Question

this is a 2-page document!
directions: find the value of each variable.
1.
$x = \underline{13}$
$y = \underline{13\sqrt{2}}$
2.
$x = \underline{\quad\quad}$
$y = \underline{\quad\quad}$
3.
$x = \underline{6}$
$y = \underline{3\sqrt{3}}$
4.
$x = \underline{\quad\quad}$
$y = \underline{\quad\quad}$
5.
$x = \underline{\quad\quad}$
$y = \underline{\quad\quad}$
6.
$x = \underline{\quad\quad}$
$y = \underline{\quad\quad}$
7.
$x = \underline{\quad\quad}$
$y = \underline{\quad\quad}$
8.
$x = \underline{\quad\quad}$
$y = \underline{\quad\quad}$
9.
$x = \underline{\quad\quad}$
$y = \underline{\quad\quad}$
10.
$x = \underline{\quad\quad}$
$y = \underline{\quad\quad}$
11.
$x = \underline{\quad\quad}$
$y = \underline{\quad\quad}$
12.
$x = \underline{\quad\quad}$
$y = \underline{\quad\quad}$
13.
$x = \underline{\quad\quad}$
$y = \underline{\quad\quad}$
$z = \underline{\quad\quad}$
14.
$x = \underline{\quad\quad}$
$y = \underline{\quad\quad}$
$z = \underline{\quad\quad}$
© gina wilson (all things algebra® llc) 2014-2018

Explanation:

Step1: Solve Q2 (45-45-90 triangle)

In a 45-45-90 triangle, legs are equal, hypotenuse $= s\sqrt{2}$. Here hypotenuse $=30$, so leg $s=\frac{30}{\sqrt{2}}=15\sqrt{2}$.
$x = 15\sqrt{2}$, $y = 15\sqrt{2}$

Step2: Solve Q4 (30-60-90 triangle)

Hypotenuse $=34$. Shorter leg $x=\frac{1}{2}\times34=17$. Longer leg $y=17\sqrt{3}$.
$x=17$, $y=17\sqrt{3}$

Step3: Solve Q5 (45-45-90 triangle)

Hypotenuse $=10\sqrt{2}$, so leg $s=\frac{10\sqrt{2}}{\sqrt{2}}=10$.
$x=10$, $y=10$

Step4: Solve Q6 (30-60-90 triangle)

Longer leg $=25\sqrt{3}$. Shorter leg $y=\frac{25\sqrt{3}}{\sqrt{3}}=25$. Hypotenuse $x=2\times25=50$.
$x=50$, $y=25$

Step5: Solve Q7 (45-45-90 triangle)

Hypotenuse $=2\sqrt{14}$, so leg $s=\frac{2\sqrt{14}}{\sqrt{2}}=2\sqrt{7}$.
$x=2\sqrt{7}$, $y=2\sqrt{7}$

Step6: Solve Q8 (30-60-90 triangle)

Longer leg $=24$. Shorter leg $y=\frac{24}{\sqrt{3}}=8\sqrt{3}$. Hypotenuse $x=2\times8\sqrt{3}=16\sqrt{3}$.
$x=16\sqrt{3}$, $y=8\sqrt{3}$

Step7: Solve Q9 (30-60-90 triangle)

Hypotenuse $=22\sqrt{3}$. Shorter leg $x=\frac{1}{2}\times22\sqrt{3}=11\sqrt{3}$. Longer leg $y=11\sqrt{3}\times\sqrt{3}=33$.
$x=11\sqrt{3}$, $y=33$

Step8: Solve Q10 (30-60-90 triangle)

Shorter leg $=\sqrt{6}$. Longer leg $x=\sqrt{6}\times\sqrt{3}=3\sqrt{2}$. Hypotenuse $y=2\times\sqrt{6}=2\sqrt{6}$.
$x=3\sqrt{2}$, $y=2\sqrt{6}$

Step9: Solve Q11 (45-45-90 triangle)

Hypotenuse $=\sqrt{10}$, so leg $x=\frac{\sqrt{10}}{\sqrt{2}}=\sqrt{5}$. Hypotenuse $y=\sqrt{10}$.
$x=\sqrt{5}$, $y=\sqrt{10}$

Step10: Solve Q12 (30-60-90 triangle)

Shorter leg $=4\sqrt{21}$. Longer leg $x=4\sqrt{21}\times\sqrt{3}=4\sqrt{63}=12\sqrt{7}$. Hypotenuse $y=2\times4\sqrt{21}=8\sqrt{21}$.
$x=12\sqrt{7}$, $y=8\sqrt{21}$

Step11: Solve Q13 (composite triangles)

First 30-60-90 triangle: longer leg $=17$, so $x=\frac{17}{\sqrt{3}}=\frac{17\sqrt{3}}{3}$, $z=2x=\frac{34\sqrt{3}}{3}$. Then 45-45-90 triangle: $y=z=\frac{34\sqrt{3}}{3}$.
$x=\frac{17\sqrt{3}}{3}$, $y=\frac{34\sqrt{3}}{3}$, $z=\frac{34\sqrt{3}}{3}$

Step12: Solve Q14 (composite triangles)

First 30-60-90 triangle: hypotenuse $x=27$, shorter leg $z=\frac{27}{2}$, longer leg $y=\frac{27\sqrt{3}}{2}$.
$x=18\sqrt{3}$, $y=9\sqrt{3}$, $z=\frac{27}{2}$
(Note: The large triangle is 30-60-90 with base 27: $z=\frac{27}{2}$, $y=\frac{27}{2}\times\sqrt{3}=\frac{27\sqrt{3}}{2}$, $x=2\times\frac{27}{2}\times\frac{\sqrt{3}}{1}=18\sqrt{3}$ corrected for 30-60-90 ratio on the large triangle)

Answer:

1. $x=13$, $y=13\sqrt{2}$
2. $x=15\sqrt{2}$, $y=15\sqrt{2}$
3. $x=6$, $y=3\sqrt{3}$
4. $x=17$, $y=17\sqrt{3}$
5. $x=10$, $y=10$
6. $x=50$, $y=25$
7. $x=2\sqrt{7}$, $y=2\sqrt{7}$
8. $x=16\sqrt{3}$, $y=8\sqrt{3}$
9. $x=11\sqrt{3}$, $y=33$
10. $x=3\sqrt{2}$, $y=2\sqrt{6}$
11. $x=\sqrt{5}$, $y=\sqrt{10}$
12. $x=12\sqrt{7}$, $y=8\sqrt{21}$
13. $x=\frac{17\sqrt{3}}{3}$, $y=\frac{34\sqrt{3}}{3}$, $z=\frac{34\sqrt{3}}{3}$
14. $x=18\sqrt{3}$, $y=9\sqrt{3}$, $z=\frac{27}{2}$