Question

find the missing side lengths. leave your answers as radicals in simplest form.
34)
a) $x = 4$, $y = 2\sqrt{3}$
b) $x = \frac{8\sqrt{3}}{3}$, $y = \sqrt{3}$
c) $x = \frac{8\sqrt{3}}{3}$, $y = 2\sqrt{3}$
d) $x = 4$, $y = \sqrt{3}$
35)
a) $x = \frac{3\sqrt{2}}{2}$, $y = \frac{\sqrt{6}}{4}$
b) $x = \frac{\sqrt{6}}{2}$, $y = \frac{\sqrt{6}}{4}$
c) $x = \frac{3\sqrt{2}}{2}$, $y = \frac{3\sqrt{2}}{4}$
d) $x = \frac{\sqrt{6}}{2}$, $y = \frac{3\sqrt{2}}{4}$
36)
a) $u = 3$, $v = 3$
b) $u = \sqrt{3}$, $v = \sqrt{3}$
c) $u = \frac{3\sqrt{3}}{2}$, $v = \frac{3\sqrt{3}}{2}$
d) $u = \frac{3\sqrt{2}}{2}$, $v = \frac{3\sqrt{2}}{2}$
37)
a) $x = 10$, $y = 5\sqrt{2}$
b) $x = 10\sqrt{2}$, $y = 10$
c) $x = 5\sqrt{2}$, $y = 10\sqrt{2}$
d) $x = 10$, $y = 10\sqrt{2}$
find the coordinates of the vertices of each figure after the given transformation.

Explanation:

Problem 34

Step1: Identify triangle type

It's a 30-60-90 right triangle. The side opposite 30° is 2 (let's assume the shorter leg is 2). In a 30-60-90 triangle, hypotenuse \( x = 2 \times \text{shorter leg} \), longer leg \( y = \text{shorter leg} \times \sqrt{3} \).

Step2: Calculate \( x \) and \( y \)

Shorter leg = 2, so \( x = 2\times2 = 4 \), \( y = 2\times\sqrt{3}=2\sqrt{3} \).

Step1: Identify triangle type

It's a 30-60-90 triangle? Wait, no, angle is 30°? Wait, the leg adjacent to 30° is \( \frac{3\sqrt{6}}{4} \)? Wait, no, let's check 30-60-90 ratios. Wait, maybe 45-45-90? No, angle is 30°. Wait, in 30-60-90, sides are in ratio \( 1:\sqrt{3}:2 \). Wait, the given leg: let's see option C: \( x=\frac{3\sqrt{2}}{2} \), \( y=\frac{3\sqrt{2}}{4} \)? No, wait, maybe I made a mistake. Wait, the triangle has a 30° angle, so hypotenuse \( x \), shorter leg \( y \), longer leg \( \frac{3\sqrt{6}}{4} \). In 30-60-90, longer leg = shorter leg \( \times \sqrt{3} \), hypotenuse = 2 \( \times \) shorter leg. Let's solve: Let shorter leg be \( y \), longer leg \( \frac{3\sqrt{6}}{4} = y\sqrt{3} \), so \( y = \frac{3\sqrt{6}}{4\sqrt{3}}=\frac{3\sqrt{2}}{4} \)? No, that's not matching. Wait, maybe it's a 45-45-90? No, angle is 30°. Wait, option C: \( x=\frac{3\sqrt{2}}{2} \), \( y=\frac{3\sqrt{2}}{4} \)? Wait, no, let's check the ratios again. Wait, maybe the triangle is 30-60-90, and the given side is the longer leg. So longer leg \( = \frac{3\sqrt{6}}{4} \), shorter leg \( y = \frac{\text{longer leg}}{\sqrt{3}}=\frac{3\sqrt{6}}{4\sqrt{3}}=\frac{3\sqrt{2}}{4} \)? No, that's not matching. Wait, option C: \( x=\frac{3\sqrt{2}}{2} \) (hypotenuse), \( y=\frac{3\sqrt{2}}{4} \) (shorter leg). Hypotenuse should be 2 \( \times \) shorter leg: \( 2\times\frac{3\sqrt{2}}{4}=\frac{3\sqrt{2}}{2} \), which matches \( x \). And longer leg \( = y\sqrt{3}=\frac{3\sqrt{2}}{4}\times\sqrt{3}=\frac{3\sqrt{6}}{4} \), which matches the given leg. So yes, option C.

Step2: Verify

Shorter leg \( y=\frac{3\sqrt{2}}{4} \), hypotenuse \( x=2y=\frac{3\sqrt{2}}{2} \), longer leg \( y\sqrt{3}=\frac{3\sqrt{6}}{4} \), which matches the given leg.

Step1: Identify triangle type

It's a 45-45-90 right triangle (angle 45°), so legs \( u \) and \( v \) are equal, hypotenuse \( 3\sqrt{2} \). In 45-45-90, hypotenuse \( = \text{leg} \times \sqrt{2} \).

Step2: Calculate legs

Let leg be \( u = v \), then \( 3\sqrt{2}=u\sqrt{2} \), so \( u = v = 3 \)? No, wait, \( 3\sqrt{2}=u\sqrt{2} \) implies \( u = 3 \)? But option D: \( u=\frac{3\sqrt{2}}{2} \), \( v=\frac{3\sqrt{2}}{2} \). Wait, wait, hypotenuse is \( 3\sqrt{2} \), so leg \( = \frac{\text{hypotenuse}}{\sqrt{2}}=\frac{3\sqrt{2}}{\sqrt{2}} = 3 \)? But option D is \( \frac{3\sqrt{2}}{2} \). Wait, I must have messed up. Wait, the hypotenuse is \( 3\sqrt{2} \), so leg \( = \frac{3\sqrt{2}}{\sqrt{2}} = 3 \), but option A is \( u=3 \), \( v=3 \). Wait, but option D is \( \frac{3\sqrt{2}}{2} \). Wait, maybe the triangle is 30-60-90? No, angle is 45°. Wait, hypotenuse \( 3\sqrt{2} \), so legs are \( \frac{3\sqrt{2}}{\sqrt{2}} = 3 \), so option A? But option D is \( \frac{3\sqrt{2}}{2} \). Wait, maybe the hypotenuse is \( 3\sqrt{2} \), so legs are \( \frac{3\sqrt{2}}{\sqrt{2}} = 3 \), so option A: \( u=3 \), \( v=3 \). But wait, let's check option D: \( u=\frac{3\sqrt{2}}{2} \), \( v=\frac{3\sqrt{2}}{2} \), hypotenuse would be \( \frac{3\sqrt{2}}{2}\times\sqrt{2}=3 \), but the hypotenuse is \( 3\sqrt{2} \), so that's wrong. So option A is correct? Wait, no, maybe I misread the hypotenuse. Wait, the hypotenuse is \( 3\sqrt{2} \), so legs are \( 3 \), so option A: \( u=3 \), \( v=3 \). But wait, the options: A) \( u=3 \), \( v=3 \); D) \( u=\frac{3\sqrt{2}}{2} \), \( v=\frac{3\sqrt{2}}{2} \). Wait, if hypotenuse is \( 3\sqrt{2} \), then legs are \( 3 \), so option A. But maybe the hypotenuse is \( 3\sqrt{2} \), so legs are \( 3 \), so answer A.

Answer:

A. \( x = 4 \), \( y = 2\sqrt{3} \)

Problem 35