The perimeter of a certain isosceles right triangle is 16+16*sqrroot2. What is the hypotenuse of the triangle?
a. 8
b. 16
c. 4*sqroot2
d. 8*sqroot2
e. 16*sqroot2
help please!
thanks
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Brent@GMATPrepNow
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Here's one approach:AJ2009 wrote:The perimeter of a certain isosceles right triangle is 16+16*sqrroot2. What is the hypotenuse of the triangle?
a. 8
b. 16
c. 4*sqroot2
d. 8*sqroot2
e. 16*sqroot2
help please!
thanks
As soon as you see the sqrt2 and the two 16's, you should be thinking that this triangle might be in the 45-45-90 family or right triangles.
The ratio of the sides of a 45-45-90 triangle is 1:1:sqrt2
The perimeter 16+16sqrt2 has an integer component and a root component.
To get 16sqrt2, we could multiply each side of the base 1:1:sqrt2 triangle by 16, but this would make the sides 16:16:16sqrt2 and the perimeter 32+16sqrt (nope)
At this point, it helps to create a second "base" triangle by multiplying each side of the 1:1:sqrt2 triangle by sqrt 2.
We get sides sqrt2:sqrt2:2 (this is a useful ratio to keep in your pocket)
Now we might see that if we multiply each side by 8 we get the desired dimensions. We get 8sqrt2:8sqrt2:16.
The perimeter is 16+16sqrt2, and so the hypotenuse is 16
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bmlaud
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For a isosceles right triangle with sides a and b
2a+b = 16+16sqrt2 ------1
2a^2 = b^2 or b = a sqrt2, putting the value in 1
2a + a sqrt2 = 16(1+sqrt2)
or a sqrt2(1+ sqrt2) = 16 (1+sqrt2)
or a = 16/sqrt2 = 8 sqrt2 ..D
2a+b = 16+16sqrt2 ------1
2a^2 = b^2 or b = a sqrt2, putting the value in 1
2a + a sqrt2 = 16(1+sqrt2)
or a sqrt2(1+ sqrt2) = 16 (1+sqrt2)
or a = 16/sqrt2 = 8 sqrt2 ..D
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bravotalks
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bmlaud, your approach seems to be correct until the last step.
You deduced a = 8 sqrt2
But the question is asking for the hypotenuse, which in your case is variable 'b' (and not variable 'a')
i.e., b = a * sqrt2 = 8 * sqrt2 * sqrt 2 = 16.
i.e., Ans B
OA?
You deduced a = 8 sqrt2
But the question is asking for the hypotenuse, which in your case is variable 'b' (and not variable 'a')
i.e., b = a * sqrt2 = 8 * sqrt2 * sqrt 2 = 16.
i.e., Ans B
OA?
Let x be the length of the isosceles sides, therefore the hypotenuse = x.sqrt(2)
Since 2x + x.sqrt(2) = 16 + 16sqrt(2)
If we compare the rational(2x) with the rational (16) and the irrational (x.sqrt(2)) with the irrational (16sqrt(2)), we cannot get a solution. Therefore we try to compare the rationals with the irrationals, i.e.
2x = 16sqrt(2), and
x.sqrt(2) = 16
From both equations, we obtain x = 8sqrt(2). Since the hypotenuse is x.sqrt(2) = 16, therefore the answer should be B.
Since 2x + x.sqrt(2) = 16 + 16sqrt(2)
If we compare the rational(2x) with the rational (16) and the irrational (x.sqrt(2)) with the irrational (16sqrt(2)), we cannot get a solution. Therefore we try to compare the rationals with the irrationals, i.e.
2x = 16sqrt(2), and
x.sqrt(2) = 16
From both equations, we obtain x = 8sqrt(2). Since the hypotenuse is x.sqrt(2) = 16, therefore the answer should be B.
