Answer :


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.


Now,


AM = DC =10cm


MB = (AB-Am)


= (20-10) = 10 cm


Also,


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


There fore


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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