Let ABCD 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 $\triangle \mathrm{BCD}$ is equal to :

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Vector Algebra

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 :

Answer: (C)

For a tetrahedron with three mutually perpendicular edges meeting at vertex A, there is a formula relating the areas:

Area (△ B C D)^{2} = Area (△ A B C)^{2} + Area (△ A C D)^{2} + Area (△ A D B)^{2}

This is analogous to the 3D Pythagorean theorem for areas.

Substituting the given values:

Area (△ B C D) = 5^{2} + 6^{2} + 7^{2} = 25 + 36 + 49 = 110

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Single correct option