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## Comment on

Rectangle’s height## Hi Brent,

Please help me with the solution to this question.

Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to the x-axis. The x- and y-coordinates of P, Q and R are to be integers that satisfy the inequalities -4 ≤ x ≤ 5 and 6 ≤ y ≤ 16. How many different triangles with these properties could be constructed ?

(A) 110

(B) 1,100

(C) 9,900

(D) 10,000

(E) 12,100

## This is more of a counting

This is more of a counting question than a geometry question. That said, it's a question I've answered many times on the Beat The GMAT forum.

My detailed solution can be found here: http://www.beatthegmat.com/og-228-t284238.html

## Hey thanks for all your help

For this I made a ratio of a smaller triangle with sides then multipled to get the height

smaller triangle sides = 1, 2, sq(5)

multiply ratio

sq(60)*(2/sq(5)) = x

x = 4sq(3)

Does this way always work as well?

## Nice!

Nice!

That approach will work for all analogous questions.

Cheers,

Brent

## Hi Brent,

Could you please help me with a simple solution to this Q : https://gmatclub.com/forum/if-a-rectangular-region-has-perimeter-p-inches-and-area-a-sq-160255.html

Thanks & Regards,

Abhirup

## You bet!

You bet!

Here's my full solution: https://gmatclub.com/forum/if-a-rectangular-region-has-perimeter-p-inche...

Cheers,

Brent

## Thanks Brent for that simple

## Sorry, I'm not exactly sure

Sorry, I'm not exactly sure what you mean by: "In Statement 1, can we logically deduce that P = 4 * sqrt(A) * sqrt(A)/3 ?"

For statement 1, my rationale was as follows:

I already concluded that, if the region is a square then P = 4√A.

Statement 1 says P = (4/3)(A)

Since (4/3)(A) looks nothing like 4√A, I asked myself "Is it possible for (4/3)(A) to equal 4√A?

So, I wrote: (4/3)(A) = 4√A

When I solved this equation, I got A = 9

So, when A = 9, (4/3)(A) = 4√A

In other words, when A = 9, then the region IS a square.

Does that help?

Cheers,

Brent

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