Question

corresponding sides of two similar triangles are 3cm and 4cms. If the area of larger triangle is 48cm.square, then find the area of smaller triangle

Answer 1

The formula states that if both figures are similar, then:

$\dfrac{\text{Area 1}}{\text{Area 2}} = \bigg ( \dfrac{\text{Length 1}}{\text{Length 2}} \bigg)^2$

Since we know that the two lengths of the two triangle are 3 cm  and 4 cm

⇒ The ratio is 3 : 4

Find the smaller area:

$\dfrac{\text{Area 1}}{48} = \bigg ( \dfrac{3}{4} \bigg)^2$

$\dfrac{\text{Area 1}}{48} = \dfrac{9}{16}$

$16 \times \text {Area 1} = 9 \times 48$

$16 \times \text {Area 1} = 432$

$\text {Area 1} = 432 \div 16$

$\text {Area 1} = 27 \text { cm}^2$

Answer: The area of the smaller triangle is 27 cm²

Answer 2

Hi ,

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We know that ,

The ratio of the areas of two similar

triangles is equal to the square of the

ratio of their corresponding sides .

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Let A1 , A2 are areas of two triangles

and

s, a are two their two corresponding

sides ,

it is given that ,

s= 3 cm ,

a = 4 cm ,

A1 = ?,

A2 = 48 cm²

A1/A2 = s²/a²

=> A1/48 = 3²/4²

=> A1/48 = 9/16

A1 = ( 48 × 9 )/16

A1 = 27 cm²

I hope this helps you.

: )