See below for the proof that the areas of the lune and the isosceles triangle are equal
How to prove the areas?
The area of the isosceles triangle is:
[tex]A_1 = \frac 12r^2\sin(\theta)[/tex]
Where r represents the radius.
From the figure, we have:
[tex]\theta = 90[/tex]
So, the equation becomes
[tex]A_1 = \frac 12r^2\sin(90)[/tex]
Evaluate
[tex]A_1 = \frac 12r^2[/tex]
Next, we calculate the length (L) of the chord as follows:
[tex]\sin(45) = \frac{\frac 12L}{r}[/tex]
Multiply both sides by r
[tex]r\sin(45) = \frac 12L[/tex]
Multiply by 2
[tex]L = 2r\sin(45)[/tex]
This gives
[tex]L = 2r\times \frac{\sqrt 2}{2}[/tex]
[tex]L = r\sqrt 2[/tex]
The area of the semicircle is then calculated as:
[tex]A_2 = \frac 12 \pi (\frac{L}{2})^2[/tex]
This gives
[tex]A_2 = \frac 12 \pi (\frac{r\sqrt 2}{2})^2[/tex]
Evaluate the square
[tex]A_2 = \frac 12 \pi (\frac{2r^2}{4})[/tex]
Divide
[tex]A_2 = \frac{\pi r^2}{4}[/tex]
Next, calculate the area of the chord using
[tex]A_3 = \frac 12r^2(\theta - \sin(\theta))[/tex]
Recall that:
[tex]\theta = 90[/tex]
Convert to radians
[tex]\theta = \frac{\pi}{2}[/tex]
So, we have:
[tex]A_3 = \frac 12r^2(\frac{\pi}{2} - \sin(\frac{\pi}{2}))[/tex]
This gives
[tex]A_3 = \frac 12r^2(\frac{\pi}{2} - 1)[/tex]
The area of the lune is then calculated as:
[tex]A = A_2 - A_3[/tex]
This gives
[tex]A = \frac{\pi r^2}{4} - \frac 12r^2(\frac{\pi}{2} - 1)[/tex]
Expand
[tex]A = \frac{\pi r^2}{4} - \frac{\pi r^2}{4} + \frac 12r^2[/tex]
Evaluate the difference
[tex]A = \frac 12r^2[/tex]
Recall that the area of the isosceles triangle is
[tex]A_1 = \frac 12r^2[/tex]
By comparison, we have:
[tex]A = A_1 = \frac 12r^2[/tex]
This means that the areas of the lune and the isosceles triangle are equal
Read more about areas at:
https://brainly.com/question/27683633
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