Question

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

Answer 1

Answer:

Question :-

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

$\bf \large\underbrace\red{Answer:}$

Given :-

• A field is in the shape of trapezium ABCD whose parallel sides are AB & CD are 25 m and 10 m. The non parallel sides are AD & BC are 14 m and 13 m.

To Find :-

• Area of the field.

Construction:-

• Through C, draw a line parallel CE || AD, intersects AB at E and let CF, h be the height.

Solution :-

ABCD is a trapezium with AB || CD

AB = 25 m

CD = 10 m

AD = 14 m

BC = 13 m

Now, AB || CD & CE || AD

So, it implies, ADCE is a parallelogram.

So, AD = CE = 14 m

Now, CD = AE = 10 m

Therefore, EB = AB - AE = 25 - 10 = 15 m

Consider triangle BEC, we have with

EB (a) = 15 m

BC (b) = 13 m

EC (c) = 14 m

Perimeter of Triangle BEC = a + b + c = 15 + 13 + 14 = 42 m

$\underline{\boxed{\sf Semi \ perimeter \: (s)=\dfrac{a + b + c}{2}}}$

So, semi perimeter, s = 42/2 = 21 m.

Area of triangle, BEC

$\underline{\boxed{\sf Area \ of \ triangle=\sqrt{s(s-a)(s-b)(s-c)} }}$

$\bf\implies \: \sqrt{21(21 - 15)(21 - 13)(21 - 14)}$

$\bf\implies \: \sqrt{21 \times 6 \times 7 \times 8}$

$\bf\implies \: \sqrt{7 \times 3 \times 3 \times 2 \times 7 \times 2 \times 2 \times 2}$

$\bf\implies \:7 \times 3 \times 2 \times 2 = 84 \: {m}^{2}$

Also, Area of triangle BEC = 1/2 × BE × h

where, h is the height of triangle BEC to base BE.

$\bf\implies \:84 = \dfrac{1}{2} \times 15 \times h$

$\bf\implies \:h = \dfrac{168}{15} \: m$

$\bf\implies \:Area \: of \: trapezium \: = \dfrac{1}{2} (AB + CD) \times h$

$\bf\implies \:Area \: of \: trapezium = \dfrac{1}{2} \times (25 + 10) \times \dfrac{168}{15}$

$\bf\implies \:Area \: of \: trapezium = 196 \: {m}^{2}$