Length of an isoscles triangle

gmattesttaker2 wrote:Hello,

Can you please assist with this:

Which of the following could be the area of an isosceles triangle with perimeter l8 and one side
of length 8 ?

(A) 6
(B) 12
(C) 14
(D) 16
(E) 18

Hey Sri,

Don't worry about this question. It requires us to use something called Heron's formula (sometimes called Hero's formula). More here: https://www.mathsisfun.com/geometry/herons-formula.html

Knowledge of this formula is NOT REQUIRED on the GMAT.

What's the source of this question?

Cheers,
Brent

gmattesttaker2 wrote:Which of the following could be the area of an isosceles triangle with perimeter l8 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.

Answer: B