Let $A B C D$ be a tetrahedron such that the edges $\mathrm{AB}, \mathrm{AC}$ and AD 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 $ riangle \mathrm{BCD}$ is equal to :

Question

Let $A B C D$ be a tetrahedron such that the edges $\mathrm{AB}, \mathrm{AC}$ and AD 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 $	riangle \mathrm{BCD}$ is equal to :

Aicharya · Accepted Answer

$\operatorname{Ar}(	riangle B C D)$ $=\sqrt{(\operatorname{Ar}(	riangle \mathrm{ABC}))^2+(\mathrm{Ar}(\mathrm{ACD}))^2+(\operatorname{Ar}(	riangle \mathrm{ADB}))^2}$ $=\sqrt{5^2+6^2+7^2}$ $=\sqrt{110}$ ^

Let $A B C D$ be a tetrahedron such that the edges $\mathrm{AB}, \mathrm{AC}$ and AD are mutually…

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