# Find the area and perimeter of a composite figure?

If someone would please tell me if I'm doing this correctly, I'd really appreciate it. Note that our instructor does not require us to calculate pi; just "set up" the problem.

Composite figure of a square and semi circle. Square measures 14cm on the left side, 20cm on the bottom. Half circle attached to the right of the square has a radius of 7. For area I did:

l * w + pi R^2 / 2

14(20) + pi (7)^2 / 2

280+24.5pi (all of our answers to composite figure problems have been left like this)

Even though the area portion of the problem is technically not complete, did I still at least get it right?

I did not do the perimeter yet. The instructor said our answer should be 54 + 7pi. What formula was used? I'd like to see step-by-step how he got to this.

Sorry this whole thing is weird; I think it makes it even more confusing since we aren't asked to completely calculate the problem. Thanks in advance to those who are willing to help.

### 2 Answers

- GreywolfLv 67 years agoBest Answer
Area: you are spot on.

for the perimeter, normally you would add all four sides for the rectangle, but one of the 14" sides is no longer part of the perimeter: 20 + 14 + 20 = 54

you aslo have half of a circle's circumference as part of the perimeter. normally, C=2*pi*r, but since you only have half of that, you only need to add pi*r = 7pi

- galeanoLv 43 years ago
enable L = length; w=width Perimeter = 2L + 2w = 40 4 Or L + W = 22 (dividing perimeter by 2) enable W = 22 - L Substituting for W and L contained in the realm function (L)(22 - L) = 40 22 L - L^2 = 40 Rewrite into commonly used quadratic variety L^2 - 22L + 40 = 0 Factoring by utilizing FOIL technique (L - 20)(L - 2) = 0 So, L = 20 or 2 If L = 20, W = 22 - L = 2 So, W = 2, L = 20 If we tried L = 2 W = 22 - L W = 2 So W = 20; L = 2 by convention, the longer area is denoted by L so L = 20, W = 2 both way, the dimensions of rectangle are 2 yards x 20 yards