Let A B C D be a tetrahedron such that edges AB , AC and AD are mutually perpendicular. Let areas of

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Let $A B C D$ 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 $\triangle BCD$ is equal to :

A$\sqrt{340}$

B$12$

C$\sqrt{110}$

D$7 \sqrt{3}$

(JEE MAIN-2025)

Solution

$ Ar (\triangle BCD )$
$=\sqrt{( Ar (\triangle ABC ))^2+( Ar ( ACD ))^2+( Ar (\triangle ADB ))^2}$
$=\sqrt{5^2+6^2+7^2}$
$=\sqrt{110}$

Standard 11

Mathematics

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