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Let A B C D be a tetrahedron such that ...

Let `A B C D` be a tetrahedron such that the edges `A B ,A Ca n dA D` are mutually perpendicular. Let the area of triangles `A B C ,A C Da n dA D B` be 3, 4 and 5sq. units, respectively. Then the area of triangle `B C D` is `5sqrt(2)` b. `5` c. `(sqrt(5))/2` d. `5/2`

A

`5sqrt2`

B

5

C

`sqrt5/2`

D

`5/2`

Text Solution

Verified by Experts

The correct Answer is:
a

Area of `triangleBCD = 1/2 |vec(BC)xx vec(BD)|`
` 1/2 (bhati -chatj) xx (bhati - dhatk)`
`=1/2 |bd hatj +bc hatk + dc hati|`
`1/2 sqrt(b^(2)c^(2)+c^(2)d^(2)+d^(2)b^(2))`
Now 6= bc, 8 =cd, 10= bd
`b^(2)c^(2) +c^(2)d^(2)+d^(2)b^(2) = 200`
substituting the value in (i) , we get
` A = 1/2 sqrt200 = 5sqrt2`
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