Question

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

पर्याय

• 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,

${\displaystyle A=\sqrt{s(s-a)(s-b)(s-c)}}$, where

${\displaystyle s=\frac{a+b+c}{2}}$

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

${\displaystyle s=\frac{a+b+c}{2}}$

${\displaystyle s=\frac{11+60+61}{2}}$

${\displaystyle s=\frac{132}{2}}$

 s = 66 m 

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

${\displaystyle A=\sqrt{66\left(55\right)\left(6\right)\left(5\right)}}$

${\displaystyle A=\sqrt{108900}}$

A = 330 m² 

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

${\displaystyle \text{Area}\left(A\right)=\frac{1}{2}(\text{Base}\times \text{Height})}$

 ${\displaystyle \text{Area}\left(A\right)=\frac{1}{2}(AC\times p)}$

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

AC = 11 m

${\displaystyle 330=\frac{1}{2}(11\times p)}$

${\displaystyle 330\times 2=(11\times p)}$

          ` p = (330xx2)/11

   p = 60 m