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Question

The parallel sides of a trapezium are 25 cm and 11 cm, while its nonparallel sides are 15 cm and 13 cm. Find the area of the trapezium.

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Solution



Let ABCD be the trapezium in which ABDC, AB=25 cm, CD=11 cm, AD=13 cm and BC=15 cm.Draw CLAB and CMDA meeting AB at L and M, respectively.Clearly, AMCD is a parallelogram.Now,MC=AD=13 cmAM=DC=11 cmMB=AB-AM
=25-11 cm=14 cm

Thus, in CMB, we have:CM=13 cmMB=14 cm BC=15 cm s=1213+14+15 cm =1242 cm =21 cms-a=21-13 cm =8 cms-b=21-14 cm =7 cms-c=21-15cm =6 cm Area of CMB=ss-as-bs-c
=21×8×7×6 cm2=84 cm2

12×MB×CL=84 cm212×14×CL=84 cm2
CL=847CL=12 cm
Area of the trapezium=12×AB+DC×CL
=12×25+11×12 cm2=12×36×12 cm2=18×12 cm2=216 cm2

Hence, the area of the trapezium is 216 cm2.

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