+129852 (a) We have two trapezoids at each end of this figure.....their combined perimeters = 2 [ 2*9 + 7 + 13 ] = 76 mm
The figure in the middle is a sector of a circle...its perimeter is :
2 pi * radius * ( 44/360) =
2 * pi * 9 * (44 / 360) ≈ 6.91 mm
So...the perimeter is [ 76 + 6.91 ] mm = 82.91 mm
(b) The two circular areas have a combined perimeter of 2 * pi * 4 * (240 / 360) ≈ 16.76 cm
The area of the equilateral triangle in the middle is (1/2)* 4^2 * √3 / 2 ≈ 6.93 cm
So...the approximate perimeter is [ 16.76 + 6.93 ] cm = 23.69 cm
+118677 Why do you say these are wrong Cassie?
You need to give a reason.
For example.
"Thanks for you time CPhill but your answers do not match with the answers in my text book.
The answers in my book are ...... "
ALWAYS THANK PEOPLE, EVEN IF IT IS JUST FOR THEIR TIME. (which they give to your for free)
+225
When I put the answer in it came out wrong. I do appreciate CPhill for answering my questions. I was telling him that it came out wrong; in a nice way not mean.
+118677 "both came out wrong" is neither nice nor mean. It is certainly not polite though.
Plus, I had not idea what you were talking about. At least you have now explained better why you think it was wrong.
Maybe it is your program that has the wrong answer, did you ever consider that?
I'll take (a) look at a for you.
P = 9+9+13+13+7+7+ the arc
= 58mm + the arc
The arc is from a circle with radius9 and subtended from an angle of 44 degrees
so
\(length \;\;of \;\;arc\;=\frac{44}{360}*2*\pi*9\\ length \;\;of \;\;arc\;=\frac{22}{10}*\pi\\ length \;\;of \;\;arc\;=2.2\pi \approx 6.9mm \)
\(Perimeter=(58+2.2\pi)mm \approx 64.9mm\)
I think CPhill misinterpreted what a composite figure is but he did all the hard bits, you should have been able to figure out why his answer was not correct yourself. Did you even try to?
