Question

If a plane cuts off intercepts OA = a, OB = b, OC = c from the co-ordinate axes, then the area of the triangle ABC=

• A.
  
• B.
  
• C.
  
• D.
  

Answer 1

The correct option is A

Length of sides are $\sqrt{a^2+b^2},\sqrt{b^2+c^2},\sqrt{c^2+a^2}$ respectively.
Now use $\Delta =\frac{1}{2}\sqrt{s(s-a)(s-b)(s-c)}$.
Trick : Put a = 2, b = 2, c = 2, then sides will be $2\sqrt{2},2\sqrt{2}$ and $2\sqrt{2}$ i.e., equilateral triangle. So area of this triangle will be $\Delta =\frac{\sqrt{3}}{4}\times (2\sqrt{2}{)}^2=2\sqrt{3}sq.units$
Now option (a) $\Rightarrow \Delta =\frac{1}{2}\sqrt{16+16+16}=\frac{1}{2}\times 4\sqrt{3}=2\sqrt{3}$. Hence the result.