Question

Please it looks like a lot but this is due in 3 hours and I spent the entire night doing work and I just want to sleep a little before my class starts. I already did 1 I just need 2 and 3

Answer 1

2)

Given:

The diameter of larger circle, D=8.

Hence, in the given figure, the diameter of the smaller semicircle, D=8.

The diameter of smaller circle, d=4.

Hence, the diameter of the two smaller semicirlces, d=4.

The radius of the smaller semicircle is,

$\begin{gathered} r=\frac{d}{2} \\ =\frac{4}{2} \\ =2 \end{gathered}$

Now, the area of the smaller semicircle can be calculated as,

$\begin{gathered} A=\frac{\pi r^2}{2} \\ =\frac{\pi\times2^2}{2} \\ =2\pi \end{gathered}$

The radius of the larger semicircle is half its diameter. So, the radius of the larger semicircle is R=4.

Now, the area of the large semicircle is,

$\begin{gathered} A^{\prime}=\frac{\pi R^2}{2} \\ =\frac{\pi\times4^2}{2} \\ =\frac{16\pi}{2} \\ =8\pi \end{gathered}$

The two smaller semicircles are darkly shaded.

Hence, the area of the lightly shaded region can be found as,

$\begin{gathered} A_S=A^{\prime}-2A \\ =8\pi-2\times2\pi \\ =8\pi-4\pi \\ =4\pi \\ =12.56 \end{gathered}$

Therefore, the area of the lightly shaded region is 4π square units or 12.56 square units.

3)

Given:

The central angle is, θ=73 °.

From part (2), the radius of larger circle, R=4.

The area of the sector with radius R=4 and central angle, θ=73° is,

$\begin{gathered} A_1=\frac{\theta}{360\degree}\times\pi R^2 \\ =\frac{73\degree}{360\degree}\times3.14\times4^2 \\ =10.19 \end{gathered}$

From part 2, the radius of the smaller circle is, r=2.

The area of the sector with radius r=2 and central angle, θ=73° is,

$\begin{gathered} A_2=\frac{\theta}{360\degree}\times\pi r^2 \\ =\frac{73\degree}{360\degree}\times3.14\times2^2 \\ =2.54 \end{gathered}$

Now, the area of the lightly shaded region is,

$\begin{gathered} A=A_1-A_2 \\ A=10.19-2.54 \\ A=7.65 \end{gathered}$

Therefore, the area of the lightly shaded region is 7.65 square units.