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Ratio of Areas: What is the area ratio between ABCD and XYZ? A B C D9 10 Y X Z 12 8 One way of determining the ratio of the areas of two figures is to.

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Presentation on theme: "Ratio of Areas: What is the area ratio between ABCD and XYZ? A B C D9 10 Y X Z 12 8 One way of determining the ratio of the areas of two figures is to."— Presentation transcript:

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2 Ratio of Areas: What is the area ratio between ABCD and XYZ? A B C D9 10 Y X Z 12 8 One way of determining the ratio of the areas of two figures is to calculate the quotient of the two areas.

3 Steps: 1.Set up fraction 2.Write formulas 3.Plug in numbers 4.Solve and label with units 1. Ratio A A 2. A = b 1 h 1 A = 1 / 2 b 2 h 2 3. = / = =15 8

4 Find the ratio of ABD to CBD CA D B When AB = 5 and BC = 2 2. ABC = 1 / 2 b 1 h CBD = 1 / 2 b 2 h 3. = ½(5)h ½(2)h 1. Ratio ABD CBD 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 Given ∆ PQR ∆WXY Find the ratio of the area. First find the ratio of the sides. Q PR 6 X Y W 4 QP = 6 XW 4 =3=3 2

7 Q PR 6 X Y W 4 Ratio of area: A PQR = 1 / 2 b 1 h 1 A WXY 1 / 2 b 2 h 2 = b 1 h 1 b2h2b2h2 = = 9 4 Because they are similar triangles the ratios of the sides and heights are the same.

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) A 1 = S 1 2 A 2 S 2 When A 1 and A 2 are areas and S 1 and S 2 are measures of corresponding segments.

9 Given the similar pentagons shown, find the ratio of their areas. S 1 = 12 S 2 9 = 4 3 A 1 = 4 2 A 2 3 = 16 9

10 If the ratio of the areas of two similar parallelograms is 49:121, find the ratio of their bases. 49cm cm 2 A 1 = S 1 2 A 2 S 2 49 = S S 2 7 = S 1 11 S 2

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

12 A C M B 1. Altitude from A is congruent for both triangles. Label it X. 2.BM = MC because AM is a median. Let y = BM and MC. A ∆ABM = 1 / 2 b 1 h 1 A ∆ACM 1 / 2 b 2 h 2 = 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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