Three Dimensional Geometry Question 262
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 $ =
Options:
A) $ \frac{1}{2}\sqrt{b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}} $
B) $ \frac{1}{2}(bc+ca+ab) $
C) $ \frac{1}{2}abc $
D) $ \frac{1}{2}\sqrt{{{(b-c)}^{2}}+{{(c-a)}^{2}}+{{(a-b)}^{2}}} $
Show Answer
Answer:
Correct Answer: A
Solution:
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 $ ,\Rightarrow \Delta =\frac{1}{2}\sqrt{16+16+16}= $ $ \frac{1}{2}\times 4\sqrt{3} $ $ =2\sqrt{3} $ . Hence the result.
BETA
