# Sat

• May 5th 2006, 12:47 PM
sara
Sat
hi,

my name is sara ...

I,m 15

and i has a sat test today

help me in this question

>>>>>>>>>>>>>>>><<<<<<<<<<<<<<<<<<<<<

in the figure at the right.
the legs of right tringle ACB are diameters of the two semicircles. If AB=4 ,What is the sum of the areas of the semicircles?

(a) 3.14 (b) 2 x 3.14 (c) 4 x 3.14 (d) 8 x 3.14 (c) 16 x 3.14

• May 5th 2006, 01:18 PM
MathGuru
Assume it is an isosceles triangle and therefore D^2 + D^2 = 4^2

solve for D and you will find one leg of the triangle or the diameter of the semicircle.

The area of the semi circle will be

1/2*(Pi*(D/2)^2) and since you have two of them you will find the sum of the area is:

Pi*(D/2)^2
• May 5th 2006, 01:21 PM
sara

(( sorrry i am not good in english because english is not my mother language))
• May 5th 2006, 01:23 PM
MathGuru
D^2 + D^2 = 4^2

2(D^2) = 16

D^2 = 8

D = Sqrt(8)

can you do the rest?
• May 5th 2006, 01:26 PM
MathGuru
Here is the rest:

Given:
Area= Pi*(D/2)^2 and D=sqrt(8)

Area= Pi*(sqrt(8)/2)^2

Area= Pi*(8/4)

Area= 2*Pi

• May 5th 2006, 01:28 PM
sara
Thank you

i hope i get a good mark in SAt text
• May 5th 2006, 01:30 PM
MathGuru

Cheers,
• May 5th 2006, 01:31 PM
sara
how do you know that i am from Qatar?
• May 5th 2006, 01:51 PM
CaptainBlack
Quote:

Originally Posted by MathGuru
Assume it is an isosceles triangle and therefore D^2 + D^2 = 4^2

solve for D and you will find one leg of the triangle or the diameter of the semicircle.

The area of the semi circle will be

1/2*(Pi*(D/2)^2) and since you have two of them you will find the sum of the area is:

Pi*(D/2)^2

No assumptions are needed the areas of the semi-circles obey a semi-circle
version of Pythagoras's theorem.

The area of the semi-circle on the hypotenuse of a right triangle is equal
to the sum of the areas of the semi-circles on the other two sides.

Hence the sum of the areas of the two semi-circles shown is equal to
the area of a semi-circle on the hypotenuse, which is:

$\pi.2^2/2=2.\pi$

RonL
• May 5th 2006, 02:39 PM
MathGuru
Thanks Ron for the correction.

Sara I checked your IP address which show that you are in Qatar.