Question

$\large \pink{\underbrace{\bf{ \red{ \: \: \: \: \: \: \: \: \: \: Question \: \: \: \: \: \: \: \: \: \: }}}}$
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Two isosceles triangles have equal vertical angles and their areas in the ratio 25: 36. Find the ratio of their corresponding heights.
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Options:-
A-4:5
B-5:6
C-6:7
D-5:7
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Need answer with explanation​

Answer 1

Answer:

${\huge{\underline{\small{\mathbb{\pink{REFER \ TO \ THE \ ATTACHMENT}}}}}}$

Δ ABC ≅ Δ PQR (SAS)

$\sf \therefore \: \frac{ar( \triangle \: abc)}{ ar \: ( \triangle \: pqr)} = \frac{ {ab}^{2} }{ {pq}^{2} } \rightarrow \: eq \: (i)$

$\sf \frac{ {ab}^{2} }{ {pq}^{2} } = \frac{25}{26} \\ \\ \therefore \: \sf\frac{ab}{pq} = \frac{5}{6}$

In Δ ABD and Δ PQS

∠B= ∠Q (Δ ABC ≅ Δ PQR)

∠ADB=∠PSQ( Each 90°)

Δ ABD Δ PQS (AA)

$\sf \therefore \: \frac{ab}{pq} = \frac{ad}{ps} \\ \\ \frac{5}{6} = \frac{ad}{ps}$

Final answer:-

$\boxed{5 \ratio \: 6}$

$\rule{200pt}{2.5pt}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

Two isosceles triangles have equal vertical angles and their areas in the ratio 25: 36.

Let assume that $\triangle$ABC and $\triangle$PQR be two isosceles triangle such that

$\rm \: \angle \: A \: = \: \angle \:P$

$\rm \: AB \: = \: AC$

$\rm \: PQ \: = \: PR$

So,

$\rm \: \dfrac{AB}{PQ} = \dfrac{AC}{PR}$

Let assume that AD and PM are perpendiculars drawn from vertex A and P on side BC and QR respectively.

Now, In $\triangle$ABC and $\triangle$PQR

$\rm \: \dfrac{AB}{PQ} = \dfrac{AC}{PR}$

$\rm \: \angle \: A \: = \: \angle \:P$

$\rm\implies \:\triangle \: ABC \: \sim \: \triangle \: PQR \: \: \: \{SAS \: Similarity \}$

So, By CPST, we have

$\rm\implies \:\dfrac{AB}{PQ} = \dfrac{BC}{QR} - - - (1)$

And

$\rm\implies \:\rm \: \angle \:B \: = \: \angle \:Q$

Consider,

In $\triangle$ABD and $\triangle$PQM

$\rm \: \angle \:B \: = \: \angle \:Q \: \: \: \: \{Proved \: above \}$

$\rm \: \angle \:D \: = \: \angle \:M \: \: \: \: \{each \: 90 \degree \}$

$\rm\implies \:\triangle \: ABD \: \sim \: \triangle \: PQM \: \: \: \{AA \: Similarity \}$

So, By CPST, we have

$\rm\implies \:\dfrac{AB}{PQ} = \dfrac{AD}{PM} - - - (2)$

From equation (1) and (2), we concluded that

$\rm\implies \:\dfrac{BC}{QR} = \dfrac{AD}{PM} - - - (3)$

Now, Further given that

$\rm \: \dfrac{ar(\triangle \: ABC)}{ar(\triangle \:PQR)} = \dfrac{25}{36}$

$\rm \: \dfrac{\dfrac{1}{2} \times BC \times AD}{\dfrac{1}{2} \times QR \times PM} = \dfrac{25}{36}$

$\rm \: \dfrac{BC}{QR} \times \dfrac{AD}{PM} = \dfrac{25}{36}$

Using equation (3), we have

$\rm \: \dfrac{AD}{PM} \times \dfrac{AD}{PM} = \dfrac{25}{36}$

$\rm \: {\bigg(\dfrac{AD}{PM}\bigg) }^{2} = \dfrac{25}{36}$

$\rm\implies \:\dfrac{AD}{PM} \: = \: \dfrac{5}{6}$

$\rm\implies \:AD : PM\: = \: 5 : 6$

So, Option (B) is correct.

$\rule{190pt}{2pt}$

Additional Information :-

1. $Pythagoras Theorem$:-

This theorem states that : In a right-angled triangle, the square of the longest side is equal to sum of the squares of remaining sides.

2. Converse of Pythagoras Theorem :-

This theorem states that : If the square of the longest side is equal to sum of the squares of remaining two sides, angle opposite to longest side is right angle.

3. Area Ratio Theorem :-

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

4. Basic Proportionality Theorem,

This theorem states that :- If a line is drawn parallel to one side of a triangle, intersects the other two lines in distinct points, then the other two sides are divided in the same ratio.