use p = 3.14 13. A gardener working in a park, trims the plants into different shapes. The gardener trims one 7. Two tanks, one in the shape of a cylinder and the particular plant in the shape of a cone. The base other in the shape of a cuboid are standing next diameter of the plant is 2 m and the height is 2 to one another. The size of the cylinder is 28 m m. If the plant gives 10 gm of oxygen per day for diameter and 21 m height and the size of the every square meter of its leaves then determine cuboid tank is 28 m long, 21 m wide and 28 m the amount of oxygen released by the plant. high. The tanks are to be painted on all outer Assume that there is no gap between the leaves surfaces and the cost of painting curved surfaces of the plant after the gardener trims the plant. is Rs. 5 per sq m and cost of painting flat surface is Rs. 4 per sq m. Determine the cost of painting (Use 5 = 2.2 ) the tanks and find out the painting cost of which tank is more. Use . 135

PRACTICE SHEET - 2 (PS-2) 14. Inavillagetherewerehutswhosetopwasmadeof grass. During the rainy season, the huts leaked and the villagers would use tarpaulin over the roof. If 50 m2 of tarpaulin was needed to cover the roof and the diameter and height of the roof were in the ratio of 6:4 then find the area under the roof. Determine the size of the roof. 15. A student wanted to make a globe for a model exhibition. He takes a world map of size 25 cm × 25 cm from a shop and pastes it on the globe. 2% of the paper is wasted during cutting. What is the diameter of the globe that the student prepared? (Use 4p = 12.5). 16. A company installs huge dome over its roof to allow more light into the building. If the radius of the dome is 21 m then find the number of glass panes required to make the dome if the size of a individual glass pane is 0.5 m × 0.5 m. 17. Sonia is a manager of an orange juice manufacturing company. She gets a special request from one medical company asking for dried orange peel. Sonia’s company uses around 15000 oranges per day and to dry the peel under sun, she needed 231 m2 of area in front of her plant. Determine the size of oranges. 18. In order to make a ball a company requires 1256 cm2 of a special rubber sheet. What would be the size of the ball? (Use p = 3.14) 19. If the diameter of the hemisphere is reduced to half, find the reduction of the total surface area. 136