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Ratios of Areas Lesson 11.7

Ratios of Areas Lesson 11.7. Y. B. A. 10. 8. Z. X. 12. C. D. 9. Ratio of Areas:. What is the area ratio between ABCD and XYZ?. One way of determining the ratio of the areas of two figures is to calculate the quotient of the two areas. 1. Ratio A.

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Ratios of Areas Lesson 11.7

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  1. Ratios of AreasLesson 11.7

  2. Y B A 10 8 Z X 12 C D 9 Ratio of Areas: What is the area ratio between ABCD and XYZ? One way of determining the ratio of the areas of two figures is to calculate the quotient of the two areas.

  3. 1. Ratio A Steps: Set up fraction Write formulas Plug in numbers Solve and label with units A 2. A = b1h1 A = 1/2b2h2 3. = 9•10 1/2 • 8 •12 = 90 48 =15 8

  4. D C A B Find the ratio of ABD to CBD When AB = 5 and BC = 2 1. Ratio ABD CBD 2. ABC = 1/2b1h CBD = 1/2b2h 3. = ½(5)h ½(2)h 4. 5:2 or 5/2

  5. Similar triangles: Ratio of any pair of corresponding , altitudes, medians, angle bisectors, equals the ratio of their corresponding sides.

  6. Q X 6 4 Y W P R Given ∆ PQR ∆WXY Find the ratio of the area. First find the ratio of the sides. QP = 6 XW 4 =3 2

  7. Q X 6 4 Y W P R A PQR = 1/2 b1h1 Ratio of area: AWXY 1/2 b2h2 = b1h1 b2h2 Because they are similar triangles the ratios of the sides and heights are the same. = 3•3 2•2 = 9 4

  8. Area ratio is the sides ratio squared! Theorem 109: If 2 figures are similar, then the ratio of their areas equals the square of the ratio of the corresponding segments. (similar-figures Theorem) A1 = S12 When A1 and A2 are areas and S1 and S2 are measures of corresponding segments. A2 S2

  9. Given the similar pentagons shown, find the ratio of their areas. S1 = 12 S2 9 = 4 3 A1 = 42 = 16 9 A2 3

  10. If the ratio of the areas of two similar parallelograms is 49:121, find the ratio of their bases. 121 cm2 49cm2 A1 = S12 A2 S2 49 = S12 7 = S1 121 S2 11 S2

  11. A C M B Corresponding Segments include: Sides, altitudes, medians, diagonals, and radii. Ex. AM is the median of ∆ABC. Find the ratio of A ∆ ABM : A ∆ACM Notice these are not similar figures!

  12. A C M B 1. Altitude from A is congruent for both triangles. Label it X. BM = MC because AM is a median. Let y = BM and MC. A∆ABM = 1/2 b1h1 A ∆ACM 1/2 b2h2 = 1/2 xy 1/2 xy = 1 Therefore the ratio is 1:1 They are equal !

  13. Theorem 110:The median of a triangle divides the triangle into two triangles with equal area.

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