Question

Find the area of the entire region round two decimal places

Answer 1

SOLUTION.

The area of the entire region will be to find the area of the isosceles triangle and the semi-circle.

`\begin{gathered} \text{Area of the triangle is } \\ =(1)/(2)* a* b*\sin \theta \\ \text{Where } \\ \text{where a and b are two sides and }\theta=angle\text{ betwe}en\text{ the two sides } \end{gathered}`

Then

`\begin{gathered} =(1)/(2)*8*8*\sin 72.6 \\ =32\sin 72.6 \\ =30.54\text{unit}^2 \end{gathered}`

The area of the isosceles triangle is 30.354 square units

We need to obtain the diameter of the semi-circle using the cosine rule

We need to obtain the value of x,

`\begin{gathered} x^2=8^2+8^2-2(8)(8)\cos 72,6 \\ x^2=128-128\cos 72.6 \\ x^2=128-38.27 \\ x^2=89.72 \end{gathered}`

Take square root of both sides, we have

`x=\sqrt[]{89.72}=9.74`

Hence, the diameter is 9.74

Then the radius will be

`r=(9.74)/(2)=4.87`

The area of the semi-circle is

`\begin{gathered} =(1)/(2)\pi r^2 \\ =(1)/(2)*3.14*4.87^2 \\ =37.24\text{unit}^2 \end{gathered}`

The area of the semi circle is 37.42 square units

Therefore the area of the region will be

`\begin{gathered} =\text{Area ot triangle + Area of semi circle } \\ =30.54+37.42 \\ =67.96uniits^2 \end{gathered}`

Thus

The area of the region is 67.96 units²