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Exam 8: Areas of Polygons and Circles

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Exam 1: Line and Angle Relationships13 Questionsadd course
Exam 2: Parallel Lines13 Questionsadd course
Exam 3: Triangles16 Questionsadd course
Exam 4: Quadrilaterals14 Questionsadd course
Exam 5: Similar Triangles12 Questionsadd course
Exam 6: Circles10 Questionsadd course
Exam 7: Locus and Concurrence4 Questionsadd course
Exam 8: Areas of Polygons and Circles5 Questionsadd course
Exam 9: Surfaces and Solids4 Questionsadd course
Exam 10: Analytical Geometry8 Questionsadd course
Exam 11: Introduction to Trigonometry4 Questionsadd course
Exam 12: Geometry Problems: Complementary Angles, Collinear Points, and Similar Triangles916 Questionsadd course
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-Use the drawing provided to explain the following theorem. The area of any quadrilateral with perpendicular diagonals of lengths and is given by . Given: Quadrilateral with at point F; and Prove: -Use the drawing provided to explain the following theorem. "The area of any quadrilateral with perpendicular diagonals of lengths -Use the drawing provided to explain the following theorem. The area of any quadrilateral with perpendicular diagonals of lengths and is given by . Given: Quadrilateral with at point F; and Prove: and -Use the drawing provided to explain the following theorem. The area of any quadrilateral with perpendicular diagonals of lengths and is given by . Given: Quadrilateral with at point F; and Prove: is given by -Use the drawing provided to explain the following theorem. The area of any quadrilateral with perpendicular diagonals of lengths and is given by . Given: Quadrilateral with at point F; and Prove: ." Given: Quadrilateral -Use the drawing provided to explain the following theorem. The area of any quadrilateral with perpendicular diagonals of lengths and is given by . Given: Quadrilateral with at point F; and Prove: with -Use the drawing provided to explain the following theorem. The area of any quadrilateral with perpendicular diagonals of lengths and is given by . Given: Quadrilateral with at point F; and Prove: at point F; -Use the drawing provided to explain the following theorem. The area of any quadrilateral with perpendicular diagonals of lengths and is given by . Given: Quadrilateral with at point F; and Prove: and -Use the drawing provided to explain the following theorem. The area of any quadrilateral with perpendicular diagonals of lengths and is given by . Given: Quadrilateral with at point F; and Prove: Prove: -Use the drawing provided to explain the following theorem. The area of any quadrilateral with perpendicular diagonals of lengths and is given by . Given: Quadrilateral with at point F; and Prove:

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To "box" the quadrilateral To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . , we draw auxiliary lines as follows:
through point D, we draw To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . ; through point B, we draw To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . ;
through point A, we draw To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . ; and through point C, we draw To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . .
The quadrilateral formed is a parallelogram that can be shown to have a right angle;
this follows from the fact that To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . is a parallelogram that contains a right angle at
vertex F . . . so the opposite angle (at vertex R) must also be a right angle.
Because To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . is a diagonal of To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . (actually rectangle To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . , To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . ;
that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . .
Similarly, To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . , To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . , and To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . . Thus, the area
of quadrilateral To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . is one half of that of rectangle To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . .
But the area of To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . is To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . , so To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . is given by To box the quadrilateral , we draw auxiliary lines as follows: through point D, we draw ; through point B, we draw ; through point A, we draw ; and through point C, we draw . The quadrilateral formed is a parallelogram that can be shown to have a right angle; this follows from the fact that is a parallelogram that contains a right angle at vertex F . . . so the opposite angle (at vertex R) must also be a right angle. Because is a diagonal of (actually rectangle , ; that is, a diagonal of a parallelogram separates the parallelogram into 2 congruent . Similarly, , , and . Thus, the area of quadrilateral is one half of that of rectangle . But the area of is , so is given by . .

Consider a circle with diameter length d, radius length r, and circumference C. Given that Consider a circle with diameter length d, radius length r, and circumference C. Given that , explain why the formula for the circumference of a circle is given by . , explain why the formula for the circumference of a circle is given by Consider a circle with diameter length d, radius length r, and circumference C. Given that , explain why the formula for the circumference of a circle is given by . .

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Given that Given that , we use the Multiplication Property of Equality to obtain . Because the length of the diameter of a circle is twice that of a radius, . By substitution, or . , we use the Multiplication Property of Equality to obtain Given that , we use the Multiplication Property of Equality to obtain . Because the length of the diameter of a circle is twice that of a radius, . By substitution, or . .
Because the length of the diameter of a circle is twice that of a radius, Given that , we use the Multiplication Property of Equality to obtain . Because the length of the diameter of a circle is twice that of a radius, . By substitution, or . . By
substitution, Given that , we use the Multiplication Property of Equality to obtain . Because the length of the diameter of a circle is twice that of a radius, . By substitution, or . or Given that , we use the Multiplication Property of Equality to obtain . Because the length of the diameter of a circle is twice that of a radius, . By substitution, or . .

-Use the drawing provided to explain the following theorem. The area A of a regular polygon whose apothem has length a and whose perimeter is P is given by . Given: Regular polygon with center O and length s for each side; apothem so that Prove: -Use the drawing provided to explain the following theorem. "The area A of a regular polygon whose apothem has length a and whose perimeter is P is given by -Use the drawing provided to explain the following theorem. The area A of a regular polygon whose apothem has length a and whose perimeter is P is given by . Given: Regular polygon with center O and length s for each side; apothem so that Prove: ." Given: Regular polygon -Use the drawing provided to explain the following theorem. The area A of a regular polygon whose apothem has length a and whose perimeter is P is given by . Given: Regular polygon with center O and length s for each side; apothem so that Prove: with center O and length s for each side; apothem -Use the drawing provided to explain the following theorem. The area A of a regular polygon whose apothem has length a and whose perimeter is P is given by . Given: Regular polygon with center O and length s for each side; apothem so that Prove: so that -Use the drawing provided to explain the following theorem. The area A of a regular polygon whose apothem has length a and whose perimeter is P is given by . Given: Regular polygon with center O and length s for each side; apothem so that Prove: Prove: -Use the drawing provided to explain the following theorem. The area A of a regular polygon whose apothem has length a and whose perimeter is P is given by . Given: Regular polygon with center O and length s for each side; apothem so that Prove:

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From center O, we draw radii From center O, we draw radii , , , , and . Because the radii are congruent to each other and the sides of the regular polygon are all congruent to each other as well, , , , , and are all congruent to each other by SSS. Each of the congruent triangles has an altitude length of . Further, the length of each base of a triangle is s, the length of side of the polygon. Therefore, the area of the regular polygon is Because the sum of the sides equals perimeter P, we have . , From center O, we draw radii , , , , and . Because the radii are congruent to each other and the sides of the regular polygon are all congruent to each other as well, , , , , and are all congruent to each other by SSS. Each of the congruent triangles has an altitude length of . Further, the length of each base of a triangle is s, the length of side of the polygon. Therefore, the area of the regular polygon is Because the sum of the sides equals perimeter P, we have . , From center O, we draw radii , , , , and . Because the radii are congruent to each other and the sides of the regular polygon are all congruent to each other as well, , , , , and are all congruent to each other by SSS. Each of the congruent triangles has an altitude length of . Further, the length of each base of a triangle is s, the length of side of the polygon. Therefore, the area of the regular polygon is Because the sum of the sides equals perimeter P, we have . , From center O, we draw radii , , , , and . Because the radii are congruent to each other and the sides of the regular polygon are all congruent to each other as well, , , , , and are all congruent to each other by SSS. Each of the congruent triangles has an altitude length of . Further, the length of each base of a triangle is s, the length of side of the polygon. Therefore, the area of the regular polygon is Because the sum of the sides equals perimeter P, we have . , and From center O, we draw radii , , , , and . Because the radii are congruent to each other and the sides of the regular polygon are all congruent to each other as well, , , , , and are all congruent to each other by SSS. Each of the congruent triangles has an altitude length of . Further, the length of each base of a triangle is s, the length of side of the polygon. Therefore, the area of the regular polygon is Because the sum of the sides equals perimeter P, we have . . Because the radii are congruent to each other and the sides of the regular polygon are all congruent to each other as well, From center O, we draw radii , , , , and . Because the radii are congruent to each other and the sides of the regular polygon are all congruent to each other as well, , , , , and are all congruent to each other by SSS. Each of the congruent triangles has an altitude length of . Further, the length of each base of a triangle is s, the length of side of the polygon. Therefore, the area of the regular polygon is Because the sum of the sides equals perimeter P, we have . , From center O, we draw radii , , , , and . Because the radii are congruent to each other and the sides of the regular polygon are all congruent to each other as well, , , , , and are all congruent to each other by SSS. Each of the congruent triangles has an altitude length of . Further, the length of each base of a triangle is s, the length of side of the polygon. Therefore, the area of the regular polygon is Because the sum of the sides equals perimeter P, we have . , From center O, we draw radii , , , , and . Because the radii are congruent to each other and the sides of the regular polygon are all congruent to each other as well, , , , , and are all congruent to each other by SSS. Each of the congruent triangles has an altitude length of . Further, the length of each base of a triangle is s, the length of side of the polygon. Therefore, the area of the regular polygon is Because the sum of the sides equals perimeter P, we have . , From center O, we draw radii , , , , and . Because the radii are congruent to each other and the sides of the regular polygon are all congruent to each other as well, , , , , and are all congruent to each other by SSS. Each of the congruent triangles has an altitude length of . Further, the length of each base of a triangle is s, the length of side of the polygon. Therefore, the area of the regular polygon is Because the sum of the sides equals perimeter P, we have . , and From center O, we draw radii , , , , and . Because the radii are congruent to each other and the sides of the regular polygon are all congruent to each other as well, , , , , and are all congruent to each other by SSS. Each of the congruent triangles has an altitude length of . Further, the length of each base of a triangle is s, the length of side of the polygon. Therefore, the area of the regular polygon is Because the sum of the sides equals perimeter P, we have . are all congruent to each other by SSS.
Each of the congruent triangles has an altitude length of From center O, we draw radii , , , , and . Because the radii are congruent to each other and the sides of the regular polygon are all congruent to each other as well, , , , , and are all congruent to each other by SSS. Each of the congruent triangles has an altitude length of . Further, the length of each base of a triangle is s, the length of side of the polygon. Therefore, the area of the regular polygon is Because the sum of the sides equals perimeter P, we have . . Further, the length of each
base of a triangle is s, the length of side of the polygon. Therefore, the area of the regular polygon is From center O, we draw radii , , , , and . Because the radii are congruent to each other and the sides of the regular polygon are all congruent to each other as well, , , , , and are all congruent to each other by SSS. Each of the congruent triangles has an altitude length of . Further, the length of each base of a triangle is s, the length of side of the polygon. Therefore, the area of the regular polygon is Because the sum of the sides equals perimeter P, we have . From center O, we draw radii , , , , and . Because the radii are congruent to each other and the sides of the regular polygon are all congruent to each other as well, , , , , and are all congruent to each other by SSS. Each of the congruent triangles has an altitude length of . Further, the length of each base of a triangle is s, the length of side of the polygon. Therefore, the area of the regular polygon is Because the sum of the sides equals perimeter P, we have . Because the sum of the sides equals perimeter P, we have From center O, we draw radii , , , , and . Because the radii are congruent to each other and the sides of the regular polygon are all congruent to each other as well, , , , , and are all congruent to each other by SSS. Each of the congruent triangles has an altitude length of . Further, the length of each base of a triangle is s, the length of side of the polygon. Therefore, the area of the regular polygon is Because the sum of the sides equals perimeter P, we have . .

-Where is the degree measure for the arc of a sector of a circle, the ratio of the area of the sector to that of the area of the circle is given by . Use this ratio to explain why the area of the sector is given by . [Note: In the figure, the sector with arc measure is bounded by radii , , and .] -Where -Where is the degree measure for the arc of a sector of a circle, the ratio of the area of the sector to that of the area of the circle is given by . Use this ratio to explain why the area of the sector is given by . [Note: In the figure, the sector with arc measure is bounded by radii , , and .] is the degree measure for the arc of a sector of a circle, the ratio of the area of the sector to that of the area of the circle is given by -Where is the degree measure for the arc of a sector of a circle, the ratio of the area of the sector to that of the area of the circle is given by . Use this ratio to explain why the area of the sector is given by . [Note: In the figure, the sector with arc measure is bounded by radii , , and .] . Use this ratio to explain why the area of the sector is given by -Where is the degree measure for the arc of a sector of a circle, the ratio of the area of the sector to that of the area of the circle is given by . Use this ratio to explain why the area of the sector is given by . [Note: In the figure, the sector with arc measure is bounded by radii , , and .] . [Note: In the figure, the sector with arc measure -Where is the degree measure for the arc of a sector of a circle, the ratio of the area of the sector to that of the area of the circle is given by . Use this ratio to explain why the area of the sector is given by . [Note: In the figure, the sector with arc measure is bounded by radii , , and .] is bounded by radii -Where is the degree measure for the arc of a sector of a circle, the ratio of the area of the sector to that of the area of the circle is given by . Use this ratio to explain why the area of the sector is given by . [Note: In the figure, the sector with arc measure is bounded by radii , , and .] , -Where is the degree measure for the arc of a sector of a circle, the ratio of the area of the sector to that of the area of the circle is given by . Use this ratio to explain why the area of the sector is given by . [Note: In the figure, the sector with arc measure is bounded by radii , , and .] , and -Where is the degree measure for the arc of a sector of a circle, the ratio of the area of the sector to that of the area of the circle is given by . Use this ratio to explain why the area of the sector is given by . [Note: In the figure, the sector with arc measure is bounded by radii , , and .] .]

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-Using the drawing provided and fact that the area of a parallelogram is given by , show that the area of a triangle is given by . -Using the drawing provided and fact that the area of a parallelogram is given by -Using the drawing provided and fact that the area of a parallelogram is given by , show that the area of a triangle is given by . , show that the area of a triangle is given by -Using the drawing provided and fact that the area of a parallelogram is given by , show that the area of a triangle is given by . .

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