Question

•→A circle field of diameter 42 m needs to be cleaned. if the cost of cleaning the field is Rs. 2.35 per m², find the cost of cleaning the field?​

Answer 1

${\orange{\bigstar}} \ {\underline{\green{\textsf{\textbf{Given :-}}}}}$

Diameter of the circle = 42 m

Cost of cleaning per m² = ₹2.35

${\blue{\bigstar}} \ {\underline{\pink{\textsf{\textbf{To Find :-}}}}}$

Cost of cleaning the whole field

${\red{\bigstar}} \ {\underline{\purple{\textsf{\textbf{Formula Used :-}}}}}$

${\boxed{\green{\textsf{\textbf{Area of a circle = }}} {\blue{\sf{\pi r^2}}}}}$

where,

r = Radius

${\sf{\pi = \dfrac{22}{7}}}$

${\orange{\bigstar}} \ {\underline{\blue{\textsf{\textbf{Solution :-}}}}}$

$Radius = {\sf{\dfrac{Diameter}{2}}}$

$\longmapsto {\sf{\dfrac{42}{2}}}$

$\longmapsto {\sf{21 \ m}}$

${\pink{\textsf{\textbf{Radius = 21 m}}}}$

According to the question by using the formula of Area of a Circle, we get,

$\dashrightarrow \ {\green{\sf{Area \ of \ circular \ field = (\pi \times 21^2) \ m^2}}}$

Solving the above equation,

$: \ \Longrightarrow \ {\sf{ \bigg ( \dfrac{22}{7} \times 21^2 \bigg ) \ m^2}}$

$: \ \Longrightarrow \ {\sf{ \bigg ( \dfrac{22}{7} \times 21 \times 21 \bigg ) \ m^2}}$

$: \ \Longrightarrow \ {\sf{ \bigg ( \dfrac{22}{7} \times 441 \bigg ) \ m^2}}$

$: \ \Longrightarrow \ {\sf{ \bigg ( \dfrac{22}{{\cancel{7}}^{ \ 1}} \times {\cancel{441}}^{ \ 63} \bigg ) \ m^2}}$

$: \ \Longrightarrow \ {\sf{(22 \times 63) \ m^2}}$

$: \ \Longrightarrow \ {\sf{1,386 \ m^2}}$

${\blue{\textsf{\textbf{Area of the circular field = 1,386 sq. m.}}}}$

Cost of cleaning the field = ₹ (1,386 × 2.35)

$: \ \Longrightarrow \ {\sf{\purple{Rs. \ 3257.1}}}$

${\boxed{\orange{\textsf{\textbf{Cost of cleaning the field is Rs. 3257.1}}}}}$

Answer 2

Radius = {\sf{\dfrac{Diameter}{2}}}

2

Diameter

\longmapsto {\sf{\dfrac{42}{2}}}⟼

2

42

\longmapsto {\sf{21 \ m}}⟼21 m

{\pink{\textsf{\textbf{Radius = 21 m}}}}Radius = 21 m

According to the question by using the formula of Area of a Circle, we get,

\dashrightarrow \ {\green{\sf{Area \ of \ circular \ field = (\pi \times 21^2) \ m^2}}}⇢ Area of circular field=(π×21

2

) m

2

Solving the above equation,

: \ \Longrightarrow \ {\sf{ \bigg ( \dfrac{22}{7} \times 21^2 \bigg ) \ m^2}}: ⟹ (

7

22

×21

2

) m

2

: \ \Longrightarrow \ {\sf{ \bigg ( \dfrac{22}{7} \times 21 \times 21 \bigg ) \ m^2}}: ⟹ (

7

22

×21×21) m

2

: \ \Longrightarrow \ {\sf{ \bigg ( \dfrac{22}{7} \times 441 \bigg ) \ m^2}}: ⟹ (

7

22

×441) m

2

: \ \Longrightarrow \ {\sf{ \bigg ( \dfrac{22}{{\cancel{7}}^{ \ 1}} \times {\cancel{441}}^{ \ 63} \bigg ) \ m^2}}: ⟹ (

7

1

22

×

441

63

) m

2

: \ \Longrightarrow \ {\sf{(22 \times 63) \ m^2}}: ⟹ (22×63) m

2

: \ \Longrightarrow \ {\sf{1,386 \ m^2}}: ⟹ 1,386 m

2

{\blue{\textsf{\textbf{Area of the circular field = 1,386 sq. m.}}}}Area of the circular field = 1,386 sq. m.

Cost of cleaning the field = ₹ (1,386 × 2.35)

: \ \Longrightarrow \ {\sf{\purple{Rs. \ 3257.1}}}: ⟹ Rs. 3257.1