# Thread: Area and Circumference, homework for 2moro!

1. ## Area and Circumference, homework for 2moro!

Hello, please can you help me solve these math problems i am having some trouble.
Q1. a circle has an area of 98mm(squared).
Calculate the circumference of the circle.

The picture for question 2 is attached to this message sorry about the bad drawing.

I really appreciate anyone who can help me with this thanks in advance.

2. Originally Posted by sammyfox07
Hello, please can you help me solve these math problems i am having some trouble.
Q1. a circle has an area of 98mm(squared).
Calculate the circumference of the circle.
The area of a circle is $A=\pi r^2$, so in this case we have:

$A=\pi r^2=98$ so $r=\sqrt{98/\pi}$

The circumference of a circle is $C=2 \pi r$, so in this case we have:

$
C=2 \pi r=2 \pi \sqrt{98/\pi}=2 \sqrt{98 \pi}=14 \sqrt{2 \pi} \rm{\ mm}
$

RonL

3. Originally Posted by sammyfox07
The picture for question 2 is attached to this message sorry about the bad drawing.

I really appreciate anyone who can help me with this thanks in advance.
I assume the shaded areais the area left in a square of side 12 cm when 4 quarter circles of radius 6 cm are removed from the corners.

So the shaded area is the difference between the area of the square and that
of a circle of radius 6 cm.

$A=12^2-\pi 6^2 \rm{\ cm^2}$

RonL

4. Q1. a circle has an area of 98mm(squared).
Calculate the circumference of the circle.
If you want baby steps... read here:

Can you solve a system of equations? That's what this method is going to be.

So the area of a circle is: $\pi r^2$

Then we know: [1] $98 = \pi r^2$

The circumference of a circle is: [2] $\pi 2r$

So to find the circumference, we need to know the r.

So we solve equation [1] for r: $\frac{98}{\pi} = r^2 \Rightarrow r = \frac{7\sqrt{2}}{\sqrt{\pi}}$

Now you have r, so solve [2] by subbing our new r value in the equation:

$\pi 2\frac{7\sqrt{2}}{\sqrt{\pi}} = \frac{14\pi \sqrt{2}}{\sqrt{\pi}}$

We're not allowed to have a sqrt on the bottom, so we multiply the top and bottom by $\sqrt{\pi}$:

$\frac{14\pi\sqrt{2\pi}}{\pi} = \boxed{\frac{14}{\sqrt{2\pi}}\,mm}$