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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

## Hi Brent,

ca we say the half triangle is 30-60-90 since x:2x:hyp and can we put values based on that?

If yes we have the folloeing x:2x:sqroot(60)

can you give a solution using this approach as im getting confused with square root 60 being the side opposite 90.

Regards,

Rohan

## That's a great idea, Rohan.

That's a great idea, Rohan.

Unfortunately, the scenario doesn't create a 30-60-90 right triangle.

In a base 30-60-90 right triangle, the two LEGS have lengths 1 and √3, and the HYPOTENUSE has length 2 (in other words, the HYPOTENUSE is twice the length of one of the LEGS.)

In the triangle in the question, one LEG is twice the length as the other LEG.

Does that help?

Cheers,

Brent

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