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There is a point P inside an equilateral...

There is a point `P` inside an equilateral ` A B C` of side `a` whose distances from vertices `A , Ba n dCa r e3,4a n d5,` respectively. Rotate the triangle and `P` through `60^0` about `Cdot` Let `A` go to `A^(prime)a n dPtoP^(prime)dot` Then the area of `P A P^(prime)` (in sq. units) is 8 (b) 12 (c) 16 (d) 6

A

8

B

12

C

16

D

6

Text Solution

Verified by Experts

The correct Answer is:
D

Clearly, `AP = A'P' = 3, BP = AP' = 4 and PC = P' C = 5`
So, `angle CPP' = angle PP'C = 60^(@)`
So, PCp' is equilateral and thus PP' = 5
So, `angle PAP' = 90^(@) " " [ :' AP = 3, PP' = 5]`
`:.` Area of `DeltaAPP' = (1)/(2) (3) (4) = 6`
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