QUESTION IMAGE
Question
find the missing side lengths. leave your answers as radicals in simplest form.
1)
a) $m = \frac{3\sqrt{2}}{2}$, $n = \frac{3\sqrt{2}}{2}$
b) $m = \sqrt{6}$, $n = \sqrt{6}$
c) $m = \frac{3}{2}$, $n = \frac{3}{2}$
d) $m = 3\sqrt{2}$, $n = 3\sqrt{2}$
2)
a) $u = \frac{4\sqrt{10}}{5}$, $v = \frac{2\sqrt{10}}{5}$
b) $u = \frac{8\sqrt{5}}{5}$, $v = \frac{4\sqrt{5}}{5}$
c) $u = \frac{4\sqrt{10}}{5}$, $v = \frac{4\sqrt{5}}{5}$
d) $u = \frac{8\sqrt{5}}{5}$, $v = \frac{2\sqrt{10}}{5}$
3)
a) $u = 3$, $v = \frac{3\sqrt{2}}{2}$
b) $u = 3$, $v = \frac{3\sqrt{2}}{4}$
c) $u = \frac{3\sqrt{2}}{2}$, $v = 3\sqrt{3}$
d) $u = 3\sqrt{3}$, $v = \frac{3\sqrt{2}}{2}$
4)
a) $a = 5$, $b = 5$
b) $a = 5\sqrt{2}$, $b = 5\sqrt{3}$
c) $a = 5\sqrt{3}$, $b = 5\sqrt{2}$
d) $a = 5\sqrt{2}$, $b = 5$
5)
a) $x = 4\sqrt{2}$, $y = 4$
b) $x = 4\sqrt{2}$, $y = 8$
c) $x = 4\sqrt{6}$, $y = 8$
d) $x = 4$, $y = 4\sqrt{6}$
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Problem 1
Step1: Identify triangle type
This is a 45-45-90 right triangle, so legs are equal, hypotenuse $= s\sqrt{2}$ where $s$ is leg length.
Step2: Solve for leg length
Let $m=n=s$, hypotenuse $3 = s\sqrt{2}$. Solve for $s$:
$$s = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$$
So $m = \frac{3\sqrt{2}}{2}$, $n = \frac{3\sqrt{2}}{2}$
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Problem 2
Step1: Identify triangle type
45-45-90 right triangle, legs are equal: $v = \frac{4\sqrt{5}}{5}$
Step2: Calculate hypotenuse $u$
Hypotenuse $= s\sqrt{2}$, where $s=\frac{4\sqrt{5}}{5}$:
$$u = \frac{4\sqrt{5}}{5} \times \sqrt{2} = \frac{4\sqrt{10}}{5}$$
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Problem 3
Step1: Identify triangle type
45-45-90 right triangle, legs are equal: $v = \frac{3\sqrt{2}}{2}$
Step2: Calculate hypotenuse $u$
Hypotenuse $= s\sqrt{2}$, where $s=\frac{3\sqrt{2}}{2}$:
$$u = \frac{3\sqrt{2}}{2} \times \sqrt{2} = \frac{3\times2}{2} = 3$$
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Problem 4
Step1: Identify triangle type
45-45-90 right triangle, legs are equal: $b = 5$
Step2: Calculate hypotenuse $a$
Hypotenuse $= s\sqrt{2}$, where $s=5$:
$$a = 5\sqrt{2}$$
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Problem 5
Step1: Identify triangle type
45-45-90 right triangle, legs are equal: $y = 4$
Step2: Calculate hypotenuse $x$
Hypotenuse $= s\sqrt{2}$, where $s=4$:
$$x = 4\sqrt{2}$$
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- A) $m=\frac{3\sqrt{2}}{2}, n=\frac{3\sqrt{2}}{2}$
- C) $u=\frac{4\sqrt{10}}{5}, v=\frac{4\sqrt{5}}{5}$
- A) $u=3, v=\frac{3\sqrt{2}}{2}$
- D) $a=5\sqrt{2}, b=5$
- A) $x=4\sqrt{2}, y=4$