Question

A triangle is placed in a semicircle with a radius of
9yd,
as shown below. Find the area of the shaded region.

Use the value 3.14 for
π , and do not round your answer.
Be sure to include the correct unit in your answer.

Answer 1

Answer to Question #84565 – Math – Geometry

Question

A triangle is placed in a semicircle with a radius of 9 yd, as shown below. Find the area of the shaded region. Use the value 3.14 for $\pi$, and do not round your answer. Be sure to include the correct unit in your answer.

Solution

The shaded figure is:

Any triangle inscribed inside a semicircle must be a right triangle.

ABC is a right angled triangle inscribed in the semicircle, with sides $AC = BC = a$

AB is the hypotenuse of the triangle ABC.

The radius of the circle is given as 9 yd and 0 is the centre of the semicircle.

The diameter of the semicircle is $AB = 2 \times OB = 2 \times 9 = 18$ yd

Now by Pythagoras theorem, in the right triangle ABC we have:

$$AC^2 + BC^2 = AB^2$$ $$a^2 + a^2 = 18^2$$ $$2a^2 = 18^2$$ $$a^2 = \frac{18^2}{2}$$ $$a ^ {2} = 18 \times 9$$ $$a = \sqrt{2 \times 9 \times 9}$$ $$a = 9 \sqrt{2}$$

Then the area of the triangle ABC is:

$$\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}$$ $$\text{Area} = \frac{1}{2} \times 9 \sqrt{2} \times 9 \sqrt{2}$$ $$\text{Area} = 81 \, \text{yd}^2$$

We see that the area of the circle is:

$$A = \pi \times \text{Radius}^2 = \pi \times 9^2 = 81\pi$$

Therefore the area of the semicircle is:

$$= \frac{1}{2} \times 81\pi = 40.5\pi = 40.5 \times 3.14 = 127.17 \, \text{yd}^2 \ (\pi = 3.14 \, \text{given})$$

We see that,

- Area of the shaded region

- $= \text{Area of semicircle} - \text{Area of triangle} \, \text{ABC}$

- $= 127.17 - 81$

- $= 46.17 \, \text{yd}^2$

Hence the area of the shaded region is 46.17 sq yd.

Answer provided by https://www.AssignmentExpert.com