Area and Perimeter of Shaded Region (Answer Confirmation)

**haftakhan** · August 17th 2010, 04:52 AM

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

Q1)Area(Shaded)=41
Perimeter(Shaded)=25

Q2)Area(shaded)=114
Perimeter(shaded)=6.44

Q3)Area(shaded)=490
Perimeter(shaded)=75

Q4)Area(shaded)=88.9
Perimeter(shaded)=

Q5)Area(shaded)=55
Perimeter(shaded)=14

Q6)Area(shaded)=123
Perimeter(shaded)=21

Q7)Area(shaded)=14
Perimeter(shaded)=15.5

Q8)Area(shaded)=
Perimeter(shaded)=37.5

**masters** · August 17th 2010, 08:35 AM

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.

**bjhopper** · August 17th 2010, 08:50 AM

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

**Soroban** · August 17th 2010, 09:43 AM

Hello, haftakhan!

Sorry, only a few of your answers are correct.

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

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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!

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

**masters** · August 17th 2010, 11:09 AM

Soroban,

Check your perimeter final answer in (2).

**haftakhan** · August 18th 2010, 05:24 AM

i have identified my mistakes thanks to your answers.

Are the last three answers right ?

**haftakhan** · August 18th 2010, 05:32 AM

Also i want to know why Perimeter of both are added to get the Perimeter of the shaded region ?

**bjhopper** · August 18th 2010, 09:29 AM

Hi haftakhan,
Show your work for questions 6,7,8 and I will help you if I think they are wrong

bjh

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