Question

A field in the shape of trapezium whose parallel sides are 25m and 10m and non parallel sides a 14 and 13 the area of the field?​

Answer 1

Answer:

ANSWER

Let ABCD be a trapezium with,

AB∥CD

AB=25m

CD=10m

BC=14m

AD=13m

Draw CE∥DA. So, ADCE is a parallelogram with,

CD=AE=10m

CE=AD=13m

BE=AB−AE=25−10=15m

In ΔBCE, the semi perimeter will be,

s=

2

a+b+c

s=

2

14+13+15

s=21m

Area of ΔBCE,

A=

s(s−a)(s−b)(s−c)

=

21(21−14)(21−13)(21−15)

=

21(7)(8)(6)

=

7056

=84m

2

Also, area of ΔBCE is,

A=

2

1

×base×height

84=

2

1

×15×CL

15

84×2

=CL

CL=

5

56

m

Now, the area of trapezium is,

A=

2

1

(sumofparallelsides)(height)

A=

2

1

×(25+10)(

5

56

)

A=196m

2

Therefore, the area of the trapezium is 196m

2

.

Answer 2

$\bf\underline{\underline{\pink{Question:-}}}$

★ A feild in the shape of trapezium whose parallel sides are 25m and 10m and non parallel sides a 14 and 13 find the area of the field.

$\bf\underline{\underline{\blue{Solution:-}}}$

Let ABCD be the field in the form of trapezium in which AB||CD such that

AB = 25m, BC = 13m

CD = 10m, DA = 14m

Draw CE||DA and CF⊥EB

Clearly, ABCD is a Parallelogram.

∴ CE = DA = 14m and AE = CE = 10m

∴ EB = AB – AE = (25 – 10)m = 15m

In ∆EBC, we have

EB = 15m, BC = 13m and CE = 14m

∴ a = 15m, b= 13m, c = 14m

∴ s = ½(15+13+14)m = 21m

∴ (s–a) = (21–15)m = 6m, (s–b) = (21–13)m = 8m, (s–c) = (21–14)m = 7m

∴ Area of ∆EBC = √s(s–a)(s–b)(s–c)

= √21×6×8×7 m²

= √7×3×3×2×2×2×2×7 m²

= (7×3×2×2)m²

= 84 m²

Also, area of ∆EBC = (½×EB×CF)

= (½×15×CF)

∴ ½×15×CF = 84m² ==> CF = $\frac{84×2}{15}$m = $\frac{56}{5}$m = 11.2 m.

∴CF = 11.2m

Area of trapezium ABCD = ½ (AB + CD)×CF

= {½(25+10)×11.2}m²

= (35×5.6)m² = 196m²

Hence, the area of trapezium ABCD is 196 m²

$\bf\underline{\underline{\green{Extra\:Formulas:-}}}$

❥Area of a Triangle = (½×base×height)sq. units

❥Heron's Formula

Let a, b, c be the sides of triangle. Then,

semi-perimeter, s = ½(a+b+c);

Area = √s(s–a)(s–b)(s–c) sq units.

❥Let each side of an equilateral triangle be a. Then,

area = ( $\frac{√3}{4}$×a²) sq units, and height = ( $\frac{√3}{2}$×a)

❥Consider an isosceles triangle having base = b and each of equal side = a Then,

area = ( $\frac{b}{4}$×√4a²–b²)sq units