Question

SOMEONE FIND THE ANSWER, ILL GIVE BRAINLIEST.

Answer 1

AREA:

SQUARE IN THE MIDDLE

This is the easiest part: a square with side length s has area s^2, so the area of the square in the middle is

$A_s = 12^2=144$

TRIANGLE ON THE LEFT

The base of this triangle is one side of the square, so it is 12 inches long. The height is given, and it is 8 inches long. The area of a triangle is half the product of width and height, and so we have

$A_t = \dfrac{12\cdot 8}{2}=48$

SEMICIRCLE ON THE RIGHT

The diameter of the semicircle is one side of the square, so it is 12 inches long. The radius is half the diameter, so it's 6 inches long. The area of a circle is $\pi r^2$, so the area of a semicircle will be half this quantity:

$A_c = \dfrac{\pi r^2}{2}\approx \dfrac{3.14\cdot 36}{2}=56.52$

The total area is the sum of these areas:

$A=A_s+A_t+A_c=144+48+56.52=248.52$

PERIMETER:

As for the perimeter, we have:

• The two sides of the triangle, which are 10 inches each.
• The top and bottom sides of the square
• Half the circumference of the circle

Since the circumference of a circle is $2\pi r$, the semicircumference will be $\pi r$

So, the perimeter is

$10+10+12+12+6\pi \approx 44+6\cdot 3.14=62.84$