Question

Two cross-roads 5m wide run parallel to the sides of a rectangular garden of length 45 m and width
40 m. Find the area of the cross-roads and also the cost of constructing these cross-roads at the rate
of 100 per square metre​

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)

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

$\fbox \orange{ Cost of cleaning the field is Rs. 3257.1 }$

Answer 2

Answer:

Solution:

$\huge \color{green} \tt{ verified✔︎ }$

Let ABCD be the rectangular park then EFGH and IJKL are the two rectangular roads with width 5m.

Given that length of rectangular park = 70m

Breadth of rectangular park = 45m

Area of the rectangular park = Length x Breadth

= 70 m x 45 m

= 3150 m²

Area of the road EFGH = 70 m x 5 m

= 350 m²

Now, Area of the road JKLI = 45 m x 5 m

= 225 m²

From the figure, it is clear that area of MNOP is common to the two roads.

Thus, Area of MNOP = 5 m x 5 m = 25 m²

Therefore,

Area of the roads = Area of EFGH + Area of JKLI – Area of MNOP

= (350 + 225) m2– 25 m²

= 550 m²

Again, it is given that the cost of constructing the roads = Rs. 105 per m²

Therefore,

Cost of constructing 550 m² area of the roads

= Rs. (105 x 550)

= Rs. 57750

$\huge\mathfrak\pink{❥︎ Sanchuu}$