Question

A field is in the shape of a trapezium whose parallel
sides are 25 m and 10 m. The non-parallel sides
are 14 m and 13 m. Find the area of the field.​

Answer 1

Given :

A trapezium such that whose parallel sides are 25 m and 10 m. And non parallel sides are 14 m and 13 m.

Find :

Area of the field (trapezium).

Construction :

Draw a perpendicular from B on CE.

Assume :

Let ABCD a trapezium such that AD = 14 m, CD = AB = CD = 10 m, DE = 25 mand BE = 13 m respectively.

Solution :

CE = DE - DC

CE = 25 - 10 = 15 m

Semi-perimeter of ΔBCE = Sum of sides/2

s = (a + b + c)/2

Here.. a = BC, b = CE and c = EB

s = (BC + CE + EB)/2

s = (14 + 15 + 13)/2

s = 42/2

s = 21 m

From Heron's formula

Area of ΔBEC = $\sqrt{s(s - a)(s - b)(s - c)}$

= $\sqrt{21(21 - 14)(21 - 15)(21 - 13)}$

= $\sqrt{21(7)(6)(8)}$

= 84 m² _____ (eq 1)

Area of ΔBCE

= 84 = 1/2 × b × h

= 84 = 1/2 × 15 × h

= 84 × 2 = 15h

= h = 11.2 m

(h = BF)

Area of parallelogram ABCD = b × h

=> 10 × 11.2

=> 112 m² ____ (eq 2)

Area of trapezium AEBD = (eq 1) + (eq 2)

=> 84 + 112

=> 196 m²

∴ Area of trapezium is 196 m².

Answer 2

Answer:

Let ABCD be a trapezium with,

AB∥CD

AB=25m

CD=10m

BC=14m

AD=13m

Draw CE∥DA. So, ADCE is a parallelogram with,

CD=AE=10m

CE=AD=13m

BE=AB−AE=25−10=15m

In ΔBCE, the semi perimeter will be,

$s = \frac{a + b + c}{2}$

$s = \frac{13 + 14 + 15}{2}$

s = 21

Area of ΔBCE,

$A = \sqrt{(s−a)(s−b)(s−c)</p><p>} </p><p></p><p></p><p>$

$\sqrt{21(21−14)(21−13)(21−15)}$

$\sqrt{</p><p>21(7)(8)(6)</p><p>}$

$\sqrt{</p><p>7056}$

= 84m²

Also, area of ΔBCE is

$A= \frac{1}{2} ×base×height$

$84= \frac{1}{2} ×15×CL$

$CL= \frac{56}{5}m$

Now, the area of trapezium is,

$A= \frac{1}{2} (sum \: of \: parallel \: sides)(height)$

$A= \frac{1}{2} ×(25+10)(556)$

=> A=196m²

Therefore, the area of the trapezium is 196m².