Question

special right triangles
date
find the missing side lengths. leave your answers as radicals in simplest form.
1)
2)
3)
4)
5)
6)
7)
8)

Explanation:

Step1: Solve Q3: Identify leg length

In a 45-45-90 triangle, legs are equal.
$y = \frac{3\sqrt{2}}{2}$

Step2: Solve Q3: Find hypotenuse

Hypotenuse = leg $\times \sqrt{2}$
$x = \frac{3\sqrt{2}}{2} \times \sqrt{2} = \frac{3 \times 2}{2} = 3$

Step3: Solve Q4: Identify leg length

In a 45-45-90 triangle, legs are equal.
$y = 3\sqrt{2}$

Step4: Solve Q4: Find hypotenuse

Hypotenuse = leg $\times \sqrt{2}$
$x = 3\sqrt{2} \times \sqrt{2} = 3 \times 2 = 6$

Step5: Solve Q6: Find leg lengths

In a 45-45-90 triangle, leg = $\frac{\text{hypotenuse}}{\sqrt{2}}$
$x = y = \frac{2\sqrt{6}}{\sqrt{2}} = 2\sqrt{3}$

Step6: Solve Q8: Find hypotenuse

In a 30-60-90 triangle, hypotenuse = $2 \times$ shorter leg
$u = 2 \times 2 = 4$

Step7: Solve Q8: Find longer leg

Longer leg = shorter leg $\times \sqrt{3}$
$v = 2 \times \sqrt{3} = 2\sqrt{3}$

Answer:

1. $x=3$, $y=\frac{3\sqrt{2}}{2}$
2. $x=6$, $y=3\sqrt{2}$
3. $x=2\sqrt{3}$, $y=2\sqrt{3}$
4. $u=4$, $v=2\sqrt{3}$