Question

IN the figure, ABCD is a trapezium in Which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm.
Find the value fo x and area of trapezium ABCD.

Answer 1

In the trapezium ABCD,
AB = 7 cm
AL = BM = 4 m
AD = BC = 5 cm

$InRight△ADL,A{D}^2=A{L}^2+L{D}^2(5{)}^2=(4{)}^2+L{D}^2\Rightarrow 25=16+L{D}^2\Rightarrow L{D}^2=25-16=9=(3{)}^2\therefore LD=3cmsimilarly,MC=LD=3cmandLM=AB=7cm\{\therefore ALandBMareperpendicularonCD\}\therefore x=DL+LM+MC=3+7+3=13cmNow,areaoftrapeziumABCD,=\frac{1}{2}(a+b)\times h=\frac{1}{2}(7+13)\times =\frac{1}{2}\times 20\times 4=40c{m}^2$