Q. 114.0( 53 Votes )

# The parallel side

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:

CL^{2} = (CM2 – ML^{2})

CL^{2} = (132 – 5^{2})

CL^{2} = (169 – 25)

CL^{2} = 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 cm^{2}.

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