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8-1 Geometric Mean p. 537 You used proportional relationships of corresponding angle bisectors, altitudes, and medians of similar triangles. Find the geometric.

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Presentation on theme: "8-1 Geometric Mean p. 537 You used proportional relationships of corresponding angle bisectors, altitudes, and medians of similar triangles. Find the geometric."β€” Presentation transcript:

1 8-1 Geometric Mean p. 537 You used proportional relationships of corresponding angle bisectors, altitudes, and medians of similar triangles. Find the geometric mean between two numbers. Solve problems involving relationships between parts of a right triangle and the altitude to its hypotenuse.

2 Geometric mean – when the means of a proportion are the same number.
𝒂 𝒙 = 𝒙 𝒅 extreme mean The geometric mean between two numbers is the positive square room of their product.

3 p. 537

4 Find the geometric mean between 2 and 50.
Let x represent the geometric mean. Definition of geometric mean Cross products Take the positive square root of each side. Simplify. Answer: The geometric mean is 10.

5 A. Find the geometric mean between 3 and 12.

6 Geometric Means in Right Triangles
In a right triangle, an altitude drawn from the vertex of the right angle to the hypotenuse forms two additional right triangles. The two triangles formed are similar to the original triangle and to each other.

7 p. 538

8 Write a similarity statement identifying the three similar triangles in the figure.
Separate the triangles into two triangles along the altitude. Then sketch the three triangles, reorienting the smaller ones so that their corresponding angles and sides are in the same position as the original triangle. Answer: So, by Theorem 8.1, Ξ”EGF ~ Ξ”FGH ~ Ξ”EFH.

9 Write a similarity statement identifying the three similar triangles in the figure.
A. Ξ”LNM ~ Ξ”MLO ~ Ξ”NMO B. Ξ”NML ~ Ξ”LOM ~ Ξ”MNO C. Ξ”LMN ~ Ξ”LOM ~ Ξ”MON D. Ξ”LMN ~ Ξ”LMO ~ Ξ”MNO

10 π‘ β„Žπ‘œπ‘Ÿπ‘‘π‘’π‘Ÿ 𝑙𝑒𝑔 π‘™π‘œπ‘›π‘”π‘’π‘Ÿ 𝑙𝑒𝑔 = 𝑏 π‘Ž = π‘₯ β„Ž = β„Ž 𝑦
C β„Žπ‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ π‘ β„Žπ‘œπ‘Ÿπ‘‘π‘’π‘Ÿ 𝑙𝑒𝑔 = 𝑐 𝑏 = 𝑏 π‘₯ = π‘Ž β„Ž β„Žπ‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ π‘™π‘œπ‘›π‘”π‘’π‘Ÿ 𝑙𝑒𝑔 = 𝑐 π‘Ž = 𝑏 β„Ž = π‘Ž 𝑦 a b h A B x D y c D x h p. 538 C a A D C b b y h A B C B c a

11 β„Ž= π‘₯𝑦 𝑏= π‘₯𝑐 π‘Ž= 𝑦𝑐 p.539

12 Find c, d, and e. Since e is the measure of the altitude drawn to the hypotenuse of right Ξ”JKL, e is the geometric mean of the lengths of the two segments that make up the hypotenuse, JM and ML. Geometric Mean (Altitude) Theorem Substitution Simplify. Since d is the measure of leg JK, d is the geometric mean of JM, the measure of the segment adjacent to this leg, and the measure of the hypotenuse JL. Geometric Mean (Leg) Theorem Substitution Use a calculator to simplify.

13 Since c is the measure of leg KL, c is the geometric mean of ML, the measure of the segment adjacent to KL, and the measure of the hypotenuse JL. Geometric Mean (Leg) Theorem Substitution Use a calculator to simplify. Answer: e = 12, d β‰ˆ 13.4, c β‰ˆ 26.8

14 8-1 Assignment Page 541, 8-19


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