JEE Main 2025 Apr 2 Shift 1, Mathematics Q18: The Plane

JEE Main2025Apr 2, Shift 1Mathematics

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 $A B C$ , $A C D$ and $A D B$ be 5, 6 and 7 square units respectively. Then the area (in square units) of the $△ B C D$ is equal to :

A
$340$
B
12
C
$110$
D
$73$

Solution

For a tetrahedron whose three edges at a vertex are mutually perpendicular, the square of the area of the face opposite that vertex equals the sum of the squares of the areas of the other three faces.

With the right-angle vertex at $A$ :

$[B C D]^{2} = [A B C]^{2} + [A C D]^{2} + [A D B]^{2}$

$[B C D]^{2} = 5^{2} + 6^{2} + 7^{2} = 25 + 36 + 49 = 110$

$[B C D] = 110$

$∴$ Option (3) is correct.

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JEE Main 2025 Apr 2 Shift 1, Mathematics Q18: The Plane

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