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m_a_m_a [10]
2 years ago
11

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

Mathematics
1 answer:
Ugo [173]2 years ago
7 0

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 = <u>4.8</u>

4. x ≈ <u>14.48</u>

5. x ≈ <u>11.37</u>

6. G.M. = <u>12·√3</u>

7. G.M. = <u>6·√5</u>

<u />

<h3>What condition guarantees the similarity of the right triangles?</h3>

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

  • <u>ΔLJK is similar to ΔKJM</u> by Angle–Angle, AA, similarity postulate

∠JLK ≅ ∠JLK by reflexive property

  • <u>ΔLJK is similar to ΔLKM</u> by AA similarity

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

  • <u>ΔKJM ~ ΔLKM</u>

2. ∠YWZ ≅ ∠YWZ by reflexive property

∠WXZ ≅ ∠YZW all 90° angle are congruent

  • <u>ΔYWZ is similar to ΔZWX</u>, by AA similarity postulate

∠XYZ ≅ ∠WYZ by reflexive property

∠YXZ ≅ ∠YZW all 90° are congruent

  • <u>ΔYWZ is similar to ΔYZW</u> by AA similarity postulate

Therefore;

  • <u>ΔZWX ~ ΔYZW</u>

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 <u>12·√3</u>

<u />

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 <u>6·√5</u>

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

brainly.com/question/12002948

brainly.com/question/12457640

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Answer:

\log_{10}(147) = 2.1673

Step-by-step explanation:

Given

\log_{10} 3 = 0.4771

\log_{10} 5 = 0.6990

\log_{10} 7= 0.8451

\log_{10} 11 = 1.0414

Required

Evaluate \log_{10}(147)

Expand

\log_{10}(147) = \log_{10}(49 * 3)

Further expand

\log_{10}(147) = \log_{10}(7 * 7 * 3)

Apply product rule of logarithm

\log_{10}(147) = \log_{10}(7) + \log_{10}(7) + \log_{10}(3)

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