Question: Rectangle’s height

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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
gmat-admin's picture

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:

Hey thanks for all your help so far!

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?

gmat-admin's picture

That approach will work for all analogous questions.


Hi Brent,

Could you please help me with a simple solution to this Q :

Thanks & Regards,
gmat-admin's picture

You bet!
Here's my full solution:


Thanks Brent for that simple yet effective solution. One query though. In Statement 1, can we logically deduce that P = 4 * sqrt(A) * sqrt(A)/3 Hence we get YES as an answer to rephrased target Q when sqrt(A)=3 and NO when sqrt(A) is not equals to 3. Is this thought process correct instead of picking numbers?
gmat-admin's picture

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?


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