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;
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;
3. The ratio of corresponding sides in similar triangles are equal
From the similar triangles, we have;
[tex]\dfrac{8}{10} = \mathbf{ \dfrac{x}{6}}[/tex]
8 × 6 = 10 × x
48 = 10·x
- [tex]x = \dfrac{48}{10} = \underline{4.8}[/tex]
3. From the similar triangles, we have;
[tex]\mathbf{\dfrac{20}{29}} = \dfrac{x}{21}[/tex]
20 × 21 = x × 29
420 = 29·x
- [tex]x = \dfrac{420}{29 } \approx \underline{14.48}[/tex]
4. From the similar triangles, we have;
[tex]\mathbf{\dfrac{20}{52}} = \dfrac{x}{48}[/tex]
20 × 48 = 52 × x
- [tex]x = \dfrac{20 \times 48}{52} = \dfrac{240}{13} \approx \underline{18.46}[/tex]
5. From the similar triangles, we have;
[tex]\mathbf{\dfrac{13.2}{26}} = \dfrac{x}{22.4}[/tex]
13.2 × 22.4 = 26 × x
- [tex]x = \dfrac{13.2 \times 22.4}{26} \approx \underline{ 11.37}[/tex]
6. The geometric mean, G.M. is given by the formula;
[tex]G.M. = \mathbf{\sqrt[n]{x_1 \times x_2 \times x_3 ... x_n}}[/tex]
The geometric mean of 16 and 27 is therefore;
- [tex]G.M. = \sqrt[2]{16 \times 27} = \sqrt[2]{432} = \sqrt[2]{144 \times 3} = \mathbf{12 \cdot \sqrt{3}}[/tex]
- The geometric mean of 16 and 27 is 12·√3
7. The geometric mean of 5 and 36 is found as follows;
[tex]G.M. = \sqrt[2]{5 \times 36} = \sqrt[2]{180} = \sqrt[2]{36 \times 5} = \mathbf{ 6 \cdot \sqrt{5}}[/tex]
- The geometric mean of 5 and 36 is 6·√5
Learn more about the AA similarity postulate and geometric mean here:
https://brainly.com/question/12002948
https://brainly.com/question/12457640
