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A triangle of area 30 is formed by the line x/c + y/(c+7) - 1 = 0 (whe

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A triangle of area 30 is formed by the line x/c + y/(c+7) - 1 = 0 (whe  [#permalink]

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New post 05 Mar 2019, 08:42
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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

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Re: A triangle of area 30 is formed by the line x/c + y/(c+7) - 1 = 0 (whe  [#permalink]

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New post 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. :tongue_opt1

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.
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A triangle of area 30 is formed by the line x/c + y/(c+7) - 1 = 0 (whe  [#permalink]

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New post 05 Mar 2019, 09:23
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fskilnik wrote:
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


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.

.
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Re: A triangle of area 30 is formed by the line x/c + y/(c+7) - 1 = 0 (whe  [#permalink]

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New post 05 Mar 2019, 09:42
GMATGuruNY wrote:
fskilnik wrote:
GMATH practice exercise (Quant Class 20)
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.
.


Nice insight. Keeping in mind this you are able to cut 30 seconds off at least. Need to check this part. +1 Kudos!
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Re: A triangle of area 30 is formed by the line x/c + y/(c+7) - 1 = 0 (whe  [#permalink]

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New post 05 Mar 2019, 14:33
fskilnik wrote:
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

Yes... 5-12-13 is an important GMAT Pythagorean Triple...


\(?\,\, = \,\,\Delta \,\,{\rm{perimeter}}\)

\({S_\Delta } = 30\,\,\,\left( * \right)\)


\({x \over c} + {y \over {c + 7}} = 1\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{
\,x{\rm{ - intercept}} = c \hfill \cr
\,y{\rm{ - intercept}} = c + 7 \hfill \cr} \right.\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,\,30 = {{c\left( {c + 7} \right)} \over 2}\,\,\,\)

\(c\left( {c + 7} \right) = 60\,\,\,\left[ { = 5 \cdot 12 = \left( { - 12} \right)\left( { - 5} \right)} \right]\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,c = 5\,\,\,{\rm{or}}\,\,\,c = - 12\,\,\,\,\,\mathop \Rightarrow \limits^{{\rm{stem}}} \,\,\,\,\,c = 5\)


\(\left\{ \matrix{
\,{\rm{right}}\,\,\Delta \hfill \cr
\,{\rm{legs}}\,\,5,12 \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{hyp}}\,\, = 13\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,? = 5 + 12 + 13 = 30\)


The correct answer is (C).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: A triangle of area 30 is formed by the line x/c + y/(c+7) - 1 = 0 (whe   [#permalink] 05 Mar 2019, 14:33
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