Question

Unit 8 right triangles and trigonometry homework 3 similar right triangles and geometric mean

Answer 1

The right triangles that have an altitude which forms two right triangles

are similar to the two right triangles formed.

Responses:

1. ΔLJK ~ ΔKJM

ΔLJK ~ ΔLKM

ΔKJM ~ ΔLKM

2. ΔYWZ ~ ΔZWX

ΔYWZ ~ ΔYZW

ΔZWX ~ ΔYZW

3. x = 4.8

4. x ≈ 14.48

5. x ≈ 11.37

6. G.M. = 12·√3

7. G.M. = 6·√5

What condition guarantees the similarity of the right triangles?

1. ∠LMK = 90° given

∠JMK + ∠LMK  = 180° linear pair angles

∠JMK = 180° - 90° = 90°

∠JKL ≅ ∠JMK All 90° angles are congruent

∠LJK ≅ ∠LJK reflexive property

• ΔLJK is similar to ΔKJM by Angle–Angle, AA, similarity postulate

∠JLK ≅ ∠JLK by reflexive property

• ΔLJK is similar to ΔLKM by AA similarity

By the property of equality for triangles that have equal interior angles, we have;

• ΔKJM ~ ΔLKM

2. ∠YWZ ≅ ∠YWZ by reflexive property

∠WXZ ≅ ∠YZW all 90° angle are congruent

• ΔYWZ is similar to ΔZWX, by AA similarity postulate

∠XYZ ≅ ∠WYZ by reflexive property

∠YXZ ≅ ∠YZW all 90° are congruent

• ΔYWZ is similar to ΔYZW by AA similarity postulate

Therefore;

• ΔZWX ~ ΔYZW

3. The ratio of corresponding sides in similar triangles are equal

From the similar triangles, we have;

$\dfrac{8}{10} = \mathbf{ \dfrac{x}{6}}$

8 × 6 = 10 × x

48 = 10·x

• $x = \dfrac{48}{10} = \underline{4.8}$

3. From the similar triangles, we have;

$\mathbf{\dfrac{20}{29}} = \dfrac{x}{21}$

20 × 21 = x × 29

420 = 29·x

• $x = \dfrac{420}{29 } \approx \underline{14.48}$

4. From the similar triangles, we have;

$\mathbf{\dfrac{20}{52}} = \dfrac{x}{48}$

20 × 48 = 52 × x

• $x = \dfrac{20 \times 48}{52} = \dfrac{240}{13} \approx \underline{18.46}$

5. From the similar triangles, we have;

$\mathbf{\dfrac{13.2}{26}} = \dfrac{x}{22.4}$

13.2 × 22.4 = 26 × x

• $x = \dfrac{13.2 \times 22.4}{26} \approx \underline{ 11.37}$

6. The geometric mean, G.M. is given by the formula;

$G.M. = \mathbf{\sqrt[n]{x_1 \times x_2 \times x_3 ... x_n}}$

The geometric mean of 16 and 27 is therefore;

• $G.M. = \sqrt[2]{16 \times 27} = \sqrt[2]{432} = \sqrt[2]{144 \times 3} = \mathbf{12 \cdot \sqrt{3}}$

• The geometric mean of 16 and 27 is 12·√3

7. The geometric mean of 5 and 36 is found as follows;

$G.M. = \sqrt[2]{5 \times 36} = \sqrt[2]{180} = \sqrt[2]{36 \times 5} = \mathbf{ 6 \cdot \sqrt{5}}$

• The geometric mean of 5 and 36 is 6·√5

Learn more about the AA similarity postulate and geometric mean here:

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