Another Math Destroyer Question!

[pandalove89]

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It asks:

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

 

[Dental2000]

good question....

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

 

[radmazindds]

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)

It asks:

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

 

[Dental2000]

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

Just follow the above posts answer and ull be fine

 

[pandalove89]

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)

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