the area of the shaded region of the rectangle is p/q (p,q are co prime natural numbers ) then p+q =
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see figure,
area of shaded region = (area of ASOP - area of ∆MPO - area of ∆NSO) + (area of RCQO - area of ∆KQO - area of ∆RLO )
=( 4 × 5/2 - 1/2 × 4 × 3/2 - 1/2 × 3 × 5/2) + (4 × 5/2 - 1/2 × 3 × 5/2 - 1/2 × 4 × 3/2 )
= (10 - 3 - 15/4) + (10 - 3 - 15/4)
= 14 - 15/2
= 13/2
hence, area of shaded region = 13/2 = p/q
p = 13 and q = 2
so, p + q = 13 + 2 = 15
area of shaded region = (area of ASOP - area of ∆MPO - area of ∆NSO) + (area of RCQO - area of ∆KQO - area of ∆RLO )
=( 4 × 5/2 - 1/2 × 4 × 3/2 - 1/2 × 3 × 5/2) + (4 × 5/2 - 1/2 × 3 × 5/2 - 1/2 × 4 × 3/2 )
= (10 - 3 - 15/4) + (10 - 3 - 15/4)
= 14 - 15/2
= 13/2
hence, area of shaded region = 13/2 = p/q
p = 13 and q = 2
so, p + q = 13 + 2 = 15
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Answer:is it from vmc
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