JEE Advance - Mathematics (2009 - Paper 2 Offline - No. 11)
Let ABC and ABC' be two non-congruent triangles with sides AB = 4, AC = AC' = 2$$\sqrt2$$ and angle B = 30$$^\circ$$. The absolute value of the difference between the areas of these triangles is ___________.
Answer
4
Explanation
We have,
$$\cos \beta = {{{a^2} + 16 - 8} \over {2 \times a \times 4}}$$
$$ \Rightarrow {{\sqrt 3 } \over 2} = {{{a^2} + 8} \over {8a}}$$
$$ \Rightarrow {a^2} - 4\sqrt 3 a + 8 = 0$$
$$ \Rightarrow {a_1} + {a_2} = 4\sqrt 3 ,{a_1}{a_2} = 8$$
$$ \Rightarrow |{a_1} - {a_2}| = 4$$
$$ \Rightarrow |{\Delta _1} - {\Delta _2}| = {1 \over 2} \times 4\sin 30^\circ \times 4 = 4$$
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