Question

2. A cow is tethered with a rope of length 35 m at the

centre of the rectangular field of length 76 m and

breadth 60 m. Find the area of the land that the cow

cannot graze?​

Answer 1

GivEn:

• Length of rope = 35 m
• Length of Rectangular field = 76 m
• Breadth of Rectangular field = 60 m

To find:

• Area of the land that the cow cannot graze?

Solution:

Here,

• Shape of field is Rectangular.

• Area of grazed land = Area of circle.

★ According to the Question:

★ Area of land that the cow can't graze = Area of rectangular field - Area of circular grazed land

→ (length × breadth) - πr²

Here,

• Length of field = 76 m
• Breadth of field = 60 cm
• Length of rope (Radius) = 35 cm

Therefore,

→ (76 × 60) - 22/7 × 35 × 35

→ 4560 - 22 × 5 × 35

→ 4560 - 3850

→ 710 m²

Hence, area of the land that the cow cannot graze is 710 m².

Answer 2

Answer:

$\huge \bf \: solution$

• Length of rope = 35 m
• Length of rectangular field = 76 m
• Breadth of rectangle = 60 m

For this we have to find the area of rectangle as well as area of circular rope.

Formulas :-

$\huge \bf \: area \: of \: field = l \times b$

$\huge \bf \: area \: of \: rope = \pi \: {r}^{2}$

First we will find area of field

_______________________

$\sf \implies \: area \: of \: field = l \times b$

$\sf \implies\: area \: of \: field \: = 76 \times 60$

$\sf \implies \: area \: of \: field \: = 4560 \: m$

______________________

$\sf \implies \: area \: of \: rope = \pi \: {r}^{2}$

$\sf \implies \: area \: of \: rope \: = \frac{22}{7} \times 35 \times 35$

$\sf \implies \: area \: of \: rope \: = 22 \times 5 \times 35$

$\sf \implies \: area \: of \: rope \: = 3850 \: m$

_____________________

Remaining area = area of field - area of rope

$\sf \implies \: remaining \: area \: = 4560 - 3850$

$\sf \implies \: remaining \: area \: = 710 \: m$