Multi-Step Special Right Triangles worksheet for finding missing side lengths in geometric problems.

This is a worksheet on "Multi-Step Special Right Triangles" from Kuta Software. The task is to find the missing side lengths, leaving answers in simplest radical form. The problems involve 45°-45°-90° triangles and 30°-60°-90° triangles.

I will solve each problem one by one, explaining the solution for each.

---

Problem 1

This is a 45°-45°-90° triangle with a hypotenuse of 10.
In a 45°-45°-90° triangle, the legs are equal, and the hypotenuse is $ \text{leg} \times \sqrt{2} $.
So, if the hypotenuse is 10, then each leg is $ \frac{10}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2} $.
The missing side $ x $ is a leg, so $ x = 5\sqrt{2} $.

---

Problem 2

This is a 45°-45°-90° triangle with a leg of 7.
In a 45°-45°-90° triangle, the hypotenuse is $ \text{leg} \times \sqrt{2} $.
So, if a leg is 7, then the hypotenuse $ x = 7\sqrt{2} $.

---

Problem 3

This is a 45°-45°-90° triangle with a leg of 9.
In a 45°-45°-90° triangle, the hypotenuse is $ \text{leg} \times \sqrt{2} $.
So, if a leg is 9, then the hypotenuse $ x = 9\sqrt{2} $.

---

Problem 4

This is a 45°-45°-90° triangle with a hypotenuse of 9.
In a 45°-45°-90° triangle, the legs are equal, and the hypotenuse is $ \text{leg} \times \sqrt{2} $.
So, if the hypotenuse is 9, then each leg is $ \frac{9}{\sqrt{2}} = \frac{9\sqrt{2}}{2} $.
The missing side $ x $ is a leg, so $ x = \frac{9\sqrt{2}}{2} $.

---

Problem 5

This is a 45°-45°-90° triangle with a hypotenuse of $ 6\sqrt{2} $.
In a 45°-45°-90° triangle, the legs are equal, and the hypotenuse is $ \text{leg} \times \sqrt{2} $.
So, if the hypotenuse is $ 6\sqrt{2} $, then each leg is $ \frac{6\sqrt{2}}{\sqrt{2}} = 6 $.
The missing side $ x $ is a leg, so $ x = 6 $.

---

Problem 6

This is a 45°-45°-90° triangle with a leg of $ 9\sqrt{2} $.
In a 45°-45°-90° triangle, the hypotenuse is $ \text{leg} \times \sqrt{2} $.
So, if a leg is $ 9\sqrt{2} $, then the hypotenuse $ x = 9\sqrt{2} \times \sqrt{2} = 9 \times 2 = 18 $.

---

Problem 7

This is a 30°-60°-90° triangle with the side opposite the 30° angle being 9.
In a 30°-60°-90° triangle, the sides are in the ratio $ 1 : \sqrt{3} : 2 $, where the side opposite 30° is the shortest, the side opposite 60° is $ \sqrt{3} $ times the shortest, and the hypotenuse is twice the shortest.
Here, the side opposite 30° is 9, so the hypotenuse is $ 2 \times 9 = 18 $, and the side opposite 60° is $ 9\sqrt{3} $.
The missing side $ x $ is the side opposite 60°, so $ x = 9\sqrt{3} $.

---

Problem 8

This is a 30°-60°-90° triangle with the side opposite the 60° angle being 5.
In a 30°-60°-90° triangle, the sides are in the ratio $ 1 : \sqrt{3} : 2 $.
If the side opposite 60° is 5, then the side opposite 30° is $ \frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3} $, and the hypotenuse is $ 2 \times \frac{5}{\sqrt{3}} = \frac{10}{\sqrt{3}} = \frac{10\sqrt{3}}{3} $.
The missing side $ x $ is the hypotenuse, so $ x = \frac{10\sqrt{3}}{3} $.

---

Final Answers:

1) $ x = 5\sqrt{2} $
2) $ x = 7\sqrt{2} $
3) $ x = 9\sqrt{2} $
4) $ x = \frac{9\sqrt{2}}{2} $
5) $ x = 6 $
6) $ x = 18 $
7) $ x = 9\sqrt{3} $
8) $ x = \frac{10\sqrt{3}}{3} $

These are the solutions to the worksheet problems.