The perimeter of a certain isosceles right triangle is 16 +

[agha79]

GMAT prep exam one question. I was unable to find this in the forms. I searched using first few words of the question as a search.

Struggling with calculating the actual value for this one.

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

OA: B

[tim]

okay, with an isosceles right triangle the sides are x, x, and x(root2). they give the perimeter, so we can set up an equation:

x + x + x(root2) = 16 + 16(root2)

factor out the x to get:

x(2+root2) = 16+16(root2)

divide both sides by (2+root2) and x = (16+16root2)/(2+root2). now the tricky part comes in rationalizing the denominator, which requires you to multiply the top and bottom of this fraction by 2-root2. once you've done this everything should cancel and leave you with 4root2 for x, which you must multiply by root2 to get the hypotenuse. alternately, you can plug in 1.4 as an approximation for root2 and get a value that is close enough to the correct answer..

[agha79]

Hi Tim - I am struggling with simplifying "x = (16+16root2)/(2+root2) x (2-root2) /(2-root2). Can you please help me step by step how everything will cancel out for us to get to x = 4 root2

[sfbay]

answer is 16, no?

from above math
after multiplying by conjugate to rationalize denominator........

s= (16 *rt2) / 2
s= 8 * rt2

hypotenuse is s * rt2 = 8 rt2 * rt2 = 16

[agha79]

Yes the answer is 16. can you please help me understand step by step how you simplified "x = (16+16root2)/(2+root2) x (2-root2) /(2-root2)" s

[sfbay]

s=((16 + 16rt2)/(2+rt2)) * ((2-rt2)/(2-rt2))

use foil to multiply top and bottom

s=(32-16rt2+32rt2-32)/ (4-2rt2+2rt2-2)

s=16rt2/2

side= 8rt2

hyp is side*rt2

=8rt2 * rt2
=8*2
=16

[david.khoy]

I think you can solve this one really quickly without doing any complicated calculation.

We know the triangle is an isoceles right triangle, so we know the ratio of the sides of the triangles is:

x : x : x * root(2)

The perimeter of an isocelese right triangle is: 2x + x * root(2)

The perimeter is 16 + 16 * root(2).

Now, let's pretend that 16 * root(2) is the hypotenuse. That would mean the other sides are 16 and 16. That does not work since the perimeter would be 32 + 16 * root(2)

Now, let's suppose that 16 is the hypothenuse. That would mean that the other sides are 8 * root(2) and 8 * root(2). (Because 8 * root(2) * root(2) = 16) The perimeter of an isoceles triangle with an hypothenuse of lenght 16 is :
16 + 8 * root(2) + 8 * root(2) = 16 + 16 * root(2).
It works. The correct answer is 16.

[jnelson0612]

Agreed David, thank you.

[tim]

hi everyone,
thanks for catching my mistake. you're right, x is 8root2 rather than 4root2, giving a correct answer of 16. sorry if i confused anyone..

[Oogway]

Let the sides be a, a and h. Where a are the equal sides of iso triangle and h is hypotenuse

a^2 + a^2 = h^2 ---> Pytho theorem
=> 2(a^2) = h^2
=> a^2 = (h^2)/2
=> a = sq rt (h^2) / sq rt 2 = h / sq rt 2

Perimeter = sum of 3 sides = a + a + h = 16 + 16(sq rt 2) which is given in quest and we have already evaluated a above

a + a + h = 16 + 16(sq rt 2)
2a + h = 16 + 16(sq rt 2)
2( h / sq rt 2) = 16 + 16 (sq rt 2)
solve this to get the value of h as 16

If someone wants, I have a scanned paper solution of the same.. Please let me know.. would love to share..

[RonPurewal]

please search the forum:
post36393.html

this thread is now locked. please post any further questions over there, not here.
thanks.