Question

please help?
WILL GIVE MEDAL

Answer 1

@whpalmer4

Answer 2

Can you find the area of the semicircle?

Answer 3

2 is the radius, isnt it?
the whole circle is 12.56 (i think) so /2 = 6.28

Answer 4

(i hope, lol)

Answer 5

Yes, \(r = 2\) so area of entire circle is \(A = \pi r^2 = \pi(2)^2 = 4\pi\) so area of the semicircle is \(2\pi = 6.28\) for this problem's purposes.

Now do you agree that the shaded area is the area of the semicircle - the area of the triangle inscribed?

Answer 6

Can you find the area of that triangle?

Answer 7

i agree the shaded part of the circle is part of the semicircle, yes. (i dont understand the triangle part)

I'm getting 4. Idk if that's right... the base is 4 and it states the height is 2

Answer 8

The area of the shaded region is what's left after you take the semi-circle and cut away the triangle, right?

The area of the triangle is \[A=\frac{1}{2}b*h = \frac{1}{2}2*4 = 4\]
So the area of the shaded region is area of the semi-circle - the area of the triangle =

Answer 9

I'm still confused...... 4 doesn't make sense to me.

Answer 10

If you look at the triangle you should be able to see that if you take the one on the right, rotate it a bit and nestle it up with the one on the left, hypotenuse to hypotenuse, you'll get a square with side length 2 and thus area 4.