Let $A B C D$ be a tetrahedron such that the edges $A B,$ $A C$ and $A D$ are mutually perpendicular. Let the area of triangles $A B C, A C D$ and $A D B$ be $1,$ $3$ and $\sqrt{90}$ sq. units respectively. Then the area of triangle $B C D$ is sq. units.

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Solution

Area of $Δ A B C = \frac{1}{2} a b = 1$
Area of $Δ A C D = \frac{1}{2} b c = 3$
Area of $Δ A D B = \frac{1}{2} a c = \sqrt{90}$
Area of $Δ B C D$
$= \frac{1}{2} \sqrt{a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}}$
$= \frac{1}{2} \times 2 \sqrt{1 + 9 + 90} = 10$

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Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the area of triangles ABC,ACD and ADB be 1, 3 and √90 sq. units respectively. Then the area of triangle BCD is sq. units.

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Let $A B C D$ be a tetrahedron such that the edges $A B,$ $A C$ and $A D$ are mutually perpendicular. Let the area of triangles $A B C, A C D$ and $A D B$ be $1,$ $3$ and $\sqrt{90}$ sq. units respectively. Then the area of triangle $B C D$ is sq. units.

Area of $Δ A B C = \frac{1}{2} a b = 1$
Area of $Δ A C D = \frac{1}{2} b c = 3$
Area of $Δ A D B = \frac{1}{2} a c = \sqrt{90}$
Area of $Δ B C D$
$= \frac{1}{2} \sqrt{a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}}$
$= \frac{1}{2} \times 2 \sqrt{1 + 9 + 90} = 10$