Question

The sides of a triangle are 11 m, 60 m and 61 m. The altitude to the smallest side is

Options

• 11 m

• 66 m

•  50 m

• 60 m

Answer 1

The area of a triangle having sides a, b, c and s as semi-perimeter is given by,

`A = sqrt(s(s-a)(s-b)(s-c))`, where

`s = (a+b+c)/2`

We need to find the altitude to the smallest side
 

Therefore the area of a triangle having sides 11 m, 60 m and 61 m is given by

a = 11 m ; b = 60 m ; c = 61 m

`s = (a+b+c)/2`

`s = (11+60+61)/2`

`s = 132/2`

 s = 66 m 

`A = sqrt(66(66-11)(66-60)(66-61))`

`A = sqrt(66(55)(6)(5))`

`A = sqrt(108900)`

A = 330 m² 

The area of a triangle having base AC and height p is given by

`" Area " (A) = 1/2 ("Base" xx "Height")`

 `"Area " (A) = 1/2 (ACxx p ) `

We have to find the height p corresponding to the smallest side of the triangle. Here smallest side is 11 m 

AC = 11 m

`330 = 1/2 (11 xx p)`

`330xx2=(11xxp)`

          ` p = (330xx2)/11

   p = 60 m