NTSE Stage 2 SAT 2018 Exam Question Paper

Question : 53 of 100

Marks: +1, -0

Let l be the length of each equal side of an isosceles triangle. If the length of each equal side is doubled, keeping its height unchanged, then the difference of the squares of bases of the new triangle and the given triangle is

0
$4l^2$
$9l^2$
$12l^2$

Validate

Solution:

Let A be the base of isosceles triangle whose equal side have length l and B be the base of changed isoscleles triangle whose equal side are of length (2l).

{AB}^{2}=4(ℓ^{2}-{h}^{2}){PQ}^{2}=4(4ℓ^{2}-{h}^{2})

{PQ}^{2}-{AB}^{2}=4(4^{2}-{h}^{2})-4(-{h}^{2})

=4[4^{2}-h^{2}-^{2}+h^{2}]=4[3^{2}]=12^{2}

Go to Question:

Prev Question

Next Question