Question

Team Brainly, a small help needed!

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.
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NCERT IXth
[Heron’s Formula]

Good quality answer required :)

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Answer 1

196m'2 is the answer

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Answer 2

Given:-

☞ Field is in the shape of Trapezium.

☞ Parallel sides are of 25 m and 10 m.

☞ Non parallel sides are of 14 m and 13 m.

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To Find:-

✯ Area of Trapezium

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Construction:-

✯ CF || AD

✯ CG perpendicular to AB

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Solution:-

In quadrilateral ADCF,

CF || AD (By construction)

CD || AF (ABCD is a trapezium)

As Quadilateral ADCF had two pairs of parallel sides,

Therefore, ADCF is a parallelogram.

As we know that opposite sides of parallelogram are equal,

Therefore, AD = CF = 13 m and CD = AF = 10 m

As, AB = AF + BF

Therefore,

BF = AB - AF

BF = 25 - 10

$\boxed {BF = 15 \: m}$

Here, The sides of the triangle BCF are:-

✯ a = CF

☞ a = 13 m

✯ b = CB

☞ b =14 m

✯ c = BF

☞ c = 15 m

Therefore, semi-perimeter of the triangle is:-

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

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

s = $\frac{42}{2}$

$\boxed {s = 21 \: m}$

Therefore, by using Heron's Formula, Area of triangle BCF:-

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

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

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

Area = $\sqrt{7056}$

$\boxed {Area = 84 \: m²}$

We can also write Area of triangle as:-

Area = $\frac{1}{2}\times Base\times Height$

Area = $\frac{1}{2}\times BF\times CG$

84 m² = $\frac{1}{2}\times BF\times CG$

84 m² = $\frac{1}{2}\times 15\times CG$

CG = $\frac{84\times2}{15}$

$\boxed {CG= 11.2 \: m}$

Therefore, area of the trapezium ABCD is:-

Area = $\frac{1}{2} \times (AB+CD)\times CG$

Area = $\frac{1}{2} \times (25+10)\times 11.2$

Area = $35\times5.6$

$\boxed {Area = 196 \: m²}$

Therefore, area of the trapezium field is 196 m².