Isosceles right triangle with perimeter 16+16(2^0.5). What

Question

Isosceles right triangle with perimeter 16+16(2^0.5). What is the length of the hypothenuse?

I know this is a special triangle and therefore the sides are in a ratio of 1:1:2^0.5 but I am having a hard time setting this up. The official answer is 16...how do you get there?

Thak you very much,

Primeminister

SimaQ · Answer

You must remember that the sum of two sides of the triangle cannot be larger than the length of one of the sides....

Therefore, if the perimeter of isoscles triangle is 16+16(2^0.5)... hypotenuse must be equal 16, because the other two sides are equal in legth and the sum of those two sides must be larger than the remaining side....

primeminister · Answer

Thanks SimaQ...i follow the triangle logic of what you said...not sure how that makes the hypothenuse 16 just by looking at the 16+16(2^0.5)??? Perhaps you can set it up in an equation so I can see it. 

I can do the following:
x+x+x(2^0.5)=16+16(2^0.5) Solving for x is very easy if you have a calculator and it gets the right answer. A bit tougher doing it just with a pencil.

Any more thoughts?

SimaQ · Answer

I meant the length of one side cannot be larger than the sum of the other two sides of the triangle.... Sorry for that mistake.... Now you should see what i REALLY meant

SimaQ · Answer

Perimeter: 16+16(2^0.5).....
From stem we know that the two sides are eqaul....
From the rule above we know that those two sides cannot be equall to 8+8=16, because the remaining side, that is hypotenuse, would be equal to more than 16, that is 16(2^0.5)....

Therefore the sum of the other two sides must be equal to 16(2^0.5)
And the remaining side, hypotenuse, 16....

Damager · Answer

if x is side hypo should be x*2^0.5

x + x + x*2^0.5 = 16 + 16*2^0.5
x(2 + 2^0.5) = 16( 1 + 2^0.5)
x(2 ^0.5 * 2^0.5 +2^0.5)=16 (1 + 2^0.5)
x*2^0.5 (2^0.5 + 1)=16( 1 + 2^0.5)

if u cancel out (1+2^0.5) from both sides

x*2^0.5=16

girikorat · Answer

if x is side hypo should be x*2^0.5

x + x + x*2^0.5 = 16 + 16*2^0.5 
x(2 + 2^0.5) = 16( 1 + 2^0.5) 
x(2 ^0.5 * 2^0.5 +2^0.5)=16 (1 + 2^0.5) 
x*2^0.5 (2^0.5 + 1)=16( 1 + 2^0.5)

if u cancel out (1+2^0.5) from both sides

x*2^0.5=16
 

Very good explaination

yogeshsheth · Answer

Thanks for the explanation damager was really lost on this one..
Guys this from Gmat Prep!!

vmalaiya · Answer

I approached it in the following way and reached a different value....

given,
2x + x*2^0.5 = 16 + 16^0.5

then,
x(2 + 2^0.5) = 8 (2 + 2^0.5)

therefore,
x = 8

and hypotneous is 8^0.5 which is not 16.

what went wrong in my approach. Damager's method works but what if I started with the above approach .... ??? :???

asaf · Answer

Isosceles right triangle with perimeter 16+16(2^0.5). What is the length of the hypothenuse? 

since a + b + c = 16 + 16*sqrt(2)
 but isoscles triangle so a = b

2a + c = 16 + 16 * sqrt(2) --- A

Also, its a right angle triangle

c^2 = 2a^2 (c^2 = a^2 + b^2 ; b = a)

a = c/(sqrt(2)) --- B

plugging in value of B in A

2c/sqrt(2) + c = 16 + 16 * sqrt(2)

solving, c = 16

HongHu · Answer

Just multiply sqrt(2) to your special triangle:
1:1:sqrt(2)
You get
sqrt(2):sqrt(2):2 
which gives you perimeter 2+2sqrt(2)
Now you know the ratio is 2:2+2aqrt(2) or 16:16+16sqrt(2).

lan583 · Answer

can someone explain how do one get the bold part?

Damager · Answer

Bold part explanation

2 = SqRoot 2 * SqRoot 2

Isosceles right triangle with perimeter 16+16(2^0.5). What

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