Name Date lass Eploring Angles of Polygons INV X 3 You have learned to identify interior and eterior angles in polygons. Now you will determine angle measures in regular polygons. Interior Angles Sum of the Interior Angles of a Polygon (n )180, where n is the number of sides of the polygon. Eample: Determine the sum of the interior angles of an octagon. Step 1: How many sides are in an octagon? 8 Step : Substitute and solve the formula. (n )180 (8 )180 6 180 1080 Interior Angle Measure of a Regular Polygon (n )180 n, where n is the number of sides of the polygon. Eample: Determine the interior angle measure of a regular octagon. Step 1: How many sides are in an octagon? 8 Step : Substitute and solve the formula. (n )180 (8 )180 n 8 6 180 8 135 omplete the statements to solve each problem. 1. What is the sum of the interior angles of a regular 15-gon? How many sides are in a 15-gon? 15 Substitute and simplify the formula. (n )180 (15 )180 13 180 30. What is the interior angle measure of a regular 1-gon? How many sides are in a 1-gon? 1 Substitute and simplify the formula. (n )180 n ( 1 )180 1 10 180 1 150 3. The sum of the measures of the interior angles of a regular 1-gon is 160.. The measure of an interior angle of a regular 30-gon is 168. 5. The sum of the measures of the interior angles of a regular 18-gon is 880. Saon. All rights reserved. 65 Saon Geometry
continued INV 3 Eterior and entral Angles Eterior Angle Measure of a Regular Polygon The measure of each eterior angle 360 n, where n is the number of sides in the polygon. Eample: Determine the measure of each eterior angle of a regular 1-gon. Step 1: How many sides are in a 1-gon? 1 Step : Substitute and solve the formula. 360 n 360 1 30 entral Angle Measure of a Regular Polygon The measure of each central angle 360 n, where n is the number of sides in the polygon. Eample: Determine the central angle of a regular -gon. Step 1: How many sides are in a -gon? Step : Substitute and solve the formula. 360 n 360 15 omplete the statements to solve each problem. 6. Determine the measure of each eterior angle of a regular 15-gon. How many sides are in a 15-gon? 15 360 n 360 15 7. Determine the central angle of a regular 30-gon. How many sides are in a 30-gon? 30 360 n 360 30 1 8. Determine the measure of each eterior angle for a regular 0-gon. 18 9. Determine the central angle of a regular 36-gon. 10 10. Determine the measure of each eterior angle of a regular 0-gon. 9 11. Determine the central angle of a regular 7-gon. 5 Saon. All rights reserved. 66 Saon Geometry
Name Date lass Flowchart and Paragraph Proofs 31 You have worked with triangle congruence (ASA and AAS). Now you will write flowchart and paragraph proofs. Flowchart Proofs You can use the flowchart proof to prove conjectures are true. Flowchart Proofs use boes and arrows. Steps go from top to bottom, as shown by arrows. The justification or reason for each step is written below the bo. Given: 1 and are right angles. Prove: 1 1 Flowchart Proof 1 and are right angles. Given m 1 = 90 m = 90 m 1 = m 1 Definition of right angle Transitive Property of Equality Definition of congruent angles 1. omplete the flowchart proof for the given two-column proof. Given: V is the midpoint of SW, and W is the midpoint of VT. Prove: SV is congruent to WT. Two-olumn Proof Flowchart Proof Statements Reasons 1. V is the midpoint of SW. 1. Given V is the midpoint of SW. Given W is the midpoint of VT. Given. W is the midpoint of VT.. Given 3. SV is congruent to VW. VW is congruent to WT.. SV is congruent to WT. 3. Definition of midpoint. Transitive Property of Equality SV VW Definition of midpoint SV WT Transitive Property of Equality VW WT Definition of midpoint. S V W T Saon. All rights reserved. 67 Saon Geometry
continued 31 Paragraph Proofs To write a paragraph proof, use sentences to write a paragraph that presents the statements and reasons. Use the two-column proof to write a paragraph proof. Given: A and DE. Prove: A DE Two-olumn Proof Statements Reasons 1. A 1. Given. DE. Given 3. A and DE 3. Definition of congruent segments. A DE. Transitive Property of Equality 5. A DE 5. Definition of congruent segments Paragraph Proof It is given that A and DE, so A and DE by the definition of congruent segments. A DE by the Transitive Property of Equality. Thus, A DE by the definition of congruent segments. A D E. Use the given two-column proof to complete the paragraph proof. Given: KL is a right angle. 1 Prove: 1 and are complementary angles. K Two-olumn Proof: Statements Reasons 1. KL is a right angle. 1. Given. m KL 90. Definition of right angle 3. m KL m 1 m 3. Angle Addition Postulate. 90 m 1 m. Substitution 5. 1 and are complementary angles. 5. Definition of complementary angles Paragraph Proof Since it is given that KL is a right angle, m KL 90 by the definition of a right angle. y the Angle Addition Postulate, m KL m 1 m. Using substitution, 90 m 1 m. Thus, by the definition of complementary angles, 1 and are complementary angles. Saon. All rights reserved. 68 Saon Geometry L
Name Date lass Altitudes and Medians of Triangles 3 You have written flowchart and paragraph proofs. Now you will solve problems involving altitudes and medians of triangles. Medians of Triangles In the picture, AH,, and G are medians of a triangle. They each join a verte and the midpoint of the opposite side. The point of intersection of the medians is called the centroid of A. entroid Theorem The centroid of a triangle is located of the distance from each 3 verte to the midpoint of the opposite side. Eample: Given: AH,, and G G are medians of A. N onclusion: AN AH, N G, N 3 3 3 A In A, suppose AH 18 and N 10. Use the entroid Theorem to find AN and. AN AH 3 entroid Theorem N 3 entroid Theorem AN 3 18 Substitute 18 for AH. 10 3 Substitute 10 for N. AN 1 Simplify. 15 Simplify. H In QRS, RX 8, and QW 30. Fill in the steps to find each length. 1. QZ. RW R OW QZ RW RX 3 3 30 QZ RW 8 3 3 5 QZ RW 3 O Y W X Z S In HK, HD 1, and K 18. Find each length. 3. H. D 1 7 D 5. K 6. 7 9 H E K Saon. All rights reserved. 69 Saon Geometry
continued 3 Finding a entroid on the oordinate Plane Find the centroid of KL with vertices at (0, ), K(6, ) and L(, 6) Step 1: Find the midpoint of two of the sides of the triangle. Since the three medians meet at the same point, the intersection of any two will give the location of the centroid. The midpoint of K is (3, 3). The midpoint of is LK is (5, ). Step : Graph the two medians and find the equation of each median. is a horizontal line, so the equation is y. To find the equation of LA, use the slope and the coordinates of one endpoint. The slope of is LA is 3. O 6 A L 6 K y y 1 m( 1 ) Point-slope Formula y 6 3( ) Substitute. y 3 6 Simplify. Step 3: Solve the two equations as a system: y and y 3 6. The y-coordinate is. Solve for the -coordinate: 3 6 3.3 The -coorinate is 3.3. The centroid is located at (3.3, ). Triangle DEF has coordinates D 1, 7, E 0, 3, and F 5, 5. omplete the steps to find the centroid of the triangle. 7. The midpoint of DE is ( 0.5, 5 ), and the midpoint of DF is ( 3, 6 ). The equation of the median from ( 0.5, 5) to F is y 5. The slope of the line from ( 3, 6) to E is 1. The equation of the line from ( 3, 6) to E is y 3. To find the coordinates of the centroid, solve the system of equations: y 5, and y 3. The coordinates of the centroid are (, 5 ). D 6 F E -6 - - O Triangle MNP has coordinates M( 3, 3), N( 1, 6), and P(1, 1). Find the centroid of the triangle. 8. centroid ( 1,.7) M N 6 5 3 1 y --3- -1-3 P Saon. All rights reserved. 70 Saon Geometry
Name Date lass onverse of the Pythagorean Theorem 33 You have worked with medians of triangles. Now you will solve problems that use the Pythagorean theorem. The Pythagorean Theorem The Pythagorean theorem states the following relationship among the lengths of the legs, a and b, and the length of the hypotenuse, c, of any right triangle. a c a b c b Use the Pythagorean theorem to find the value of in the triangle. Substitute 6 for a, for b, and 1 for c in the equation a b c and solve for. a b c 6 1 Substitute 36 1 Simplify 108 108 36 3 6 3 The length of the unknown side is 6 3. Pythagorean theorem Subtract Square root Simplify radical 6 1 omplete the steps to find the value of. 1. 5 8 5 6 5 5 6 5 39 39 8 5 Find the value of. Give your answer in simplest radical form.. 3. 0 16 5 1 9 Saon. All rights reserved. 71 Saon Geometry
Pythagorean Inequality Theorem You can use the following theorem to classify triangles by their angles if you know their side lengths. Always use the length of the longest side for c. Pythagorean Inequality Theorem continued 33 A A b m > 90 c a m < 90 b a c If a b c, then A is obtuse. If a b c, then A is acute. Determine if A is an obtuse, acute, or right triangle. Step 1: Let c 13. The other two lengths, a and b, will equal 7 and 9, respectively. Step : Substitute the lengths into a b c. a b c 7 9 13 9 81 169 A 9 13 7 130 169 Since a b c, the triangle is an obtuse triangle. omplete the steps to determine if the given side lengths form an acute, obtuse, or right triangle.. 10, 11, 1 5. 3,, 5 10 11 1 3 5 100 11 1 9 16 5 1 1 5 5 The triangle is a/an acute triangle. The triangle is a/an right triangle. Determine whether the triangle with the given side lengths is an obtuse, an acute, or a right triangle. 6. 8, 13, 16 7., 7, 8 obtuse obtuse 8. 8, 8, 11 9. 9, 1, 15 acute right Saon. All rights reserved. 7 Saon Geometry
Name Date lass Properties of Parallelograms 3 All parallelograms have the following properties. F G H FG H GH F F G H F H G Opposite sides are congruent. Opposite angles are congruent. Prove opposite sides of a parallelogram are congruent. Given: WXYZ is a parallelogram. Z W Prove: WZ XY, WX ZY Solution Y X Statements Reasons 1. WXYZ is a parallelogram. 1. Given. WZ XY, WZ XY. Definition of parallelogram 3. ZX ZX 3. Refleive property. WXZ XZY, ZXY WZX. orresponding Angles Postulate 5. WZX YXZ 5. ASA triangle congruence. 6. WZ XY, WX ZY 6. orresponding Parts of ongruent Triangles are ongruent. omplete the paragraph proof to prove that opposite angles of a parallelogram are congruent. F G 7. Given: FGH is a parallelogram. Prove: G, F H It is given that FGH is a parallelogram. y the definition of a parallelogram, GH F, and FG H. Since same-side interior angles are supplementary, m G m H 180, and m F m 180. For the same reason m G m F 180 and m H m 180. Using substitution, m H m m G m H. y the Subtraction Property of Equality, m H m m H m G m H m H. Simplifying, m m G. Using substitution, m F m m H m. y the Subtraction Property of Equality, m F m m m H m m. Simplifyig, m F m H. y the definition of congruent angles, G and F H. Saon. All rights reserved. 73 Saon Geometry H
continued 3 More Properties of Parallelograms F G H m F m G 180 m G m H 180 m H m 180 m m F 180 F G P H FP HP GP P onsecutive angles are supplementary. The diagonals bisect each other. You can use the properties of parallelograms to find the measures of angles and sides of parallelograms. WXYZ is a parallelogram. Find m X. Step 1: Since W and X are consecutive angles, they are supplementary. Thus, m W m X 180. Step : Substitute m W (7 15) and m X () into m W m X 180 and solve for. X (7 + 15) W Y Z m W m X 180 onsecutive angles are supplementary. (7 15) 180 Substitute the given values. 11 15 180 ombine like terms. 11 165 Subtract 15 from both sides. 15 Divide both sides by 11. Step 3: Substitute 15 into the epression for m X. m X () [ (15) ] 60 DEF is a parallelogram. omplete the steps to find the length of D. 8. D EF Opposite sides are congruent D EF ongruent sides have the same length. w 8 5w 1 Substitute. 8 w 1 Subtract w from both sides. 7 w Subtract 1 from both sides. D w 8 7 8 36 Find each measure for the parallelogram DEF. 9. EF 10. m F 36 8 D w + 8 E (9z - 1) 5w + 1 3z F Saon. All rights reserved. 7 Saon Geometry
Name Date lass You have worked with properties of parallelograms. Now you will find arc lengths and the area of sectors. Finding Areas of Sectors and Arc Length 35 Arc Length Arc length is the distance along a part of the circumference of a circle measured in linear units. The arc length of a circle is given by the formula L r m 360. A The variable m stands for the degree measure of the arc. Find the arc length of K. Give the answer in terms of and rounded to the nearest hundredth. From the circle, r 9 cm, and m 8. Substitute the values into the formula. L r m 360 Formula for arc length. 9 cm 8 360 Substitute 9 cm and 8. 1 cm Solve for the answer in terms of. 5 8 K r 9 cm m 13.19 cm Round to the nearest hundredth. omplete the steps to find the arc length. Give your answer in terms of and rounded to the nearest hundredth. 1. A A 110. WX W 90 18 in. 1 cm X L r m 360 18 in. 110 L r m 360 360 1 cm 90 360 11 in. 3.56 in. 7 cm 1.99 cm Saon. All rights reserved. 75 Saon Geometry
continued 35 Sector Area A sector of a circle is a region bounded by two radii of the circle and their intercepted arc. The area of a sector of a circle is given by the formula A r m 360. Find the area of the sector DE. Give the answer in terms of and rounded to the nearest hundredth. From the circle, r 6 cm, and m 70. Substitute the values into the formula. A r m 360 Formula for sector area. 6 cm 70 360 Substitute 6 cm and 70. 6 cm D 36 cm 70 360 Simplify. A r m 70 E 7 cm 1.99 cm Solve for the answer in terms of. Round to the nearest hundredth. omplete the steps to find the area of each sector. Give your answer in terms of and rounded to the nearest hundredth. 5. sector QRS 6. sector HG G Q 9 in. 10 R in. H S A r m 360 9 in. 81 in 10 360 A r m 360 10 360 in. 7 in in in 90 360 90 360 8.8 in 3.1 in Saon. All rights reserved. 76 Saon Geometry
Name Date lass Right Triangle ongruence Theorems 36 Hypotenuse-Angle ongruence Theorem: If the hypotenuse and an acute angle of one right triangle are congruent to the hypotenuse and an acute angle of another right triangle, then the triangles are congruent. Use the Hypotenuse-Angle ongruence Theorem to prove that A KL. Statement Reason 1. A and KL 1. Given are right triangles.. A. Given 3. A K 3. Given. A KL. HA ongruence Theorem A K L 1. Use the Hypotenuse- Angle ongruence Theorem to show that GHI MNP. Statement Reason G P N 1. GHI and MNP are right triangles.. GI MP 1. Given. Given H I M 3. G M 3. Given. GHI MNP. HA ongruence Theorem. Use the Hypotenuse-Angle ongruence Theorem to prove that HK XYZ. Statement Reason 1. HK and XZY are 1. Given right triangles.. H XZ. Given 3. Z 3. Given. HK XZY. HA ongruence Theorem H K Y X Z Saon. All rights reserved. 77 Saon Geometry
continued 36 Hypotenuse-Leg ongruence Theorem: If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of another right triangle, then the triangles are congruent. U Use the Hypotenuse-Leg ongruence Theorem to prove that RST UVX. R RST and UVX are both given as right triangles. Leg RT is congruent to leg UX as given. Also, hypotenuse RS is congruent to X V hypotenuse UV as given. Therefore, RST UVX are congruent by the HL ongruence Theorem. T S 3. Use the Hypotenuse-Leg ongruence Theorem to prove that PQR DE. Statement 1. PQR and DE are right triangles.. PR E and PQ D 1. Given. Given Reason 3. GHI MNP. HA ongruence Theorem P R Q E D. Use the Hypotenuse-Leg ongruence Theorem to prove that A VWX. Possible answer: A and VWX are both right triangles. Leg A is congruent to leg VX as given. Also, hypotenuse A is congruent to hypotenuse VW as given. Therefore, A VWX by HL ongruencetheorem. 5. Use the Hypotenuse-Leg ongruence Theorem to prove that EFG MNP. Statement Reason 1. EFG and MNP 1. Given are right triangles.. EG MP and EF MN. Given 3. EFG MNP 3. HA ongruence Theorem G E P F N M Saon. All rights reserved. 78 Saon Geometry
Name Date lass Writing Equations of Parallel and Perpendicular Lines 37 Find the slopes of lines parallel and perpendicular to line k. First, find the slope of line k. m y 1 y 1 hoose two points on line k to substitute into the equation. ( 1, y 1 ) (0, ) and (, y ) ( 3, 0) 0 m 0 ( 3) 3 Since parallel lines have the same slope, the slope of a line parallel to line k is 3. Since the slopes of two perpendicular lines are opposite reciprocals, the slope of a line perpendicular to line k is 3. y k (0, ) O - - (-3, 0) - - omplete the steps to find the slopes of lines parallel and perpendicular to each line. 1. line a: m a Slope of line parallel to line a: ( 1) 0 0 1 5 Slope of line perpendicular to line a: 5 5 1 5. line b: m b 3 0 0 ( 6) Slope of line parallel to line b: Slope of line perpendicular to line b: Find the slopes of the lines parallel and perpendicular to each line. 1 3. line g 1 Slope of line parallel to line g: Slope of line perpendicular to line g: y b (0, -1) O (3, 0) (5, 0) - - a - - -6 (0, -6) y (0, ) (, 3) g O - - - -. line h Slope of line parallel to line h: 5 Slope of line perpendicular to line h: 5 Saon. All rights reserved. 79 Saon Geometry
continued 37 Graph the line that is perpendicular to the line y 1 3 and passes through (, 1). The slope of the perpendicular line is the opposite reciprocal of 1 3, so m 3. y y 1 m( 1 ) Point-slope formula y 1 3( ) Substitute. y 3 7 Solve. y 6 O - - - omplete the steps to graph the line that is perpendicular to y = 1 and passes through (, 3). 6 y 5. y y 1 m( 1 ) y 3 1 y 1 ( ) ( ) - - O - omplete the steps to graph the line that is parallel to y 3 and passes through (1, ). y 6. y y 1 m ( 1 ) y 3 ( 1 ) y 3 5 O - - - - Find and graph each line. 7. Graph the line that is perpendicular to 8. Graph the line that is parallel to y 1 and passes through (, 5). y 1 3 and passes through (, 1). y y - - O - - O - - - - Saon. All rights reserved. 80 Saon Geometry
Name Date lass Perpendicular and Angle isectors of Triangles 38 In the diagram, XZ bisects X. According to the Triangle Angle isector Theorem, WZ and ZY are proportional to WX and XY. The proportion is WX XY WZ ZY. Find the length of WX if WZ 9, XY, and ZY 10. WX XY Triangle Angle isector Theorem WZ ZY W X n 9 10 10n 16 n 1.6 Substitute the values for WZ, XY, and ZY. ross multiply. Solve. Z Y So, WX 1.6. 1. Find the length of TV if RS 18, ST 8 and RV 0. 18 8 0 n 18 n 160 S T R n 8. 8 V. Find the length of RV if RS 7, ST 1, and TV 11. 7 n 1 11 1 n 97 A n.75 3. Find the length of if AD 15, A 1, and D 9. 8. D. Find the length of AD if A 30, 1, and D 15. 37.5 5. Find the length of D if AD 1, A 5, and 15. 1.6 6. Find the length of A if AD 8, D 8, and 6. 1 Saon. All rights reserved. 81 Saon Geometry
continued 38 The circumcenter of a triangle is the point at which all the perpendicular bisectors of the sides intersect. The circumcenter is equidistant from all the vertices of the triangle. Find the circumcenter of the triangle with vertices at A(3, 3), (9, 3), and (6, 7). To find the circumcenter, first choose two sides to use. Then, find the midpoints of the two sides. The midpoint of A is (6, 3), and the midpoint of A is (.5, 5). Net, find the slope of the two sides. The slope of A is 0, and the slope of A is 3. Since the circumcenter is the point at which the perpendicular bisectors intersect, find the equations of the lines perpendicular to A and A. The equation of the line perpendicular to A through (6, 3) is 6, and the equation of the line perpendicular to A through (.5, 5) is y 3 67 8. Finally, solve the system of equations to find the point of intersection of the perpendicular bisectors: 6 and y 3 67 8. The solution is (6, 3.875). This is the circumcenter of triangle A. 10 8 6 y Z (1, 7) Y (1, 3) (5, 5) X O 6 8 10 omplete the steps to find the circumcenter of the triangle. 7. XYZ with vertices at X(5, 5), Y(1, 3), and Z(1, 7). The midpoint of XY is 3,. The midpoint of YZ is 1, 5. The slope of XY is 1. The slope of is YZ is undefined. 10 8 6 y Z (1, 7) Y (1, 3) (5, 5) X O 6 8 10 The equation for the perpendicular bisector of XY is y 10. The equation for the perpendicular bisector of YZ is y 5. The circumcenter for XYZ is 5, 5. 8. Find the circumcenter of the triangle with vertices at A(, ), (8, ), and (6, 10). 5, 6. 3 Saon. All rights reserved. 8 Saon Geometry
Name Date lass You learned that the sum of the angles inside a triangle is 180. Now you will relate the angle measures to the side lengths of a triangle. Inequalities in a Trianglee 39 A triangle s largest side is always opposite its largest angle, and its smallest side is always opposite its smallest angle. Order the angle measures of A from least to greatest. The shortest side length is A, so the smallest angle in the triangle is. The longest side is, so A is the largest angle. Finally, A is the middle side length, so is the middle angle measure. Therefore, the order of the angle measures from least to greatest is,, A. A 8 in. 1 in. 0 in. Fill in the blanks to put the angle measures in order from least to greatest for FGH. 1. The shortest side is The longest side is FH, so the smallest angle is G. GH, so the largest angle is F. The remaining angle, H, is the middle angle measure. F 16 cm 7 cm G H 5 cm Therefore, the order of the angles from least to greatest measure is G, H, F. Fill in the blanks to put the angle measures in order from least to greatest for XYZ.. The measure of Y is 100. Therefore, XZ is the longest side. The angle with the smallest measure is Z, so the shortest side length is XY. So the middle side length is YZ. 71 17 17 8 8 5 5 5 5 = 1.75 6 Therefore, the order of the side lengths from least to greatest is XY, YZ, XZ. Saon. All rights reserved. 83 Saon Geometry
continued 39 Triangle Inequality Theorem: The sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Determine whether each set of side lengths could form a valid triangle. (, 5, 9) and (9, 11, 1) For the first set of side lengths, the sum of and 5 is equal to 9. This is not greater than the third side, so these side lengths cannot be a triangle. For the second set of side lengths, the sum of any pair of sides is greater than the third side, so these lengths do form a triangle. What is the range for the values of n in the given triangle? Using the Triangle Inequality Theorem, the sum of any two sides must be greater than the third side. First, write inequalities for the sum of each pair of sides. 10 3 n 10 n 3 3 n 10 13 n n 7 n 7 Then, eliminate any side lengths that are not possible and find the solution. Side lengths cannot be negative, so the solution is 7 n 13. 10 3 n Fill in the blanks to find the range for the values of. 3. 5 1 1 5 19 9 5 1 9 5 1 The range of values for is 9 19. Determine whether each set of side lengths could form a triangle.. 6, 10, 5 yes 5., 8, 15 no Give the range for the values of in the given triangles. 6. 7. 6 16 1 13 1 0 7 Saon. All rights reserved. 8 Saon Geometry
Name Date lass Finding Perimeters and Areas of omposite Figures 0 You have learned how to find the perimeter and area of polygons and circles. Now you will find the perimeter and area of composite figures. Area of omposite Figures Find the area of this composite figure. The figure is made up of a triangle and two rectangles. First, find the area of each shape in the figure. The area of the triangle is A 1 1 bh 1 (8)() 8 cm. The area of the large rectangle is A bh (8)(1) 96 cm. The area of the small rectangle is A 3 bh ()(5) 10 cm. Add all the areas. y the Angle Addition Postulate, the area of the composite figure is A 8 96 10 11 cm. cm 8 cm 1 cm cm 5 cm omplete the steps to find the area of each figure. 1. A 1 bh 7 7 9 in 7 in. A 1 bh 1 7 6 1 in A 9 1 70 in 7 in. 6 in.. A bh 3 9 7 m 9 m A bh 1 7 8 m 3 m 3 m A 7 8 111 m 7 m Find the area of each composite figure. When necessary, write your answer in terms of. 3.. 16 ft 5 in. 10 in. 16 ft 10 in. in. 56 3 ft 15 in Saon. All rights reserved. 85 Saon Geometry
Area of Shaded Regions Find the area of the shaded region. Leave your answer in terms of π. The figure is made up of a semicircle inside a rectangle. First, find the area of the whole rectangle. continued 0 A 1 bh (1)(1) 168 yd Then, find the area of the semicircle. A 1 r (6) 18 yd To find the area of the shaded region, we need to know the area of the rectangle without the area of the triangle. If A is the area of the shaded region, then A 1 A A. So, 168 A 18π, and A 168 18π yd. 1 yd 1 yd Fill in the blanks to find the area of the shaded region for each figure. 5. A 1 bh 15 1 180 mm A 1 bh 1 8 15 60 mm 180 A 60 mm A 10 mm 8 mm 15 mm 1 mm 6. A 1 bh 11 11 11 ft A 1 π r 1 π 8 π ft 11 A 8 ft A 11 8 ft Find the area of the shaded region. When necessary, write your answer in terms of. 11 ft 11 ft 8 ft 7. 1 in. 8. 1 in. 7 in. 6 cm 1 cm 18 in. Diameter = 1 cm 31 in 9 cm Saon. All rights reserved. 86 Saon Geometry