Question ID: #600
Let $ABCD$ 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:
- (1) $\sqrt{340}$
- (2) 12
- (3) $\sqrt{110}$
- (4) $7\sqrt{3}$
Solution:
For a tetrahedron with three mutually perpendicular edges meeting at a vertex (rectangular tetrahedron), the square of the area of the face opposite to that vertex is equal to the sum of the squares of the areas of the other three faces.
Given:
Area($\triangle ABC$) = 5
Area($\triangle ACD$) = 6
Area($\triangle ADB$) = 7
Using the property:
$$ (\text{Area of } \triangle BCD)^2 = (\text{Area of } \triangle ABC)^2 + (\text{Area of } \triangle ACD)^2 + (\text{Area of } \triangle ADB)^2 $$
$$ (\text{Area of } \triangle BCD)^2 = 5^2 + 6^2 + 7^2 $$
$$ (\text{Area of } \triangle BCD)^2 = 25 + 36 + 49 $$
$$ (\text{Area of } \triangle BCD)^2 = 110 $$
$$ \text{Area of } \triangle BCD = \sqrt{110} $$
Ans. (3)
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