Question

name marina cambre pacheco id: 1 date 10/01/2025 period 4 all special right triangles 13) 14) 15) 16) find the missing side lengths. leave your answers as radicals in simplest form and round to the 100ths place. 17) 18) 19) 20) 21) 22)

Explanation:

Step1: Recall special - right - triangle ratios

For a 30 - 60 - 90 triangle, the side - length ratios are $1:\sqrt{3}:2$ (shorter leg : longer leg : hypotenuse). For a 45 - 45 - 90 triangle, the side - length ratios are $1:1:\sqrt{2}$ (legs : hypotenuse).

Step2: Solve problem 17

This is a 45 - 45 - 90 triangle. If one leg $a=\frac{5\sqrt{3}}{8}$, then the other leg $x = \frac{5\sqrt{3}}{8}$ (since legs are equal in a 45 - 45 - 90 triangle), and the hypotenuse $h=\frac{5\sqrt{3}}{8}\sqrt{2}=\frac{5\sqrt{6}}{8}\approx1.53$.

Step3: Solve problem 18

This is a 45 - 45 - 90 triangle. If one leg $a = \frac{5}{8}$, then the other leg $x=\frac{5}{8}$, and the hypotenuse $h=\frac{5}{8}\sqrt{2}=\frac{5\sqrt{2}}{8}\approx0.88$.

Step4: Solve problem 19

We have a right - triangle with angles 45 and 60 degrees. First, consider the 30 - 60 - 90 part. If the side opposite the 60 - degree angle is $6\sqrt{5}$, then the side opposite the 30 - degree angle is $\frac{6\sqrt{5}}{\sqrt{3}}=\frac{6\sqrt{5}\times\sqrt{3}}{3}=2\sqrt{15}$. Then, considering the 45 - 45 - 90 part, the hypotenuse $x$ of the large triangle is $2\sqrt{15}\sqrt{2}=2\sqrt{30}\approx10.95$.

Step5: Solve problem 20

This is a 30 - 60 - 90 triangle. If the side opposite the 60 - degree angle is $7\sqrt{3}$, then the side opposite the 30 - degree angle $x=\frac{7\sqrt{3}}{\sqrt{3}} = 7$.

Step6: Solve problem 21

We have a triangle with 45 and 60 - degree angles. Let's assume the side opposite the 45 - degree angle is $a$. If the side adjacent to the 60 - degree angle is 6. In the 30 - 60 - 90 part, the side opposite the 30 - degree angle is $\frac{6}{\sqrt{3}} = 2\sqrt{3}$. Then in the 45 - 45 - 90 part, $x = 2\sqrt{3}$ and $y=2\sqrt{3}\sqrt{2}=2\sqrt{6}\approx4.90$.

Step7: Solve problem 22

We have a triangle with 45 and 60 - degree angles. If one side is $\frac{8}{9}$. In the 45 - 45 - 90 part, if one leg is $\frac{8}{9}$, then $y=\frac{8}{9}\sqrt{2}$. In the 30 - 60 - 90 part, $x=\frac{y}{\sqrt{3}}=\frac{8\sqrt{2}}{9\sqrt{3}}=\frac{8\sqrt{6}}{27}\approx0.73$.

Answer:

17: Leg $x=\frac{5\sqrt{3}}{8}\approx1.08$, Hypotenuse $\frac{5\sqrt{6}}{8}\approx1.53$;
18: Leg $x = \frac{5}{8}$, Hypotenuse $\frac{5\sqrt{2}}{8}\approx0.88$;
19: $x = 2\sqrt{30}\approx10.95$;
20: $x = 7$;
21: $x = 2\sqrt{3}\approx3.46$, $y=2\sqrt{6}\approx4.90$;
22: $x=\frac{8\sqrt{6}}{27}\approx0.73$, $y=\frac{8\sqrt{2}}{9}\approx1.26$