JEE Mains · Maths · STD 11 - 9. straight line
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
- C \(\sqrt{110}\)
- D \(7 \sqrt{3}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{110}\)
Step-by-step Solution
Detailed explanation
\(\operatorname{Ar}(\triangle B C D)\) \(=\sqrt{(\operatorname{Ar}(\triangle \mathrm{ABC}))^2+(\mathrm{Ar}(\mathrm{ACD}))^2+(\operatorname{Ar}(\triangle \mathrm{ADB}))^2}\) \(=\sqrt{5^2+6^2+7^2}\) \(=\sqrt{110}\)
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