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Q.

Let A(0,0), B(5,0), C(5,3) and D(0,3) be the vertices of rectangle ABCD, If P is a variable point lying inside the rectangle ABCD and d (P, L) denotes perpendicular distance of point P from line LIf d(P,AB)≤min{d(P,BC),d(P,CD),d(P,AD)} then area of the region in which P lies is (in sq.units)If d(P,AB)≥maxd(P,BC),d(P,CD),d(P,AD) then the area of the region in which P lies is (in sq.units)If d(P,AB)−322+(d(P,AD))2≥1, then area of region in which P lies is (in sq.units)

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a

174

b

194

c

214

d

234

e

1

f

12

g

34

h

14

i

15-2π

j

10-π2

k

15-π

l

15-π2

answer is , , .

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Detailed Solution

d(P,AB)≤d(P,BC)d(P,AB)≤d(P,CD)d(P,AB)≤d(P,AD)⇒P lies in region as shownArea=12×32×7=214d(P,AB)≥d(P,BC)d(P,AB)≥d(P,CD)d(P,AB)≥d(P,AD)⇒P lies in region as shownArea==12×1×12=14x2+y−322≥1P lies in region as shownArea =5×3−π(1)22=15−π2
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