<p>What is the ratio of the diagonal of a square to the hypotenuse of the isosceles right traingle having the same area?
a) 1:2
b)1: V2
c) 1: 1
d) V2:1
E) 2:1</p>
<p>What is V2?</p>
<p>Without loss of generality:</p>
<p>Let the area of the square and triangle be 1.</p>
<p>Each side of the square would be root(1) = 1. The diagonal would then be root(2).</p>
<p>Each side of the triangle would be root(2). The hypotenuse would be 2 root(2).</p>
<p>The ratio would be root(2)/2 root(2) = 1:2</p>
<p>A</p>
<p>You can do this with algebra if you want. I find it easier to plug in.</p>
<p>I think it’s B.</p>
<p>Create a square that has the same area as an isosceles right triangle. For me, this would be a square with side square root 2 (area = 2) and an isosceles right triangle with side 2 (area = 2)</p>
<p>Diagonal of square = side x square root of 2 = 2</p>
<p>Hypotenuse of isosceles right triangle = leg of triangle x square root of 2 = 2sqrt 2</p>
<p>Diagonal of square : hypotenuse = 2: 2sqrt 2 = 1:sqrt 2</p>
<p>I’m sure this problem can be worked without drawing an example, but this is normally how I solve questions like these.</p>
<p>
missed the sqrt(2)</p>
<p>LEt x = side of a square with diagonal x(V2) and area x^2
let y = 1 leg of a isoceles triangle with hypotenuse y(V2)</p>
<p>the areas are the same, so (y^2)/2 = x^2
multiply both sides by 2</p>
<p>y^2= 2x^2 (I hate fractions and I just thought it would be easier to get everything in terms of Y)</p>
<p>take the sqrt of both sides
y = x(V2)</p>
<p>the hyp of the triange is y(V2), so plug in the in-terms-of-X version of Y we just got</p>
<p>hyp= x(V2)(V2), or 2x.
the diagonal of the square is x(V2)
divide both sides by X</p>
<p>and get a 2:1 relationship, hyp:square
invert it-- square:hyp = 1:2 choice A</p>
<p>EDIT: after seeing lolcake’s solution, do I always do it the hard way?</p>
<p>^and thats the algebraic way. Usually for the SAT you don’t have to, even tho that the most widely recognized way.</p>
<p>Using a triangle and square that both have an area of 8,</p>
<p>triangle base = 4, height = 4, hypotenuse = 4 sqrt 2</p>
<p>square side = sqrt8, diagonal = 4</p>
<p>diagonal: hypotenuse = 4: 4sqrt 2 = 1:sqrt2</p>
<p>What am I missing? This is the second time I have gotten B…</p>
<p>the hypotenuse is shared by the two identical isos triangle and is equal in length to the diagonal of the square the two triangles make up.</p>
<p>for an isos to be as large as a square in area, its hyp has to be twice as long as the diagonal that serves as the hyp for the two smaller isos that make up the square</p>
<p>wow that is a lot easier than I had initially made it out to be</p>
<p>Using a square and an isosceles triangle both with areas of 1, I’m getting B.</p>
<p>The diagonal of the square is sqrt(2).</p>
<p>The sides of the triangle are sqrt(2); the hypotenuse is 2.</p>
<p>Ratio is sqrt(2):2, which reduces to 1:sqrt(2), which is B.</p>
<p>
</p>
<p>I don’t think so. Using the Pythagorean Theorem to calculate the length of the hypotenuse:</p>
<p>Hypotenuse^2=Side^2+Side^2</p>
<p>Hypotenuse^2=(sqrt(2))^2+(sqrt(2))^2</p>
<p>Hypotenuse^2=2+2</p>
<p>Hypotenuse^2=4</p>
<p>Hypotenuse=sqrt(4)=2</p>
<p>^At least I’m not alone anymore :)</p>
<p>^^oh your right. I accidentally multiplied by 2 instead of sqrt(2)</p>
<p>oh wait my algebriac solution was correct until the end when my wires crossed and i thought that 2 was 2 times as great as sqrt2</p>
<p>it’s b folks!</p>
<p>Can someone explain to me where I erred?</p>
<p>Area of square and isosceles triangle = 1
Length of side of square=1
Diagonal =1xV2=V2
Length of leg of isosceles triangle=V2
Hypotenuse=V2xV2=2
Ratio=V2:2, which is not an option.</p>
<p>What did I do wrong?</p>
<p>Haha. Ignore my last post; I forgot that V2:2 simplifies to 1:V2</p>
<p>The question can also be thought of this way: If a half-square (the isos rt triangle) has the same area as a whole square, how many times longer are the half-square’s sides?</p>
<p>For this to work, the square that was cut in half must have started out with twice the area of the smaller square. Then, since area varies with the square of length, to double area you need to multiply length by v2.</p>