Two sectors of a circle of radius 12 overlap as shown, with P and R as the centers of the respective circles. Determine the area of the shaded region.

Guest Jan 13, 2022

#2**0 **

My approach is to get the area of half the shaded area,

then double that value for the area of the entire shaded area.

Call the point where the two circles intersect, label it X.

Call the point where the two diameters meet, label it Y.

Draw a line from X to Y. Draw a line from X to P.

The line from X to P and the line from Y to P mark off a quarter of circle P.

Since the area of the whole circle P is πr^{2} (144π), the quarter of the circle is **36π**.

There is a right triangle XYP with legs equal to 12 apiece,

so using Pythagoras Theorem, the hypotenuse is **sqrt(288)**.

Subtract the area of the right triangle from the area of the quarter circle.

36π – sqrt(288) gives you that little piece at the perimeter of the quarter circle.

The other circle's little piece has the same amount

so the entire shaded area is 2 • (36π – sqrt(288) )

Let's simplify that expression. The sqrt(288) is sqrt(2•144) = 12sqrt(2)

This makes the expression 2 • (36π – 12sqrt(2) ) so let's take 12 outside the parentheses

2 • (12 • (3π – sqrt(2) ) = **24 • (3π – sqrt(2) )**

I've double & quadruple checked my figures, and I don't think I tangled them up anywhere.

If I did make a mistake, then I hope somebody will show me how and explain it.

I don't like it when my answer differs from another answer...

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Guest Jan 14, 2022

#3**0 **

**ARRRRRGH I did make a mistake.** Too late to edit.

I called the hypotenuse the area of the triangle.

We don't' even care what the hypotenuse is. The area of the triangle is one-half 12 • 12

Quarter circle is 36π Triangle is 72

Sliver is 36π – 72 Thus the total area is **2 • (36π – 72)** __Just like the first Guest said.__

I will take 36 outside the parentheses **72 • (π – 2)**

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Guest Jan 14, 2022