An Isosceles Right Triangle With a Leg of Length "a" And Perimeter "p" GMAT Problem Solving

Question: 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?

p/2
p−a
(p+a)/2
2a
p−a√2

“An isosceles right triangle with a leg of length "a" and perimeter "p"- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:

Approach Solution 1:

Let us see the diagram below

Looking at fig 1
The hypotenuse = \(\sqrt{a^2+a^2}\)= a\(\sqrt{2}\)
It is already given that, a + a√2 = p
2a+a√2 = p..... (1)
Looking at fig 2
Let us Divide Fig I in two parts
now,
Two new equal size isosceles right triangles will be formed,
So.
Height and the base of small isosceles right triangle = \(\frac{1}{2}\)a\(\sqrt{2}\) = \(\frac{a}{\sqrt{2}}\)
Now ,Perimeter of small isosceles right triangle
a +\(\frac{a}{\sqrt{2}}\)+ \(\frac{a}{\sqrt{2}}\)
a + a\(\sqrt{2}\)
2a +a\(\sqrt{2}\) - a

Therefore p - a

Correct Answer: B

Approach Solution 2:

There is another approach to answering this question which is Pretty simple
It is Given that the isosceles right triangle with leg of length a.
The hypotenuse will be √2a using pythagorean theorem.
It is given a+a+√2a = p
p =2a + √2a
The length of the altitude, let's say L, is what you drop to create two smaller, identical triangles.
So,
Area of the big triangle = (1/2) ∗ a ∗ a =(1/2) ∗ √2a ∗ L
L = a / \(\sqrt{2}\)
Now Perimeter of small triangle
= a+L+2\(\sqrt{\frac{a}{2}}\)
= a + a / \(\sqrt{2}\) + 2\(\sqrt{\frac{a}{2}}\)
= a + \(\sqrt{2a}\)
But since we need the perimeter in terms of p too.
Lets evaluate
p = 2a + \(\sqrt{2a}\)
So,
\(\sqrt{2a}\)= p - 2a
Perimeter of small triangle
=a + p − 2a
= p−a