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Question

The parallel sides of a trapezium is 77 cm and 60 cm and its non-parallel sides are 26 cm, and 25 cm. The area of the trapezium is equal to

A
1600 cm2
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B
1666 cm2
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C
1644 cm2
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D
1500 cm2
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Solution

The correct option is C 1644 cm2
Let ABCD be a trapezium where AB//DC
AB=60cm
DC=77cm
AD=26cm
BC=25cm
Draw ADDC and BQDC
PQ=60cm and AP=BQ=h
Let DP=x and QC=y
In right ΔAPD,
Using Pythagoras theorem, we have
AP2+DP2=AD2
h2+x2=(26)2
h2=262x2 ………..(1)
In right ΔBQC,
Using Pythagoras theorem, we have
BQ2+QC2=BC2
h2+y2=252
h2=252y2 …………..(2)
Equating equations (1) and (2), we have
262x2=252y2
x2y2=262252
x2y2=676625
x2y2=51 ……….(3)
Also, we have, x+60+y=77
x=7760y
x=17y
Substituting value of x in (3) we get
(17y)2y2=51
172+y234yy2=51
28951=34y
y=23834
y=7cm
Thus, x=17y=177=10cm
Substituting values of x and y in (1) we get
h2=262(10)2
=676100
=576
h=576
h=24cm
Area of rectangle ABQP=AB×BQ
=60×24
=1440cm2
Area of ΔAPD=12×x×h
=12×10×24
=120cm2
Area of ΔBQC=12×y×h
=12×7×24
=84cm2
Hence area of the trapezium ABCD=(1440+120+84)cm2
=1644cm2.

1210290_1314699_ans_c9029815ec384b1d9d5ad6f5fa2f2ec0.jpg

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