Question

If a plane cuts off intercepts $O C = c$ from the co-ordinate axes, then the area of the triangle $A B C$=

• A. $\frac { 1 } { 2 } \sqrt { b ^ { 2 } c ^ { 2 } + c ^ { 2 } a ^ { 2 } + a ^ { 2 } b ^ { 2 } }$
• B. $\frac { 1 } { 2 } ( b c + c a + a b )$
• C. $\frac { 1 } { 2 } a b c$
• D. $\frac { 1 } { 2 } \sqrt { ( b - c ) ^ { 2 } + ( c - a ) ^ { 2 } + ( a - b ) ^ { 2 } }$

Answer 1

Answer: (C)

$\frac { 1 } { 2 } a b c$

Answer 2

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 (1) $\Rightarrow \Delta = \frac { 1 } { 2 } \sqrt { 16 + 16 + 16 } =$ $\frac { 1 } { 2 } \times 4 \sqrt { 3 }$ $= 2 \sqrt { 3 }$ . Hence the result.