Question

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directions: find the value of each variable.
1.
$x = 13$
$y = 13\sqrt{2}$
2.
$x = 15\sqrt{2}$
$y = 15\sqrt{2}$
3.
$x = 3$
$y = 3\sqrt{2}$
4.
$x = 17$
$y = 17\sqrt{2}$
5.
$x = 10$
$y = 10$
6.
$x = $
$y = $
7.
$x = 2\sqrt{7}$
$y = 2\sqrt{7}$
8.
$x = $
$y = $
9.
$x = $
$y = $
10.
$x = $
$y = $
11.
$x = \sqrt{10}$
$y = 2\sqrt{5}$
12.
$x = $
$y = $
13.
$x = \frac{17\sqrt{3}}{3}$
$y = \frac{34\sqrt{3}}{3}$
$z = \frac{34\sqrt{3}}{3}$
14.
$x = $
$y = $
$z = $

Explanation:

Step 1: Solve Triangle 6 (30-60-90)

Given long leg $=25\sqrt{3}$. For 30-60-90 triangles:
Short leg $y = \frac{\text{long leg}}{\sqrt{3}} = \frac{25\sqrt{3}}{\sqrt{3}} = 25$
Hypotenuse $x = 2 \times \text{short leg} = 2 \times 25 = 50$

Step 2: Solve Triangle 8 (30-60-90)

Given long leg $=24$.
Short leg $y = \frac{24}{\sqrt{3}} = 8\sqrt{3}$
Hypotenuse $x = 2 \times 8\sqrt{3} = 16\sqrt{3}$

Step 3: Solve Triangle 9 (30-60-90)

Given hypotenuse $=22\sqrt{3}$.
Short leg $x = \frac{22\sqrt{3}}{2} = 11\sqrt{3}$
Long leg $y = 11\sqrt{3} \times \sqrt{3} = 33$

Step 4: Solve Triangle 10 (30-60-90)

Given short leg $=\sqrt{6}$.
Long leg $x = \sqrt{6} \times \sqrt{3} = 3\sqrt{2}$
Hypotenuse $y = 2 \times \sqrt{6} = 2\sqrt{6}$

Step 5: Solve Triangle 12 (30-60-90)

Given short leg $=4\sqrt{21}$.
Long leg $x = 4\sqrt{21} \times \sqrt{3} = 12\sqrt{7}$
Hypotenuse $y = 2 \times 4\sqrt{21} = 8\sqrt{21}$

Step 6: Solve Triangle 14 (30-60-90)

Given hypotenuse of large triangle $=27$.
Short leg $z = \frac{27}{2} = 13.5$
Long leg of large triangle $= \frac{27\sqrt{3}}{2}$
For the small 30-60-90 triangle:
$y = \frac{1}{2} \times \frac{27\sqrt{3}}{2} = \frac{27\sqrt{3}}{4}$
$x = \frac{\sqrt{3}}{2} \times \frac{27\sqrt{3}}{2} = \frac{81}{4} = 20.25$

Answer:

Triangle 6: $x=50$, $y=25$
Triangle 8: $x=16\sqrt{3}$, $y=8\sqrt{3}$
Triangle 9: $x=11\sqrt{3}$, $y=33$
Triangle 10: $x=3\sqrt{2}$, $y=2\sqrt{6}$
Triangle 12: $x=12\sqrt{7}$, $y=8\sqrt{21}$
Triangle 14: $x=\frac{81}{4}$, $y=\frac{27\sqrt{3}}{4}$, $z=\frac{27}{2}$