Question

Drow
2 pappendiculars flom somdk on PQ
As U is the midit. and VCIIPA
Using midpt theotwe we have Cis afso a mid
So is ( D ). Srmilarily
( sin Delta operatorname{suc} ) and ( Delta operatorname{sen} A )
( s c=frac{1}{2} s A )
midpt thesenn ( y )
( therefore ) Ala of ( s v c=frac{1}{4} )
Similarly Ara of ( R D V=frac{1}{4} ) Alla ( oint R B varphi )
( left.operatorname{Ratio}=frac{frac{3}{4}left(f frac{1}{2}(p A x s A)right)+frac{1}{2}(A B x S A)+frac{3}{4}left(frac{1}{2}(B Q x S A)right)}{frac{1}{4}left(frac{1}{2}(P A times S A)+frac{1}{2}(S R times S A)+frac{1}{L}left(frac{1}{2}(B Q times S A)right.right.}right) )
Since ( S R=A B )
[
frac{frac{3}{8}(P++B varphi)+frac{1}{2} A B}{frac{1}{8}(P A+B Q)+frac{1}{2} A B} quadleft(begin{array}{c}
A B=11
P A+B varphi=15-11
end{array}right)
]
( =frac{frac{3}{2}+frac{11}{2}}{frac{1}{2}+frac{11}{2}}=7 / 6 )
( operatorname{Ans} frac{1}{7} )

# 71. In the given figure, PQRS is a trapezium with PQ|| SR and PQ = 15 cm and SR = 11 cm. If U is the mid point of PS and UV|| PQ, then the ratio of areas of PQVU and that of UVRS is equal to S PL 471 11:15 (2) 7:6 (3) 13:15 (4) 8:7

Solution