# Another Math Destroyer Question!

Discussion in 'DAT Discussions' started by pandalove89, Jul 31, 2011.

1. ### pandalove89 7+ Year Member

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What is the length of one side of an isosceles right triangle whose area is 49 square cm?

The solution says: (1/2)s^2 = 49 therefore, s= 7*(square root of 2)

I'm confused. I know that the area of a triangle is 1/2bh, but how could you assume that the base is equal to one of the lengths? Doesn't make any sense to me

2. ### Dental2000 7+ Year Member

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good question....

The only way i can answer it is that they have assumed the isosceles triangle to be an Isosceles right triangle...

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3. ### radmazindds 7+ Year Member

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an isosceles right triangle is a 45/45/90 triangle. Therefore base = height.

(BxH)/2 = (BxB)/2 = 49

49x2 = 98 = B^2

sqrt (98) = B = 7sqrt(2)

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4. ### Dental2000 7+ Year Member

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ok disregard my previous post, dint read the question....it lookds like they HAVE given the triangle to be an isosceles RIGHT triangle

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5. OP

### pandalove89 7+ Year Member

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wow i didn't even realize the right triangle part. thanks!

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