Question

The sides of a triangle are 11cm, 60 cm and 61 cm. Find the altitude of the smallest side.

Answer 1

Answer:

60 cm

Step-by-step explanation:

Let the triangle be ΔABC, where,

AB = 60 cm

BC = 11 cm

AC = 61 cm

And altitude be AD

First, lets find the area of the triangle

S=(11+60+61)/2

S=132/2=66 cm

A = $\sqrt{66(66-11)(66-60)(66-61}$

A = $\sqrt{66 X 55 X 6 X 5\}$

A = $\sqrt{6X11X5X11X6X5}$

A = 6 × 11 × 5

A = 330 cm²

Now, smallest side is 11 cm. And we know that altitude is a perpendicular dropped to a side from the opposite vertex (i.e. the height of the triangle)

∴ The altitude of the smallest side is AD dropped to BC from vertex A.

base is 11 cm

height is unknown

We know that,

A = $\frac{1}{2}$×b×h

∴330=$\frac{1}{2}$×11×h

11h=660                               ...(transposing 1/2 to L.H.S)

h=60 cm

Hope it helps!