What is the value of y
1 answer:
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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>
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<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;
8 × 6 = 10 × x
48 = 10·x
3. From the similar triangles, we have;
20 × 21 = x × 29
420 = 29·x
4. From the similar triangles, we have;
20 × 48 = 52 × x
5. From the similar triangles, we have;
13.2 × 22.4 = 26 × x
6. The geometric mean, G.M. is given by the formula;
The geometric mean of 16 and 27 is therefore;
- The geometric mean of 16 and 27 is <u>12·√3</u>
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7. The geometric mean of 5 and 36 is found as follows;
- The geometric mean of 5 and 36 is <u>6·√5</u>
Learn more about the AA similarity postulate and geometric mean here:
Answer:
(a) There are 113,400 ways
(b) There are 138,600 ways
Step-by-step explanation:
The number of ways to from k groups of n1, n2, ... and nk elements from a group of n elements is calculated using the following equation:
Where n is equal to:
n=n1+n2+...+nk
If each team has two students, we can form 5 groups with 2 students each one. Then, k is equal to 5, n is equal to 10 and n1, n2, n3, n4 and n5 are equal to 2. So the number of ways to form teams are:
For part b, we can form 5 groups with 2 students or 2 groups with 2 students and 2 groups with 3 students. We already know that for the first case there are 113,400 ways to form group, so we need to calculate the number of ways for the second case as:
Replacing k by 4, n by 10, n1 and n2 by 2 and n3 and n4 by 3, we get:
So, If each team has either two or three students, The number of ways form teams are:
113,400 + 25,200 = 138,600