# Ratio of heights of an isosceles triangle

There were two isosceles triangles whose angles are same. The ratio of their areas is $16:25$. What is the ratio of their heights?

This is a question given in the aptitude section of a company written. I was confused and can't understand where to start. Can anyone please help me

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Hints:

=== In similar triangles, the similarity ratio is the same between corresponding sides, corresponding medians, corresponding heights and corresponding angle bisectors (and others, like midsegments and etc ).

=== In similar triangles, the ratio between areas equals the similarity ratio squared.

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Pretend they asked you about squares instead of isosceles triangles. Work it out for that easy case, should give you insight.

--For a square, you know the exact formula for the area vs length. For an iscosceles triangle you don't. Does it matter?

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so 4:5 is the answer – sai kiran grandhi Sep 27 '13 at 5:12
In mathematics, the answer is always an argument, not a number ^_^ But you found the correct last step in the argument. – DanielV Sep 27 '13 at 5:24

Hey as given there area is 16:25

We know as area are h1^2/h2^2=16/25 --> h1/h2=4/5

describe below how to calcuate...

Lets assume that heights are h1 and h2 and base are b1 and b2

so are are (b1h1)/2 and b2h2/2

becase is same angle is @ than as in traingle tan@=h/a/2=2h/a 16/25=(2h1*h1/(tan@*2))/(2h2*h2/(tan@*2))

h1/h2=4/5 ans

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