Find the area of a trapezium whose parallel sides `25 c m ,13 c m` and other sides are `15 cm and 15 cmdot`

Explanation

ADCE is a parallelogram ( AD || CE & AE || CD).
∴ `AE = DC = 13` cm ( Opposite side of parallelogram are equal)
`BE = AB - AE`
`BE = 25 - 13`
`BE = 12` cm
In `triangle BCE`
`S = a + b + c/2`
`S = 15 + 15 + 12 /2`
`S = 21`
Area of triangle `BCE = sq(s( s - a)(s - b)( s - c ))`
Area of triangle `BCE = sq(21(21-15)(21-15)(21-12))`
Area of triangle `BCE = sq(21 xx6 xx 6 xx 9)`
Area of triangle `BCE = 18 sq(21)cm^2` -----`1`
`h` is the height of triangle ` BCE`
Area of `BCE = 1/2 ( Base xx Height )`
`= 1/2(12)(h)`
`= 6h ----2`
From `1` & `2`
`6h = 18 sq(21)`
`=> h = 3 sq(21)` cm
The height of trapezium `ABCD` is equal to height of triangle ` BCE`.
Area of trapezium `= 1/2 ( AB + CD ) xx h`
`= 1/2 (25 + 13) xx 3 sq(21)cm^2`
`= 57sq(21) cm^2`