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

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It is well known that a triangle’s area is √(p(p-a)(p-b)(p-c)), when  [#permalink]

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New post Updated on: 18 Jan 2016, 03:17
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Question Stats:

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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 two days.

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Originally posted by MathRevolution on 13 Jan 2016, 18:17.
Last edited by Bunuel on 18 Jan 2016, 03:17, edited 1 time in total.
Added the OA.
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Re: It is well known that a triangle’s area is √(p(p-a)(p-b)(p-c)), when  [#permalink]

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New post 13 Jan 2016, 19:52
MathRevolution 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 two days.


Hi,
how can this Q be of 700+ difficulty..
Its not worth sub-600..
All values given, even formula given.. Just put values and get the answer
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Re: It is well known that a triangle’s area is √(p(p-a)(p-b)(p-c)), when  [#permalink]

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New post 13 Jan 2016, 22:26
2
MathRevolution 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 two 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

It is a 600 level question (in my opinion) as per GMAT standard
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Re: It is well known that a triangle’s area is √(p(p-a)(p-b)(p-c)), when  [#permalink]

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New post 18 Jan 2016, 03:11
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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Re: It is well known that a triangle’s area is √(p(p-a)(p-b)(p-c)), when  [#permalink]

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New post 18 Jan 2016, 03:44
Umm, don't wish to labour this point...but its a very basic question involving tedious calculation. Does it deserve a place on this forum?
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Re: It is well known that a triangle’s area is √(p(p-a)(p-b)(p-c)), when  [#permalink]

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New post 18 Jan 2016, 04:43
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Horatio wrote:
Umm, don't wish to labour this point...but its a very basic question involving tedious calculation. Does it deserve a place on this forum?


Yes it deserves a place in this forum but not by the method used by Heron's formula.

The formula √(p(p-a)(p-b)(p-c)) can never be expected out of Test takers by GMAC. The other method of solving this question is already posted by me. Here it is again.


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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Re: It is well known that a triangle’s area is √(p(p-a)(p-b)(p-c)), when  [#permalink]

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Re: It is well known that a triangle’s area is √(p(p-a)(p-b)(p-c)), when &nbs [#permalink] 02 Nov 2017, 09:43
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