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14 Oct 2008, 10:37
Kudos
Bookmarks
Hello,
What is the value of the hypotenuse of an isosceles triangle with a perimeter equal to 16 + 16√2?
The answer choices are:
a. 8
b. 16
c. 8√2
d. 16√2
e. 32
Please help. I know it's pretty simple but I can't seem to wrap my head around it. Also, the GMAT practice test from GMAC says it's 16 but 8√2 is the only answer that makes sense to me.
Nevertheless, please teach me how to solve for this.
Thanks,
Elizabeth
What is the value of the hypotenuse of an isosceles triangle with a perimeter equal to 16 + 16√2?
The answer choices are:
a. 8
b. 16
c. 8√2
d. 16√2
e. 32
Please help. I know it's pretty simple but I can't seem to wrap my head around it. Also, the GMAT practice test from GMAC says it's 16 but 8√2 is the only answer that makes sense to me.
Nevertheless, please teach me how to solve for this.
Thanks,
Elizabeth
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14 Oct 2008, 11:11
Kudos
Bookmarks
Answer B 
Do you remember following rule?
isoscelss triangle rule
= 1:1: √2
= x : x : x √2
= x√2 : x√2 : 2x
then perimeter (P)
= x√2 + x√2 + 2x
= 2x√2 + 2x
Now, apply 16√2 + 16 = 2x√2 + 2x, and solve for x, which would be 8
Now you have all sides of triagles: = 8√2 : 8√2 : 16
In an isosceles right triangle, hypontenuse is the longest of sides
(or in x√2 : x√2 : 2x, whatever is equivalent to 2x), which is 16
hypotenuse is 16
(To double check)
if x = 8,
= x√2 + x√2 + 2x
= 8√2 + 8√2 + 16
= 16√2 + 16
Give me Kudos, Kudos, Kudos!!
Do you remember following rule?
isoscelss triangle rule
= 1:1: √2
= x : x : x √2
= x√2 : x√2 : 2x
then perimeter (P)
= x√2 + x√2 + 2x
= 2x√2 + 2x
Now, apply 16√2 + 16 = 2x√2 + 2x, and solve for x, which would be 8
Now you have all sides of triagles: = 8√2 : 8√2 : 16
In an isosceles right triangle, hypontenuse is the longest of sides
(or in x√2 : x√2 : 2x, whatever is equivalent to 2x), which is 16
hypotenuse is 16
(To double check)
if x = 8,
= x√2 + x√2 + 2x
= 8√2 + 8√2 + 16
= 16√2 + 16
Give me Kudos, Kudos, Kudos!!
14 Oct 2008, 11:20
Kudos
Bookmarks
since the question says hypotenuse, must mean that it is a right isoceles there for perimeter is x + x + xsqrt2
so 2x + xsqrt2 = 16 + 16sqrt2
x(2 + sqrt2) = 16 + 16sqrt2
x= 16 + 16sqrt2/2 + sqrt2
x=16(1+sqrt2)/2(1+sqrt2/2)
x=8(1+sqrt2)/1+sqrt2/2
x=8(1+sqrt2)*2/1+sqrt2
x=16
so 2x + xsqrt2 = 16 + 16sqrt2
x(2 + sqrt2) = 16 + 16sqrt2
x= 16 + 16sqrt2/2 + sqrt2
x=16(1+sqrt2)/2(1+sqrt2/2)
x=8(1+sqrt2)/1+sqrt2/2
x=8(1+sqrt2)*2/1+sqrt2
x=16
