Question

The sides of a triangle are 11 cm, 15 cm and 16 cm. The altitude to the largest side is

Options

• \[30\sqrt{7} cm\]

   

• \[\frac{15\sqrt{7}}{2}cm\]

   

• \[\frac{15\sqrt{7}}{4}cm\]

   

• 30 cm

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 corresponding to the longest side
 

Therefore the area of a triangle having sides 11 cm, 15 cm and 16 cm is given by

a = 11 m ; b = 15 cm ; c = 16 cm

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

`s =(11+15+16)/2`

`s = 42/2`

s = 21 cm 

`A = sqrt(21(21-11)(21-15)(21-6))`

`A = sqrt(21(10)(6)(5)`

`A = sqrt(6300)`

`A = 30 sqrt(7)   cm^2` 

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 (AC xx p)`

We have to find the height p corresponding to the longest side of the triangle.Here longest side is 16 cm, that is AC=16 cm 

`30 sqrt(7) = 1/2 (16 xx p)`

`30 sqrt(7) xx 2 = (16 xx p)`

                    ` p = (30sqrt(7) xx 2) /16` 

`p = (15 sqrt(7) )/4 cm `