How to do approach this one! OE not satisfactory

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

GmatKiss wrote:How to do approach this one! OE not satisfactory x=y+2 => x-y =2 -(1) 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

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

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 In red !

damn,what's happened to me!! thnks again neelgandham

shankar.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 Nice catch! thanks

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.

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 .