Q49 / 75 · JEE Main 2025 (2 Apr Shift 1) · Mathematics

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

  • A
    $\sqrt{340}$
  • B
    12
  • Correct answer
    C
    $\sqrt{110}$
  • D
    $7 \sqrt{3}$

Solution

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

Hence, the correct answer is option: 'C'.

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