CAT 2001 — QA Question 39

The third side of the triangle with sides 10 and x - 3 is $\sqrt{100-{(x-3)}^2}$

∴ In the large triangle,

(x - 3)² + (x + 4)² = ${\left[x+\sqrt{100-{(x-3)}^2}\right]}^2$

Only x = 11 satisfies this equation

Hence, option (b).

Alternatively,

x – 3 and x + 4 form part of a Pythagorean triplet. x = 12 and x = 10 does not give a Pythagorean triplet. With x = 11, we have 8, 15, 17.

This means that the large triangle has a hypotenuse of 17.

Therefore, one side of the smallest triangle is 17 – x = 17 – 11 = 6

As 6, 8, and 10 form a Pythagorean triplet, x is definitely equal to 11.