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perimeter

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by gmatmachoman » Thu May 27, 2010 9:22 pm
What is the perimeter of the triangle bounded by the line 8x – 15y = 120 and the x and y intercepts?


(1) 60 (2) 20 (3) 40 (4) 36

OA : 40
Last edited by gmatmachoman on Thu May 27, 2010 9:57 pm, edited 1 time in total.
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by albatross86 » Thu May 27, 2010 9:32 pm
I'm assuming the equation is 8x - 15y = 120

Plugging in x = 0 gives y = -8
Plugging in y = 0 gives x = 15

So x and y intercepts are (0,-8) and (15,0)
The third point is the origin (0,0)

This gives us lengths of 8, 15 and root(15^2 + 8^2) = 17

So perimeter is : 8 + 15 + 17 = 40

Pick (3)
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by Rahul@gurome » Thu May 27, 2010 9:37 pm
The coordinates of the triangle are (0, 0), (15, 0) and (0, -8)
Length between (15, 0) and (0, -8) is √(15)^2 + (-8)^2 = 17
Perimeter = 15 + 8 + 17 = 40

The correct answer is (3).
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by selango » Thu May 27, 2010 9:38 pm
The equation is 8x+15y=120

x intercept (ie)y=0 gives,x=15

y intercept (ie)x=0 gives ,y=-8

Distance is origin and x and y

Distance between 0 and x =15

Distance between 0 and y= 8

-->Third side = sqrt(225+64)=17

Perimeter =15+8+17=40
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by liferocks » Thu May 27, 2010 9:39 pm
8x-15y=120

in intercept form it becomes

x/(15)+y/(-8)=1..so the sides of the triangle are 8,15,17

hence perimeter=8+15+17=40

Ans option3
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