Thread: Square inscribed in triangle

A square is inscribed in a right triangle whose short sides are in the ratio of 1:2. What is the length of the side of the square in terms of the length of the shortest side of the circumscribed triangle?

I tried assuming that the shortest sides were 1 and 2, which makes the hypotenuse sq. root 5

Then I solved for the area of the whole triangle and I got 1

So then I made 1 = (1/2)(1-w)(2-w)

But for some reason I got w = 3 or 0 which doesn't work out...

i may be making up some BS but does this look correct/true?

Originally Posted by realintegerz

A square is inscribed in a right triangle whose short sides are in the ratio of 1:2. What is the length of the side of the square in terms of the length of the shortest side of the circumscribed triangle?

I tried assuming that the shortest sides were 1 and 2, which makes the hypotenuse sq. root 5

Then I solved for the area of the whole triangle and I got 1

So then I made 1 = (1/2)(1-w)(2-w)

But for some reason I got w = 3 or 0 which doesn't work out...

I will try to explain my solution without a picture... I somehow cannot upload a picture from my school...

Draw a right triangle, B = 90 degrees AB = 1 BC = 2
Draw inside the triangle a square label the square EBFG with E on AB en F on BC and G on AC

EB = BF = x
Then AE = 1 - x and FC = 2 - x

You've got two similar triangles AEG and GFC there is a ratio between the two triangles and because of that

AE : GF = EG : FC

but also AE x FC = GF x EG and



and x = 2/3

so we get the ratio 2/3 : 1 : 2
[in my school we have to multiply by 3 ..... 2 : 3 : 6]