Question

find the area of trapezium whose parallel sides are 25 cm and other sides are 15 cm and 15 cm

Answer 1

hii Here area of trapezium = Area of parallelogram + area of triangle

Also area of triangle = (1/2) × base × height

Hence height of parallelogram is also 12 cm
Area of parallelogram = base × height
= 11 × 12 = 132 sq cm
Hence area of trapezium = 84 + 132 = 216 sq cm mark me brainlist

Answer 2

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