Question

a field is in the shape of a trapezium whose parallel sides are 25 meter and 10 meter . the non parallel sides are 14 meter and 13 meter . 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

•°• $\sf{\underline {Area\:of\:trapezium \:is\:196\:m^2}}$

Answer 2

$\huge\star{\tt{\underline{\underline{\red{Answer}}}}}\star$

See the attachment for figure

In the parallel sides

The Longer one will be 25cm(CD)

while the Shorter one will be 10cm(AB)

and we can select other sides as any of the two left

AC=13cm

BD=14cm

$\large{\orange{\underline{\tt{Formula:}}}}$

Divide the Trapezium in triangle and a Parallelogram

By drawing a Line parallel to AC We could do this

(See the 2nd attachment)

and name the Point as E

Now,

Area of Trapezium = Area of ∆ BED + Area of ABCE

Area of ∆ BED

By Heron's Formula

S= $\large{\frac{BE + ED+ BD}{2}}$

Solution :

Semi Perimeter = $\large{\frac{13 + (25-10) + 14}{2}}$

$\implies$ = $\large{\frac{42}{2}}$

$\implies$ S= 21cm

Now,

$\large{\sqrt{S(S-BE)(S-ED)(S-BD)}}$

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

$\implies$ $\large{\sqrt{21 x 8 x 6 x 7}}$

$\implies$ $\large{\sqrt{7056}}$

$\implies$ $\large{\boxed{\boxed{84 m{}^{2}}}}$✰1

For height of the Trapezium

84= ½ x b x h

$\implies$ h= 168 / 15

$\implies$ ${\boxed{h=11.2cm}}$

Area of Parallelogram ABCE

Area = Base x Height

= 10 x 11.2

$\implies$ $\large{\boxed{\boxed{112m{}^{2}}}}$✰2

Area of Trapezium

✰1 + ✰2

=84 + 112

= $\huge{\boxed{\boxed{196m{}^{2}}}}$

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