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Posts: 246
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
61% (01:59) correct
39%
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wrong
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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) \(16\sqrt{2}\)
(A) \(8\)
(B) \(16\)
(C) \(4\sqrt{2}\)
(D) \(8\sqrt{2}\)
(E) \(16\sqrt{2}\)
Posts: 6,727
Quote:
An IMPORTANT point to remember is that, in any isosceles right triangle, the sides have length x, x, and x√2 for some positive value of x.
Note: x√2 is the length of the hypotenuse, so our goal is to find the value of x√2
From here, we can see that the perimeter will be x + x + x√2
In the question, the perimeter is 16 + 16√2, so we can create the following equation:
x + x + x√2 = 16 + 16√2,
Simplify: 2x + x√2 = 16 + 16√2
IMPORTANT: Factor x√2 from the left side to get : x√2(√2 + 1) = 16 + 16√2
Now factor 16 from the right side to get: x√2(√2 + 1) = 16(1 + √2)
Divide both sides by (1 + √2) to get: x√2 = 16
Answer = B
Cheers,
Brent
It's B
16+16(rt2)=2a+a(rt2) (where a is the lenght of each leg)
16+16(rt2)=a(2+rt2)
(16+16(rt2))/(2+rt2)=a
Hypoteneuse of isosceles rigth triangle is a(rt2)
H=a(rt2)=((16+16(rt2)(rt2))/(2+rt2)
H=(16(rt2)+16x2))/(2+(rt2))
H=16(rt2+2)/(rt2+2)=16
16+16(rt2)=2a+a(rt2) (where a is the lenght of each leg)
16+16(rt2)=a(2+rt2)
(16+16(rt2))/(2+rt2)=a
Hypoteneuse of isosceles rigth triangle is a(rt2)
H=a(rt2)=((16+16(rt2)(rt2))/(2+rt2)
H=(16(rt2)+16x2))/(2+(rt2))
H=16(rt2+2)/(rt2+2)=16
As mentioned in the subject, this is a 45:45:90 isosceles triangle. As per properties of 45:45:90 triangle, the sides are in the ratio of 1:1: √2 (check out OG, or you can try this with couple of numbers)
So, let the sides be x, x, and x√2 (where x is the length of legs and x√2 is hypotenuse). We are supposed to find length of hypotenuse, i.e. x√2
Perimeter =
2x + x√2 = 16 +16√2
or, x√2(√2 + 1) = 16(1 + √2)
or, x√2 = 16
B is correct answer
So, let the sides be x, x, and x√2 (where x is the length of legs and x√2 is hypotenuse). We are supposed to find length of hypotenuse, i.e. x√2
Perimeter =
2x + x√2 = 16 +16√2
or, x√2(√2 + 1) = 16(1 + √2)
or, x√2 = 16
B is correct answer
General Discussion
Let the length of the opposite and adjacent = a
Length of hypotenuse = h = sqrt(a^2+a^2) = sqrt(2) * a
So a = h/sqrt(2) = (sqrt(2)/2) * h
Perimeter = 2a + h
= 2 * (sqrt(2)/2) * h + h
= h (sqrt (2) + 1 )
Also given
Perimeter = 16 + 16 sqrt (2) = 16 (sqrt (2) + 1 )
So h = 16
Length of hypotenuse = h = sqrt(a^2+a^2) = sqrt(2) * a
So a = h/sqrt(2) = (sqrt(2)/2) * h
Perimeter = 2a + h
= 2 * (sqrt(2)/2) * h + h
= h (sqrt (2) + 1 )
Also given
Perimeter = 16 + 16 sqrt (2) = 16 (sqrt (2) + 1 )
So h = 16
I also get B
If the length of one of the two equal sides is x, then the other side is also x and the hypotenuse is xrt2.
we know that P= 16+16rt2
to get the perimeter you add the lengths of the three sides so either the equal sides are 16/2 = 8 and the other side is 8rt2 (which it isn't)
or the equal sides are 16rt2/2 = 8rt2 and the hypotenus is 16.
2(8rt2) = 16rt2 so it's pretty obvious that the hypotenuse is 16.
If the length of one of the two equal sides is x, then the other side is also x and the hypotenuse is xrt2.
we know that P= 16+16rt2
to get the perimeter you add the lengths of the three sides so either the equal sides are 16/2 = 8 and the other side is 8rt2 (which it isn't)
or the equal sides are 16rt2/2 = 8rt2 and the hypotenus is 16.
2(8rt2) = 16rt2 so it's pretty obvious that the hypotenuse is 16.
y^2 = x^2+x^2 => x = y/sqrt(2)
2x+y=16+16sqrt(2)
ysrrt(2)+y=16+16sqrt(2)
y=16
2x+y=16+16sqrt(2)
ysrrt(2)+y=16+16sqrt(2)
y=16
gmatacer
I looked at it in a more simple manner....
Simply kept in mind that two sides must be equal and that the sum of the two sides must be more than the length of the remaining side....
Put each sides to be X
Then hyp = xsqrt(2)
Perimeter = x + x + x*sqrt(2)
Solving for x, we get x = 8Sqrt(2)
Hence, Hyp = 16
Then hyp = xsqrt(2)
Perimeter = x + x + x*sqrt(2)
Solving for x, we get x = 8Sqrt(2)
Hence, Hyp = 16
This is actually more simple than you think....
Remember the rule that the sum of any two sides of the triangular must not be less than the length of the remaining side....
Thus isosceles triangular has two sides equal in length... so the sum of those two sides must be more than the remaining side...
You are given a perimeter, you can trace down the answer quickly.... and without calculations...
Remember the rule that the sum of any two sides of the triangular must not be less than the length of the remaining side....
Thus isosceles triangular has two sides equal in length... so the sum of those two sides must be more than the remaining side...
You are given a perimeter, you can trace down the answer quickly.... and without calculations...
My method for this is simple....
two equal sides
hypotenuse
Perimeter=16+16sqrt2
the sum of the two equal sides cannot be less than the hypotenuse.
Thus hypotenuse must be less than the sum of the other two equal sides:
Perimeter=16+16sqrt2.... the lower value in this equation is the hypotenuse, thus answer is 16
B
two equal sides
hypotenuse
Perimeter=16+16sqrt2
the sum of the two equal sides cannot be less than the hypotenuse.
Thus hypotenuse must be less than the sum of the other two equal sides:
Perimeter=16+16sqrt2.... the lower value in this equation is the hypotenuse, thus answer is 16
B
We know that an isocele right triangle has the following sides dimension: x, x and xsqrt(2) and we are looking for xsqrt(2).
Lets calculate x:
x+x+xsqrt(2) = 16 + 16sqrt(2)
x (2+sqrt(2)) = 8 (2+2sqrt(2))
x = 8 (2+2sqrt(2)) / (2+sqrt(2))
Lets caculte xsqrt(2):
xsqrt(2) = 8 (2+2sqrt(2)) sqrt(2) / (2+sqrt(2))
xsqrt(2) = 8 (2sqrt(2)+4) / (2+sqrt(2)) = 16
Lets calculate x:
x+x+xsqrt(2) = 16 + 16sqrt(2)
x (2+sqrt(2)) = 8 (2+2sqrt(2))
x = 8 (2+2sqrt(2)) / (2+sqrt(2))
Lets caculte xsqrt(2):
xsqrt(2) = 8 (2+2sqrt(2)) sqrt(2) / (2+sqrt(2))
xsqrt(2) = 8 (2sqrt(2)+4) / (2+sqrt(2)) = 16
above720
x + x + x√2 = 16 + 16√2
2x + x√2 = 16 + 16√2
x√2 (1 + √2) = 16 (1 + √2)
x√2 = 16
x = 16/√2 = 8√2
h^2 = x^2 + x^2
h^2 = 2x^2
h^2 = 2 (8√2)^2
h^2 = 2x2x8x8
h = 2 x 8 = 16
plaguerabbit
Guys, there is a simpler approach. On this board, with all the practice that everyone's doing, we are all so focused on the various nuances of the GMAT, so this should jump out at you.
We know that the triangle has to be x to x to xroot2. But when we try to make it work, it simply doesn't make sense. I mean, if the sides were an integer and the hypotenuse were the same integer times root 2, then the perimeter would have to just be 2x + xroot2. But it's not that. It's x + xroot2. So something's wrong.
You should instantly think - maybe the hypotenuse is the integer. It's the only other way the GMAT has ever really made these things hard.
So now we can eliminate C, D, and E. And since we're left with just 8 or 16, in this case, plugging in isn't so tough, and we get to 16 in about 31 seconds.
This is essentially what Squirrel was saying. Realize it's backwards ahead of time. But then just get the right answer. Remember, the GMAT doesn't award points for slickness of the math, it awards points for right answers in the shortest amount of time.
lahoosaher
Joined: 16 Jan 2009
Last visit: 20 Apr 2018
Posts: 236
Concentration: Technology, Marketing
Schools: Ross '16 (D) Tepper '16 (D) McCombs '16 (D) ISB '15 (D) Marshall '16 (WL) McDonough '16 (WL) Goizueta '16 (WL) Kelley '16 (D) Kenan-Flagler '16 (D) Rotman '16 (S)
GMAT 1: 700 Q50 V34
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Schools: Ross '16 (D) Tepper '16 (D) McCombs '16 (D) ISB '15 (D) Marshall '16 (WL) McDonough '16 (WL) Goizueta '16 (WL) Kelley '16 (D) Kenan-Flagler '16 (D) Rotman '16 (S)
GMAT 1: 700 Q50 V34
Posts: 236
Perimeter =
2x + x√2 = 16 +16√2
or, x√2(√2 + 1) = 16(1 + √2)
or, x√2 = 16
B
2x + x√2 = 16 +16√2
or, x√2(√2 + 1) = 16(1 + √2)
or, x√2 = 16
B
Himalayan
much easier to solve, at least saves few seconds:
if x=8√2, => 2x=2*8√2=16√2, Perimeter-2x= hypotenuse, or 16 + 16√2-16√2=16
whiplash2411
Joined: 09 Jun 2010
Last visit: 02 Mar 2015
Posts: 1,760
Status:Three Down.
Concentration: General Management, Nonprofit
Schools: HBS '17 (M) Stanford '17 (A)
Schools: HBS '17 (M) Stanford '17 (A)
Posts: 1,760
Updated on: 02 Jul 2010, 19:33
Originally posted by whiplash2411 on 02 Jul 2010, 19:30.
Last edited by whiplash2411 on 02 Jul 2010, 19:33, edited 3 times in total.
Last edited by whiplash2411 on 02 Jul 2010, 19:33, edited 3 times in total.
Okay, so since this is an isosceles triangle, let's assume the equal sides are x and x and the hypotenuse is
\(\sqrt{x^2 + x^2}\) = \(\sqrt{2}x\)
Perimeter = \(x + x + \sqrt{2}x\) = \(2x+\sqrt{2}x\) = \(\sqrt{2}x (\sqrt{2} + 1)\)
This is equal to 16 + 16 \(\sqrt{2}\)
So equating we get:
\(\sqrt{2}x (\sqrt{2} + 1)\) = 16 + 16 \(\sqrt{2}\) = 16\((\sqrt{2} + 1)\)
Solving this and canceling the \((\sqrt{2} + 1)\) on both sides, we get:
\(\sqrt{2}x=16\)
So we have \(x = \frac{16}{\sqrt{2}}\)
From the first step, we know that the hypotenuse is:\(\sqrt{2}x\) = \(\frac{16}{\sqrt{2}}*\sqrt{2}\) = 16
So I'd say the answer is B.
Hope this helps.
\(\sqrt{x^2 + x^2}\) = \(\sqrt{2}x\)
Perimeter = \(x + x + \sqrt{2}x\) = \(2x+\sqrt{2}x\) = \(\sqrt{2}x (\sqrt{2} + 1)\)
This is equal to 16 + 16 \(\sqrt{2}\)
So equating we get:
\(\sqrt{2}x (\sqrt{2} + 1)\) = 16 + 16 \(\sqrt{2}\) = 16\((\sqrt{2} + 1)\)
Solving this and canceling the \((\sqrt{2} + 1)\) on both sides, we get:
\(\sqrt{2}x=16\)
So we have \(x = \frac{16}{\sqrt{2}}\)
From the first step, we know that the hypotenuse is:\(\sqrt{2}x\) = \(\frac{16}{\sqrt{2}}*\sqrt{2}\) = 16
So I'd say the answer is B.
Hope this helps.
Posts: 416
Hey guys,
Nice solution - just one thing I like to point out on these:
Think like the testmaker!
If I were writing the test and knew that everyone studies the 45-45-90 ratio as: 1, 1, sqrt 2
I would make a living off of making the shorter sides a multiple of sqrt 2 so that the long side is an integer. People aren't looking for that! And they often won't trust themselves enough to calculate correctly...they'll look at the answer choices and see that 3 of them are Integer*sqrt 2, and they'll think they screwed up somehow because the right answer "should" have a sqrt 2 on the end.
So...keep in mind that with the Triangle Ratios:
45-45-90
x: x: x*sqrt 2
30-60-90
x : x*sqrt3 : 2x
One of the easiest tricks up the GMAT author's sleeve is to make x equal to a multiple of the radical so that the radical appears on the side you're not expecting and the integer shows up where you think it shouldn't!
Also, as you go through questions like these, ask yourslf "how could they make that question a little harder" or "how could they test this concept in a way that I wouldn't be looking for it". Training yourself to look out for unique cases, from the testmaker's perspective, helps you to get a real mastery of the GMAT from a high level.
Nice solution - just one thing I like to point out on these:
Think like the testmaker!
If I were writing the test and knew that everyone studies the 45-45-90 ratio as: 1, 1, sqrt 2
I would make a living off of making the shorter sides a multiple of sqrt 2 so that the long side is an integer. People aren't looking for that! And they often won't trust themselves enough to calculate correctly...they'll look at the answer choices and see that 3 of them are Integer*sqrt 2, and they'll think they screwed up somehow because the right answer "should" have a sqrt 2 on the end.
So...keep in mind that with the Triangle Ratios:
45-45-90
x: x: x*sqrt 2
30-60-90
x : x*sqrt3 : 2x
One of the easiest tricks up the GMAT author's sleeve is to make x equal to a multiple of the radical so that the radical appears on the side you're not expecting and the integer shows up where you think it shouldn't!
Also, as you go through questions like these, ask yourslf "how could they make that question a little harder" or "how could they test this concept in a way that I wouldn't be looking for it". Training yourself to look out for unique cases, from the testmaker's perspective, helps you to get a real mastery of the GMAT from a high level.
So we know that the proportions of a isosceles right triangle are x:x:x\sqrt{2}, since the perimeter is 16+16*sqrt(2), then
2x+x*sqrt(2)=16+16\sqrt{2}. we can solve for x and use that information to figure out what the hypotenuse is
First, let's isolate x, so we get x=16+16*sqrt(2)/(2+sqrt(2))
we can eliminate the root from the denominator by multiplying both the numerator and the denominator by 2-sqrt(2), and we end up with:
16*sqrt(2)/2=x or x=8*sqrt(2)
This is one of the options, because the GMAT is trying to trick you into picking it. However, we are looking for the hypotenuse and this is one of the sides, so we have to multiply it by sqrt(2) and we get 16 (B)
2x+x*sqrt(2)=16+16\sqrt{2}. we can solve for x and use that information to figure out what the hypotenuse is
First, let's isolate x, so we get x=16+16*sqrt(2)/(2+sqrt(2))
we can eliminate the root from the denominator by multiplying both the numerator and the denominator by 2-sqrt(2), and we end up with:
16*sqrt(2)/2=x or x=8*sqrt(2)
This is one of the options, because the GMAT is trying to trick you into picking it. However, we are looking for the hypotenuse and this is one of the sides, so we have to multiply it by sqrt(2) and we get 16 (B)
Flexxice
Firstly you have missed out onright angle in "certain isosceles triangle"..
secondly..
16 +16\(\sqrt{2}\) should tell us that the two equal sides are either 8 or 8\(\sqrt{2}\)..
If it is 8, perimeter = 8+8+8\(\sqrt{2}\), which is not the case here..
if it is 8\(\sqrt{2}\), P = 8\(\sqrt{2}\)+8\(\sqrt{2}\)+8\(\sqrt{2}*\sqrt{2}\) = 16 + 16\(\sqrt{2}\).. which is the P
so side is 8\(\sqrt{2}\) and HYP = 16
B
OA is being edited accordingly
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