gmat_guy666 wrote:An isosceles right triangle with a leg of length "a" and perimeter "p" is further divided into two similar triangles of equal area. Which of the following represents the perimeter of one of these smaller triangles?
A. p/2
B. p−a
C. (p+a)/2
D. 2a
E. p-a* root2
OAB
In square ABCD, ∆ABE is 1/2 of isosceles right triangle ∆ABD.
∆ABE:
As the figure shows, ∆ABE is itself an isosceles right triangle.
The sides of an isosceles triangle are in the following ratio:
x : x : x√2.
Let AE=2, BE=2, and AB = 2√2.
∆ABD:
a = the length of each leg = 2√2.
p = 2√2 + 2√2 + 2 + 2 = 4 + 4√2.
The question stem asks for the perimeter of ∆ABE.
The perimeter of ∆ABE = 2 + 2 + 2√2 = 4 + 2√2. This is our target.
Now plug into the answers a = 2√2 and p = 4 + 4√2 to see which answer yields our target of 4 + 2√2.
Only
B works:
p-a = (4 + 4√2) - 2√2 = 4 + 2√2.
The correct answer is
B.
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