Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the ACD is equal to : [2025]

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Solution

Q.
Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the $△$ BCD is equal to : [2025]

1 $\sqrt{110}$

2 $7 \sqrt{3}$

3 $\sqrt{340}$

4 12

Ans.
(1)

We have, ar( $△$ ABC) = 5

$\Rightarrow \frac{1}{2} \times b \times c = 5 \Rightarrow b c = 10$ ,

ar( $△$ ACD) = 6

$\Rightarrow \frac{1}{2} \times c \times d = 6 \Rightarrow c d = 12$ ,

$a r (△ A B D) = 7$

$\Rightarrow \frac{1}{2} \times b \times d = 7 \Rightarrow b d = 14$

$\vec{B C} = - b \hat{i} + c \hat{j} and \vec{B D} = - b \hat{i} + d \hat{k}$

$\vec{B C} \times \vec{B D} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - b & c & 0 \\ - b & 0 & d \end{matrix} |$ $= c d \hat{i} + b d \hat{j} + b c \hat{k}$

$| \vec{B C} \times \vec{B D} |$ $= \sqrt{c^{2} d^{2} + b^{2} d^{2} + b^{2} c^{2}}$

$= \sqrt{12^{2} + 14^{2} + 10^{2}} = \sqrt{440}$

$∴ A r e a (△ B C D) = \frac{1}{2}$ $| \overset{—}{B C} \times \overset{—}{B D} |$

$\Rightarrow \frac{1}{2}$ $| \sqrt{440} |$ $= \sqrt{110} sq . units$

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