Question

Two isosceles triangles have equal angles and their areas are in the ratio 16:25. Find the ratio of their corresponding altitudes.​

Answer 1

Answer:

The ratio of their corresponding heights is 4 : 5.

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Answer 2

Let the two triangles be ∆ABC & ∆DEF

IN ∆ABC

AB = AC [Given]

$\implies{\mathsf{ \frac{AB }{AC} = 1 }}$ ...(1)

Similiary in traingle DEF,

$\implies{\mathsf{\frac{DE}{DF}=1}}$ ...(2)

By equating both the equations (1) & (2)

$\implies{\mathsf{ \frac{ AB }{AC } = \frac{ DE }{DF }}}$

Now, This gives us

In ∆ABC & ∆DEF we get ∠A = ∠D

By using S.A.S (Side - angle - side) Congruency/similarity both triangles will be congruent.

$\boxed{\bold{\mathsf{S.A.S \: similarity}}}$

$\implies{\mathsf{}}$ Any two triangles will be similar, if any one pair of córresponding sides are proportional and the angles between them are equal.

∆ABC ~ ∆DEF

By using an basic property that is

$\bigstar$ Area's of any two similar triangle are in the ratio of the square of their corresponding altitudes.

As AX & DY are the altitudes of the ∆ABC & ∆DEF respectively

$\implies{\mathsf{ \frac{ ar(ABC)}{ar(DEF } = \frac{ {AX}^{2} }{{DY}^{2} } }}$

$\implies{\mathsf{ \frac{ 16}{ 25 } = \frac{ {AX}^{2} }{{DY}^{2} } }}$

$\implies{\mathsf{\frac{ {4}^{2} }{ {5}^{2}} =\frac{ {AX}^{2} }{{DY}^{2} } }}$

$\huge{\boxed{\implies{\mathsf{\frac{ {AX} }{{DY} } = \frac{4 }{ 5 } }}}}$

So , The ratio of their corresponding heights is 4 : 5