Q. 114.0( 53 Votes )
The parallel side
Let ABCD be the given trapezium in which AB|| DC,
AB = 20 cm, DC = 10 cm and AD=BC=13cm
Draw CL AB and CM || DA meeting AB at L and M, respectively.
Clearly, AMCD is a parallelogram.
AM = DC =10cm
MB = (AB-Am)
= (20-10) = 10 cm
CM = DA = 13cm
Therefore, CMB is an isosceles triangle and CL MB.
And L is midpoint of B.
⇒ML = LB = = = 5 cm
From right CLM, we have:
CL2 = (CM2 – ML2)
CL2 = (132 – 52)
CL2 = (169 – 25)
CL2 = 144
CL = 12
Therefore length of CL is 12 cm that is height of trapezium is 12 cm
We know that area of trapezium is × (sum of parallel sides) × height
Therefore Area of trapezium = × (20 + 10) × 12 = 180 cm2.
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