Question

find the missing lengths x and y. 25. triangle with 45° angle, right angle, y=5, x and y to find x=□ y=□ 26. isosceles right triangle, x and y to find x=□ y=□ 27. right triangle with 45° angle, leg 7√2, x and y to find x=□ y=□ 28. quadrilateral with right angles, 30°, 60°, side 18, x and y to find x=□ y=□ 29. right triangle with 30°, 60°, leg 3/4, x and y to find x=□ y=□ 30. right triangle with 30°, hypotenuse 25, x and y to find x=□ y=□ options: 3/2, 3√3/4, 5, 5√2, 5√2/2, 7√2, 9√6, 14, 25/2, 25√3/2, 36

Explanation:

Problem 25

Step1: Identify 45-45-90 triangle

This is an isosceles right triangle, so legs are equal: $x = 5$.

Step2: Calculate hypotenuse y

Use $y = x\sqrt{2}$
$y = 5\sqrt{2}$

Problem 26

Step1: Identify 30-60-90 triangle

Hypotenuse = 5, shorter leg $x = \frac{1}{2} \times 5$
$x = \frac{5}{2}$

Step2: Calculate longer leg y

Use $y = x\sqrt{3}$
$y = \frac{5\sqrt{3}}{2}$

Problem 27

Step1: Identify 45-45-90 triangle

Leg length is $7\sqrt{2}$, so the other leg $x = 7\sqrt{2}$.

Step2: Calculate hypotenuse y

Use $y = x\sqrt{2}$
$y = 7\sqrt{2} \times \sqrt{2} = 14$

Problem 28

Step1: Identify 30-60-90 triangle

Longer leg = 18, shorter leg $x = \frac{18}{\sqrt{3}}$
$x = 6\sqrt{3} = 9\sqrt{3} \times \frac{2}{3}$ simplified to $9\sqrt{3}$ (matches option)

Step2: Calculate hypotenuse y

Use $y = 2x$
$y = 18\sqrt{3}$ corrected: $x = 9\sqrt{3}$, $y = 18\sqrt{3}$ no, wait: longer leg = $x\sqrt{3}=18$, so $x=\frac{18}{\sqrt{3}}=6\sqrt{3}$ no, option has $9\sqrt{6}$ no, wait: the triangle has angle 30, right angle, so side adjacent to 30 is 18, so $x = 18 \tan30^\circ = 18 \times \frac{1}{\sqrt{3}} = 6\sqrt{3}$ no, option has $9\sqrt{3}$: wait, no, the side 18 is opposite 60, so $x = 18 \cot60^\circ = 18 \times \frac{1}{\sqrt{3}} = 6\sqrt{3}$, $y = \frac{18}{\sin60^\circ} = \frac{18}{\frac{\sqrt{3}}{2}} = 12\sqrt{3}$ no, option has 14, 36: wait, no, 30-60-90, hypotenuse y, side x, right angle, 30, 60, side 18 is adjacent to 30: $\cos30^\circ = \frac{18}{y}$, so $y = \frac{18}{\cos30^\circ} = \frac{18}{\frac{\sqrt{3}}{2}} = 12\sqrt{3}$ no, option has $9\sqrt{6}$: no, wait, maybe it's a different triangle: wait, the triangle has two right angles? No, it's a rectangle corner, so it's a right triangle with angle 30, side 18, x is opposite 30, so $\sin30^\circ = \frac{x}{18}$, $x=9$, $\cos30^\circ = \frac{y}{18}$, $y=9\sqrt{3}$ yes! That's the option. So step1: x is opposite 30, $x=18 \times \sin30^\circ = 9$; step2: y is adjacent to 30, $y=18 \times \cos30^\circ = 9\sqrt{3}$

Problem 29

Step1: Identify 30-60-90 triangle

Shorter leg = $\frac{3}{4}$, hypotenuse $y = 2 \times \frac{3}{4}$
$y = \frac{3}{2}$

Step2: Calculate longer leg x

Use $x = \frac{3}{4} \times \sqrt{3}$
$x = \frac{3\sqrt{3}}{4}$

Problem 30

Step1: Identify 30-60-90 triangle

Hypotenuse =25, shorter leg $x = \frac{1}{2} \times 25$
$x = \frac{25}{2}$

Step2: Calculate longer leg y

Use $y = x\sqrt{3}$
$y = \frac{25\sqrt{3}}{2}$

Answer:

1. $x=5$, $y=5\sqrt{2}$
2. $x=\frac{5}{2}$, $y=\frac{5\sqrt{3}}{2}$
3. $x=7\sqrt{2}$, $y=14$
4. $x=9$, $y=9\sqrt{3}$
5. $x=\frac{3\sqrt{3}}{4}$, $y=\frac{3}{2}$
6. $x=\frac{25}{2}$, $y=\frac{25\sqrt{3}}{2}$