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GMAC paper based tests

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Re: GMAC paper based tests

by gabriel » Sat May 12, 2007 7:34 am
Cybermusings wrote:OA coming in after explanations
… we need to find x to find the area … point to remember … the perpendicular to the unequal side from the opposite angle will divide the unequal into 2 equal parts ..

so accordingly we have from the Pythagoras theorem … (2x-2)^2=x^2+((3x-2)/2)^2 … solving this we get the quadratic eqn .. 3x^2-20x+12 = 0 .. solving the quadratic eqn we get 2 possible values for x 2/3 and 6 .. but x cant be 2/3 coz then the base wuld be equal to 0 … so taking the value of x as 6 we get the base is equal to 16 and the perpendicular equal to 6 .. so the area is equal to 16*6/2 = 48 .. so the answer is C …
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by ns88 » Sat May 12, 2007 9:03 am
since isosceles triangle

2x-2=3x-8

x=6


area =(3x-2) (x)1/2 =48

ans=c
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by bww » Sat May 12, 2007 11:07 am
A=(1/2)(3x-2)(x)
=(3x^2-2x)/2

2x-2=3x-8
x=6

(3(3)^2-2(6))/2
=48 or C
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