Question

the
area of two similar triangles are 81cm2

. and 49 cm2

respectively.

If altitude of bigger triangle is 4.5 cm. Find the corresponding altitude

of smaller triangle.​

Answer 1

$\Large{\underline{\underline{\mathfrak{\bf{Question}}}}}$

The area of two similar triangles are 81 cm² and 49 cm² respectively. If altitude of bigger triangle is 4.5 cm. Find the corresponding altitude of smaller triangle.

$\Large{\underline{\underline{\mathfrak{\bf{Solution}}}}}$

$\Large{\underline{\mathfrak{\bf{Given}}}}$

• Area of big triangle = 81 cm²
• Area of small triangle = 49 cm²
• Altitude of bigger triangle = 4.5 cm

$\Large{\underline{\mathfrak{\bf{Find}}}}$

• Altitude of small triangle

$\Large{\underline{\underline{\mathfrak{\bf{Explanation}}}}}$

Let, Altitude of small triangle = x cm

As we know that,

Area's of two similar triangle are in the ration of the squares of the corresponding altitudes .

$\mapsto\small{\sf{\:\dfrac{Area\:of\:big\:triangle}{Area\:of\:small\:triangle}\:=\:\dfrac{square\:of\:Altitude\:of\:big\:triangle}{square\:of\:Altitude\:of\:small\:triangle}}}\\ \\ \mapsto\sf{\:\dfrac{81}{49}\:=\:\dfrac{4.5^2}{x^2}}\\ \\ \mapsto\sf{\:x^2\:=\:\dfrac{49\times 20.25}{81}}\\ \\ \mapsto\sf{\:x^2\:=\:\dfrac{992.25}{81}}\\ \\ \mapsto\sf{\:x^2\:=\:12.25}\\ \\ \mapsto\sf{\:x\:=\:\sqrt{12.25}}\\ \\ \mapsto\mathfrak{\bf{\:x\:=\:3.5}}$

$\Large{\underline{\mathfrak{\bf{Thus}}}}$

• Altitude of small triangle = 3.5 cm

$\Large{\underline{\underline{\mathfrak{\bf{Answer\:Verification}}}}}$

A/c to this,

$\mapsto\small{\sf{\:\dfrac{Area\:of\:big\:triangle}{Area\:of\:small\:triangle}\:=\:\dfrac{square\:of\:Altitude\:of\:big\:triangle}{square\:of\:Altitude\:of\:small\:triangle}}} \\ \\ \mapsto\sf{\:\dfrac{81}{49}\:=\:\dfrac{4.5^2}{3.5^2}} \\ \\ \mapsto\sf{\:1.65\:=\:\dfrac{20.25}{12.25}} \\ \\ \mapsto\sf{\:1.65\:=\:1.65} \\ \\ \:\:\mathfrak{\bf{\:L.H.S.\:=\:R.H.S.}}$

That's proved.

Answer 2

Given :

The area of big triangle = 81cm²

The area of small triangle = 49cm²

Altitude of bigger triangle = 4.5 cm

To Find :

The corresponding altitude of smaller triangle = ?

Solution :

Let altitude of smaller triangle be X cm.

$\sf \pink{ \frac{area \: of \: bigger \triangle}{area \: of \: smaller \triangle} = \frac{altitude \: of \: bigger \triangle}{altitude \: of \: smaller \triangle} }$

$\implies\sf \frac{81}{49} = \frac{4. {5}^{2} }{ {x}^{2} } \\ \\ \implies \sf{ {x}^{2} = (20 \times 25 \times \frac{49}{81} }) \\ \\ \sf \implies \: {x}^{2} = 12.25 \\ \\ \sf \implies \: x = \sqrt{12.25} \\ \\ \sf \: x = 3.5 \: cm$

$\bold{hence \: the \:corresponding \: altitude } \\ \bold{ \: of \: the \: smaller \: triangle \: is \: {\boxed {\rm {\red {\underline {\underline{3.5 \: cm }}}}}}}$

corres