A field is in the shape of an isosceles trapezium with parallel sides measuring $20 m and 8m$ and its non-parallel sides measuring $10 m$ . Find its area ___ (in $c m^{2}$ ).

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Solution

Let ABCD be the trapezium given to us with $AB = 8 cm, CD = 20 cm, AD = 10 cm and BC = 10 cm$ .
If we draw a line BQ parallel to AD where Q lies on DC
such that, ABQD is a parallelogram
Hence, BQ = AD = 10cm ( opposite sides of parallelogram)
DQ = AB = 8 cm ( opposite sides of parallelogram)
Also, BQC is an isosceles triangle
As, BQ = BC
Now, drop a perpendicular BL , where L lies on QC
Such that ,
QL =LC = $\frac{Q C}{2}$
= $\frac{20 - 8}{2}$
= $\frac{20 - 8}{2}$
= 6 cm ( BL is the perpendicular bisector of the triangle)
Now, In triangle BLC
using pythagoras theorem:
$(L C)^{2} + (B L)^{2} = (B C)^{2}$
$(6)^{2} + (B L)^{2} = (10)^{2}$
$(B L)^{2} = (10)^{2} - (6)^{2}$
$(B L)^{2} = 100 - 36$
$(B L) = s q r t 64$
$(B L) = 8$

area of trapezium = $\frac{1}{2} \times (s u m o f p a r a l l e l s i d e s) \times h e i g h t$
= $\frac{1}{2} \times (A B + D C) \times B L$
= $\frac{1}{2} \times (8 + 20) \times 8$
= 112 $c m^{2}$

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A field is in the shape of an isosceles trapezium with parallel sides measuring 20 m and 8m and its non-parallel sides measuring 10 m. Find its area ___ (in cm2).

A field is in the shape of an isosceles trapezium with parallel sides measuring $20 m and 8m$ and its non-parallel sides measuring $10 m$ . Find its area ___ (in $c m^{2}$ ).