Question

If an isoscles triangle ABC,AB=AC=6cm is inscribed in a circle of radius9cm,find the area of the circle

Answer 1

If an isosceles triangle ABC, AB=AC=6cm is inscribed in a circle of radius 9cm.

Find the area of the triangle.

Solution:

$$S = \frac {1}{2} B C \times A H = H B \times A H$$

Theorem: The angle subtended at the centre of a circle by an arc is twice any angle at the circumference standing on the same arc.

$$\text{So } \angle AOC = 2\angle ABC$$

$\Delta AOC$ is an isosceles triangle (OA and OC are radiuses)

So if OM is a perpendicular to AC, then AM=MC and $\angle AOM = \angle MOC = \frac{1}{2}\angle AOC = \angle ABC$

$\Delta AOM$ and $\Delta ABH$ are similar triangle ($\angle AOM = \angle ABC$, $\angle AMO$ and $\angle AHB$ are right -AA)

$$\text{So } \frac{AM}{AO} = \frac{AH}{AB}$$ $$AM = \frac {1}{2} AC = 3 \text{ cm} \quad AO = R = 9 \text{ cm} \quad AB = 6 \text{ cm}$$ $$AH = \frac{AM \times AB}{AO} = \frac{3 \times 6}{9} = 2 \text{ cm}$$

Using Pythagoras' theorem

$$HB = \sqrt{AB^2 - AH^2} = \sqrt{36 - 4} = 4\sqrt{2} \text{ cm}$$ $$S = 4\sqrt{2} \times 2 = 8\sqrt{2} \text{ cm}^2$$

Answer: the area of the triangle is $8\sqrt{2} \text{ cm}^2$, the area of the circle is $S_e = \pi R^2 = 81\pi \text{ cm}^2$