find the area of a trapezium whose parallel sides are 25cm, 13cm and other sides are 15cm, 15cm.
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Given:- ABCD Is a trapezium
AB = 25 cm
DC = 13 cm
AD & BC = 15 cm
Construction:- Draw CE || AD
To Find :- Area of trapezium ABCD
Solution :- 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 ∆ BCE
S = a + b + c/2
S = 15 + 15 + 12 /2
S = 21
Area of ∆ BCE = √ s( s - a)(s - b)( s - c )
Area of ∆ BCE = √ 21(21-15)(21-15)(21-12)
Area of ∆ BCE = √ 21 × 6 × 6 × 9
Area of ∆ BCE = 18√21 cm^2 -----1
h is the height of ∆ BCE
Area of BCE = 1/2 ( Base × Height )
= 1/2(12)(h)
= 6h -----2
From 1 & 2
6h = 18√21
=> h = 3√21 cm
The height of trapezium ABCD is equal to height of ∆ BCE.
Area of trapezium = 1/2 ( AB + CD ) × h
= 1/2 (25 + 13) × 3√21cm^2
= 57√21 cm^2
