Another geometry problem!

[tbroker]

The areas of two similiar trianles are 25 and 16. If the perimeter of the first is 15 find the perimeter of the second. Goodluck

[ekonomiks]

The area is 15*Sqrt[16/25] = 12?

[asquare]

Are you giving us your homework as "brain teasers?"

[tbroker]

no its from simmons precalc in a nushell, thought it would be fun. I know it is 12 but where does the sqrt of the perimeters come on

[snappythecrab]

I know it is 12 but where does the sqrt of the perimeters come on

It is an immediate (almost) result from the definition of similar triangles and from the formula for the area of a triangle.

You may also have a little bit more luck getting people to bite by posting this in the GRE forum. These seem more up their alley.

[ekonomiks]

Let subscripts 2 and 1 denote the big and small triangles respectively. We have
h1/h2 = b1/b2 = r
A1/A2 = h1*b1/(h2*b2) = r^2 = 16/25 -> r = 4/5
Then, P1 = r*P2 = 4/5*(15) = 12

Notations: h = height, b = base, A = area.

[snigai]

Let S1 and S2 denote areas of triangles 1 and 2 respectively and P1 and P2 their perimeters.
The solution is trivial, since we have the formula:
S1/S2=P1^2/P2^2
5/4=15/P2 >> P2=60/5=12
I am quite sure that every 8th-9th grader in my country would solve this problem without thinking too much.

[tbroker]

there is really no need for the cocky attitude. I have had no dealings with geometry for 7 years. I bet the 8-9 graders in your country can do it so easily and I am happy for them

[reactor]

Please do NOT post math problems in the PhD Economics - Admissions Forum.

[KBTA]

This problem has helped me augment my conception in similitude. If two traingles are equal, then in all cases the ratio of two similar linear dimensions would be same.a1/a2=b1/b2=c1/c2=h1/h2=p1/p2=k (constant)Here, a,b,c are arms and h is height and p is perimeter.