Question

A triangle with integral sides has perimeter $8cm$. Then the area of the triangle is

• A. $2c{m}^2$
• B. $4c{m}^2$
• C. $2\sqrt{2}c{m}^2$
• D. $Noneofthese$

Answer 1

The correct option is C $2\sqrt{2}c{m}^2$

Let the integral sides are $p,q,r$

and ATQ: $p+q+r=8cm$

$p+q=8-r$

Now,

Sum of two sides

$p+q=8-1=7$

$p+q=8-2=6$

$p+q=8-3=5$

$p+q$ can't be $1,2,3,4$,

Hence the sum of two sides is greater than the third side

out of them only the sum $5$ or $6$ possible, because if the sum, is $7$ sides $p,q,r$ are $(6,1,1),(4,3,1),(5,2,1)$, which isn't possible , as the sum of two sides is greater than third side.

If the sum is $6$ then possible values of $p,q,r$ are $(3,3,2),(5,1,2),(4,2,2)$. and $(4,2,2),(5,1,2)$ arn't exist. as the sum of two sides is greater than third side.

If the sum is $5$ then $p,q,r$ to be $(4,1,3),(3,2,3)$, first Will not be exist. and not the other. as the sum of two sides is greater than third side.

finally the sides of triangle are $3,3,2$

and it will be an isosceles triangle with sides and base $3,2$ respectively.

So, perpendicular length from the vertex to the base $=\sqrt{3^2-1^2}=\sqrt{8}$

Now, the area $=\frac{1}{2}\times base\times perpendicular=\frac{1}{2}\times 2\times \sqrt{8}=2\sqrt{2}c{m}^2$