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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



See, dividing the trapezium into two triangle and one rectangle,
rectangle having one side =6 cm
triangle's sides are 5cm and 4cm
To find the third side or the perpendicular distance between both parallel lines,
Apply pythagoras' theorem.
height =√((5)² -(4)²)
           = 3 cm
so, area = (1/2)×(sum of parallel side)×(distance between them)
              = (1/2)×(6+ 14)×3
              = 30 cm²
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The Brainliest Answer!
Let the trapezium be ABCD where AB and CD are the parallel sides.
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
=36 sq.cm
Hence, area of ABCD is=12+36 i.e. 48 sq. cm.
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