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GRE Math Challenge #34-the area of an isosceles triangle [#permalink]
Expert's post 00:00

Question Stats: 53% (01:10) correct 46% (01:03) wrong based on 13 sessions
Which of the folllowing could be the area of an isosceles triangle with perimeter 18 and one side of length 8 ?

A.6
B.12
C.14
D.16
E.18
[Reveal] Spoiler: OA

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Re: GRE Math Challenge #34-the area of an isosceles triangle [#permalink]
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Expert's post
sandy wrote:
Which of the folllowing could be the area of an isosceles triangle with perimeter 18 and one side of length 8 ?

A.6
B.12
C.14
D.16
E.18

Since the isosceles triangle has a perimeter of 18 and a side length of 8, we could have the following sides:

1) 8, 8, 2

or

2) 8, 5, 5

In option 1, the base is 2 and the legs (the sides that have equal length) are 8 each. The height of this triangle, h, satisfies the Pythagorean theorem in the form (b/2)^2 + h^2 = l^2 where b is the base and l is a leg of the isosceles triangle. Thus:

(2/2)^2 + h^2 = 8^2

1 + h^2 = 64

h^2 = 63

h = √63

Recall that the area of a triangle is (b x h)/2, so the area of the triangle is (2 x √63)/2 = √63. However, this is not one of the answer choices. Thus, we must consider option 2.

In option 2, the base is 8 and the legs are 5 each. So, we have:

(b/2)^2 + h^2 = l^2

(8/2)^2 + h^2 = 5^2

16 + h^2 = 25

h^2 = 9

h = √9 = 3

So, the area of the triangle is (8 x 3)/2 = 24/2 = 12.

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500+ lessons 3000+ practice problems 800+ HD solutions Re: GRE Math Challenge #34-the area of an isosceles triangle   [#permalink] 03 Jan 2018, 07:19
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