The area of a region is the amount of space the region occupies.
The area of the shaded region is [tex]\frac{2}{3}(16\pi - 24\sqrt3)[/tex]
First, we calculate the measure of [tex]\angle BCQ[/tex] using cosine ratio
[tex]\cos(\angle BCQ) = \frac{QC}{CB}[/tex]
Where:
[tex]CB = BA = 4cm[/tex]
[tex]QC = \frac 12 \times AC = \frac 12 \times 4cm = 2cm[/tex]
So, we have:
[tex]\cos(\angle BCQ) = \frac{2cm}{4cm}[/tex]
[tex]\cos(\angle BCQ) = \frac{1}{2}[/tex]
Take arc cos of both sides
[tex]\angle BCQ = \cos^{-1}(\frac{1}{2})[/tex]
[tex]\angle BCQ = 60^o[/tex]
Convert to radians
[tex]\angle BCQ = 60 \times \frac{\pi}{180}[/tex]
[tex]\angle BCQ = \frac{\pi}{3}[/tex]
Calculate the measure of [tex]\angle BCD[/tex]
[tex]\angle BCD = 2 \times \angle BCQ[/tex]
[tex]\angle BCD = 2 \times \frac{\pi}{3}[/tex]
[tex]\angle BCD = \frac{2\pi}{3}[/tex]
Calculate the area of sector BCD using:
[tex]Area = \frac 12 r^2 \theta[/tex]
So, we have:
[tex]A_1 = \frac 12 \times 4^2 \times \frac{2\pi}{3}[/tex]
[tex]A_1 = 16 \times \frac{\pi}{3}[/tex]
[tex]A_1 = \frac{16\pi}{3}[/tex]
Calculate the area of the complete sector
[tex]Area = 2 \times A_1[/tex]
[tex]Area =2 \times \frac{16\pi}{3}[/tex]
[tex]Area =\frac{32\pi}{3}[/tex]
Next, calculate the area of the rhombus
First, calculate side length QB using Pythagoras theorem
[tex]BC^2 = QB^2 + QC^2[/tex]
[tex]2^2 = QB^2+ 4^2[/tex]
[tex]4 = QB^2 + 16[/tex]
Collect like terms
[tex]QB^2 = 16 - 4[/tex]
[tex]QB^2 = 12[/tex]
Take square roots
[tex]QB = \sqrt{12[/tex]
Multiply by 2 to calculate side length DB
[tex]DB = 2 \times \sqrt{12[/tex]
Expand
[tex]DB = 2 \times \sqrt{4 \times 3[/tex]
Split
[tex]DB = 2 \times \sqrt{4} \times \sqrt{3[/tex]
[tex]DB = 2 \times 2 \times \sqrt{3[/tex]
[tex]DB = 4\sqrt{3[/tex]
So, the area of the rhombus is:
[tex]Area = DB \times AC[/tex]
[tex]Area = 4\sqrt3 \times 4[/tex]
[tex]Area = 16\sqrt3[/tex]
The area of the shaded region is the area of the rhombus subtracted from the area of the complete sector
[tex]Area =\frac{32\pi}{3}[/tex] ---- Sector
[tex]Area = 16\sqrt3[/tex] ---- Rhombus
So, we have:
[tex]A = \frac{32\pi}{3}- 16\sqrt3[/tex]
Take square roots
[tex]A = \frac{32\pi - 48\sqrt3}{3}[/tex]
Factor out 2
[tex]A = \frac{2(16\pi - 24\sqrt3)}{3}[/tex]
Rewrite as:
[tex]A = \frac{2}{3}(16\pi - 24\sqrt3)[/tex]
Hence, the area of the shaded region has been proven to be[tex]\frac{2}{3}(16\pi - 24\sqrt3)[/tex]
Read more about areas at:
https://brainly.com/question/16418397
