In the figure above, the perimeter of triangle I is 16 feet greater th

in triangle 1 apply Pythagoras theorem; we get hypotenuse = 15y
similarly, apply Pythagoras thm in trianlge 2; we get PQ = 17y

Since, we now have all three sides of both triangles, we can use the first condition of question to determine 'y'
now perimeter(triangle1) - perimeter(triangle2) = 16
40y-36y=16
y=4
so PQ = 17y = 68...C

Hello guys

This question belongs again to the great Pythagoras

https://en.wikipedia.org/wiki/Pythagoras

Let's begin

\(a^2 + b^2 = c^2\)

\((9y)^2 + (12y)^2 =c^2\)

\(81y^2 + 144y^2 = c^2\)

\(225y^2 = c^2\)

\(c = 15y\)

The hypotenuse of triangle II is a leg of triangle I

Do Pythagoras again

\(a^2 + b^2 = c^2\)

\((8y)^2 + (15y)^2 =c^2\)

\(64y^2 + 225y^2 = c^2\)

\(289y^2 = c^2\)

Here it's really useful if you know the squares through 20 but at least be familiar with them so the number stands out at you a bit

\(c = 17y\)

The perimeter of triangle I = 8y + 15y + 17y = 40y

The perimeter of triangle II = 9y + 12y + 15y = 36y

We are told that the perimeter of triangle I is 16ft greater than the perimeter of triangle II

40y = 36y + 16

4y = 16

y = 4

We want to find the length of PQ, which is 17y

17(4) = 68

The answer is (C)

Right, I missed Y's from the values.

In triangle II: \(12y^2 + 9y^2 = (3rd side)^2\)

=> \(144y^2 + 81y^2 = 225y^2 = (3rd side)^2\)

=> (3rd side) = 15y

Perimeter of II triangle: 12y + 9y + 15y = 36y

In triangle I: \(15y^2 + 8y^2 = (PQ)^2\)

=> \(225y^2 + 64y^2 = 289y^2 = (PQ)^2\)

=> (PQ) = 17y

Perimeter of I triangle: 15y + 8y + 17y = 40y


=> 40y - 36y = 16

=> 4y = 16

=> y = 4

=> PQ = 17y = 17 * 4 = 68

Answer C