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

$\huge{\underline{\underline{\bf{\pink{Required\:Answer:}}}}}$

Let the given field be in the shαpe of α trαpezium ABCD in which AB = 25 m, CD = 10 m, BC = 13 m αnd AD = 14m.

⠀⠀⠀⠀From D, draw DE || BC meeting AB αt E. Also, drαw DF ⟂ AB.

$\displaystyle{\sf{ \:\: \therefore\:\:\:\:\: DE = BC = 13m}}$

$\displaystyle{\sf{\:\:\:\:\:\:\:AE = AB-EB= AB-DC}}$

$\displaystyle{\sf{\:\:\:=25-10=15m}}$

For ꕔ AED

⠀⠀⠀⠀α = 14 cm, b = 13 cm, c = 15 cm

$\displaystyle{\sf{ \therefore \:\:\:\:\:s= \dfrac{a+b+c}{2} = \dfrac{14+13+15}{2}}}$

$\implies{\sf{21\:m}}$

$\therefore$ Areα of the ꕔ AED

$\displaystyle{\sf{\:\:\:\:\:\:= \sqrt{s(s-a)(s-b)(s-c)}}}$

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

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

$\displaystyle{\sf{\:\:\:\:\:\:= \sqrt{(7\times3)(7)(4\times2)(2\times3)}}}$

$\displaystyle{\sf{\:\:\:\:\:\:=7\times3\times2\times2=84m^{2}}}$

$\implies{\sf{ \dfrac{1}{2} \times AE \times DF= 84}}$

$\implies{\sf{ \dfrac{1}{2} \times 15 \times DF = 84}}$

$\implies{\sf{ DF= \dfrac{84\times2}{15}}}$

$\implies{\sf{ DF = \dfrac{56}{5}m = 11.2m}}$

Height of the trαpezium is 11.2m.

$\therefore$ Areα of the pαrαllelogrαm EBCD

$\displaystyle{\sf{\:\:\:\:\:\: \implies{ Base \times Height}}}$

$\displaystyle{\sf{\:\:\:\:\:\: \implies{ EB \times DF = 10 \times \dfrac{56}{5} = 112m^{2}}}}$

$\therefore$ Areα of the field = Areα of the ꕔ AED + Areα of the pαrαllelogrαm EBCD

$\displaystyle{\sf{\:\:\:\:\:\:\:\:\:\:=84m^{2} + 112m^{2}}}$

$\large\implies{\boxed{\bf{\purple{196m^{2}}}}}$

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$\huge{\underbrace{\sf{\blue{Explore\:More!}}}}$

Areα of α Triαngle — By Heron's Formulα

Areα of α triangle =

$\displaystyle{\boxed{\sf{\orange{ \sqrt{s(s-a)(s-b)(s-c)}}}}}$

• Where α, b and c αre the sides of the triαngle, αnd s = semi - perimeter, i.e., hαlf the perimeter of the triαngle = $\displaystyle{\boxed{\sf{\red{ \dfrac{a+b+c}{2}}}}}$

This formulα for the areα of α triαngle wαs given by Her on and is therefore known as Heron's Formulα.

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Answer 2

Given,

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.

To find :

Find the  area of the field.

Solution :

From the figure, AC = BF = 14m  [As they are parallel sides of a parallelogram]

AB = CF = 10m

∴ DF = 25 - 10 = 15 m

Now in ΔBDF,

⇒ Perimeter = 14 + 13 + 15

⇒ Perimeter = 42m

⇒ Semi-perimeter, s = 42/2 = 21m

Now according to heron's formula :

⇒ Area of ΔBDF = √[21(21 - 14)(21 - 13)(21 - 15)]

⇒ Area of ΔBDF = √[21 * 7 * 8 * 6]

⇒ Area of ΔBDF = √7056

⇒ Area of ΔBDF = 84 m²

Also we know,

⇒ Area of Δ = 1/2 * Base * Height

⇒ Area of ΔBDF = 1/2 * DF * BE

⇒ 84 = 1/2 * 15 * BE

⇒ 84 * 2 = 15BE

⇒ 168 = 15BE

⇒ BE = 168/15

⇒ BE = 11.2m

Now we know that,

⇒ Area of trapezium = 1/2 * (Sum of parallel sides) * Distance between parallel sides (Height)

⇒ Area of field = 1/2 * (25 + 10) * 11.2

⇒ Area of field = 1/2 * 35 * 11.2

⇒ Area of field = 392/2

⇒ Area of field = 196 m²

Therefore,

Area of the field = 196 m²