A) The area of an equilateral triangle is 53 extsqrt(3) cm2. Find its perimeter. b) In a right-angled triangle ABC, AC = 6 cm, BC = 8 cm. Find the length of the hypotenuse AB and the length of the alt

The area $A$ of an equilateral triangle with side length $s$ is given by:

$$A = \frac{\sqrt{3}}{4} s^2$$

We are given $A = 53\sqrt{3} \text{ cm}^2$.

$$53\sqrt{3} = \frac{\sqrt{3}}{4} s^2$$

Divide both sides by $\sqrt{3}$:

$$53 = \frac{1}{4} s^2$$

Multiply both sides by 4:

$$s^2 = 53 \times 4$$ $$s^2 = 212$$

Take the square root of both sides:

$$s = \sqrt{212}$$

Simplify the radical:

$$s = \sqrt{4 \times 53}$$ $$s = 2\sqrt{53} \text{ cm}$$

The perimeter $P$ of an equilateral triangle is $3s$.

$$P = 3 \times (2\sqrt{53})$$ $$P = 6\sqrt{53} \text{ cm}$$

The perimeter of the equilateral triangle is $\boxed{\mathbf{6\sqrt{53} \text{ cm}}}$.

5. b)

Given $|AC| = 6 \text{ cm}$ and $|BC| = 8 \text{ cm}$.

$$AB^2 = AC^2 + BC^2$$ $$AB^2 = 6^2 + 8^2$$ $$AB^2 = 36 + 64$$ $$AB^2 = 100$$ $$AB = \sqrt{100}$$ $$AB = 10 \text{ cm}$$

In a right-angled triangle, the altitude to the hypotenuse creates two smaller triangles that are similar to the original triangle and to each other. Thus, $\triangle ADC \sim \triangle ACB$.
From the similarity, the ratio of corresponding sides is equal:

$$\frac{AD}{AC} = \frac{AC}{AB}$$

Substitute the known values:

$$\frac{x}{6} = \frac{6}{10}$$

Multiply both sides by 6:

$$x = \frac{6 \times 6}{10}$$ $$x = \frac{36}{10}$$ $$x = 3.6 \text{ cm}$$

The value of $x$ is $\boxed{\mathbf{3.6 \text{ cm}}}$.