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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