The perimeter of a certain isosceles right triangle is 16 + 16*2^(1/2)

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.

If the legs are x and x, then Hypotenuse = \(\sqrt{2}x\)
\(x+x+\sqrt{2}x=16+16 \sqrt{2}\)
\(x(2+\sqrt{2})=16(1+\sqrt{2})\)
\(x(\sqrt{2}*\sqrt{2} +\sqrt{2})=16(1+\sqrt{2})\)
\(x\sqrt{2}(\sqrt{2} +1)=16(1+\sqrt{2})\)
=>\(\sqrt{2}*x =16\)

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?

First of all, we have a linear (first degrees) equation and it cannot give two solutions. Next, \(x=8\) does not satisfy \(x(2+\sqrt{2})=8(2+2\sqrt{2})\).

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




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

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?

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.

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.

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.

If algebra gets tricky, find a way to play with the numbers.

Solution is attached.

Hope it is helpful.

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

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)

great, thanks for explaining! now I get it