Two parallel sides of an isosceles trapezium are 6 cm and 14 cm respectively. if the length of each non parallel side is 5 cm, find the area of the trapezium.
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Let the trapezium be ABCD where AB and CD are the parallel sides.
Now,
AB=6 cm
CD=14 cm
AD=BC=5 cm
Construct a line parallel to BC from A to a point E on DC.
NOw, ar(ADE)= 12 sq. cm (By Heron's Formula)
ar(ADE)= 1/2*b*h
Therefore the height is 6 cm.
ar(ABCE)= b*h
=6cm*6cm
=36 sq.cm
Hence, area of ABCD is=12+36 i.e. 48 sq. cm.
Now,
AB=6 cm
CD=14 cm
AD=BC=5 cm
Construct a line parallel to BC from A to a point E on DC.
NOw, ar(ADE)= 12 sq. cm (By Heron's Formula)
ar(ADE)= 1/2*b*h
Therefore the height is 6 cm.
ar(ABCE)= b*h
=6cm*6cm
=36 sq.cm
Hence, area of ABCD is=12+36 i.e. 48 sq. cm.
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10
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