Question

Find the area of trapezium whose parallel sides are 25 CM, 13 CM and the other sides are 15cm each

Answer 1

Let ABCD be a trapezium
Let E is point at which perpendicular from A meet CD
Let F is point at which perpendicular from B meet CD
as ABFE is a rectangle so, AB=FE=13 cm
Let DF and FC  be equal to x
  DF=x+x+13=25
=>x=6 cm = DE = FC
by pythagorous theorem in triangle AED
AE^2 + DE^2= AD^2
AE^2 + 6^2 =15^2
=> AE =3*sqrt(21)
area of triangle AED=1/2* base*height
                                =1/2*6*3*sqrt(21)= 9*sqrt(21)
similarly , area of triangle BFC= 9*sqrt(21)
 area of rectangle ABFE = a * b =13* 3* sqrt (21)
      Area of trapezium =area of triangle AED+ area of rectangle ABFE +area of triangle BFC
                                 =9 *sqrt (21) + 39 * sqrt (21) + 9*sqrt(21)
                                 = 57 * sqrt(21) =261.21 square cm

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