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655-705 Level|   Geometry|      
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Senthil1981
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 18 Aug 2016, 22:57
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coffeeloverfreak
The perimeter of a certain isosceles right triangle is \(16 + 16 \sqrt{2}\). What is the length of the hypotenuse of the triangle?

\((A) \hspace{5} 8\)
\((B) \hspace{5} 16\)
\((C) \hspace{5} 4\sqrt{2}\)
\((D) \hspace{5} 8\sqrt{2}\)
\((E) \hspace{5} 16\sqrt{2}\)
Show more

Hence the answer is B) 16
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 05 Nov 2017, 20:33
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I am little confuse here. given that this is a Isosceles right angle triangle which can be either 30-60-90 triangle or 45-45-90.
I don't understand,why it should be intuitive to consider 45-45-90 triangle ?

would appreciate your response. thanks !
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 05 Nov 2017, 22:55
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jyotipes21@gmail.com
I am little confuse here. given that this is a Isosceles right angle triangle which can be either 30-60-90 triangle or 45-45-90.
I don't understand,why it should be intuitive to consider 45-45-90 triangle ?

would appreciate your response. thanks !
Show more

An isosceles triangle is a triangle with at least two sides of the same length. An isosceles triangle also has at least two angles of the same measure; namely, the angles opposite to the two sides of the same length. So, 30-60-90 is NOT an isosceles triangle.

You should go through the basics once more.

Basics to Brush-Up Fundamentals



23. Geometry




24. Coordinate Geometry




25. Triangles




26. Polygons




27. Circles




28. Rectangular Solids and Cylinders




29. Graphs and Illustrations



For more check Ultimate GMAT Quantitative Megathread
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 08 Mar 2018, 17:46
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coffeeloverfreak
The perimeter of a certain isosceles right triangle is \(16 + 16 \sqrt{2}\). What is the length of the hypotenuse of the triangle?

\((A) \hspace{5} 8\)

\((B) \hspace{5} 16\)

\((C) \hspace{5} 4\sqrt{2}\)

\((D) \hspace{5} 8\sqrt{2}\)

\((E) \hspace{5} 16\sqrt{2}\)
Show more
Main Idea: Equate the expression for the sum of the sides to the value given

Details: Let the equal sides be x. So hypotenuse is x*sqrt(2)

Perimeter = 2x+x*sqrt(2)= 16 + 16 *sqrt(2)

Simplifying we have x= 16(1+ sqrt(2)) /(2+sqrt(2))

So x*sqrt(2) = 16.

Hence B.
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 09 Apr 2019, 03:28
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coffeeloverfreak
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) \(16\sqrt{2}\)
Show more
If the legs are x and x, then Hypotenuse = \(\sqrt{2}x\)
\(x+x+\sqrt{2}x=16+16 \sqrt{2}\)
\(x(2+\sqrt{2})=16(1+\sqrt{2})\)
\(x(\sqrt{2}*\sqrt{2} +\sqrt{2})=16(1+\sqrt{2})\)
\(x\sqrt{2}(\sqrt{2} +1)=16(1+\sqrt{2})\)
=>\(\sqrt{2}*x =16\)
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 19 Sep 2019, 00:02
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coffeeloverfreak
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) \(16\sqrt{2}\)
Show more
guy please clear my doubt: i solved this way, I don't know what I did wrong. Please help me:-
x+x+\(x\sqrt{2}\)=16+\(16\sqrt{2}\)
2x+\(x\sqrt{2}\)= 16+\(16\sqrt{2}\)
x(2+\(\sqrt{2}\))=8(2+\(2\sqrt{2}\))
x=8 and x=\(8\sqrt{2}\)
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 19 Sep 2019, 01:22
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coffeeloverfreak
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) \(16\sqrt{2}\)
guy please clear my doubt: i solved this way, I don't know what I did wrong. Please help me:-
x+x+\(x\sqrt{2}\)=16+\(16\sqrt{2}\)
2x+\(x\sqrt{2}\)= 16+\(16\sqrt{2}\)
x(2+\(\sqrt{2}\))=8(2+\(2\sqrt{2}\))
x=8 and x=\(8\sqrt{2}\)
Show more


\(x\sqrt{2}+x+x=16+16\sqrt{2}\);

\(x\sqrt{2}+2x=16+16\sqrt{2}\);

Factor out \(x\sqrt{2}\) from the left hand side: \(x\sqrt{2}(1+\sqrt{2})=16+16\sqrt{2}\);

Factor out 16 from the right hand side: \(x\sqrt{2}(1+\sqrt{2})=16(1+\sqrt{2})\);

Reduce by \(1+\sqrt{2}\): \(x\sqrt{2}=16\);

Hope it's clear.
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 19 Sep 2019, 01:26
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rishabhmishra1993
coffeeloverfreak
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) \(16\sqrt{2}\)
guy please clear my doubt: i solved this way, I don't know what I did wrong. Please help me:-
x+x+\(x\sqrt{2}\)=16+\(16\sqrt{2}\)
2x+\(x\sqrt{2}\)= 16+\(16\sqrt{2}\)
x(2+\(\sqrt{2}\))=8(2+\(2\sqrt{2}\))
x=8 and x=\(8\sqrt{2}\)


\(x\sqrt{2}+x+x=16+16\sqrt{2}\);

\(x\sqrt{2}+2x=16+16\sqrt{2}\);

Factor out \(x\sqrt{2}\) from the left hand side: \(x\sqrt{2}(1+\sqrt{2})=16+16\sqrt{2}\);

Factor out 16 from the right hand side: \(x\sqrt{2}(1+\sqrt{2})=16(1+\sqrt{2})\);

Reduce by \(1+\sqrt{2}\): \(x\sqrt{2}=16\);

Hope it's clear.
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but what's wrong with what i did in this question?
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 19 Sep 2019, 01:40
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Bunuel
rishabhmishra1993
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) \(16\sqrt{2}\)

guy please clear my doubt: i solved this way, I don't know what I did wrong. Please help me:-
x+x+\(x\sqrt{2}\)=16+\(16\sqrt{2}\)
2x+\(x\sqrt{2}\)= 16+\(16\sqrt{2}\)
x(2+\(\sqrt{2}\))=8(2+\(2\sqrt{2}\))
x=8 and x=\(8\sqrt{2}\)


\(x\sqrt{2}+x+x=16+16\sqrt{2}\);

\(x\sqrt{2}+2x=16+16\sqrt{2}\);

Factor out \(x\sqrt{2}\) from the left hand side: \(x\sqrt{2}(1+\sqrt{2})=16+16\sqrt{2}\);

Factor out 16 from the right hand side: \(x\sqrt{2}(1+\sqrt{2})=16(1+\sqrt{2})\);

Reduce by \(1+\sqrt{2}\): \(x\sqrt{2}=16\);

Hope it's clear.
but what's wrong with what i did in this question?
Show more

First of all, we have a linear (first degrees) equation and it cannot give two solutions. Next, \(x=8\) does not satisfy \(x(2+\sqrt{2})=8(2+2\sqrt{2})\).
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 07 Jun 2020, 12:48
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coffeeloverfreak
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) \(16\sqrt{2}\)
Show more


Perimeter = 2x + x√2 = 16 + 16√2


JeffTargetTestPrep EducationAisle BrentGMATPrepNow

Can't we just compare LHS and RHS since there's only one expression with square root on both sides? I mean why can't we do this?
x√2 = 16√2 => Gives x = 16
and
2x = 16 => Gives x = 8

I'm confused why we can't do that here just like it's done for this Q:

https://gmatclub.com/forum/if-the-sum-o ... l#p1958772

JeffTargetTestPrep
Bunuel
If the sum of the square roots of two integers is \(\sqrt{9+6\sqrt{2}}\), what is the sum of the squares of these two integers?

(A) 40
(B) 43
(C) 45
(D) 48
(C) 52

We can let a = the first integer and b = the second integer. Thus:

√a + √b = √(9 + 6√2)

We are asked to find a^2 + b^2.

Let’s square both sides of the equation above.

(√a + √b)^2 = [√(9 + 6√2)]^2

a + 2√ab + b = 9 + 6√2

Since a and b are integers, we must have:

a + b = 9 and 2√ab = 6√2

If we square both sides of a + b = 9, we have:

a^2 + 2ab + b^2 = 81

If we square both sides of 2√ab = 6√2, we have:

4ab = 36(2)

2ab = 36

We can now substitute 36 for 2ab in a^2 + 2ab + b^2 = 81 to obtain:

a^2 + 36 + b^2 = 81

a^2 + b^2 = 45

Answer: C
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Thanks!
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 08 Jun 2020, 05:42
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coffeeloverfreak
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) \(16\sqrt{2}\)


Perimeter = 2x + x√2 = 16 + 16√2


JeffTargetTestPrep EducationAisle BrentGMATPrepNow

Can't we just compare LHS and RHS since there's only one expression with square root on both sides? I mean why can't we do this?
x√2 = 16√2 => Gives x = 16
and
2x = 16 => Gives x = 8

I'm confused why we can't do that here just like it's done for this Q:

https://gmatclub.com/forum/if-the-sum-o ... l#p1958772

Answer: C
Thanks!
Show more

Question is asking us to find THE value of x.
Your solution suggests there are TWO values of x.

Here is an analogous example: 2x + 3x = 2(1) + 3(6)
When we compare 2x and 2(1), we conclude that x = 1
When we compare 3x and 3(6), we conclude that x = 6
However, the actual (one) solution to the equation is x = 4

Cheers,
Brent
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 08 Jun 2020, 07:19
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dabaobao
coffeeloverfreak
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) \(16\sqrt{2}\)


Perimeter = 2x + x√2 = 16 + 16√2


JeffTargetTestPrep EducationAisle BrentGMATPrepNow

Can't we just compare LHS and RHS since there's only one expression with square root on both sides? I mean why can't we do this?
x√2 = 16√2 => Gives x = 16
and
2x = 16 => Gives x = 8

I'm confused why we can't do that here just like it's done for this Q:

https://gmatclub.com/forum/if-the-sum-o ... l#p1958772

Answer: C
Thanks!

Question is asking us to find THE value of x.
Your solution suggests there are TWO values of x.

Here is an analogous example: 2x + 3x = 2(1) + 3(6)
When we compare 2x and 2(1), we conclude that x = 1
When we compare 3x and 3(6), we conclude that x = 6
However, the actual (one) solution to the equation is x = 4

Cheers,
Brent
Show more

Thanks BrentGMATPrepNow for the reply.

Sorry, I get it that my solution is wrong since it has 2 answers. I want to understand the difference between these 2 Qs: Why can't we equate expression with square root on LHS with the one on RHS?

1) Reduction on both sides help:
2x + x√2 = 16 + 16√2 => x√2 (√2 + 1) = 16 (1 + √2) => Cross out (1 + √2) from both sides => x√2 = 16

2) Reduction on both sides doesn't help. Why? I'm confused
a + b + 2√(ab) = 3(3 + 2√2) => Can't do anything with this reduced expression.


My theory: "In an equation with square roots, match the expression with square root on LHS with expression with square root on RHS. Similarly, do the same for expressions without square root." => Works for second and not for first. What's the rule/theory behind this since my theory doesn't seem to work for the first question?
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 08 Jun 2020, 09:52
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Thanks BrentGMATPrepNow for the reply.

Sorry, I get it that my solution is wrong since it has 2 answers. I want to understand the difference between these 2 Qs: Why can't we equate expression with square root on LHS with the one on RHS?

1) Reduction on both sides help:
2x + x√2 = 16 + 16√2 => x√2 (√2 + 1) = 16 (1 + √2) => Cross out (1 + √2) from both sides => x√2 = 16

2) Reduction on both sides doesn't help. Why? I'm confused
a + b + 2√(ab) = 3(3 + 2√2) => Can't do anything with this reduced expression.


My theory: "In an equation with square roots, match the expression with square root on LHS with expression with square root on RHS. Similarly, do the same for expressions without square root." => Works for second and not for first. What's the rule/theory behind this since my theory doesn't seem to work for the first question?
Show more

It may not be possible to find an all-encompassing strategy for solving all equations with square roots.

It's important to note that there are considerable differences between the two equations you're referring to.
The first equation, 2x + x√2 = 16 + 16√2, features one variable, while the second equation, a + b + 2√(ab) = 3(3 + 2√2), features two variables (AND the product of two variables).
Given these significant differences, we can't expect to solve them using the same approach.
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 19 Nov 2020, 02:22
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Quote:
The Perimeter of a certain isosceles right triangle is 16 + 16√2. What is the length of the hypotenuse of the triangle?

A) 8
B) 16
C) 4√2
D) 8√2
E) 16√2

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
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Hi BrentGMATPrepNow, thank you so much for spelling this answer out in a very elegant manner. Although I understand the solution now, I'm still unclear why the solution does not occur to me more intuitively (let's say on a thought-process level). For instance, while solving the question the first time around, I considered the isosceles right triangle to be (x, x, x sqrt(2)), then the total perimeter = 2x+x sqrt(2). I equated the two equations and assumed that x=16. Hence, x sqrt(2) = 16 sqrt(2) as the hypotenuse.

Where do you think I made the mistake while thinking about this problem (in addition to not noticing the equations properly, since x can't take 16 and 8 with my approach)? Or what can I do to ensure I do not fall into this trap while solving problems?

Bunuel, VeritasKarishma, chetan2u. Any suggestion is much appreciated. Thank you.
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 19 Nov 2020, 04:23
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BrentGMATPrepNow
Quote:
The Perimeter of a certain isosceles right triangle is 16 + 16√2. What is the length of the hypotenuse of the triangle?

A) 8
B) 16
C) 4√2
D) 8√2
E) 16√2

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

Hi BrentGMATPrepNow, thank you so much for spelling this answer out in a very elegant manner. Although I understand the solution now, I'm still unclear why the solution does not occur to me more intuitively (let's say on a thought-process level). For instance, while solving the question the first time around, I considered the isosceles right triangle to be (x, x, x sqrt(2)), then the total perimeter = 2x+x sqrt(2). I equated the two equations and assumed that x=16. Hence, x sqrt(2) = 16 sqrt(2) as the hypotenuse.

Where do you think I made the mistake while thinking about this problem (in addition to not noticing the equations properly, since x can't take 16 and 8 with my approach)? Or what can I do to ensure I do not fall into this trap while solving problems?

Bunuel, VeritasKarishma, chetan2u. Any suggestion is much appreciated. Thank you.
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The error you made was careless, not strategic. So you understood the concept it was testing but got lost in the twist of the numbers. Mind you, careless mistakes are harder to recover from. A tutor can explain you the concept but after that it is upto you to ensure you do not make them.

\(2x + x\sqrt{2} = 16 + 16\sqrt{2}\)

x = 16 does not satisfy this. So you need to solve it further. Take out common.

\(x\sqrt{2}(\sqrt{2} + 1) = 16(1 + \sqrt{2})\)

Now it matches so x*sqrt(2) is 16. That is precisely the value you need.
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AntonioGalindo
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 30 May 2022, 15:11
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If algebra gets tricky, find a way to play with the numbers.

Solution is attached.

Hope it is helpful.
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RastogiSarthak99
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 17 Sep 2022, 01:20
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solvable through 45 45 90 triangle rule

If its right angled isosceles then sides will be x x sq root 2 x

2x + sq root 2 x = 16 + 16 sq root 2

solve for x. x = 8 sq root 2. But hypotenuse is sq root 2 x so 8 sq root 2 * sq root 2 gives us 16
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coffeeplease123
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 11 Jan 2024, 15:18
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Can someone tell me please why the hypothenuse IS NOT 16√2 (answer E)? I have the same calculations as you guys, and know that the hypothenuse is x√2 which is equal to 16√2. So why not to keep it this way?
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KarishmaB
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 12 Jan 2024, 02:15
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coffeeplease123
Can someone tell me please why the hypothenuse IS NOT 16√2 (answer E)? I have the same calculations as you guys, and know that the hypothenuse is x√2 which is equal to 16√2. So why not to keep it this way?
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Consider an isosceles right triangle with hypotenuse as 16√2. What are the two legs then? 16 each.
What is the perimeter then? 16 + 16 + 16√2 = 32 + 16√2
Does it match the given perimeter? No.

Given perimeter is 16 + 16√2 = x + x + x√2
Note that x is not 16 here.
But if we put x = 8√2, then we get 8√2 + 8√2 + 8√2*√2 = 16√2 + 16 (which works)
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coffeeplease123
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The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2) 12 Jan 2024, 09:12
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KarishmaB
coffeeplease123
Can someone tell me please why the hypothenuse IS NOT 16√2 (answer E)? I have the same calculations as you guys, and know that the hypothenuse is x√2 which is equal to 16√2. So why not to keep it this way?

Consider an isosceles right triangle with hypotenuse as 16√2. What are the two legs then? 16 each.
What is the perimeter then? 16 + 16 + 16√2 = 32 + 16√2
Does it match the given perimeter? No.

Given perimeter is 16 + 16√2 = x + x + x√2
Note that x is not 16 here.
But if we put x = 8√2, then we get 8√2 + 8√2 + 8√2*√2 = 16√2 + 16 (which works)
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great, thanks for explaining! now I get it
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