Bunuel wrote:

enigma123 wrote:

The perimeter of a right isoscles triangle is 32 + 32 \(\sqrt{2}\). What is the length of the hypotenues of the triangle?

A) 16

B) 8 \(\sqrt{3}\)

C) 32

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

E) 16\(\sqrt{2}\)

How come answer is E ?

This is how I am trying to solve

Isoscles Right angle triangle mean has x:x:x\(\sqrt{2}\) as its sides. Therefore, x+x+x\(\sqrt{2}\) = 32 + 32\(\sqrt{2}\)

2x+x\(\sqrt{2}\) = 32 + 32\(\sqrt{2}\). I am stuck after this.

In a right triangle where the angles are 45°, 45°, and 90° the sides are always in the ratio \(1 : 1 : \sqrt{2}\). With the \(\sqrt{2}\) being the hypotenuse (longest side). This can be derived from Pythagoras' Theorem. Because the base angles are the same (both 45°) the two legs are equal and so the triangle is also isosceles.

Now, if we proceed the way you suggest:

Let the side of this triangle be \(x\) then, \(P=32+32\sqrt{2}=x+x+\sqrt{2}x\) --> \(32+32\sqrt{2}=x(2+\sqrt{2})\) -->\(x=\frac{32+32\sqrt{2}}{2+\sqrt{2}}\) --> muliptly both nominatro and denominator by \(2-\sqrt{2}\) (to rationalize the denominator): \(x=\frac{(32+32\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}=\frac{64+64\sqrt{2}-32\sqrt{2}-64}{4-2}=\frac{32\sqrt{2}}{2}=16\sqrt{2}\).

So, hypotenuse is \(16\sqrt{2}*\sqrt{2}=32\)

Answer: C.

P.S. Of course you can also solve this problem with approximation method or plug-in method using the answer choices.Good tip. I solved it the long way, but now that i think about it, we can arrive at a much quicker solution by estimating.

Perimeter is \(32+32\sqrt{2}\)

\(\sqrt{2}\) is approx 3/2

so perimeter is approx: 32 + 32(3/2) = 80

If we take the second largest choice, option E, the hypothenuse is approx 16(3/2)=24. The other 2 legs must be less than the hypothenuse, so the max possible perimeter is less than 24+24+24=72. This leaves option C as the only sensible choice.

Or another way...

The sides of the triangle are in the ratio \(x:x:x\sqrt{2}\)

so perimeter, in terms of x, is approx \(x+x+\frac{3}{2}x=\frac{7}{2}x\)

given perimeter is approx 80,

\(\frac{7}{2}x=80\)

since we need to find the value of (3/2)x

\(\frac{7}{2}x*\frac{2}{7}*\frac{3}{2}=80*\frac{2}{7}*\frac{3}{2}\) = approx 30