Question

find the area of trapezium whose parallel sides 25 cm ,13 cm and other sides are 15 cm and 15 cm

Answer 1

Let AB = 25 cm , CD = 13 cm and BC = CD = AD cm

Draw perpendiculars from C and D to AB meeting AB at F and E respectively

Now A E = F B = 6 cm

In triangle AED,

AE2+DE2=AD2AE2+DE2=AD2

DE2=AD2−AE2DE2=AD2−AE2

=102−62=64=102−62=64

DE=8cmDE=8cm

Area of the Trapezium = 1/2*h(a+b) sq cm

Area of the Trapezium = 1/2*8(25+13)sq cm =152 sq 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