Let ABCD be a tetrahedron such that the edges AB, AC and, AD are mutually perpendicular. Let the area of triangles ABC, ACD and ADB be 3,4 and 5 sq. units, respectively. Then the area of triangle BCD is

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$\frac{5}{2}$

$\frac{\sqrt{5}}{2}$

$5 \sqrt{2}$

answer is A.

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Detailed Solution

$\begin{matrix} Area of Δ B C D = \frac{1}{2} | \vec{B C} \times \vec{B D} | \\ = \frac{1}{2} | (b \hat{i} - c \hat{j}) \times (\hat{b} \hat{i} - d \hat{k}) | \\ = \frac{1}{2} | b d \hat{j} + b c \hat{k} + d c \hat{i} | \\ = \frac{1}{2} \sqrt{b^{2} c^{2} + c^{2} d^{2} + d^{2} b^{2}} - - - (i) \end{matrix}$

$\begin{aligned} now, 6 = bc; 8 = cd; 10 = bd \\ b^{2} c^{2} + c^{2} d^{2} + d^{2} b^{2} = 200 \end{aligned}$

Substituting the value in (i), we get
$A = \frac{1}{2} \sqrt{200} = 5 \sqrt{2}$

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