Solved Trigonometry Prerequisite: Special Right Triangles | Chegg.com - Free Printable

Here are the step-by-step solutions for each problem on the worksheet.

Part 1: Find the value of x in each triangle (45°-45°-90° Triangles)

*Rule: In these triangles, the two legs are equal. The hypotenuse is the leg multiplied by $\sqrt{2}$.*

1.
The legs are both 8. We need to find the hypotenuse ($x$).
$$x = 8 \cdot \sqrt{2}$$
Answer: $8\sqrt{2}$

2.
The hypotenuse is 14. We need to find a leg ($x$).
$$x = \frac{14}{\sqrt{2}}$$
To simplify, multiply top and bottom by $\sqrt{2}$:
$$x = \frac{14\sqrt{2}}{2} = 7\sqrt{2}$$
Answer: $7\sqrt{2}$

3.
This is a square cut in half. The side length is 18. The diagonal is the hypotenuse ($x$).
$$x = 18 \cdot \sqrt{2}$$
Answer: $18\sqrt{2}$

4.
The hypotenuse is 18. We need to find the leg ($x$).
$$x = \frac{18}{\sqrt{2}}$$
Simplify:
$$x = \frac{18\sqrt{2}}{2} = 9\sqrt{2}$$
Answer: $9\sqrt{2}$

5.
The hypotenuse is 8. We need to find the leg ($x$).
$$x = \frac{8}{\sqrt{2}}$$
Simplify:
$$x = \frac{8\sqrt{2}}{2} = 4\sqrt{2}$$
Answer: $4\sqrt{2}$

6.
The diagonal (hypotenuse) is 10. We need to find the side ($x$).
$$x = \frac{10}{\sqrt{2}}$$
Simplify:
$$x = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$$
Answer: $5\sqrt{2}$

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Part 2: Sketch and Solve Word Problems

7. The perimeter of a square is 48 meters. Find the length of a diagonal.
* First, find the side length. A square has 4 equal sides.
$$\text{Side} = \frac{48}{4} = 12 \text{ meters}$$
* Now, find the diagonal using the special right triangle rule (Side $\cdot \sqrt{2}$).
$$\text{Diagonal} = 12\sqrt{2}$$
Answer: $12\sqrt{2}$ meters

8. The perimeter of a square is 20 cm. Find the length of a diagonal.
* First, find the side length.
$$\text{Side} = \frac{20}{4} = 5 \text{ cm}$$
* Now, find the diagonal.
$$\text{Diagonal} = 5\sqrt{2}$$
Answer: $5\sqrt{2}$ cm

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Part 3: Find the value of x and y

9. (45°-45°-90° Triangle)
* Hypotenuse is 7.
* Legs ($x$ and $y$) are equal.
$$x = y = \frac{7}{\sqrt{2}} = \frac{7\sqrt{2}}{2}$$
Answer: $x = \frac{7\sqrt{2}}{2}, y = \frac{7\sqrt{2}}{2}$

10. (45°-45°-90° Triangle)
* One leg is 10. Since it's a 45-45-90 triangle, the other leg ($x$) is also 10.
* The hypotenuse ($y$) is the leg times $\sqrt{2}$.
$$y = 10\sqrt{2}$$
Answer: $x = 10, y = 10\sqrt{2}$

11. (45°-45°-90° Triangle)
* One leg is 7. The other leg ($x$) is also 7.
* The hypotenuse ($y$) is $7\sqrt{2}$.
Answer: $x = 7, y = 7\sqrt{2}$

12. (30°-60°-90° Triangle)
* *Rule: Short leg is opposite 30°. Long leg is opposite 60°. Hypotenuse is $2 \times$ short leg.*
* The side labeled 20 is opposite the 60° angle (Long Leg).
* To find the short leg ($x$): Divide the long leg by $\sqrt{3}$.
$$x = \frac{20}{\sqrt{3}} = \frac{20\sqrt{3}}{3}$$
* To find the hypotenuse ($y$): Multiply the short leg by 2.
$$y = 2 \cdot \left(\frac{20\sqrt{3}}{3}\right) = \frac{40\sqrt{3}}{3}$$
Answer: $x = \frac{20\sqrt{3}}{3}, y = \frac{40\sqrt{3}}{3}$

13. (30°-60°-90° Triangle)
* The hypotenuse is 20.
* The short leg ($x$, opposite 30°) is half the hypotenuse.
$$x = \frac{20}{2} = 10$$
* The long leg ($y$, opposite 60°) is the short leg times $\sqrt{3}$.
$$y = 10\sqrt{3}$$
Answer: $x = 10, y = 10\sqrt{3}$

14. (30°-60°-90° Triangle)
* The short leg (opposite 30°) is 20.
* The hypotenuse ($x$) is twice the short leg.
$$x = 20 \cdot 2 = 40$$
* The long leg ($y$, opposite 60°) is the short leg times $\sqrt{3}$.
$$y = 20\sqrt{3}$$
Answer: $x = 40, y = 20\sqrt{3}$

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Final Answer:
1. $8\sqrt{2}$
2. $7\sqrt{2}$
3. $18\sqrt{2}$
4. $9\sqrt{2}$
5. $4\sqrt{2}$
6. $5\sqrt{2}$
7. $12\sqrt{2}$ m
8. $5\sqrt{2}$ cm
9. $x = \frac{7\sqrt{2}}{2}, y = \frac{7\sqrt{2}}{2}$
10. $x = 10, y = 10\sqrt{2}$
11. $x = 7, y = 7\sqrt{2}$
12. $x = \frac{20\sqrt{3}}{3}, y = \frac{40\sqrt{3}}{3}$
13. $x = 10, y = 10\sqrt{3}$
14. $x = 40, y = 20\sqrt{3}$