Question

The areas of two similar triangles are 100 cm and 49 cm respectively . If the altitude of the bigger triangles is 5 cm find the corresponding altitude of other

Answer 1

As we know that, for two similar triangles
(Area)₁/(Area)₂ = (side)₁²/(side)₂²

(Area)₁ = 100 cm²
(Area)₂ = 49 cm²
(altitude)₁ = 5 cm
(altitude)₂ = x

100/49 = (5)²/x²
⇒x² = [(25)×49]/100
⇒x = (5×7)/10
      =3.5 cm

Answer 2

Answer:

3.5 cm

Step-by-step explanation:

We are given that The areas of two similar triangles are 100 cm and 49 cm respectively .

We are also given that The altitude of the bigger triangles is 5 cm

Let the altitude of smaller triangle be x

Theorem : The ratio of area of similar triangles is equal to the ratio of squares of corresponding altitudes

So, $\frac{100}{49}=\frac{5^2}{x^2}$

$x^2=\frac{5^2 \times 49}{100}$

$x=\sqrt{\frac{5^2 \times 49}{100}}$

$x=3.5$

Hence the corresponding altitude of other triangle is 3.5 cm