# 4 urgent circumference and areas

• Jul 20th 2007, 02:29 PM
sanee66
4 urgent circumference and areas
:confused:My son is almost done with this course, I just need a little more help explaining this stuff to him please!!!!!

I think the last one is 99pi.
For the question with the trapezoid, do I take 32-14=18 to try to get the areas of the shaded parts? and if so, how do i decide how long each side is?
• Jul 20th 2007, 07:08 PM
DivideBy0
Question 2

$\displaystyle A_{shaded}=\frac{1}{2} \times 11 \times (32-14)$
$\displaystyle A_{shaded}=99$

Question 3

$\displaystyle HD=DF=GF=HG=32\pi$

Then, by pythagoras' formula:

$\displaystyle \left(\frac{HG}{2} \right)^2 + \left(\frac{GF}{2} \right)^2=IJ^2$

$\displaystyle \sqrt{(16\pi)^2+(16\pi)^2}=IJ$

$\displaystyle \sqrt{512\pi^2}=IJ$

$\displaystyle IJ=16\sqrt{2}\pi$

Question 4

$\displaystyle 2\pi r_{A}=24\pi$
$\displaystyle r_{A}=12$
$\displaystyle \pi r_{A}^2=144\pi$

$\displaystyle 2\pi r_{B}=12\pi$
$\displaystyle r_{B}=6$
$\displaystyle \pi r_{B}^2=36\pi$

Difference is $\displaystyle 144\pi-36\pi=108\pi$

Sorry I can't do Q.1, are you sure there is enough information?