MATH 3 part 2

b. 51 1 º, 67 1 º 22Solutions: Let x be the measure of the third angle a. x + (46º + 82º) = 180º x + 128º = 180º x = 180º - 128º x = 52º, measure of the third angleb. x + (51 1 º + 67 1 º) = 180º P 22 x + 119º = 180º x = 180º - 119º x = 61º , measure of the third angle2. The measure of one acute angle of right triangle PQR is 33º. R Q What are the measures of the other angles?Solution:Since ΔPQR and ∠ R = 90º , and ∠ Q = 33º, then ∠ P = 33º + 90º + ∠ P = 180º 23º + ∠ P = 180º ∠ P = 180º - 123º ∠ P = 57ºAlternate Solution: Remember, the two acute angles of a right triangle are complementary, hence thesum of their measures is 90º. In other words, the sum of the measures of the two acuteangles of a right triangle is 90. You can use this knowledge in solving the given problem.33º + ∠ P = 90º A ∠ P = 90º - 33º ∠ P = 57º3. In an isosceles triangle, one base angle measures 49º. Find the measures of the other two angles.Solution: Let ΔMAN be isosceles. ∠ M ≅ ∠ N M N m ∠ M = 49º m ∠ N = m ∠ M = 49º m ∠ M + m ∠ N + m ∠ A = 180º 49º + 49º + m ∠ A = 180º 98º + m ∠ A = 180º m ∠ A = 180º - 98º m ∠ A = 82º 3

Try this outA. Given the measures of the two angles of a triangle, determine the measure of the third.1. 36º, 70º 2. 58º, 50º3. 90º, 25º 5. 64 1 º, 59º4. 35º, 35º 2B. Given the measure of one base angle of an isosceles triangle, determine the measuresof the other two angles.1. 45º 4. 37 1 º2. 50º3. 53º 2 5. 63 1 º 4C. Given the measure of the vertex angle of an isosceles triangle, find the measure of abase angle.1. 75º 4. 88º2. 100º 5. 97º3. 39ºD. In a right triangle, the measure of one acute angle is given. Find the measure of theother acute angle.1. 45º 4. 43º2. 36º 5. 61 1 º 23. 52ºE. Solve the following problems.1. The measure of one acute angle of a right triangle is twice that of the other acute angle. Find the measures of the three angles.2. Find the measures of the angles of a triangle that are in the ratio 1:2:3.3. The vertex angle of an isosceles triangle is 15 degrees more than three times the base angle. Find the measures of all the three angles of the triangle.4. n a triangle, one angle is ten degrees more than twice the smallest angle and the third angle is four less than three times the smallest angle. Find the measures of the three angles.5. What are the measures of the angles of an isosceles right triangle? 4

Lesson 2Sum of the Measures of Exterior Angles of a Triangle. Sum of the Interior Angles of a Quadrilateral To find the sum of the measures of the exterior angles of a triangle, it is necessary toapply the previous lesson; the sum of the measures of the interior angles of the triangle is180º. The phrase sum of the exterior angles of a triangle means you will use one exteriorangle from each vertex of the triangle. Thus in the given figure, 3 Ato find the sum of the exterior angles C 1of ΔABC, ∠ 1, ∠ 2 and ∠ 3 are the designatedexterior angles. Each of the exterior angle 2is adjacent to one of the interior angle of the Btriangle. Study the following procedure:m ∠ A + m ∠ B + m ∠ C = 180 Sum of the angles of a triangle is 180 m ∠ 1 + m ∠ A = 180 Definition of exterior angle of a triangle m ∠ 2 + m ∠ B = 180 Definition of exterior angle of a triangle m ∠ 3 + m ∠ C = 180 Definition of exterior angle of a trianglem ∠ 1 + m ∠ 2 + m ∠ 3 + m ∠ A + m ∠ B + m ∠ C = 540 APE by subtraction m ∠ A + m ∠ B + m ∠ C = 180m∠1 + m∠2 + m∠3 = 360 Therefore, the sum of the measures of the exterior angles of a triangle one on eachvertex is equal to 360. What about a quadrilateral? How do you get the sum of the measures of the interiorangles of a four-sided polygon? Consider quadrilateral ABCD. BWhat is the sum of the measures of∠ A, ∠ B, ∠ C and ∠ D? To determine A 12this, draw a diagonal, say BD. Twotriangles are formed, ΔABD and ΔBCD. 43 In each triangle, the sum of the Dinterior angles is 180. C 5

To determine the sum of the four angles, study the following procedures.In ΔABD, m ∠ A + m ∠ 1 + m ∠ 4 = 180 by APEIn ΔBCD, m ∠ C + m ∠ 2 + m ∠ 3 = 180m ∠ A + m ∠ 1 + m ∠ 2 + m ∠ C + m ∠ 3 + m ∠ 4 = 360But m ∠ B = m ∠ 1 + m ∠ 2 and m∠D = m∠3 + m∠4By substitution, m ∠ A + m ∠ B + m ∠ C + m ∠ D = 360. Therefore, the conclusion is that, the sum of the measures of the interior angles of aquadrilateral is 360. How about the sum of the exterior angles of a quadrilateral? Is it the same as that ofthe sum of the exterior angles of the triangle?Exploration: Given the figure at the right. Quadrilateral ABCD has one exterior angle in each vertex.Those are ∠ 1, ∠ 2, ∠ 3 and ∠ 4. Let us name the interior angles, ∠ A, ∠ B, ∠ C and ∠ D. To determine the sum of the measures of 1 B∠ 1, ∠ 2, ∠ 3 and ∠ 4, study the following steps. A 2From the previous paragraph, it is known that m ∠ A + m ∠ B + m ∠ C + m ∠ D = 360.By definition of exterior angle, 4 3Cm ∠ A + m ∠ 1 = 180 Dm ∠ B + m ∠ 2 = 180m ∠ C + m ∠ 3 = 180m ∠ D + m ∠ 4 = 180By Addition Property of Equality or APE (m ∠ A + m ∠ B + m ∠ C + m ∠ D) + (m ∠ 1 + m ∠ 2 + m ∠ 3 + m ∠ 4) = 720Subtract: (m ∠ A + m ∠ B+ m ∠ C + m ∠ D) = 360 m ∠ 1 + m ∠ 2 + m ∠ 3 + m ∠ 4 = 360Therefore, the sum of exterior angles of a quadrilateral is equal to 360.Examples:1. The measures of the three angles of a quadrilateral are 75º, 101º and 83º. Find the measure of the fourth angle. 6

Solution: Let x = the measure of the fourth angle. x + 75º + 101º + 83º = 360º x + 259º = 360º x = 360º - 259º x = 101º2. Given quadrilateral PRST. Using the figure, R S find x, m ∠ P, m ∠ R, m ∠ S and m ∠ T. 2x 3x - 23Solution: Write the representations for each angle. 2x - 10 x + 33 m ∠ P = 2x – 10 m ∠ R = 2x P T m ∠ S = 3x – 23 m ∠ T = x + 33Using the representations for each angle, substitute to the equation m ∠ P + m ∠ R + m ∠ S + m ∠ T = 360 (2x – 10) + (2x) + (3x – 23) + (x + 33) = 360 8x + 33 – 33 = 360 8x + 0 = 360 8x = 360 x = 360 8 x = 45 m ∠ P = 2x – 10 = 2(45) – 10 = 90 – 10 m ∠ P = 80 m ∠ R = 2x = 2(45) m ∠ R = 90 m ∠ S = 3x – 23 = 3(45) – 23 = 135 – 23 m ∠ S = 112To check: m ∠ T = x + 33 = 45 + 33 m ∠ T = 78 m ∠ P + m ∠ R + m ∠ S + m ∠ T = _____ 80 + 90 + 112 + 78 = 360 7

Try this outA. Use the given information to find the measure of each 45 numbered angle in the figure. ∠ 1 ≅ ∠ 5.1. m ∠ 2 = 106, m ∠ 6 = 85 1 23 62. m ∠ 1 = 43, m ∠ 7 = 1123. m ∠ 2 = 115, m ∠ 4 = 31 74. m ∠ 5 = 51, m ∠ 3 = 72B. In the figure, determine the measure of the 460 820 numbered angles. 270 4 2 4. m ∠ 1 5. m ∠ 2 1 6. m ∠ 3 7. m ∠ 4 3 390C. In quadrilateral MNOP, ∠ M ≅ ∠ O, M 2x–7 x + 16 N ∠ N ≅ ∠ P. Using the given in the figure, find the measures of P O all the angles. 32 410 5D. Use the given information to find the measures of the following angles. 580 450 750 4 1. ∠ 1 2. ∠ 2 3. ∠ 3 4. ∠ 4 5. ∠ 5 6. ∠ 6 Lesson 3Finding the Sum of the Interior Angles and Exterior Angles of Any Polygon. To determine the sum of the measures of interior angles and exterior angles of anypolygon (polygon of n sides) like the given pentagon, you have to find out the number oftriangles that can be formed by drawing the diagonals without intersecting each otherexcept at their endpoints. 8

In quad ABCD, draw diagonal AC. A B D C With the introduction of diagonal AC,two triangles are formed. The same thinghappens if diagonal BD is drawn. If thesum of the measures of the interior anglesof each triangle is 180, then for quadrilaterals,the sum of the interior angles is 360. The polygon at the right has five (5) sides. DEIf the diagonal is drawn from one vertex, say D,only two diagonals can be drawn, DF and DG .With these two diagonals, three non-overlapping Htriangles are formed, ΔDEF, ΔDFG and ΔDGH. FIf in each triangle, the sum of the measures Gof interior angles is 180, then in pentagonDEFGH, the sum of the measures of the fiveangles is equal to 3(180) or 540. This findings can be summarized using the formula Sa = (n – 2) 180, where Sq is thesum of the measures of interior angles of the polygon, n is the number of sides of thepolygon. This formula can be applied to convex polygons only. You also have to remember this: The sum of the measures of interior angles of aregular polygon with n sides is equal to Sa = (n – 2) 180.Regular polygons are polygons which are both equilateral and equiangular. A polygon is convex if and only if the lines containing the sides of the polygon do notcontain points in its interior. To get the measure of each interior angle of a regular polygon you simply rememberthis. The measure of each interior angle of a regular polygon with n sides is equal to Ia =(n − 2)180 , where, Ia is an interior angle, and n is the number of sides of the polygon. n If you can get the measure of each interior angle of a regular polygon, then you canalso compute for the measure of each exterior angle of a regular polygon. You only have toremember these formulas. From the generalizations that you had made in the earlier part of the lesson, the sumof the measures of the exterior angle of the convex polygon is equal to 360. This includesthe regular polygons too. 9

The measure of each exterior angle of a regular polygon with n sides is given by theformula: Ea = 360 , where Ea is an exterior angle of a regular polygon, and n is the number nof sides of the polygon.Examples:1. What is the sum of the measures of the interior angles of a convex polygon witha. 11 sidesb. 15 sidesSolutions: Sa = (n – 2) 180 a. = (11 – 2) 180 = 9(180) = 1620b. Sa = (n – 2) 180 = (15 – 2) 180 = 13(180) = 23402. How many sides does a convex polygon have if the sum of the measures of its interiorangles is 1440?Solution: Sa = (n – 2) 180 1440 = (n – 2) 180 1440 = 180n – 360 n = 1440 + 360 180 n = 1800 180 n = 10The polygon has 10 sides.3. Find the sum of the measures of the interior angles of a convex heptagon.Solution: Sa = (n – 2) 180 = (7 – 2) 180 = 5(180) = 900 10

4. Find the measure of each interior angle of a regular 11-sided polygon.Solution: Ia = (n − 2)180 Measure of each interior angle is n Ia = (11 − 2)(180) 11 Ia = 9(180) 11 Ia = 1620 115. How many degrees are there in each of the exterior angle of a regular hexagon.Solution: Ea = 360 n Ea = 360 6 = 606. If each exterior angle of a polygon is 36º, how many sides does the polygon have?Solution: Ea = 360 n 36 = 360 n 36n = 360 n = 360 36 n = 107. If each interior angle of a polygon is 150º, how many sides does the polygon have?Solution: m + 150 = 180 m = 180 – 150 m = 30Using the figure, with m as the exterior angle of the polygon,Ea = 360 n 11

30 = 360 m 150 n30n = 360 n = 360 30 n = 12Try this outA. Complete the table below. No. of Sum of the Polygon No. of No. of Diagonals from No of Measures of Sides Vertices one vertex triangles interior Angles 1 Quadrilateral 2 Pentagon 3 Hexagon 4 Nonagon 5 Dodecagon 6 N-gonB. Find the number of sides in each regular polygon with exterior angle of the givenmeasure. 1. 90º 2. 45º 3. 36º 4. 60ºC. How many sides does a regular polygon have if each angle measures: 5. 120º 6. 90º 7. 135º 8. 150ºD. Find the sum of the measures of the exterior angles one at each vertex, of each of thefollowing convex polygon. 9. decagon 10. hexagon 11. octagon 12. dodecagonE. The sum of the measures of the interior angles of a polygon is given. Find the number ofsides of the convex polygon. 13. 1260º 14. 1620º 15. 2520º 16. 3060º 12

F. The sum of the measures of the interior angles of a regular polygon is given. Find themeasure of each angle. 17. 1260º 18. 2520º 19. 3060º 20. 720ºG. Solve the following problems. 62021. Find y in the given figure. 680 y22. The measures of the two angles of a quadrilateral are 105º and 107º. If the remainingtwo angles are congruent, find the measure of each angle.23. In a convex quadrilateral WXYZ, the measure of ∠ X is twice the measure of ∠ Y. If∠ Y ≅ ∠ Z ≅ ∠ W, find the measure of each angle.24. The number of diagonals from a vertex in a regular polygon is 7. How many sides doesthe polygon have? What is the measure of each interior angle?25. Each interior angle of a regular polygon is twice the measure of each exterior angle.How many sides does the polygon have? M N26. In the figure, like markings indicate 420 congruent parts. Find the measure of each unknown angle. P 420 O A27. ∠ ABC is a right angle, ∠ CDB is D a right angle. If the m ∠ C = 32, find the measures of the following angles: CB ∠A ∠ ADB S ∠ ABD ∠ CBD 7 428. In the given figure, RV = UV, RU = SU, ST = TU, m ∠ RVU = 92, m ∠ STU = 106, R6 m ∠ U = 137, find the measures of all 1 the numbered angles. T V 920 1060 53 2U 13

Let’s summarize 1. The sum of the measures of the interior angles of a triangle is 180. 2. The sum of the measures of the interior angles of a quadrilateral is 360. 3. The sum of the measures of the interior angles of a convex polygon is given by the formula (n -2)180, where n is the number of sides of the polygon. 4. The sum of the measures of the exterior angle one on each vertex of a polygon is 360. 5. The measure of each interior angle of a regular polygon of n sides is given by the formula Ia = (n − 2)180 , where n is the number of sides of the polygon. n 6. The measure of each exterior angle of a regular polygon of n sides is given by the formula Ea = 360 , where n is the number of sides of the polygon. n What have you learnedAnswer each question as indicated. 1. What is the measure of each angle of a regular pentagon? 2. The sum of the measures of the two angles of a triangle is 127. What is the measure of the third angle? 3. The measure of one base angle of an isosceles triangle is 67. Find the measure of the vertex angle. AB 4. Given quadrilateral ABCD. If m ∠ C = 60, and if ∠ A ≅ ∠ B, what is the m ∠ D if its measure is 27 less than the measure of ∠ A. D C 5. What is the sum of the measures of the interior angles of an octagon? 6. If each interior angle of a polygon is 160º, how many sides does the polygon have? 14

7. Find the value of x using the given in the P Q figure 2x + 10 3x xR 2x x T S8. The angles of a triangle are in the ratio 1:3:5. Find the measures of the three angles.9. If each interior angle of a regular polygon measures 150º, how many sides does the polygon have?10. In the figure, AN ≅ XN . If ∠ A = 62º, find the measure of ∠ 1. A 1XN15

Answer KeyHow much do you know. 1. 60º 2. 50º 3. 63º 4. 120º 5. 65º 6. 360º 7. 6 8. 90 9. 24 cm 10. circle ATry this outLesson 1A. 1. 74º 2. 72º 3. 65º 4. 110º 5. 56 1 º 2B. 1. 45º, 90º 2. 50º, 80º 3. 53º, 74º 4. 37 1 º, 105º 2 5. 63 1 º, 53 1 º 42C. 1. 52 1 º 2 2. 40º 3. 70 1 º 2 4. 46º 5. 41 1 º 2 16

D. 1. 45º 2. 54º 3. 38º 4. 47º 5. 28 1 º 2E. 1. 30º, 60º, 90º 2. 30º, 60º, 90º 3. 33º, 33º, 114º 4. 29º, 68º, 83º 5. 45º, 45º, 90ºLesson 2A. 1. m ∠ 3 = 74, m ∠ 4 = 53, m ∠ 5 = 21, m ∠ 1 = 21, m ∠ 7 = 95 2. m ∠ 2 = 11, m ∠ 3 = 69, m ∠ 4 = 26, m ∠ 6 = 68, m ∠ 5 = 43 3. m ∠ 1 = 34, m ∠ 3 = 65, m ∠ 5 = 34, m ∠ 6 = 81, m ∠ 7 = 99 4. m ∠ 1 = 51, m ∠ 2 = 108, m ∠ 4 = 21, m ∠ 6 = 57, m ∠ 7 = 123B. 5. m ∠ 1 = 114 6. m ∠ 2 = 66 7. m ∠ 3 = 68 8. m ∠ 4 = 32C. m ∠ M = m ∠ O = 2(57) – 7 = 107 m ∠ P = m ∠ N = 57 + 16 = 73D. 1. m ∠ 1 = 120 2. m ∠ 2 = 19 3. m ∠ 3 = 62 4. m ∠ 4 = 15 5. m ∠ 5 = 45 6. m ∠ 6 = 60 17

Lesson 3A. Polygon No. of No. of No. of No of Sum of the Sides Vertices Diagonals triangles Measures 1 Quadrilateral from one Of interior 2 Pentagon 4 4 2 3 Hexagon 5 5 vertex 3 Angles 4 Nonagon 6 6 1 4 360 5 Dodecagon 9 9 2 7 540 6 N - gon 12 12 3 10 720 n N 6 N–2 1260 9 1800 N–3 (n – 2 )180B. 1. 4 sides 2. 8 sides 3. 10 sides 4. 6 sidesC. 5. 6 sides 6. 4 sides 7. 8 sides 8. 12 sidesD. 9. 360 10. 360 11. 360 12. 360E. 13. 9 sides 14. 11 sides 15. 16 sides 16. 19 sidesF. 17. 140º 18. 157.5º 19. 161.05º 20. 120ºG. 21. y = 130º 22. Each of the remaining angle measures 74º. 23. m ∠ W = 72, m ∠ X = 144, m ∠ Y = 72, m ∠ Z = 72 24. 10 sides, 144 25. 6 sides 26. m ∠ N = 69, m ∠ MON = 69, m ∠ P = 69, m ∠ PMO = 69 18

27. m ∠ A = 58, m ∠ ADB = 90, m ∠ ABD = 32, m ∠ CBD = 58 28. m ∠ 1 = 44, m ∠ 2 = 44, m ∠ 3 = 37, m ∠ 4 = 37, m ∠ 5 = 56, m ∠ 6 = 62 m ∠ 7 = 62What have you learned 1. 108º 2. 53º 3. 46º 4. 82º 5. 1080º 6. 18 sides 7. x = 50º 8. 20º, 60º, 100º 9. 12 sides 10. m ∠ 1 = 124 19

Module 5 Geometry of Shape and Size What this module is about This module will discuss the perimeter of the commonly used polygons in geometrysuch as triangles, quadrilaterals and others (with their corresponding formulas). In addition,this module will also expound on the circumference of the circle. Furthermore, this modulewill help you apply these concepts in solving problems associated with real life. What you are expected to learn This module is written for you to1. Recall the different plane figures and their properties which are commonly used in geometry.2. Name the properties of the sides of different polygons.3. Define perimeter and determine the different formulas of getting the perimeter of the different polygons .4. Define a circle.5. Identify the lines and segments associated with circles.6. Define circumference of a circle and determine the formula for getting the circumference of the circle.7. Use the formulas for finding perimeter and circumference in solving real life problems.How much do you knowAnswer the following questions as indicated.Find the perimeter of a regular polygon(indicated) given the length of a side.1. triangle, s = 13 cm2. square, s = 10 cm3. rhombus, s = 11.5 dm4. pentagon, s = 9.25 cm5. hexagon, s = 12.73cm

6. The circumference of a circle is 66 cm. Find the radius of the circle. Use π = 22 . 77. The length of a rectangle is 13 more than twice its width. If the perimeter is 116 cm, find the dimensions of the rectangle.8. One of the sides of a rhombus is 2x + 1. What is its perimeter?9. The diameter of a circle is 14 cm. Find its circumference. Use π = 3.14.10. If the perimeter of a regular hexagon is 69 cm, what is the length of each side? What you will do Lesson 1 Perimeters of Polygon The perimeter of a polygon is the distance around it. It can be computed by gettingthe sum of the length or measures of all the sides. If the polygon is identified as regular,then the perimeter is computed by simply multiplying the given measure of the side with thenumber of sides. Let us recall the different formulas for finding the perimeter of differentpolygons.Triangle:For a general triangle, the perimeter (P) is ab P = a + b + c , where a, b, and c c are the measures of the sides.For isosceles triangle, perimeter is mm P = m + m + n, where m is the length of one of the legs, and n is the length of the base. nFor equilateral triangle, the perimeter is s P = 3s, where s is the length of one of the equal sides. 2

For polygon of four sides or quadrilaterals, consider the given figures.Given quad ABCD, then its perimeter is x y P = w + x + y + z, where w, x, y w and z are the length of the sides zFor a square, its perimeter is s P = s + s + s + s, where s is the length of a side. Therefore, P = 4sFor a rectangle, the perimeter is w l P = l + l + w + w, where l is the length and w is the width. Therefore, w P = 2l + 2w or P = 2 (l + w) l aFor a parallelogram, the perimeter is P = a + a + b + b, where a, and b are the bb lengths of the two consecutive sides a of the parallelogram. Thus P = 2a + 2b P = 2(a + b)For other figures like pentagon and hexagon or those polygons with more than fivesides, the formula is practically the same. Get the sum of all the length of the sides. aFor the given pentagon, the perimeter isP = a + b + c + d + e, where a, b, c, bcd, and e are the measures of thesides. efFor a regular pentagon with a as the length of a side, the perimeter isP = 5a s rFor a hexagon, the perimeter is t uP = r + s + t + u + v + w, wherer, s, t, u, v, and w are the length wof the sides. v 3

For a regular hexagon whose length of side is denoted by r, the perimeter is given bythe formula P = 6rExample 1. Find the perimeter of the following figures.1. 2. 3. 3cm 5 cm 6 cm 6 cm 5 cm 4 cm4. 5. 7 cm 6. 7 cm 5 cm 3 cm 5 cm 9 cm 6 cmSolutions:1. P = (3 + 5 + 6)cm = 14 cm2. P = [2(6) + 4] cm = 12 + 4 = 16 cm3. P = 3(5 cm) = 15 cm4. P = 4(6 cm) = 24 cm5. P = [2(7cm) + 2(5cm)] = 14cm + 10cm = 24cm6. P = (3 + 7 + 5 + 9)cm = 24cmExample 2. Find the perimeter of the following regular polygons given the measure of a side (s).1. triangle, s = 10 cm 4

2. square, s = 7cm3. pentagon, s = 13.5cm4. hexagon, s = 11.25 cm5. nonagon, s = 9.3 cmSolutions:1. P = 3s = 3 (10cm) = 30 cm2. P = 4s = 4(7cm) = 28cm3. P = 5s = 5(13.5cm) = 67.5 cm4. P = 6s = 6(11.25cm) = 67.50cm5. P = 9s = 9(9.3cm) = 83.7cmExample 3 The width of a rectangular garden is 5m. If the length is 2m more than its width, how many meters of fencing materials are needed to enclose the whole garden?How much will the cost of fencing material be if the owner pays P59 per meter?Solution:Width (w) = 5mLength (l) = 2m more than the widthLength (l) = 5m + 2m = 7mGet the perimeter:P = 2(l + w) = 2(7m + 5m) = 2(12m) = 24mCost of materials = 24m (P59 /m) = P1416.00 5

Example 4. The perimeter of a regular pentagon is 120 m. What is the length of each side?Solution:Pentagon has 5 sides, and since it is a regular pentagon, the sides are equal.P = 5s 5s = 120 m s = 120m 5 s = 24m.Example 5. A saleslady is preparing to wrap a box whose dimensions are 30cm by 20cm by 7cm. If she is going to tie a ribbon around the box as in the figure, how long should the ribbon be if she will allow 25cm for the design at the top? How much will the ribbon cost if a meter is P8.25.Solution: To find the length of the ribbon, find the two perimeters.P1 = 2(30cm) + 2(7cm) = 60cm + 14cm = 74 cmP2 = 2(20cm) + 2(7cm) = 40cm + 14cm = 54 cmTotal length = P1 + P2 + 25cm = 74cm + 54cm + 25cm = 128cm = 1.28 mTotal cost = 1.28 m(P8.25) = P12.6225 ≈ P12.65Try this outA. Find the perimeter of the following using the given information in the figure.1. 2. 3. 95 7 15.3 10 10 6

4. 21 5. 12 15 10 16B. The sides of an isosceles triangle are given. Find the perimeter. 1. 8, 8, 13 2. 5, 4, 4 3. 7, 7, 5 4. 6, 7, 6 5. 1.2, 1.2, 1.3 6. 8.13, 8.14, 8.13 7. 12.75, 12.75, 10 8. 1, 2, 2C. Given are the length and width of a rectangle. Find its perimeter. 1. 4.5, 8.3 2. 12.01, 19.22 3. 18.3, 21.5 4. 2.03, 5.43 5. 9.75, 12.25 6. 3 , 5 3 7. 4 5 , 8 5 8. a, 3a 9. x + 1, 5x + 4 10. x2 + 1, x2 + 25D. Find the perimeter of the following 1. A square with side of 25cm. 2. A parallelogram whose consecutive sides are 21cm and 27.5 cm respectively. ( )3. A regular hexagon whose side is 2 3 + 7` cm. 4. A rectangle 17m by 11m. 5. A rhombus with side of 18.23dm.E. Solve the following problems. 1. The length of a rectangle is 4 less than three time its width. If the perimeter of the rectangle is 272, what are its dimensions? 7

2. The perimeter of a regular pentagon is 215 cm. Find the length of each side.3. In an isosceles trapezoid, the length of one leg is 53dm and the median is 63dm. Find its perimeter.4. James always jogs around his rectangular pool, 10m by 6m. If he jogs around it 5 times, what is the distance covered by James?5. Find the perimeter of the given quadrilateral. 4x - 9 2x + 1 3x - 46. When the side of a square is increased by 2cm, its perimeter becomes 40 cm. What is the length of the original square?7. When the side of a square is reduced by 7cm, the perimeter becomes 84. What is the perimeter of the original square?8. The length of a rectangle is 64 cm. Its width is 13cm more than half its length. What is the perimeter of the rectangle?9. The base of an isosceles triangle is 15 more than one-third of the length of the leg. If a leg measures 57 dm, find the perimeter of the triangle.10. If the side of a square is increased by 25%, by how many percent will the perimeter increase? Lesson 2Circle and its Circumference A circle is defined as the set of points equidistant from a fixed point called the center.Though the center is not a part of the circle, it is essential that every circle has a center Acircle on a plane can be represented geometrically or algebraically. In this part of thelesson, geometric circles will be the focus of our discussion. In the given circle, A is the center of the circle, ●Athus we can name the circle, circle A. There are otherlines and segments associated with a circle likeradius, chord, diameter, secant and tangent.8

In circle O, OA is a segment from the center to the B 0circle. OA therefore is a radius of the circle. Aside from OA , ACother radii are OB and OC . BC is a segment whoseendpoints are points on the circle and it passes throughthe center of the circle. Therefore, BC is called thediameter of the circle. If the endpoints of a segment arepoints on the circle, it is called chord of the circle. By this definition, the diameter is also a chord of a circle. By inspection, it is veryeasy to recognize that the length of the diameter is twice that of the radius. Therefore, in thefigure, BC = OB + OC . Other lines and segments associated with a circle DEare shown in the figure. They are BC,DE and line t. BC has ●A Cits endpoints on the circle. Thus it is an example of chord Bof the circle. It is not a diameter since it did not pass through TXthe center of the circle. DE intersects the circle at two pointsD and E. It is called a secant of the circle. Line t intersects thecircle at only one point X. Line t is called tangent of the circle.Example 1. Given circle A and points B, H, C, D and E on it.Name: B H1. 4 radii A C2. 2 chords3. a diameter E4. a secant D5. a tangent FSolutions: 1. AB, AC, AH , AE 2. EH , EC 3. EH 4. EC 5. DF Every polygon has its own perimeter. Likewise a circle has its own perimeter too. Todistinguish it from those of polygons, we call the distance around the circle circumference.There is also a formula for finding the circumference of a circle. The letter “C” will be usedto represent the circumference of the circle. The formula that we will use is 9

C = 2π r, where r is the radius of the circle and π is an irrational number whosevalue is approximated at 3.1416 or 22 . 7 An alternative formula for circumference can be used utilizing the diameter instead ofthe radius of the circle. C = π D, where D is the length of the diameter of the circle.Example 1. Find the circumference of a circle of radius 9cm.Solution: C = 2π r C = 2π (9cm) C = 18π cm If there is assigned value for π , say 22 , then 7 C = 2( 22 )(9cm) 7 C = 396 cm 7 C ≈ 56.57cm If π ≈ 3.1416, then C = 2(3.1416)(9cm) C ≈ 56.55cmExample 2. The diameter of a circle is 12.6 dm. What is its circumference? Express theanswer in terms of π .Solution: C= πD C = π (12.6dm) C = 12.6π dmExample 3. The circumference of a circle is 39.27cm. Find the radius of the circle. Useπ =3.1416.Solution: C = 2π r r= C 2π r = 39.27 2(3.1416) r = 6.25cm 10

Try this outA. Find the length of the diameter of a circle given its radius. 1. 24cm 2. 19 m 3. 3.96dm 4. .08 cm 5. 4 dm 7 6. 18 m 7. (x+2) cm 8. x km 2 9. m + n m10. 2a − b dm 4B. Use the figure and answer each of the following: BC D 1. What is the center of the circle? E F 2. Name the circle. 3. Name 5 radii A 4. Name 2 diameters HG 5. Name all the chords 6. Name a tangent of the circle 7. Name a secantC. Find the circumference of the circle given the radius of the circle. Express answers interms of π .1. 19cm2. 25dm3. 11cm4. 8cm5. 13.75cm6. 2.03cm7. 5 m 78. 5cm9. a dm10. (x + 3)cm 11

D. Find the circumference of the circle if the diameter is given. Use π = 3.1416.1. 24cm2. 50dm3. 35m4. 3 km 45. 12.56m6. 19.75cm7. 1 km 38. 18.7cm9. 25.76cm10. 10.705cm11. 2 m 512. 105.031cm13. (a+b)dm( )14. 2 +1 km15. 6 5 dm16. 3x2 cm17. (5x – 1) cm18. 7x + 4 dm19. 3.1416m20. 3 m 5+3E. Given the circumference of the circle, find the radius and the diameter of the circle. Usethe appropriate value of π .1. 36π2. 58π3. 126π4. 39π5. 101π6. 37.24π7. 65.78π8. 137.5π9. 89.93π10. 452.76π11. 14.2512. 549.7813. 298.45214. 314.16 12

15. 75.5 16. 78.54 17. 3π 18. 18456.9 19. 72 π 20. (10x – 4y) πF. Solve the following problems. 1. An artificial lake has a diameter of 34 m. If Teena jogs around it 6 times, how many meters will that be? (Use π =3.14) 2. At one point in a race, Joseph was 15 m behind Allan and 18 m ahead of Brad. Brad was trailing Nick by 30m. Allan was ahead of Nick by how many meters? 3. What is the perimeter of a square inscribed in a circle of radius 10 cm? Let’s summarize 1. Perimeter of a polygon is the distance around the polygon. 2. The general formula for finding the perimeter of a polygon is Perimeter = s1 + s2 + s3 +. . . + sn-1 + sn , where sn is the measure of the sides and n is the number of sides of the polygon. 3. For regular polygons, the perimeter is equal to the measure of a side multiplied by the number of sides. 4. A circle is the set of points equidistant from a fixed point called the center. The center of the circle defines the name of the circle. 5. Every circle must have a center. 6. Lines and segments associated with circle are the following: Radius – the segment joining the center and any point on the circle. Chord – segment joining any two points on the circle. Diameter – chord which passes through the center. Secant – a line intersecting a circle at two points. Tangent – a line intersecting a circle at only one point. 7. Circumference of a circle is the distance around a circle. 13

8. The formula for finding the circumference of a circle is C = 2π r , where r is the radius of the circle and π is an irrational number approximately equal to 3.1416 or 22 . 7What have you learned1. What is the perimeter of a regular heptagon with side of 23 cm?2. The length of a leg of an isosceles trapezoid is 15 dm. The length of the median is 35dm. What is the perimeter of the trapezoid? 3x - 73. Find the perimeter of the given rectangle. 2x + 94. The circumference of a circle is 141π . Find the radius of the circle.5. If the side of a square is increased by 4 cm, the perimeter becomes 136 cm. Find the length of the side of the original square.6. The perimeter of a regular pentagon is 17.5 m. What is the length of each side?7. What is the longest chord in a circle?8. If the diameter of a circle is 25 cm , what is the length of the radius of the circle?9. The base of an isosceles triangle is 27 dm. If the length of a leg is 11 more thanone-third of the base, find the perimeter of the triangle. D 3x - 7 C10. ABCD is a parallelogram. Using the given in thefigure, find x if the perimeter is 130 cm. 2x + 1 AB14

Answer KeyHow much do you know 1. 52 cm 2. 40 cm 3. 46 dm 4. 46.25 cm 5. 76.38 cm 6. 10.5 cm 7. length = 43, width = 15 8. 8x + 4 9. 43.96 cm 10. 11.5 cmLesson 1A. 1. 24 2. 24 3. 61.2 4. 72 5. 48B. 1. 29 2. 13 3. 19 4. 19 5. 3.7 6. 24.4 7. 35.5 8. 5C. 1. 25.6 2. 62.46 3. 79.6 4. 14.92 5. 44 6. 12 3 7. 24 5 8. 8a 9. 12x + 10 10. 4x2 + 52 15

D. 1. 100 cm 2. 97 cm ( )3. 12 3 + 42 cm 4. 56 m 5. 72.92 dmE. 1. length = 101, width = 35 2. 43 cm 3. 232 dm 4. 160 m 5. 11x - 11 6. 8 cm 7. 112 cm 8. 218 cm 9. 148 dm 10. 25%Lesson 2A. 1. 48 cm 2. 38 m 3. 7.92 dm 4. 0.16 cm 5. 8 dm 7 6. 6 2 m 7. (2x + 4)cm 8. x cm 9. 2 m + n m 10. 2a − b dm 2B. 1. E 2. circle E 3. EC, ED, EF, EG, EA 4. AD,CG 5. AB, AD,CG 6. HG 7. AB 16

C. 1. 38π cm 2. 50π dm 3. 22π cm 4. 16π cm 5. 27.5π cm 6. 4.06π cm 7. 10 π m 7 8. 2 5 cm 9. 2a dm 10. (2x + 6) cmD. 1. 75.3984 cm 2. 157.08dm 3. 109.956 m 4. 2.3562 km 5. 39.458 m 6. 62.0466 cm 7. 1.0472 km 8. 58.74797 cm 9. 80.93 cm 10. 33.63 cm 11. 1.2566 m 12. 329.96 cm 13. 3.1416(a+b) dm 14. 7.58 km 15. 42.15 dm 16. 9.4248x2cm 17. (15.708x – 3.1416) cm 18. 3.1416( 7x + 4 ) dm 19. 9.8696 m 20. 1.8 mE. r = 18, d = 36 1. r = 29, d = 58 2. r = 63, d = 126 3. r = 19.5 d = 39 4. r = 50.5, d = 101 5. r = 18.62, d = 37.24 6. r = 32.89, d = 65.78 7. r = 68.75, d = 137 8. r = 44.965, d = 89.93 9. 17

10. r = 226.38, d = 452.7611. r = 2.268, d = 4.53612. r = 87.34, d = 174.6813. r = 47.5, d =9514. r = 50, d =10015. r = 12.02, d = 24.0416. r = 12.5, d = 2517. r = 3 , d = 3 218. 2937.5, d = 587519. 3 2 , d= 6 220. 5x – 2y, d = 10x – 4yF. 1. 640.56 m 2. 3 meters 3. 40 2 cmWhat have you learned1. 161 cm2. 100 dm3. 10x + 44. 70.55. 30 cm6. 3.5 m7. diameter8. 12.5 cm9. 67 dm10. 14.2 cm 18

Module 6 Geometry of Shape and Size What this module is about This module is about areas of plane figures. In this module you will study the areas ofsquares, rectangles, parallelograms, triangles, trapezoids and circles, and learn to computethem. What you are expected to learnThis module is designed for you to 1. Apply the formulas for the measurements of the following plane figures a. square, b. rectangle, c. parallelogram, d. triangle, e. trapezoid, and f. circle. 2. Solve problems involving areas of plane figures How much do you know 1. Find the area of a square whose side is 9 cm. 2. Find the area of the triangle below. 8 cm6 cm

3. Find the area of a rectangle whose length and width are 12 cm and 5 cm respectively.4. Find the area of the trapezoid ABCD below B 6 cm C D 4 cm A 10 cm5. Find the area of a circle whose radius is 7 mm.6. Find the base of a triangle if the altitude is 4 cm and the area is 16 cm2.7. Find the area of a parallelogram with base 12 cm and height 8 cm.8. The area of the parallelogram ABCD below is 96 cm2. Find x. B C D X A 12 cm9. Find the area of the figure 9 cm 5 cm 3 cm6 cm 1 cm10. Beth’s garden is 4 meters wide and 6 meters long. Find the area of the garden. 2

What you will do Lesson 1 Areas of Rectangles and SquaresThe first figure below is a rectangle and the second is a rectangular region.Rectangle Rectangular region A rectangular region is a union of a rectangle and its interior. When you are asked tofind the area of a rectangle, you are actually asked to determine the area of a rectangularregion. The area of a region is the number of square units contained in the region. A squareunit is a square with a side 1 unit in length. I unit l unitExample 1 In the rectangle below, each small square is one unit in length. Find the area of the rectangle. The area can be determined by counting the number of small squares. Since thereare 24 small squares, therefore, the area is 24 square units. The standard units of area are square units, such as square centimeters, squaredecimeters and square meters. 3

Example 2 Find the area of the rectangle below. 12 cm 9 cm Solution: The length ( l ) of the rectangle is 12 cm and the width (w) is 9 cm. Substitute thesedata in the formula. A = lw = (12 )( 9) = 108 cm2 The area is 108 square centimeters.Example 3 Find the area of a rectangle whose length and width are 14 cm and 12 cmrespectively. Solution: Step 1. Draw and label the figure. 12 cm 14 cm Step 2. Substitute the data in the formula. The figure shows that l = 14 cm and w = 12 cm. A = lw = (14)(12) = 168 cm2 The area is 168 square centimeters.. 4

Example 4 Find the width of the rectangle with an area of 80 cm2 and length equal to 10 cm.Solution: Step 1. Draw and label the figure. 10 cm A = 80 cm2 Step 2. A = 80 cm 2 , l = 10 cm. Substitute these data in the formula. A = lw 80 = 10 w 10w = 80 w = 8 cm The width is 8 cmExample 5 The area of rectangle EFGH below is 48 cm2. Find its width. 8 cm (x + 2) cmSolution: In the figure, l= 8 cm and w = (x + 2) cm. Substitute 8 in place of l and (x + 2) inplace of w in the formula A = lw. Replace A by 48 since the area of the rectangle is given as48 cm2. Then solve the resulting equation for x. A = lw. 48 = 8 ( x + 2 ) 48 = 8x + 16 48 –16 = 8x 32 = 8x or 8x = 32 x=4 Since the width is represented by x + 2, then the width is 4 + 2 or 6 cm 5

The Area of a Square When the length and width of a rectangle are equal, the figure is a square. Theformula for finding the area of a square is A = s2 where s = length of a side. Ds C s AB Square AB = BC = CD = DAExample 1Find the area of the square ABCD.Solution: DC A = s2 = 62 s = 6 cm = (6)(6) = 36 cm2 A B The area is 36 square centimetersExample 2Find the side of a square whose area is 25 cm2.Solution:Step 1. Draw and label the figure.Step 2. Substitute the data in the formula. A = 25 cm2 s = ? A = s2 25 = s2 s2 = 25 s2 = 25 s = 5 cm The side is 5 centimeters 6

Example 3 Find the area of the shaded region. 20 cm15 cm 8 cm 8 cm Solution: Step 1. Find the area of the rectangle A = lw = (20)(15) = 300 cm2 Step 2. Find the area of the square A = s2 = 82 = (8)(8) = 64 cm2 Step 3. Find the area of the shaded region by subtracting the area of the square from the area of the rectangle. A = lw - s2 A = 300 – 64 = 236 cm2 The area of the shaded region is 236 square centimeters. Another important thing you will learn from this lesson is that areas of plane figurescan be added as long as the figures do not overlap. 7

The area of the entire figure in the following illustration is the sum of the areas of therectangle and the square.Example:Find the area of the figure below. C B 2 E F 6 4 A D4 G Notice that polygon ABCD is a rectangle and polygon DEFG is a square. The area ofthe entire figure is the sum of the areas of the rectangle and the square.Solution:Step 1. Find the area of the rectangle ABCD A1 = lw = (10)(6) = 60Step 2. Find the area of the square DEFG. A2 = s2 = 42 = (4)(4) = 16Step 3. Find the area of the entire figure by adding the area of the rectangle to the area of the square. A3 = A1 + A2 = 60 + 16 = 76 cm2The area of the entire figure is 76 square centimeters 8

Try this outSet AIn exercises 1-4, each plane figure is divided into small squares each with a side 1 unit inlength. 1. Find the area. A = _______ square units. 2. Find the area. A = ________ square units 3. Find the area. A = ________ square units 9

4. Find the area. A = _________ square unitsFind the area of each rectangle or square described below.5. 5 cm 7. 19 cm 5 cm 16 cm 8 cm 5.5 m6. 8. 16 cm 5.5 m 9. A rectangle with a length of 14 meters and a width of 10 meters.10. A square with a side of 7 mm.Set B 1. Find the area of a square whose side is 15 cm. 2. Find the area of a rectangle whose length and width are 12 m and 7 m respectively. 3. Find the area of a square whose side is 8.5 m 4. Find the area of a rectangle with length of 8 cm and width of 5 cm. 5. A side of a square is 13 cm. Find its area. 6. The length and width of a rectangle are 18 cm and 10 cm respectively. Find its area. 10

Find the area of the following figure.7. 2 cm 8. 3 cm 3 cm 3 cm2 cm 2 cm 10 5 cm 4 cmFind the area of the shaded region in the following figures9. 6 12 8 6 10 6 4 8Set CFind the area of each square1. side = 4 m 2. side = 4.5 cm 4. length = 9 cm and width = 8 cmFind the area of each rectangle 6. Area = 81 c m23. Length = 11 cm and width = 8 cmFind the length of a side of each square.5. Area = 36 cm2Find the length of each rectangle7. Area = 112 c m2 and width = 8 cm8. Area = 135 c m2 and width = 9 cm 11

Find the width of each rectangle9. Area = 176 c m2 and length = 16 cm10 Area = 216 c m2 and length = 18 cm Lesson 2 Areas of Parallelograms and Triangles The area of a parallelogram is equal to the product of the base times the height.The formula is A = bhExample 1 Find the area of a parallelogram with a base of 5 cm and a height of 3 cm. Solution: Step 1. Draw and label the figure. The base AB has length of 5 cm and the height or altitude is 3 cm. DC 3 cm A 5 cm BStep 2. Substitute the data in the formula A = bh = 5(3) = 15 cm2The area is 15 square centimetersExample 2Find the area of the parallelogram below h = 5.5 15.2 12

The figure shows that b = 15.s cm and h = 5.5 cm. Substitute these data in theformula. Solution: A = bh = (15.2)(5.5) = 83.6 cm2 The area is 83.6 square centimetersExample 3 The area of a parallelogram is 84 m2. The lengths of a base is 6 m. Find the length ofthe corresponding altitude. Solution: Step 1. Draw and label the figure.A = 84 m2 h=? 6m Step 2. Substitute the data in the formula A = bh 84 = 6h 6h = 84 h = 84/6 h = 14 m The height is 14 mExample 4 Find the height of a parallelogram that has a base of 14 cm and an area of 126 cm2Solution:Step 1. Draw and label the figure.A = 126 cm2 h=? 14 cm 13

Step 2. Substitute the data in the formula A = bh 126 = 14 h 14 h = 126 h = 126/14 h = 9 cmThe height is 9 centimetersThe Area of a TriangleThe diagonal separates the parallelogram into two congruent triangles. D CD C DAB B AB The diagonal BD divides the parallelogram ABCD into two congruent triangles. Twocongruent triangles have equal areas. The area of ∆ABD is equal to the area of ∆CDB.Since the formula for finding the area of a parallelogram is A = bh, therefore, the formula forfinding the area of a triangle is A = 1 bh. 2Example 1Find the area of the triangle shown below. h=4m 9mSolution: A = 1 bh 2 = 1(9)(4) 2 = 18 m2The area is 18 square meters 14

Example 2 Find the area of the triangle below. 12 dm 20 dmSolution A = 1 bh 2 = 1 (20) (12) 2 = 120 dm2The area is 120 square decimetersExample 3Find the area of the triangle shown below. 5.5 m 7.2 m Solution: A = 1 bh 2 = 1 ( 7.2) ( 5.5) 2 = 19.8 m2The area is 19.8 square metersExample 4 Find the area the triangle with base equal to 12 cm and altitude equal to 10 cm. 15