Area of the shaded region = 1/2 * x * y = 1/2 * y(y+2)
y(y+2) = 48
y = 6
hence x = 8
x^2 + y^2 = z^2
36 + 64 = z^2 = 100
z = 10
Option E
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neelgandham
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x=y+2 => x-y =2 -(1)GmatKiss wrote:How to do approach this one!
OE not satisfactory
area of traiangle = 0.5*x*y = 24 => xy = 48 -(2)
From 1 and 2
Z^2 = x^2 + y^2 = ((x-y)^2) +2*x*y = (2^2)+(2*48) = 100 => Z =10 Option E
Last edited by neelgandham on Tue Nov 01, 2011 4:10 am, edited 1 time in total.
Anil Gandham
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The area of a rt. triangle is given.
1/2 * b*h = 24
b*h=48.
Also given 'h' is 2 more than 'b'.
So only possibility is 6 and 8 (6*8 = 48)
Now using phy. theorem find hyp; Sqrt(36+64) = 10 IMO
1/2 * b*h = 24
b*h=48.
Also given 'h' is 2 more than 'b'.
So only possibility is 6 and 8 (6*8 = 48)
Now using phy. theorem find hyp; Sqrt(36+64) = 10 IMO
Last edited by shankar.ashwin on Tue Nov 01, 2011 4:13 am, edited 1 time in total.
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neelgandham
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In red !rijul007 wrote:Area of the shaded region = 1/2 * x * y = 1/2 * y(y+2)
1/2 * y(y+2) = 24 => y(y+2)=48 and not 24, => y =6 and x = 8 and z =10
y(y+2) = 24
y = -6 or 4
as the value cant be ngative, we take y=4
hence x = 6
x^2 + y^2 = z^2
36 + 16 = z^2 = 52
z = 2*root(13)
Option B
Anil Gandham
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Nice catch! thanksshankar.ashwin wrote:The area of a rt. triangle is given.
1/2 * b*h = 24
b*h=48.
Also given 'h' is 2 more than 'b'.
So only possibility is 6 and 8 (6*8 = 48)
Now using phy. theorem find hyp; Sqrt(36+64) = 10 IMO
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GMATGuruNY
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ALWAYS LOOK AT THE ANSWER CHOICES before you do any math.
Seeing 6, 8 and 10 in the answer choices should make us look for a 6-8-10 triangle.
y=6 and x=8 satisfy the condition that x = y+2.
If x=8 and y=6, then the area = (1/2)*6*8 = 24, satisfying the only other condition given.
Thus, the triangle is a 6-8-10 triangle with a hypotenuse of 10.
The correct answer is E.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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As a tutor, I don't simply teach you how I would approach problems.
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I agree with the approach by GMAT guru !! that seems to be the best one !!
This is what I did :
area of triangle = 1/2 B.H => 1/2 (y+2)(y) .
Then looking at the options I could easily tellit has to be 6 and 8(6+2) leading the value of z to be 10 .
This is what I did :
area of triangle = 1/2 B.H => 1/2 (y+2)(y) .
Then looking at the options I could easily tellit has to be 6 and 8(6+2) leading the value of z to be 10 .
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vaibhavgupta
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E it is!GmatKiss wrote:How to do approach this one!
OE not satisfactory
If OA is A, IMO B
If OA is B, IMO C
If OA is C, IMO D
If OA is D, IMO E
If OA is E, IMO A
FML!! :/
If OA is B, IMO C
If OA is C, IMO D
If OA is D, IMO E
If OA is E, IMO A
FML!! :/
