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Triangles Problem (geometry)

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Triangles Problem (geometry)

by vongdn » Tue Oct 05, 2010 9:31 pm
This is a geometry problem so I will try my best to describe the figure shown.

You have two triangles shown, both with equal interior angles x, y, and z. So the triangles are similar because they have the same angles. The left triangle is drawn as larger than the right triangle. The base of the smaller triangle is shows as equaling s, and the base of the bigger triangle is show as equaling S.

If the area of the triangle on the right is twice the area of the triangle on the left, then in terms of s, S = ?

Since the area of a triangle is 1/2 b * h

I come up with the equation 2*( 1/2 (s*h) ) = 1/2 (S*h)....I solve that and get S=2s

However the answer says that S = sqr_root(2)*s or (2^(1/2))s

What am I missing?
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by neerajkumar1_1 » Tue Oct 05, 2010 10:00 pm
what u r missing is that u assumed that height of the triangles is same...

2(1/2 s h) = 1/2 S H
S = 2 s h/H

since the two triangles are similar... and the ratio of their areas is 1/2 then the ratio of their sides will be in the ratio 1/root(2)

hence H= root(2) h

therefore S = 2 s h/(root(2)h)
= root(2)s
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by vongdn » Tue Oct 05, 2010 10:17 pm
neerajkumar1_1 wrote:what u r missing is that u assumed that height of the triangles is same...

2(1/2 s h) = 1/2 S H
S = 2 s h/H

since the two triangles are similar... and the ratio of their areas is 1/2 then the ratio of their sides will be in the ratio 1/root(2)

hence H= root(2) h

therefore S = 2 s h/(root(2)h)
= root(2)s
Thanks, I knew I was doing something probably dumb.

Can I ask, how you got that h:H was 1:root(2)? Was it because the ratio of areas was 1:2, and because for h and H you are dealing with length, you must then take the root(1/2), ie root(1)/root(2) kind of like going from cm^2 to cm?
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by neerajkumar1_1 » Wed Oct 06, 2010 12:47 am
vongdn wrote:
neerajkumar1_1 wrote:what u r missing is that u assumed that height of the triangles is same...

2(1/2 s h) = 1/2 S H
S = 2 s h/H

since the two triangles are similar... and the ratio of their areas is 1/2 then the ratio of their sides will be in the ratio 1/root(2)

hence H= root(2) h

therefore S = 2 s h/(root(2)h)
= root(2)s
Thanks, I knew I was doing something probably dumb.

Can I ask, how you got that h:H was 1:root(2)? Was it because the ratio of areas was 1:2, and because for h and H you are dealing with length, you must then take the root(1/2), ie root(1)/root(2) kind of like going from cm^2 to cm?
yes ... remember when the sides are similar the sides are in comparative ratio...
so since the area is give... which is side * side... then each side will be in the ratio of a under root of that factor...
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