# Math Help - Area and Perimeter of Shaded Region (Answer Confirmation)

In attachment there are 8 question and i want to confirm my answers

2. Hi haftakhan,

For i, I got this:

Area of big circle: $100 \pi$
Area of small circle: $36 \pi$

Area of shaded region: $100 \pi - 36 \pi \approx 201.06$

The perimeter of the circle with radius 10: $20 \pi \approx 62.83$
The perimeter of the circle with radius 6: $12\pi \approx 37.70$
The combined perimeter of the shaded region: $\approx 100.53$

For ii, I got this:

Area of circle: $100 \pi$
Area of Square: $14.14^2$

Area of shaded region: $100 \pi - 14.14^2 \approx 114.22$

The perimeter of the shaded region: $20\pi + 4(14.14) \approx 119.39$

For iii, the area looks good, but I don't see how you got 75 for the perimeter.

I haven't looked at the others, but if you want to post them individually and show how you got your answers, that'd be good.

3. Hi haftakhan,
I agree with masters show your work. I looked only at 8 where you had no area answer. My answer is area 72pi and 24pi for perimeter

4. Hello, haftakhan!

Are you sure of your area and perimeter formulas?
The perimeter is the amount of fencing to enclose the shaded region.

In attachment there are 8 question and i want to confirm my answers

(1) Area = 41, Perimeter = 25

We have two concentric circles with radius 6 and 10.

The area of the shaded region is the difference of the two areas:
. . $A \:=\:\pi(10^2) - \pi (6^2) \;=\;64\pi \;\approx\;201.1$

The perimeter of the shaded region is the sum of the two perimeters:
. . $P \;=\;2\pi(10) + 2\pi(6) \;=\;32\pi \;\approx\;100.5$

(2) Area = 114, Perimeter = 6.44

We have a circle with radius $10$
. . and an inscribed square with side $10\sqrt{2}$

$\text{(Area)} \;=\; \text{(Area of circle)} - \text{(Area of square)}$
. . $A \;=\;\pi(10^2) - (10\sqrt{2})^2 \;=\;100\pi - 200 \;\approx\;114.2$ .
You're right!

$\text{(Perimeter)} \;=\;\text{(Perimeter of circle)} + \text{(Perimeter of square)}$
. . $P \;=\;2\pi(10) + 4(10\sqrt{2}) \;=\;20\pi + 40\sqrt{2} \;\approx\; 119.4$

Corrected my typo . . . Thanks, masters!

(3) Area = 490. Perimeter = 75

We have a circle with radius 10 inside a circle with radius 16.

$\text{(Area)} \;=\;\text{(Area of large circle)} - \text{(Area of smaller circle)}$
. . $A \;=\;\pi(16^2) - \pi(10^2) \;=\;156\pi \;\approx\;490.1$ .
Correct!

$\text{(Perimeter)} \;=\;\text{(Sum of the two perimeters)}$
. . $P \;=\;2\pi(16) + 2\pi(10) \;=\;52\pi \;\approx\;163.4$

(4) Area = 88.9, Perimeter = ?

We have a quarter-circle with radius 40
. . and an isosceles right triangle with sides 40.

$\text{Area} \;=\;\text{(Area of quarter-circle)} - \text{(Area of triangle)}$
. . $A \;=\;\frac{1}{4}\pi(40^2) - \frac{1}{2}(40)(40) \;=\;400\pi - 800 \;\approx\;456.6$

$\text{(Perimeter)} \;=\;\text{Perimeter of quarter-circle)} + \text{Length of hypotenuse)}$
. . $P \;=\;\frac{1}{4}\cdot2\pi(40) + 40\sqrt{2} \;\approx\; 119.4$

(5) Area = 55, Perimeter = 14

We have a 16-by-16 square with two semicircles inside with radius 8.

$\text{(Area)} \;=\;\text{(Area of square)} - \text{(Area of circle)}$
. . $A \;=\;16^2 - \pi(8^2) \;=\;256 - 64\pi \;\approx\;54.9$ .
Yes!

$\text{(Perimeter)} \;=\;\text{(2 sides of the square}) + \text{(Circumference of circle)}$
. . $P \;=\;2(16) + 2\pi(8) \:=\:32 + 16\pi \;\approx\;82.3$

5. Soroban,