Question

In a triangle the sum of any two side exceeds the third side by $6$ cm, then find its area.

• A. $6\sqrt{3}$ sq. cm
• B. $9\sqrt{3}$ sq. cm
• C. $12\sqrt{3}$ sq. cm
• D. $15\sqrt{3}$ sq. cm

Answer 1

Answer: (B)

$9\sqrt{3}$ sq. cm

Let a, b, and c are sides of triangle

a/c question

$a+b-c=6$

$b+c-a=6$

$c+a-b=6$

add all equations

$a+b+c=18$

hence perimeter $\left(s\right)=18$

$s-(a+b-c)=18-6$

$a+b+c-a-b+c=12$

$2c=12$

$c=6cm$

in the same way

$b=a=6cm$

hence triangle is an equilatral

so ,

area of equilatral triangle$=\frac{\sqrt{3}}{4}(6{)}^2$

$=9\sqrt{3}c{m}^2$