Let's analyze and solve the problems shown in the image.

13. Solve for $x$:

The triangle is a right triangle. Using the Pythagorean Theorem: $(x + 3)^2 = x^2 + 12^2$

Expanding and simplifying: $x^2 + 6x + 9 = x^2 + 144$

Cancel $x^2$ from both sides: $6x + 9 = 144$

Subtract 9: $6x = 135$

Divide by 6: $x = 22.5$

14. Use the diagram to find $h$:

In the given triangle, $h$ is the height from point $C$ perpendicular to $AB$, forming two right triangles. Using the geometric mean property: $h^2 = xy$

Taking the square root: $h = \sqrt{xy}$

Correct answer: (c) $\sqrt{xy}$.

15. If the area of a right isosceles triangle is $8 \, \text{cm}^2$, then the length of the hypotenuse is:

For a right isosceles triangle, the two legs are equal ($a$), and the hypotenuse is: $\text{Hypotenuse} = a\sqrt{2}.$

The area of the triangle is: $\text{Area} = \frac{1}{2}a^2 = 8$

Solve for $a^2$: $a^2 = 16 \implies a = 4.$

Hypotenuse: $\text{Hypotenuse} = 4\sqrt{2}.$

Correct answer: (b) $4\sqrt{2} \, \text{cm}$.

16. If the short leg of a 30°, 60°, 90° triangle is $12 \, \text{cm}$, then the area of the triangle is:

For a 30°, 60°, 90° triangle:

• Short leg = $12$
• Long leg = $12\sqrt{3}$
• Hypotenuse = $24$.

The area is: $\text{Area} = \frac{1}{2} \times \text{Short leg} \times \text{Long leg}.$

Substitute: $\text{Area} = \frac{1}{2} \times 12 \times 12\sqrt{3} = 72\sqrt{3}.$

Correct answer: (c) $72\sqrt{3} \, \text{cm}^2$.

Summary of Answers:

13. $x = 22.5$
14. (c) $\sqrt{xy}$
15. (b) $4\sqrt{2} \, \text{cm}$
16. (c) $72\sqrt{3} \, \text{cm}^2$

Do you want step-by-step explanations for any of these?
Here are 5 related questions to deepen understanding:

1. What is the derivation of the Pythagorean Theorem?
2. How do you calculate the altitude of a right triangle given its sides?
3. What is the general formula for the area of an isosceles triangle?
4. How are the side ratios of a 30°, 60°, 90° triangle derived?
5. How does the area formula change for a non-right triangle?

Tip: For 30°, 60°, 90° triangles, remember that the sides are always in the ratio $1:\sqrt{3}:2$.