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It is well known that a triangle’s area is √(p(p-a)(p-b)

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by Max@Math Revolution » Sun Jan 17, 2016 8:59 pm
It is well known that a triangle's area is √(p(p-a)(p-b)(p-c)),
when p=(a+b+c)/2, such that a, b, c are the lengths of sides of the triangle. If the triangle has 360, 300, and 300 as the side's lengths, what is the triangle's area?


A. 34,200 B. 36,200 C. 38,200 D. 42,200 E. 43,200


* A solution will be posted in 2 days.
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by GMATinsight » Wed Jan 20, 2016 6:15 am
Max@Math Revolution wrote:It is well known that a triangle's area is √(p(p-a)(p-b)(p-c)),
when p=(a+b+c)/2, such that a, b, c are the lengths of sides of the triangle. If the triangle has 360, 300, and 300 as the side's lengths, what is the triangle's area?


A. 34,200 B. 36,200 C. 38,200 D. 42,200 E. 43,200


* A solution will be posted in 2 days.
Point No. 1: GMAT Takers are not supposed to know about the well know Heron's formula√(p(p-a)(p-b)(p-c))

Point No. 2: This question doesn't require application of Heron's formula √(p(p-a)(p-b)(p-c))

Make an Isosceles triangle with two sides 300 and base 360

Drop Perpendicular on Base (dividing base into two halves of 180 each) and use Pythagorus theorem to calculate height = √((300^2)-(180^2)) = 240

Area of Triangle = (1/2)*360*240 = 43200

Answer: Option E
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by Max@Math Revolution » Thu Jan 21, 2016 8:48 pm
It is well known that a triangle's area is √(p(p-a)(p-b)(p-c)),
when p=(a+b+c)/2, such that a, b, c are the lengths of sides of the triangle. If the triangle has 360, 300, and 300asthe side's lengths, what is the triangle's area?

A.34,200 B.36,200 C.38,200 D. 42,200 E.43,200

-> P=(360+300+300)/2=480, area=√(480(480-360)(480-300)(480-300))=43,200

Therefore, the answer is E.
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