(Q9) The areas of two similar triangles are 81 cm2 and 49 cm2 respectively. If the altitude of the bigger triangle is 4.5 cm. Find the corresponding altitude of the smaller triangle.

ABCDEFYX

Solution :

Given : ∆ABC ∼ ∆DEF

∆ABC = 81 cm2

∆DEF = 49 cm2

AX = 4.5 cm

To find : DY

We know that,

∆ABC
∆DEF
 =
AX2
DY2

[∵ Ratio of areas of two similar triangles is equal to ratio of the squares of their corresponding altitudes]

 =
()
∴ =
×
⇒ DY = √(
×
) =
×

∴ DY = cm