Question

in ∆ PQR , x and y are on sides of pq and pr respectively such that xy//QR if xy=4cm ,qr=10cm , ar(∆pxy) = 18 sq cm find ar(square xqry)​

Answer 1

Answer:

Let perpendicular drawn from P be intersect XY at K and QR at L. area of ∆PXY = XY*PK/2=18 => PK = 9cm

Since ∆PXY ~ ∆PQR => XY/QR = PK/PL => PL = 9*10/4=45/2cm

So KL = PL-PK = 27/2 cm

Now area of trapezium XQRY = KL*(XY+QR)/2= 14*27/4=94.5 square cm