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05 Mar 2019, 08:42
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
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Question Stats:
65% (02:36) correct
35%
(02:53)
wrong
based on 69
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GMATH practice exercise (Quant Class 20)
A triangle of area 30 is formed by the line x/c + y/(c+7) - 1 = 0 (where c is a positive constant) and the coordinate axes. What is the perimeter of this triangle?
(A) 12
(B) 24
(C) 30
(D) 36
(E) 52
A triangle of area 30 is formed by the line x/c + y/(c+7) - 1 = 0 (where c is a positive constant) and the coordinate axes. What is the perimeter of this triangle?
(A) 12
(B) 24
(C) 30
(D) 36
(E) 52
05 Mar 2019, 09:19
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This question required me a bit of time to figure out how to solve it, but probably it's just me and geometry. 
First step: find the intersections with the x-axis and y-axis.
Hence: \(y = c + 7\) when \(x = 0\).
\(x = c\) when \(y = 0\).
So, you have two points: \(A (c,0)\) and \(B (0, c + 7)\). If you join them in the Cartesian plane, you'll have a triangle with base \(c\) and height \(c + 7\).
Thus: \(A = \frac{bh}{2} = \frac{c(c+7)}{2} = 30\)
\(c^2 +7c - 60 = 0\)
\((c - 5) (c + 12) = 0\)
\(c = 5\)
Now let's use Pitagora to find the hypotenuse: \(\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} =13\)
\(5 + 13 + 12 = 30\)
I think I used a too long approach, and I'm waiting for somebody who enlighten me with a quicker method. By the way, this is my try.
First step: find the intersections with the x-axis and y-axis.
Hence: \(y = c + 7\) when \(x = 0\).
\(x = c\) when \(y = 0\).
So, you have two points: \(A (c,0)\) and \(B (0, c + 7)\). If you join them in the Cartesian plane, you'll have a triangle with base \(c\) and height \(c + 7\).
Thus: \(A = \frac{bh}{2} = \frac{c(c+7)}{2} = 30\)
\(c^2 +7c - 60 = 0\)
\((c - 5) (c + 12) = 0\)
\(c = 5\)
Now let's use Pitagora to find the hypotenuse: \(\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} =13\)
\(5 + 13 + 12 = 30\)
I think I used a too long approach, and I'm waiting for somebody who enlighten me with a quicker method. By the way, this is my try.
05 Mar 2019, 09:23
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fskilnik
Given that the area and the answer choices are all integers, the triangle is probably a Pythagorean Triple.
The only Pythagorean Triple with an area of 30 is 5-12-13.
In the given equation, test c=5, with the result that c+7 = 12:
x/5 + y/12 - 1 = 0
If x=0, then y=12, implying a y-intercept at (0, 12).
If y=0, then x=5, implying an x-intercept at (5, 0).
Implication:
In combination with the two axes, the line above will form a right triangle that has a height of 12 along the y-axis, a base of 5 along the x-axis, and a hypotenuse of 13, with the result that the area = (1/2)(12)(5) = 30.
Perimeter = 5+12+13 = 30.
.
