Question

Length of the fence of a trapezium shaped field ABCD is 120 m. If
BC=48 m, CD=17 m and AD= 40 m, find the area of this field. Side
AB is perpendicular to the parallel sides AD and BC.​

Answer 1

Step-by-step explanation:

Length of the fence = Perimeter

∴ AB + BC + CD + DA = 120

or AB + 48 + 17 + 40 = 120

or AB + 105 = 120

or AB = 120 - 105 = 15 m

∵ Area of a trapezium =

1/2×{sum of parallel lines)×height

=1/2×(AD+BC)×AB

=1/2×(40m+48m)×15m^2

=44m×15m^2

=660m^2

Answer 2

Question :-

Length of the fence of a trapezium-shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD= 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Given :-

• Length of ABCD = 120m

• BC = 48m

• CD = 17m

• AD = 40m

• AB is perpendicular to the parallel sides AD.

To find :-

• Area of field ?¿

Solution :-

$➠$ AB + BC + CD + DA = 120 m .

$➠$ BC = 48 m, CD = 17 m, AD = 40 m

$➠$ AB = 120 m – (48 m + 17 m + 40 m)

= 120 – 105 m

= 15 m

_________

Area of the trapezium ABCD

$= \: \: \: \: \: \: \: \dfrac{1}{2} × (BC + AD) × AB$

$= \: \: \: \: \: \: \: \dfrac{1}{2} × (48 + 40) × 15$

$= \: \: \: \: \: \: \dfrac{1}{2} \times 88 \times 15$

$= \: \: \: \: \: \: 44 \times 15$

$= \: \: \: \: \: 660 {m}^{2}$

hence, required area = $660 {m}^{2}$