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Hi!lind0923 wrote:Hi Ron and dmateer,
I know this is an old post...... but just referencing it and I don't understand how to simplify 16/root 2 to 8*root 2. Maybe I'm just being dense, but I would appreciate any help you can give me.
thanks!!
this is very nicely done, although the factoring on the left is certainly not obvious.minhphuoc wrote:Dear all,
I think up a solution:
As some of you said, 2a + a sqrt(2) = 16 + 16 sqrt(2)
--> a sqrt(2) x [1 + sqrt(2)] = 16 x [1 + sqrt(2)]
--> a sqrt(2) = 16, and we already know a sqrt (2) is hypotenuse.
Ron, thanks for the above. Have one quick question. when i first saw the 16 + 16sqrt2, I equated it to X + X + Xsqrt2 -- the standard sides of a 45-45-90 triangle. then, i figured the 2x must equal 16 and Xsqrt2 must equal 16sqrt2. But I realized that by equating the two i was getting conflicting answers for the sides -- at which point I got apprehensive and guessed. Now, since this a 45-45-90 why doesn't it abide by the xlunarpower wrote:two ways for you to solve this problem much more easily than with the rather insane algebra:airan wrote:The perimeter of a certain isosceles right triangle is 16 +16 sqrt(2). What is the length of the hypotenuse of the triangle
1. 8
2. 16
3. 4 sqrt(2)
4. 8 sqrt(2)
5. 16 sqrt(2)
(1) plug in the answer choices
according to the 45-45-90 template, the hypotenuse of an isosceles right triangle is √2 times each of the legs. therefore, each leg is the hypotenuse divided by √2.
if you plug in choice (b), hypotenuse = 16, you'll find that leg = 16/√2 = 8√2 (post back if you don't know how to simplify this). therefore, perimeter = 2(leg) + hypotenuse = 16√2 + 16.
this is WAY easier than going through with all the algebra.
also, you shouldn't have to do that much plugging. if you plug in (a) first, then you'll get exactly half the desired perimeter, proving that (b) is correct. and if you plug in (c), (d), or (e), then you'll get, respectively, 8 + 4√2, 16 + 8√2, or 32 + 16√2, any of which will allow you to deduce that all three are incorrect.
(2) memorize TWO templates for the 45-45-90 triangle
everyone knows the 1-1-√2 template, which has perimeter 2 + 1√2.
you can also memorize an additional template: √2-√2-2.
this template is just √2 times the original template, but the multiples aren't obvious. this template has perimeter 2 + 2√2.
if you memorize this template, then you'll immediately recognize 16 + 16√2 as exactly eight times its perimeter, so that your triangle is 8√2-8√2-16.
Xsqrt2 rule? what am I missing here??heh.san2009 wrote:Now, since this a 45-45-90 why doesn't it abide by the x:x:Xsqrt2 rule? what am I missing here??
that's a good takeaway.san2009 wrote:gotcha. so basically the main takeaway is: one has to be super careful about equating different parts of the LHS and RHS of an equation. I'd rather just do the algebra or use estimation in the future.
maybe so, but you want to keep in mind that this is actually the entire point of the whole quantitative section!san2009 wrote:hah!
that's the part the wears me out.
thanks ron!
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.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
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