Question

The sides of a triangle are 50 cm, 78 cm and 112 cm. The smallest altitude is

विकल्प

• 20 cm

• 30 cm

• 40 cm

• 50 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`

Therefore the area of a triangle, say A having sides 50 cm, 78 cm and 112 cm is given by

a = 50 cm ; b = 78 cm ;  c = 112 cm 

`s = (50+78+112)/2`

`s = 240/2`

 s = 120 cm

`A = sqrt(120 (120-50)(120-78)(120-112))`

`A = sqrt((120(70)(42)(8))`

`A = sqrt(2822400)`

A = 1680 cm²

The area of a triangle, having p as the altitude will be,

\[\text{ Area}  = \frac{1}{2} \times \text{ base }  \times \text{ height } \]

Where, A = 1680  cm²

We have to find the smallest altitude, so will substitute the value of the base AC with the length of each side one by one and find the smallest altitude distance i.e. p

Case 1 

AC = 50 cm 

1680 = `1/2 (50 xx p)`

1680 `xx 2 = 50 xx p `

`p = (1680 xx 2)/50`

`p = 67.2   cm `

Case 2

`AC = 78  cm `

`1680 = 1/2 (78 xx p)`

`1680 xx 2 = 78 xx p `

`p = (1680xx2)/78`

p = 43 cm 

Case 3

Ac = 112 cm

`1680 =1/2 (112 xx p)`

`1680 xx 2 = 112 xx p `

`p = (1680 xx 2 )/112`

 p = 30 cm