The Perimeter of a certain isosceles right triangle is 16+16*sqrt(2). What is the length of the hypotenuse of the triangle?
A)8
B)16
C)4*sqrt(2)
D)8*sqrt(2)
E)16*sqrt(2)
Answer is B
Is this question categorized as a tough one?
Relation between Perimeter and hypotenuse of triangle
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 amirhakimi
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 amirhakimi
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Here is what I've done:
Since its a right isosceles triangle, the perimeter is 2x+x*sqrt(2)
2x+x*sqrt(2)=16+16*sqrt(2)
x(2+sqrt(2))=16(1+sqrt(2)) > X=[16(1+sqrt(2))]/[2+sqrt(2)]
using FOIL to simplify denominator > X= [(16*sqrt(2))]/2 > X=8*sqrt(2)
I stuck here
Since its a right isosceles triangle, the perimeter is 2x+x*sqrt(2)
2x+x*sqrt(2)=16+16*sqrt(2)
x(2+sqrt(2))=16(1+sqrt(2)) > X=[16(1+sqrt(2))]/[2+sqrt(2)]
using FOIL to simplify denominator > X= [(16*sqrt(2))]/2 > X=8*sqrt(2)
I stuck here
 amirhakimi
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Ohh, X is one of the sides not hypotenuse, hypotenuse will be X*sqrt(2) and will be 8*sqrt(2) * sqrt(2) which is 16
silly me
silly me

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Since it is an isoceles right angled triangle its a 454590 triangle with the sides in the ratio 1:1:âˆš2amirhakimi wrote:The Perimeter of a certain isosceles right triangle is 16+16âˆš2. What is the length of the hypotenuse of the triangle?
A)8
B)16
C)4âˆš2
D)8âˆš2
E)16âˆš2
Perimeter = x + x + xâˆš2 = 2x + xâˆš2
But the perimeter is 16+16âˆš2
Equating:
2x + xâˆš2 = 16+16âˆš2
xâˆš2 (âˆš2 + 1) = 16 (1 + âˆš2)
xâˆš2 = 16
From the figure this is the hyp. thus the answer is B
P.S: Wont call it a tough problem; the apt word would be tricky, especially in time crunch situations
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An important point here is that, in any isosceles right triangle, the sides have length x, x, and xâˆš2 for some positive value of x.amirhakimi wrote:The Perimeter of a certain isosceles right triangle is 16 + 16âˆš2. What is the length of the hypotenuse of the triangle?
A)8
B)16
C)4âˆš2
D)8âˆš2
E)16âˆš2
Answer is B
Note: xâˆš2 is the length of the hypotenuse, so our goal is to find the value of xâˆš2
From here, we can see that the perimeter will be x + x + xâˆš2
In the question, the perimeter is 16 + 16âˆš2, so we can create the following equation:
x + x + xâˆš2 = 16 + 16âˆš2,
Simplify: 2x + xâˆš2 = 16 + 16âˆš2
IMPORTANT: Factor xâˆš2 from left side to get : xâˆš2(âˆš2 + 1) = 16 + 16âˆš2
Now factor 16 from right side to get: xâˆš2(âˆš2 + 1) = 16(1 + âˆš2)
Divide both sides by (1 + âˆš2) to get: xâˆš2 = 16
Answer = B
Cheers,
Brent
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Another option is to test the answer choices.amirhakimi wrote:The Perimeter of a certain isosceles right triangle is 16 + 16âˆš2. What is the length of the hypotenuse of the triangle?
A)8
B)16
C)4âˆš2
D)8âˆš2
E)16âˆš2
Answer is B
We'll use the fact that, for any isosceles right triangle, the sides have length x, x, and xâˆš2 for some positive value of x. The hypotenuse is xâˆš2
So, if x is the length of one leg of the right triangle, the perimeter = x + x + xâˆš2
Start with answer choice D.
If the hypotenuse is 8âˆš2, then x = 8 [since the hypotenuse = xâˆš2]
So, the perimeter = 8 + 8 + 8âˆš2 = 16 + 8âˆš2
NO GOOD! We're told that the perimeter is 16 + 16âˆš2, so we need the triangle to be LARGER THAN 8âˆš2.
This means we can ELIMINATE A, C and D
IMPORTANT: At this point, we can test either B or E. It doesn't matter which one we test. If B works, we're done. If B doesn't work, then we'll automatically choose E.
Let's try answer choice B.
If the hypotenuse is 16, then we know that xâˆš2 = 16
So, x = 16/âˆš2 = 8âˆš2
So, in this instance, the perimeter is 8âˆš2 + 8âˆš2 + 16 = 16âˆš2 + 16
PERFECT! We're told that the perimeter is 16 + 16âˆš2
Answer: B
Cheers,
Brent
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You were on track until 2x+x*sqrt(2)=16+16*sqrt(2).amirhakimi wrote:Here is what I've done:
Since its a right isosceles triangle, the perimeter is 2x+x*sqrt(2)
2x+x*sqrt(2)=16+16*sqrt(2)
x(2+sqrt(2))=16(1+sqrt(2)) > X=[16(1+sqrt(2))]/[2+sqrt(2)]
using FOIL to simplify denominator > X= [(16*sqrt(2))]/2 > X=8*sqrt(2)
I stuck here
After this you could have tried this way:
x(2+sqrt(2)) = 8sqrt(2) (sqrt(2)+2) => x=8sqrt(2)
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