MATH 3 part 2

B. Write parallel or perpendicular. 4. 1.2. 5.3. 6. A pair of lines of your pad paper 7. The corner of the blackboard 8. Railroad tracks 9. The grills 10. A pair of consecutive sides of a picture frameC. Are we parallel or perpendicular? 1. Two lines on a plane which do not meet. 2. Two intersecting lines which form a right angle. 12

3. KL4. AB & BD FI JG5. FG & EH6. KI & FG AE HC7. PT & TU BD8. QR & PS9. PV & TU PV Q10. BY & BO TU R S BO YD 13

1. Trapezium Lesson 4 Kinds of Quadrilaterals DR MEIf a quadrilateral has no parallel sides, then it is a trapezium. MORE is atrapezium.2. Trapezoid H O EPIf a quadrilateral has exactly a pair of parallel sides, then it is a trapezoid.If HO // EP , then HOPE is a trapezoid.3. Parallelogram S T POIf a quadrilateral has two pairs of parallel sides, then it is a parallelogram.If ST // PO and SP // PO then STOP is a parallelogram. 14

Try this outA. Identify the kind of quadrilateral:1. A quadrilateral whose opposite sides are parallel.2. A quadrilateral with no parallel sides.3. A quadrilateral with a pair of parallel sides.4. Quadrilateral WHEN. W H E N5. Quadrilateral COLD. CX O6. Quadrilateral SORE X X DX L SX O E X R P E7. Quadrilateral PEAS S A8. Quadrilateral TOME TX O X X EX M 15

9. Quadrilateral PEAK PE10. Quadrilateral LOAF A K L O A FB. Select the correct word from the set (trapezium, trapezoid, parallelogram) J KA L R N W O Q M CDE R P F SU VHT Y HG 16

1. quadrilateral DETH2. quadrilateral DANS3. quadrilateral SNLU4. quadrilateral LOVU5. quadrilateral JMRK6. quadrilateral OMPV7. quadrilateral RWCR8. quadrilateral WGFC9. quadrilateral PWGY10. quadrilateral TERYC. Draw the following figures:1. trapezium ZENY2. trapezoid BETH3. parallelogram LOVE4. parallelogram with diagonals DL & BW5. trapezoid with diagonals IE & MCTwo tents are fixed below: B AF D CE HGive me: G J H K M 6. N 7. three parallelograms L 8. 9. Two trapezoids 10. 17

Lesson 5 TRAPEZOIDS TrapezoidXX Isosceles Trapezoid 18

Parts of a Trapezoid Aside from the vertices, angles, sides and diagonals of a trapezoid, atrapezoid has the bases, the base angles, the legs, the median and the altitude.In quadrilateral BETH, BE // HT , then BETH is a trapezoid.The bases are the parallel sides B E BE and AT are the bases. O TThe base angles are the angles M including either base. ∠H and ∠T are the bases.The legs are the non parallel sides.BH and ET are the legs. HThe median of a trapezoid is a segment joining the midpoints of the non-parallel sides. MO is the median of trapezoid BETHThe altitude of a trapezoid is the segment included between the bases and perpendicular to either one of them. EP is the altitude of trapezoid BETH.Special Kind of a TrapezoidA special kind of a trapezoid Z Eis the isosceles trapezoid. NAn isosceles trapezoid is atrapezoid whose legs are congruent.If ZY ≅ EN then ZENY is an YIsosceles trapezoid. 19

Try this out TA. Match Column A with Column B. S BP R A IColumn A Column B1. quadrilateral STAR a. consecutive sides2. quadrilateral STIR b. trapezoid3. TI c. legs4. RA & ST d. base angles5. ST & SR e. midpoint 6. SA f. altitude 7. BP g. bases 8. SR & AT h. median 9. ∠R & ∠STA i. Opposite angles 10. ∠R & ∠RAT j. parallelogram k. diagonalB. Identify1. A quadrilateral with a pair of parallel sides.2. The segment joining the midpoints of the legs of a trapezoid.3. The non-parallel sides of a trapezoid.4. The parallel sides of a trapezoid.5. The segment joining the opposite vertices of a trapezoid. 20

6. A pair of sides with a common vertex.7. Two angles whose vertices are the endpoints of a side.8. Trapezoid with equal non parallel sides.9. A segment from a side ⊥ to the opposite side.10. A polygon of four sides.C. Use the figure to identify: ATXYDM P E 1. X of AD 2. AM of quadrilateral DATE 3. Quadrilateral DATE if AD ≅ TE 4. AD and TE of quadrilateral DATE 5. DE of quadrilateral DATE 6. XY of quadrilateral DATE 7. Quadrilateral ATPM 8. ∠D of quadrilateral DATE 9. AT and DE 10. AE21

Lesson 6 Parallelogram ParallelogramRectangle Rhombus SquareParts of a Parallelogram B E CIn a parallelogram, you have also Pthe bases and the altitudes. D The bases of a parallelogram areeither pair of opposite sides.The altitude is a segment includedbetween the bases and perpendicularto either one of them. A TAD or BC is a base and ET is the altitude to either base.AB or CD may also be considered as a base and BP is the altitude to either of these bases. 22

Kinds of Parallelogram There are three kinds of parallelogram. You have the rectangle, therhombus and the square.A rectangle is a parallelogram with a right angle.PURA is a rectangle. P U A S R OA rhombus is a parallelogram with apair of congruent consecutive sides. SONY is a rhombus. YNA square is a parallelogram with a right angle and a pair of congruentconsecutive sides. B O SSo, if you take a square with a right angle,it is a rectangle with a pair of congruentconsecutive sides, if you take a square witha pair of congruent consecutive sides, it isa rhombus with a right angle.BOSE is a square. EA. Identify:1. A polygon of four sides.2. A quadrilateral with two pairs of parallel sides.3. A segment included between the bases and perpendicular to either one of them.4. Either pair of opposite sides of a parallelogram.5. A segment joining any 2 opposite vertices.6. A parallelogram with a right angle.7. A parallelogram with a pair of consecutive sides congruent. 23

8. A rhombus with a right angle. 9. A rectangle with a pair of consecutive sides equal 10. A quadrilateral with four equal sides.B. Classify each statement as true or false. 1. Every parallelogram is a quadrilateral. 2. Every quadrilateral is a parallelogram. 3. If quadrilateral PONY is a parallelogram then PO // YN . 4. There exists a parallelogram with all sides congruent. 5. There exists a parallelogram with all angles congruent. 6. There exists a parallelogram ABCD where AB // AD 7. CL is a diagonal of parallelogram COLD. 8. There exists a parallelogram that is not a plane figure. 9. A square is a rectangle. 10. A rhombus is a square.C. Write the letter which corresponds to those given in the figure or parts of the figures below.AB CDw xy r E F G m v s o y1. rectangle2. parallelogram t3. trapezoid4. square 6. base5. rhombus 7. isosceles trapezoid 8. altitude 9. diagonal 10. median 24

Let’s Summarize A quadrilateral is a polygon of four sides. The parts of a quadrilateral are the vertices, the angles, the sides and thediagonals. A quadrilateral can be named by its vertices. The kinds of quadrilateral can be summarized by drawing the Family Treeof Quadrilaterals. Quadrilateral four sided polygon Trapezium Trapezoid parallelogramNo parallel sides With a pair of With 2 pairs of parallel sides parallel sides isosceles rectangle rhombus With a right With a pair of angle consecutive sides congruent square With a right angle and with a pair of consecutive sides congruent 25

Two lines are parallel if they are coplanar and do not meet. Perpendicular lines are intersecting lines which form a right angle. In a trapezoid, the non-parallel sides are the legs, the parallel sides are the bases. Base angles of a trapezoid are the two angles including a base. Themedian is a segment joining the midpoints of the non-parallel sides of atrapezoid. The altitude is the segment included between the bases andperpendicular to either one of them. In a parallelogram, the bases are either pair of opposite sides. Thealtitude is a segment included between the bases and perpendicular to one ofthem. What have you learned K R OAM SC E W X Y P Q 26

Identify by matching Column A with Column B. Write only the letter.Column A Column B1. ∠MOR & ∠REM a. quadrilateral2. quadrilateral MORE b. opposite angles3. AC c. consecutive vertices4. CE & QP d. parallelogram5. CQ & EP e. trapezium6. quadrilateral CEPQ f. square7. XY8. quadrilateral WSCQ g. rhombus9. quadrilateral CAKE h. altitude10. S & W i. Bases j. kgs k. isosceles trapezoid l. median 27

Answer KeyHow much do you know1. a 6. a2. b 7. b3. b 8. d4. d 9. c5. c 10. bTry this outLesson 1A. 1. quadrilateral B. 1. DLDC, LOGD, OCDL, DLOC 2. not COLD, OLDC, LDCO, DCOL 3. quadrilateral 4. not 2. BEAH 5. quadrilateral 3. LARY 6. quadrilateral 4. MANY 7. quadrilateral 5. SERO 8. quadrilateral 6. HOPE 9. quadrilateral 7. STOP 10. not 8. BAMY 9. ABCD 10. PQRS (Answers may vary) C. 1. ADWP 2. LVWP 3. OKUV 4. MITV 5. IFST 6. FCDS 7. MHCF 8. GYCB 9. LNRP 10. MNRQ (Answers may be in any order)Lesson 2A. 1. C & L, A & O B. 1. Q C. 1. b 2. ∠C & ∠L, ∠A & ∠O 3. CO & AL, CA & OL 2. QR 2. c 4. CL & DA 3. ∠S 3. a 4. consecutive 4. c 28

5. C & O, O & L 5. PR 5. b L & A, A & C 6. consecutive 6. d 7. A 7. a6. ∠C & ∠O, ∠O & ∠L 8. BD 8. a ∠L & ∠A, ∠A & ∠C 9. consecutive 9. b7. AC & CO, CO & OL 10. ∠C 10. d OL & LA, LA & AC8. PO, OR, EM & MP9. ∠P, ∠O, ∠E & ∠M10. PE & MOLesson 3A. 1. yes B. 1. parallel C.1. parallel 2. yes 2. perpendicular 2. perpendicular 3. no 3. perpendicular 3. perpendicular 4. no 4. parallel 4. perpendicular 5. yes 5. parallel 5. parallel 6. no 6. parallel 6. perpendicular 7. no 7. perpendicular 7. perpendicular 8. no 8. parallel 8. parallel 9. yes 9. parallel 9. parallel 10. no 10. perpendicular perpendicular 10.Lesson 4A. 1. parallelogram B. 1. parallelogram C.1-5. may vary 2. trapezium 2. trapezium 6. ABCD 3. trapezoid 3. trapezoid 7. GKLS 4. trapezium 4. parallelogram 8. JLMH 5. parallelogram 5. trapezoid 9. FBDC 6. trapezoid 6. trapezium 7. parallelogram 7. parallelogram 10. JHMN 8. parallelogram 8. trapezoid 9. trapezoid 9. trapezoid 10.trapezium 10. parallelogramLesson 5A. 1. b B. 1. trapezoid C. 1. midpoint 2. j 2. median 2. altitude 3. f 3. legs 3. isosceles 4. g 4. bases trapezoid 5. a 5. diagonal 4. legs 29

6. k 6. consecutive sides 5. base7. h 7. consecutive angles 6. median8. c 8. isosceles trapezoid 7. parallelogram9. I 9. altitude 8. base angle10. d 10. quadrilateral 9. parallel sides orLesson 6 bases 10. diagonalA. 1. quadrilateral B. 1. true 2. 2. false C. 1. B 3. parallelogram 3. true 2. C 4. 4. true 3. F 5. altitude 5. true 4. D 6. 6. false 5. E 7. bases 7. true 6. r, t, n, m 8. 8. false 7. G 9. diagonal 9. true 8. v, s, x 10. 10. false 9. w rectangle 10 o rhombus square square rhombus & squareWhat Have You Learned1. b2. d3. h4. I5. j6. k7. l8. g9. f10. c 30

Module 3 Plane Coordinate Geometry What this module is about This module will show you a different kind of proving. The properties of triangles andquadrilaterals will be verified in this module using coordinate plane and the application ofthe lessons previously discussed in the other modules. Furthermore, you will also have the chance to do analytical proof and compare it withthe geometric proof that you have been doing since the start of the year. Included in thismodule are lessons that will discuss in detail properties/relationships of circles with otherfigures in the coordinate plane. What you are expected to learn This module is written for you to 1. Define and illustrate coordinate proof. 2. Verify some properties of triangles and quadrilaterals by using coordinate proof. 3. Determine the difference between geometric proofs and coordinate proof. 4. Illustrate the general form of the equation of the circle in a coordinate plane. 5. Derive the standard form of the equation of the circle from the given general form. 6. Find the coordinate of the center of the circle and its radius given the equation . 7. Determine the equation of a circle given its a. center and radius b. radius and the point of tangency with the given line 8. Analyze and solve problems involving circles. How much do you knowAnswer the following questions Tell which of the axes placements will simplify a coordinate proof for a theoreminvolving the figure shown.

1. bc a2. a bc3. a bcGive the center and radius of the following circles: 4. x2 + y2 = 16 5. x2 + y2 – 25 = 0 6. (x – 3)2 + (y + 1)2 = 36 7. x2 + y2 – 4x + 10y + 16 = 0 8. Give the standard form of the circle whose center is at (2, -1) and a radius of 7 units.Write the equation of the circle satisfying the following conditions. 9. The line segment joining (4, -2) and (-8, -6) is a diameter. 10. The center is at (0, 5) and the circle passes through (6, 1). 2

What you will do Lesson 1 Coordinate Proof The coordinates of a point on the coordinate plane are real numbers, thus it ispossible to prove theorems on geometric figures by analytic or algebraic method. We callthis proof the coordinate proof. In writing coordinate proof, some suggestions have to be taken into considerations.First, we may choose the position of the figure in relation to the coordinate axes. If thefigure is a polygon, it is always simpler to put one of the sides on either the x-axis or the y-axis. Second, we may place one of the vertices on the origin. Third, the essential properties of the given figure should be expressed by thecoordinates of key points. The proof is accomplished by setting up and simplifyingalgebraically equations and relations involving these coordinates. Fourth, the figure should never be made special in any way so that the proof will begeneral and can be applied to all cases. Except for zero (0), numerical coordinate shouldnot be used in the proof.Example 1. Prove analytically that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and measures one-half of it.Solution: y A(a, t) X Y O(0.0) X B(b,0) 3

a. Analysis: The figure shows the triangle appropriately placed on rectangular coordinate plane. We have to show two properties here.1) XY ║ OB and (2) XY = 1 OB. 2b. Since X and Y are midpoints, then the midpoint formula can be used to get theircoordinates. The coordinates of X are  a , t  and that of Y are  a + b , t  . 2 2  2 2The slope m1 of OB = 0 − 0 = 0 . The slope m2 of XY is b−0 m2 = t −t = 0 =0 22 −b a −a+b 22 2 Since the slope of OB = 0 and that of XY = 0, then the two slopes are equal.Therefore, XY ║ OB . Therefore, the segment joining the midpoints of the two sides of atriangle is parallel to the third side.c. For the second conclusion, get the length of XY and OB.XY =  a + b − a 2 +  t − t 2  2 2 2 2 =  b 2 2 =b 2 OB = (b − 0)2 + (0 − 0)2 = b2 =b The computation showed that XY = b and OB = b. Therefore, XY = 1 OB. 22Hence, the segment joining the midpoints of the two sides of a triangle is one-half the thirdside. 4

Example 2. Prove that the diagonals of a rectangle are equal. yD(0, c) C(b, c) xA(0,0) B(b, 0)Given: ABCD is a rectangleProve: AC = BDProof: In this problem we are given not just any quadrilateral but a rectangle. Place onevertex on the origin. In the figure, it was vertex A which was in the origin. Then vertex B onthe x-axis, and vertex D on the y-axis. The following are the coordinates of the vertex : A(0, 0), B(b, 0), C(b, c) and D(0, a).AC and BD are the two diagonals. We have to determine if the length of AC is equal tothe length of BD. AC = (b − 0)2 + (c − 0)2 = b2 + c2 BD = (0 − b)2 + (c − 0)2 = (−b)2 + c2 = b + c2 Therefore, from the computed distances or lengths, AC = BD. Hence, thediagonals of rectangle are equal. 5

Example 3. Prove that the diagonals of a square are perpendicular to each other. yGiven: HOPE is a squareProve: HP ⊥ OE E(0, a) P(a, a) H(0,0) O(a, 0)Proof: To show the perpendicularity of HP and OE , then their slopes must be thenegative reciprocal of each other. Since the vertices of the square were placed on thecoordinate plane in such a way that makes proving simpler, then we are now ready to findthe slopes of both HP and OE .Computing for the two slopes , m HP = a − 0 = a = 1 a−0 a m OE = 0 − a = − a = −1 a−0 a From the results you can easily see that the reciprocals of HP and OE are 1 and – 1which are negative reciprocals. Therefore, HP and OE are perpendicular to each other. 6

Try this outA. Determine the coordinates of A, B or C.1. y MAIN is a rectangle 2. y BEAM is a parallelogram M(c,b) B M(0,b) A A(0,0) E(a,0)N(0,0) I(a, 0) y y3. LEAD is an isosceles trapezoid 4. ABC is an isosceles triangle C L(b,c) E(a-b, c) x CD(0,0) F B(0,0) A(a, 0) x A(a, 0)B. Given the following statements, do the following:a. Use the given figure on the rectangular coordinate plane.b. Write the hypothesis.c. Write what is to be proven.d. Prove analytically. 7

1. The medians to the legs of an isosceles triangle are equal. y E (a, b) D F DA (0, 0) F x C (2a, 0)2. The diagonals of an isosceles trapezoid are equal. y D (b, c) C (a-b, c) xA (0, 0) B (a, 0)3. The diagonals of a parallelogram bisect each other. y E (b, c) P (a+b, c) xH (0, 0) O (a, 0) 8

4. The midpoint of the hypotenuse of a right triangle is equidistant from the vertices. yC (0, 2b) E xA (0, 0) B (2a, 0)5. The segments joining the midpoints of the opposite sides of a quadrilateral bisect each other. y K (4d, 4e) BE (4b, 4c) C F A (4a, 0) xM(0, 0) D Lesson 2 Circles on Coordinate Plane You know from the previous chapter that the set of all points equidistant from a fixedpoint is called a circle. The fixed point is called the center. If the circle is on the Cartesiancoordinate plane, the distance formula will lead us to write an algebraic condition for acircle. 9

In the figure, if P has ycoordinates (x, y), using the distanceformula will give us (x,y) r (x − 0)2 + ( y − 0)2 = r x c (0,0) x2 + y2 = rSquaring both sides, we get x2 + y2 = r2 Thus the equation of a circle withcenter at (0, 0) and radius r in standardform is x2 + y2 = r2Example 1. Find the equation of the circle whose center is the origin and the radius isa. r = 3b. r = 1c. r = 5d. r = 2 2Solution:a. C(0, 0) r=3b. C(0, 0) The equation of the circle is x2 + y2 = 32c. C(0, 0)d. C(0, 0) x2 + y2 = 9 r=1 The equation of the circle is x2 + y2 = 12 x2 + y2 = 1 r=5 The equation of the circle is x2 + y2 = 52 x2 + y2 = 25 r=2 2 The equation of the circle is x2 + y2 = (2 2 )2 x2 + y2 = 8 10

Example 2. Determine the center and the radius of the circle whose equations are given. a. x2 + y2 = 49 b. x2 + y2 = 36 c. x2 + y2 = 18Solutions: The equations are of the form x2 + y2 = r2, hence the center is the origin. The radiusin each circle is a. r = 9 b. r = 6 c. r = 3 2 Not all circles on the rectangular coordinate plane has its center at the origin. In thegiven circle below, its center is not the origin. We will represent the center of the circlewhich is not the origin as (h, k). y (x,y) r c (h,k) x In the coordinate plane given above, center C has coordinates (h, k) and radius r. ByPythagorean Theorem, the distance from the center of the circle to a point A(x, y) on thecircle can be solved. This will also give us the standard form of the equation of a circle. AC = (x − h)2 + (y − k )2 = r (x - h)2 + (y – k)2 = r2 The standard form of the equation of the circle with center at (h, k) and radius r is (x – h)2 + (y – k)2 = r2 11

Example 1. Find the equation of a circle with center at (1, 5) and a radius of 3 units.Solution: Substitute the following values in the standard formh = 1, k = 5, r=3The equation is (x – 1)2 + (y – 5)2 = 32 (x – 1)2 + (y – 5)2 = 9Example 2. Find the equation of a circle with center at (2, -3) and radius of 5 units.Sketch the figure. h = 2, k = -3, r=5The equation is (x – 2)2 + (y + 3)2 = 52The figure is (x – 2)2 + (y + 3)2 = 25 yx r C (2,-3) The standard form of the equation of a circle with center at C(h, k) and radius r canbe presented in another form. This is done by squaring the binomials and simplifying theresults. (x – h)2 + (y – k)2 = r2 x2 – 2hx + h2 + y2 – 2yk + k2 = r2 x2 + y2 – 2hx – 2yk + h2 + k2 – r2 = 0 By assigning capital letters D, E and F to represent the constants, the equation willnow assume this general form. x2 + y2 + Dx + Ey + F = 0 12

Example 1. Find the radius and the center of the circle given its equation. x2 + y2 – 4x – 6y -12 = 0Solution: First isolate the constant term at the right side of the equal sign by applying theaddition property of equality x2 + y2 – 4x – 6y – 12 = 0 x2 + y2 – 4x – 6y = 12 Then group the terms with x together and those with y together. (x2 - 4x)+ (y2 - 6y) = 12 Complete each group like in completing the square by adding the third term of thetrinomial. Note that what you added to each group should be added to the right side of theequation also. (Application of addition property of equality) (x2 - 4x + 4 ) + (y2 – 6y + 9 ) = 12 + 4 + 9 (x2 - 4x + 4 ) + (y2 – 6y + 9 ) = 25Rewrite each perfect trinomial square into binomial factors. (x – 2)2 + (y – 3)2 = 25 (x – 2)2 + (y – 3)2 = 52 Since the equation is in center-radius form, then we can determine the coordinates ofthe center and the radius of the circle. The center is at (2, 3) and the radius is 5 units.Example 2. Find the radius and the coordinates of the center of the circle given the EquationSolution: x2 + y2 + 6x – 2y + 6 = 0 Isolate first the constant term x2 + y2 + 6x – 2y = -6 Then group the x and y together (x2 + 6x) + (y2 – 2y) = - 6 Add constants to each group by completing the square. Add to the right sideof the equation what you will add to the left side. (x2 + 6y + 9) + (y2 – 2y + 1) = -6 + 9 + 1 (x2 + 6y + 9) + (y2 – 2y + 1) = 4 13

Write each trinomial as factors or square of binomial. (x + 3)2 + (y – 1)2 = 4 x + 3)2 + (y – 1)2 = 22 The equation is in center-radius form. So the center of the circle is at (-3, 1) and itsradius is 2 units.Example 3. Tell whether the equation x2 + y2 -4x + 8y + 24 determines a circle.Solution: To solve this problem, you should be aware of these fact. The existence of a circledepends on the value of r2. In the standard form of equation of the circle, (x – h)2 + (y – k)2 = r2, one of thefollowing statements is always true if r2 > 0, the graph is a circle. r2 = 0, the graph is a point (We call this the point circle) r2 < 0, the graph or the circle does not exist In the given example, solve for the value of r2. In doing this, you simply follow theprocedure in examples 1and 2. x2 + y2 - 4x + 8y + 24 = 0 x2 + y2 – 4x + 8y = - 24 (x2 – 4x) + (y2 + 8y) = - 24 (x2 – 4x + 4 ) + (y2 + 8y + 16 ) = -24 +4 + 16 (x – 2)2 + (y + 4)2 = -4 Since r2 = -4, then the circle does not exist. The following examples discuss of problems that involve circles in the coordinateplane. Each problem is treated differently according to what is given and what is beingasked . 14

Example 4. Find the equation of a circle whose center is at (4, 2) and is tangent to y-axis.Sketch the figure. y y r C(4,2)xSolution: Since the circle is tangent to y-axis, the radius of the circle is perpendicular to y-axis.It also means that the length of the radius is also the length of the perpendicular segmentfrom the center of the circle to y –axis. From the figure, you can determine that the point oftangency is at (0, 2). To find the length of the radius, use the distance formula. r = (4 − 0)2 + (2 − 2)2 = 42 + 0 = 16 r= 4 To solve for the equation , use the coordinates of the center (4, 2) and the computedlength of radius r = 4. (x - h)2 + (y – k)2 = r2 (x – 4 )2 + (y – 2)2 = 42 x2 – 8x + 16 + y2 – 4y + 4 = 16 x2 + y2 – 8x – 4y + 4 = 0 15

Example 5. Write the equation of the circle with the given condition. (10, 8) and (4, -2) arethe endpoints of the diameter. Sketch the figure. y (10,8) ●C x (4,-2)Solution: In a circle, the radius is one-half of the diameter. Since the given are the endpointsof the diameter, then the center of the circle is the midpoint of the diameter. M 10 + 4 , 8 + (−2)  2 2 M 14 , 6   2 2 M (7,3) The next step is to get the length of the radius. Since radius is one-half of the circle,so get the distance from the center to one of the endpoint of the diameter. Any endpoint willdo. r = (10 − 7)2 + (8 − 3)2 r = 42 + 52 r = 16 + 25 r = 41 16

To find the equation of the line, use C(7, 3) and r = 41 (x – 7)2 + (y – 3)2 = ( 41 )2 x2 – 14x + 49 + y2 – 6y + 9 = 41 x2 + y2 – 14x – 6y + 17 = 0Example 6. Write the equation of the circle with center at (-8, 5) and passing through A(- 6, 4).Solution: Since the circle is passing through A, then the distance from the center to A is the length of the radius of the circle. Compute first the radius of the circle. r = [− 8 − (− 6)]2 + (5 − 4)2 .r = (−2)2 + 12 r = 4+1 C r= 5 (-8,5)The equation of the circle is (x + 8)2 + (y – 5)2 = ( 5 )2 x (-6,4)x2 + 16x + 64 + y2 – 10y + 25 = 5x2 + y2 + 16x – 10y + 84 = 0 yTry this out.A. Determine if the following are equations of a circle, a point or a circle that does not exist.1. x2 + y2 = 32. x2 + y2 - 12 = 03. x2 + y2 + 121 = 04. (x – 5)2 + y2 = 15. x2 + y2 – 10x – 8y + 41 = 0B. Give the center and the radius of each circle.1. x2 + y2 = 252. x2 + y2 - 12 = 03. x2 + (y – 3)2 = 121 17

4. (x – 7)2 + (y – 5)2 = 185. (x + 1)2 + (y – 4)2 = 36. (x – 8)2 + y2 = 497.  x + 1 2 + (y – 7)2 = 25  28. x2 + y2 + 6x + 16y – 11 = 09. x2 + y2 + 2x – 6y – 8 = 010. x2 + y2 – 4x – 12y + 30 = 0C. Write the equation of a circle in standard form with center C and radius r given.1. C(0, 0), r= 42. C(0, 0),3. C(1, 1), r= 2 34. C(-2, -5), r= 35. C(-3, 4), r = 106. C(2, -5),7. C(0, 6), r= 2 28. C(-4, 0), r= 59. C(0, -5), r= 6 r = 3.510. C(3, 0), r= 5 r= 3 3D. Solve the following problems. Sketch the figure. Show the complete solution.1. Write the equation of the circle with center at (3, -1) and tangent to the x-axis.2. Write the equation of a circle with center at (2, 5) and passing through (-2, 1).3. The line segment joining ( -2, 5) and (-2, -3) is a diameter.4. A circle is tangent to both axes and the radius at the first quadrant is 3 units.5. A circle is tangent to the line 3x – 4y = 24 and the center is at (1, 0). Lets summarize1. Proving theorems in geometry analytically is also known as coordinate proof.2. In coordinate proof, the location of one of the vertices and one of the sides of the figure can help in proving easily the theorem.3. The set of all points equidistant from a fixed point is called a circle. The fixed point is called the center.4. The standard form of equation of a circle with center at the origin and radius r is x2 + y2 = r25. The standard form of equation of a circle with center at (h, k) and radius r is (x – h)2 + (y – k)2 = r2 18

6. The general form of equation of a circle is x2 + y2 + Dx + Ey + F = 0 7. The existence of a circle depends upon the value of r2. If r2 > 0, the circle exist. If r2 = 0, point circle exist. If r2 < 0, the circle does not exist. What have you learned1. yD(0, b) CA(0, 0) B(2a, 0) x ABCD is a rectangle, What is the coordinate of C ? 2. In figure in # 1, find the length of AC . 3. In doing coordinate proof, it is always simpler to put one of the vertices of the polygon or figure on the ___________. 4. What is the standard form of the equation of a circle if the center is the origin and the radius is r units?What is the center and the radius of the following circles given their equations? 5. x2 + y2 = 64 6. x2 + (y + 1)2 = 25 7. (x – 6)2 + y2 = 1 8. (x – 3)2 + (y + 7)2 = 12 9. (x – 1)2 + (y +1)2 = 49 10. What is the equation of the circle whose center is at (-4, -1) and passing through the origin? Sketch the figure. Express the answer in general form. 19

Answer KeyHow much do you know.1. c r=42. a r=53. b r= 64. C(0,0) or origin,5. C(0, 0) or origin,6. C(3, -1),7. C(2, -5), r = 138. (x – 2)2 + (y + 1)2 = 499. x2 + y2 + 4x + 8y – 20 = 010. x2 + y2 – 10y – 27 = 0Lesson 1A. 1. A(a, b) 2. B(a + c, b) 3. C  b , c  2 2 4. C  a ,c  2 B. 1. y E (a, b) DF A(0, 0) C (2a, 0) x Given: ACE is an isosceles triangle. AF and DC are medians Prove: AF = DCSolution:1. First determine the coordinates of D and F. 20

Since AF and DC are medians, then D and F are midpoints.D  a + 0 , b + 0  2 2D  a , b  2 2F  a + 2a , b   2 2F  3a , b   2 22. After finding the coordinates of D and F, determine the length of AF and DC .AF =  3a − 02 +  b − 02 =  2  2  =  3a 2 +  b 2 =  2  2 9a2 + b2AF = 44 9a2 + b2 4 9a2 + b2 2DC =  2a − a 2 +  0 − b 2 =  2  2 =  3a 2 +  − b 2 =  2   2 9a2 + b2DC = 44 9a2 + b2 4 9a2 + b2 2 21

Based on computations, AF = DC. Therefore, the two medians are equal and we canconclude that that the medians to the legs of an isosceles triangle are equal.2. y D(b, c) C (a-b, c) A (0, 0) B (a, 0) x Given: ABCD is an isosceles trapezoid. AC and BD are its diagonals Prove: AC ≅ BDSolution: To prove that AC ≅ BD we have to show that AC = BD. AC = (a − b − 0)2 + (c − 0)2 = (a − b)2 + c2AC = a2 − 2ab + b2 + c2 BD = (a − b)2 + (0 − c)2 = a2 − 2ab + b2 + c2 The computations showed that AC = BD. Thus we can conclude that the diagonals ofan isosceles trapezoid are congruent.3. y E (b, c) P (a+b, c) x H (0,0) O (a, 0) 22

Given: HOPE is a parallelogram HP and EO are the diagonals Prove: HP and EO bisect each otherSolution: The simplest way of proving this is to show that the midpoints of the two diagonalsare one and the same.1. Find the midpoints of HP and EO and compare. M (HP)  a + b + 0 , c   2 2 M (HP)  a + b , c   2 2 M (OE)  a + b , c   2 2 The results showed that HP and EO have the same midpoint. Hence they bisecteach other. We can conclude that the diagonals of a parallelogram bisect each other. y4. C (0, 2b) E xA (0, 0) B (2a, 0) Given: ∆ABC is a right triangle. E is the midpoint of the hypotenuse BC. Prove: CE = BE = AESolutions: Since E is the midpoint of BC, then its coordinates are 23

E  2a , 2b  2 2 E(a, b)After finding the coordinates of E, find CE, BE and AE. Then compare the lengths.CE = (0 − a)2 + (b − 2b)2 = (− a)2 + (− b)2CE = a2 + b2BE = (2a − a)2 + b2BE = a2 + b2AE = (a − 0)2 + (b − 0)2AE = a2 + b2 Since the three segments CE, BE and AE have the same lengths, we can thereforeconclude that E is equidistant from the vertices of the right triangle. y5. K (4d, 4e) E (4b, 4c) B FC M (0, 0) D x A (4a, 0)Given: MAKE is a quadrilateral B, C, D and F are midpoints of the sides Prove: FC and BD bisect each otherSolutions: 1. Get the coordinates of F, B, C and D. F  4b + 0 , 4c + 0  2 2 24

F (2b,2c)B  4b + 4d , 4c + 4e 2 2B (2b + 2d,2c + 2e)C  4d + 4a , 4e   2 2C (2d + 2a,2e)D  4a , 0   2 2D (2a,0)2. Get the midpoints of FC and BD .Midpoint of FCM1  2b + 2d + 2a , 2c + 2e  2 2M1 (b + d + a,c + e)Midpoint of BDM2  2b + 2d + 2a , 2c + 2e + 0  2 2M2 (b + d + a,c + e) You can see that the midpoints of FC and BD are both (a+b+d, c+e). Therefore,we can conclude that the segments joining the midpoints of opposite sides of a quadrilateralbisect each other.Lesson 2A.1. a circle2. a circle3. the circle does not exist4. a circle5. a point circle 25

B. r= 51. C(0, 0), r=2 3 r = 112. C(0, 0), r= 3 23. C(0, 3), r= 34. C(7, 5), r= 75. C(-1, 4), r= 56. C(8, 0),7. C  − 1 ,7 , r= 6 r= 3 2 2 r = 108. C(-3, -4),9. C(-1, 3),10. C(2, 6),C. .1. x2 + y2 = 162. x2 + y2 = 12 y3. (x – 1)2 + (y – 1)2 = 9 ●4. (x + 2)2 + (y + 5)2 = 105. (x + 3)2 + (y – 4)2 = 86. (x – 2)2 + (y + 5)2 = 257. x2 + (y – 6)2 = 368. (x + 4)2 + y2 = 12.259. x2 + (y + 5) = 510. (x – 3)2 + y2 = 27D.1. C(3, -1) , tangent to x-axisSolution: Since the circle is tangent to x-axis, xthen r is ⊥ to x-axis. The distance fromthe center of the circle to the x-axis or thelength of radius r is I unit. The equation of the circle with centerat (3, -1) and r = 1 is (x – 3)2 + (y + 1) = 1 x2 – 6x + 9 + y2 + 2y + 1 = 1 x2 + y2 – 6x + 2y + 10 – 1 = 0 x2 + y2 – 6x + 2y + 9 = 0 26

2. Center at (2, 5) and passing through ( -2, 1) ySolution: Since the center is knownwhat we need is the length of theradius. To find the length of theradius, use the other point as the otherend of the radius.r = [2 − (− 2)]2 + (5 −1)2 C(2,5)r = (2 + 2)2 + 42 x (-2,1)r = 42 + 42r = 16 +16r = 32r= 4 2The equation of the circle therefore is (x – 2)2 + (y – 5)2 = (4 2 )2 x2 – 4x + 4 + y2 – 10y + 25 = 32 x2 + y2 – 4x – 10y – 3 = 03. Line segment joining (-2, 5) and (-2, -3) is a diameter.Solution: The midpoint of the segment is (-2,5)the center of the circle. Find the coordinatesof the midpoint.M  − 2 + (−2) , 5 − 3 2 2M  − 4 , 2  (-2,1)  2 2 M (-2, 1) Then find the length of the radiususing the midpoint and one of the endpointsof the diameter . r = [− 2 − (−2)]2 + (5 −1)2 = (−2 + 2)2 + 42 = 0 + 42 = 42 27

r=4The equation of the circle with center at ( -2, 1) and r = 4 is(x + 2)2 + (y – 1)2 = 42 yx2 + 4x + 4 + y2 – 2y + 1 = 16x2 + y2 + 4x – 2y - 11 = 04. Solution: Since the circle is tangent to bothaxes, therefore the center of the circle is atequal distance from both x and y axes. That distance is 3 units since the radius C(3,3)is given at 3 units. The center is also at the rfirst quadrant. The circle passes through (0, 3)and (3, 0). The center is at (3, 3) x The equation of the circle is (x – 3)2 + (y – 3) = 32x2 – 6x + 9 + y2 – 6y + 9 = 9 x2 + y2 – 6x – 6y + 9 = 05. Tangent to the line 3x – 4y = 24 with center at (1, 0). y Solution: The radius of the circle is equalto the distance of the center (1, 0) from theline 3x – 4y = 24. To find the distance from a point to a line, C(1,0) xwe use the formula r d = Ax1 + By1 + C 3x-4y=24 − A2 + B2where A and B are the coefficients of x andy, and C is the constant in the equation ofline. x1 and y1 are the coordinates of thepoint.Therefore in the given,A = 3, B = -4 and C = -24. x1 = 1, y2 = 0r = 3(1) + (−4)(0) − 24 − 32 + 42 28

= 3 + 0 − 24 − 9 +16= − 21 −5r = 21 5The equation of the circle with center at (1, 0) and radius r = 21 is (x – 1)2 + y2 =  212 = 441 5  5  25What have you learned1. C(2a, b)2. AC = 2a2 + b23. origin r=84. x2 + y2 = r2 r=55. C(0, 0), r=16. C(0, -1),7. C(6, 0),8. C(3, -7), r=2 39. C(1, -1), r=710. Solution: Find the length of r using the origin and the coordinates of the center (-4, -1). r = (−4 + 0)2 + (−1 − 0)2 y = (−4)2 + (−1)2 = 16 +1 = 17The equation is (x + 4)2 + (y + 1)2 = ( 17 )2 x x2 + 8x + 16 + y2 + 2y + 1 = 17 C(4,-1)x2 + y2 + 8x + 2y + 17 – 17 = 0 x2 + y2 + 8x + 2y = 0 29

Module 4 Geometry of Shape and Size What this module is about This module is about polygons, specifically angles of the polygon. This module willteach you how to find the sum of the interior angles of the polygon. You will also discoverother interesting facts about the interior and exterior angles of the polygon. You can alsoform your own generalization regarding the sum of the measures of the interior and exteriorangles of a convex polygon. What you are expected to learnThis module will help you: 1. Determine the sum of the measures of the interior and exterior angles of the triangle. 2. Determine the sum of the measures of the interior and exterior angles of a quadrilateral. 3. Make generalizations on the sum of the measures of the interior and exterior angles of a polygon. How much do you know 1. What is the measure of each angle of an equilateral triangle? 2. If the sum of the two angles of a triangle is 130º, what is the measure of the third angle? 3. One acute angle of a right triangle is 27º. What is the measure of the other acute angle? 4. What is the measure of an exterior angle of an equilateral triangle? 5. If the measure of the vertex angle of an isosceles triangle is 50º, what is the measure of each base angles? 6. What is the sum of the measures of the interior angles of a quadrilateral? 7. How many sides does a regular polygon have if each interior angle is 120º? 8. The measures of the four angles of a pentagon are 140º, 75º, 120º, and 115º. Find the measure of the fifth angle. 9. If the radius of the circle is 12 cm, what is the diameter of the circle? 10. What is the appropriate name of the given circle?

What you will do Lesson 1Determining the Sum of the Measures of the Interior Angles of a Triangle. In any given triangle, say ΔABC, there 6 Aare three interior angles and by observation, C5 12there are 6 exterior angles. In the given figure,∠ A, ∠ B, and ∠ C are the interior angles of ΔABC. 3Each interior angle has two angles adjacent to it. 4BFor ∠ A, ∠ 1 and ∠ 2, for ∠ B, ∠ 3 and ∠ 4, andfor ∠ C, ∠ 5 and ∠ 6. Aside from being adjacent, ∠ 1and ∠ 2 are both supplementary to ∠ A.Those six angles which are adjacent to Athe interior angles of the triangle arecalled exterior angles of ΔABC. Measuresof the exterior angles will be discussed inthe next lesson. To get the sum of the measures of ∠ A C B∠ B and ∠ C, you can use the protractor and B Athen add the sum of their measures. There isanother way of doing this. Here are the steps: C 1. Prepare a cut out of any triangle of any size. 2. Cut off the three angles as in the figure 3. On the given line l, align the 3 cut out angles with all the vertices coinciding with point O. 4. All three vertices should fit perfectly on one side of the line. 5. The measure in degree of the angles about a point on the same side of a line is 180. From the two experiment that you did, you can l Bconclude that the m ∠ A + m ∠ B + m ∠ C = 180. ACExamples: o1. What is the measure of the third angle of a triangle if the measures of the two angles are a. 46º, 82º 2