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triangles

by GmatTakerNo.1 » Mon Apr 26, 2010 10:43 am
Hey, got a problem with this one.

A right isosceles triangle fits tightly inside a rectangle that has one side that spans the triangle's hypotenuse. If the triangle's area is 144, what is the perimeter of the rectangle?

The answer is 72.

My result is 16 + 8*sqroot 3, my way as follows: I marked the hypothenuse as a and height as h.

144 = 1/2*a*h = 1/2*a*a*sqroot3/2, it follows: a=8

then h=4*sqroot3

and the perimeter of the rectangle must be: 2*8 + 2*4*sqroot3 = 16 + 8*sqroot3

What´s wrong?
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by iamseer » Mon Apr 26, 2010 11:23 am
the triangle is right isosceles. so the length of the rectangle is the hypotenuse.

Let the 2 sides of the triangle be = x
therefore 1/2*x*x=144
therefore x=12root2

therefore the breadth of the rectangle is 12 and length is 12*2 =24
so perimeter =12+24+12+12=72

Draw the figure on paper. It is much easier that way.

HTH
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by clock60 » Mon Apr 26, 2010 11:29 am
GmatTakerNo.1 wrote:Hey, got a problem with this one.

A right isosceles triangle fits tightly inside a rectangle that has one side that spans the triangle's hypotenuse. If the triangle's area is 144, what is the perimeter of the rectangle?

The answer is 72.

My result is 16 + 8*sqroot 3, my way as follows: I marked the hypothenuse as a and height as h.

144 = 1/2*a*h = 1/2*a*a*sqroot3/2, it follows: a=8

then h=4*sqroot3

and the perimeter of the rectangle must be: 2*8 + 2*4*sqroot3 = 16 + 8*sqroot3

What´s wrong?
let it be sides of triangle -a
the area of triangle=1/2*a^2=144
so a=12*2^1/2
if we drop perpendicular on the base of rectangle we will got another triangle with sides equal to the height of rectangle
let it be h
h^2+h^2=(12*2^1/2)^2
h=12
and h=1/2* base of the rectangle, it can be proven in many ways- (12*2^1/2)^2+(12*2^1/2)^2=base^2
so base=24
perimeter=2*(12+24)=72
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by moliver » Mon Apr 26, 2010 11:33 am
GmatTakerNo.1 wrote:Hey, got a problem with this one.

A right isosceles triangle fits tightly inside a rectangle that has one side that spans the triangle's hypotenuse. If the triangle's area is 144, what is the perimeter of the rectangle?

The answer is 72.

My result is 16 + 8*sqroot 3, my way as follows: I marked the hypothenuse as a and height as h.

144 = 1/2*a*h = 1/2*a*a*sqroot3/2, it follows: a=8

then h=4*sqroot3

and the perimeter of the rectangle must be: 2*8 + 2*4*sqroot3 = 16 + 8*sqroot3

What´s wrong?
Hi gmattaker1 could you please explain why you consider
144 = 1/2*a*h = 1/2*a*a*sqroot3/2, it follows: a=8
I think that the problem could be there
how are you imagine the figure? to me. the base of the triangle is one side of the rectangle and the height is the other side
if we divide the rectangle in two parts equal we have another isosceles triangle with an area of 144/2
call "a" to one of the side that are equal so the area must be a*a/2=144/2 so a = 12
then one side of the rectangle is 12 and the other is 12*2
perimeter = 24*2 +12*2 =72

please let me know if that help you
and you please put the source of the problem, thanks!!
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