JEE MAIN - Mathematics (2025 - 2nd April Morning Shift - No. 8)
Let $A B C D$ be a tetrahedron such that the edges $A B, A C$ and $A D$ are mutually perpendicular. Let the areas of the triangles $\mathrm{ABC}, \mathrm{ACD}$ and ADB be 5,6 and 7 square units respectively. Then the area (in square units) of the $\triangle B C D$ is equal to :
$\sqrt{110}$
12
$\sqrt{340}$
$7 \sqrt{3}$
Explanation
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$$\begin{aligned} & \operatorname{ar}(\triangle A B C)=5 \\ & \frac{1}{2} \times b c=5 \\ & \Rightarrow \quad b c=10 \\ & \operatorname{ar}(\triangle A C D)=6 \\ & \frac{1}{2} \times c d=6 \\ & \Rightarrow \quad c d=12 \\ & \operatorname{ar}(\triangle A B D)=7 \\ & b d=14 \end{aligned}$$
$$\begin{aligned} & \operatorname{area}(\triangle B C D)=\frac{1}{2}|\overrightarrow{B C} \times \overrightarrow{B D}| \\ & \overrightarrow{B C}=<-b, c, 0> \\ & \overrightarrow{B D}=<-b, 0, d> \\ & |\overrightarrow{B C} \times \overrightarrow{B D}|=\sqrt{c^2 d^2+b^2 d^2+b^2 c^2} \\ & =\sqrt{12^2+14^2+10^2} \\ & =\sqrt{440} \\ & \operatorname{ar}(\triangle B C D)=\sqrt{110} \end{aligned}$$
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