Question

Two isosceles triangles have equal angles and their areas are in the ratio 16 : 25. The ratio of their corresponding heights is
(a) 4 : 5
(b) 5 : 4
(c) 3 : 2
(d) 5 : 7

Answer 1

Answer:

The ratio of their corresponding heights is 4 : 5.

Among the given options option (a) is 4 : 5 is the correct answer.

Step-by-step explanation:

Given:

Two isosceles ∆s have equal vertical angles and their areas are in the ratio of 16: 25.

Let the two isosceles triangles be  ΔABC and ΔPQR with ∠A = ∠P.

Therefore,

AB/AC = PQ/PR

In ΔABC and ΔPQR,

∠A = ∠P   (given)

AB/AC = PQ/ PR (sides of a isosceles∆)

ΔABC – ΔPQR    (By SAS similarity)

Let AD and PS be the altitudes(height) of ΔABC and ΔPQR.

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding altitudes.

ar(ΔABC)/ar(ΔPQR) = (AD/PS)²

16/25 = (AD/PS)²

√16/25 = √(AD/PS)²

AD/ PS = 4/5

AD : PS = 4 : 5

Hence, the ratio of their corresponding heights is 4 : 5.

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Answer 2

ANSWER:

A)4:5

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 corresponding 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