Question

The areas of two similar triangles are 48 cm2

and 75 cm2

respectively. If the altitude of the first

triangle be 3.6 cm, find the corresponding altitude of the other​

Answer 1

$\large{\underline{\rm{\blue{\bf{Given:-}}}}}$

The areas of two similar triangles are 48 cm² and 75 cm²

Altitude of the first triangle = 3.6 cm

$\large{\underline{\rm{\blue{\bf{To \: Find:-}}}}}$

The other altitude of the other triangle.

$\large{\underline{\rm{\blue{\bf{Solution:-}}}}}$

We know that,

Ratio of areas of two similar triangle is equal to the squares of the ratio of their corresponding altitudes.

$\sf Area=(\triangle_1)=48 \: cm^{2}$

$\sf Area=(\triangle_2)=75 \: cm^{2}$

$\sf a_1=3.6 \: cm$

For similar triangles the ratio of areas is equal to the ratio of square of their altitudes.

$\implies \sf \dfrac{A_1}{A_2} =\bigg(\dfrac{h_1}{h_2} \bigg)^{2}$

Substituting their values, we get

$\implies \sf \dfrac{48}{75} =\bigg(\dfrac{3.6}{h_2 }\bigg)^{2}$

$\implies \sf \dfrac{16}{25} =\bigg( \dfrac{3.6}{h_2} \bigg)^{2}$

$\implies \sf \bigg(\dfrac{4}{5}\bigg)^{2}=\bigg(\dfrac{3.6}{h_2} \bigg)^{2}$

$\implies \sf \dfrac{4}{5} =\dfrac{3.6}{h_2}$

Now, solving $\sf h_2$

$\implies \sf h_2=\dfrac{3.6 \times 5}{4}$

$\implies \sf h_2=4.5 \: cm$

Therefore, the other corresponding height is 4.5 cm

Answer 2

Answer:

4.5 cm

Step-by-step explanation:

Ar. (△1) = 48cm2

Ar. (△2) = 75cm2

a1=3.6cm

For similar triangles the ratio of areas is equal to the ratio of square of their altitudes.

Thus, A(△2)A(△1)=(a2)2(a1)2

7548=(a2)2(3.6)2

(a2)2=4812.96×75

(a2)2=20.25

a2=4.5cm