Question

a field is in thr of a trapezium whose parallel sides are 25m and 10 m the non parallel sides are 14 m and 13m. find the area of the field

Answer 1

Here's your answer!!

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It's given that,

Length of parallel sides are 25cm and 10cm.

And length of non-parallel sides are 14 cm and 13 cm.

We have to find it's area , to find the area of Trapezium we must have sum of parallel sides and height of Trapezium.

We have parallel sides , but we have to find height.

So,

We will first drawn AE // BC
=> AE =BC= 13 cm

Now,

In ∆ ADE,
=>AD =14 cm (Given)
=>AE =13 cm

And,

=>DE=DC-EC
$= > 25 - 10 \\ = > 15cm$

Therefore,

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

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

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

$= > s = 21$

Therefore,

Area of ∆ADE =

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

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

$= > \sqrt{21 \times 7 \times 8 \times 6}$

$= > \sqrt{7 \times 3 \times 7 \times 2 \times 2 \times 2 \times 3 \times 2}$

$= > 7 \times 3 \times 2 \times 2$

$= > 84$

Hence,

Area of ∆ ADE = 84 cm^2

$= > \frac{1}{2} \times base \times height = 84 {cm}^{2}$

$= > \frac{1}{2} \times 15 \times height = 84$

So,

$= > height = \frac{84 \times 2}{15}$

$= > height = \frac{168}{15} = \frac{56}{5} = 11.2$
Hence,

Height of Trapezium= 11.2cm

Now, we can easily find the area of Trapezium.

Area of Trapezium=

$= > \frac{1}{2} \times (sum \: of \: parallel \: side) \times height$

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

$= > \frac{1}{2} \times 35 \times 11.2$

$= > 196 \: {cm}^{2}$

Hence,

Area of Trapezium is 196 cm^2

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Hope it helps you!! :)

Answer 2

Since,

there is a Trapezium,

whose one parallel side (CD) = $\textsf{10 m}$

and, another parallel side (AB) = $\textsf{25 m}$

also, one of the non parallel side (AD) = $\textsf{14 m}$

another non parallel side (BC) = $\textsf{13 m}$

$\textbf{Construction}$ => Draw CE || AD and CF $\perp$AB .

so, AECD is a parallelogram. (by Construction)

$\therefore$ AD = CE = $\textsf{14 m}$

and, CD = AE = $\textsf{10 m}$

then,
$\textsf{BE = AB - AE}$
$\textsf{BE = 25m - 10m}$
$\textsf{BE = 15m }$

and, in ∆BCE ,

let,
a = BC = 13m ,
b = CE = 14m ,
c = BE = 15m .

so,

s (semi-perimeter) = $\frac{sum\: of\: all\: sides}{2}$

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

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

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

s = $\textsf{ 21 m }$

Hence, by Heron's formula,

$= \sqrt{s(s - a)(s - b)(s - c)} \\ \\ = \sqrt{21(21 - 13)(21 - 14)(21 -15 )} \\ \\ = \sqrt{21 \times 8 \times 7 \times 6}$

= $\sqrt{ 3 \times 7 \times 2 \times 2 \times 2 \times 7 \times 3 \times 2}$

= 3 × 7 × 2 × 2

= 84 m²

we know that,

area of triangle = $\frac{1}{2}$ × base × height

=> 84 m² = $\frac{1}{2}$ × EB × CF

=> 84 m² = $\frac{1}{2}$ × 15 × CF

=> $\frac{84 \times 2}{15}$ = CF

=> CF = 11.2 m = height.

thus,

area of trapezium,

= $\frac{1}{2}$ × sum of parallel sides × height

= $\frac{1}{2}$ × ( 10 + 25 ) × 11.2

= $\frac{1}{2}$ × 35 × 11.2

= $\mathsf{ 196 \: m^2}$