Find the area of the shaded regions. Give your answer as a completely simplified exact value in terms of π (no approximations).

The area of the shaded regions would be 52π m².

Step-by-step explanation:

So we know that the line OP is 8 m long and that the line AB is the same length as the line OA.

Because the radius is the distance from the center of a circle to the edge of a circle, the line OP is the radius of the circle. The line OB is also the radius of the circle. Because the radius of a circle is the same throughout the circle, the line OP is the same length as the line OB, so the line OB is 8 m long. Because the lines OA and AB are the same length as each other, and they both lie on the line OB, OA and AB are both the same length as half of OB, or 4 m long.

To find the area of the shaded regions, we can subtract the area of the medium-sized circle from the area of the bigger circle, and then add the area of the smallest circle.

The area of a circle is:

A = πr²

where A is the area and r is the radius.

Knowing that the radius of the largest circle is 8 m, the area of the largest circle would be:

A = π(8)² = 64π m²   (m² stands for square meters)

So the area of the largest circle is 64π m².

OA would be the radius of the medium-sized circle for the same reason OP was the radius of the largest circle. The length of OA is 4 m, so the radius of the medium-sized circle would be 4 m. Knowing this, the area of the circle would be:

A = π(4)² = 16π m²

So the area of the medium-sized circle would be 16π m².

AB would be the diameter of the smallest circle. The diameter is 2 times the radius of the circle. Because the length of AB is 4 m, the diameter would be 4 m and the radius would be 2 m (we do not need to know the diameter to find the area, only the radius). Knowing this, the area of the smallest circle would be:

A = π(2)² = 4π m²

So the area of the smallest circle would be 4π m².

So now be know that the area of the largest circle is 64π m², the area of the medium-sized circle would be 16π m², and that the area of the smallest circle would be 4π m². Now we use these to find the area of the shaded regions. Remember, the area of the shaded regions would be the area of the largest circle minus the area of the medium sized circle plus the area of the smallest circle. Knowing this, the area of the shaded regions would be:

64π - 16π + 4π = 52π m²

The area of the shaded regions would be 52π m².

I hope this helps. :)

the multiples of 15 are 15, 30,45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195,   210, 225, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375,   390, 405, 420, 435, 450, 465, 480, 495, 510, 525, 540, 555, 570, 585, 600, 615, 630, 645, 660, 675, 690, 705, 720, and 735.

the factors of 15 are 1, 3, 5, and 15.

hope that , fam!

-une, your fren : )