DAV Class 7 Maths Chapter 11 Worksheet 5

$$\begin{array}{rl}\text{Radius of the circle}& =7cm\\ \text{Circumference of a circle}& =2\pi r\\ \text{Circumference}& =2\times \frac{22}{7}\times 7\\ & =2\times \frac{22}{{\cancel{7}}^1}\times {\cancel{7}}^1\\ & =2\times 22\\ & =44\text{cm}\end{array}$$

Answer The circumference of the circle $=44cm$

$$\begin{array}{rl}\text{Circumference of the circle}& =66\text{m}\\ \text{Circumference}& =\pi \times d\\ \pi \times d& =66m\\ \frac{22}{7}\times d& =66m\\ d& ={\cancel{66}}^{3}m\times \frac{7}{{\cancel{22}}^1}\\ & =3m\times 7\\ d& =21m\end{array}$$

Answer The diameter of the circle $=21m$

$$\begin{array}{rl}\text{Circumference of the circle}& =176\text{cm}\\ 2\pi r& =176\\ 2\times \frac{22}{7}\times r& =176\\ r& =\frac{176\times 7}{2\times 22}\\ \\ & =\frac{{\cancel{176}}^8\times 7}{2\times {\cancel{22}}^1}\\ \\ & =\frac{{\cancel{8}}^4\times 7}{{\cancel{2}}^1\times 1}\\ \\ & =4\times 7\\ r& =28cm\end{array}$$

Answer The radius of the circle $=28cm$

$$\begin{array}{rl}\text{Diameter of the wheel }(d)& =1.4\text{m}\\ \text{Circumference}& =\pi d\\ & =\frac{22}{{\cancel{7}}^1}\times {\cancel{1.4}}^{0.2}\\ & =22\times 0.2\\ & =4.4m\end{array}$$

Answer The circumference of the wheel $=4.4m$

$$\begin{array}{rl}\text{Let the radii of first circle be }{r}_1& =2x\\ \text{Let the radii of second circle be }{r}_2& =3x\\ \text{The circumference of the first circle }(C_1)& =2\pi (r_1)\\ & =2\pi (2x)\\ C_1& =4\pi x\\ \text{The circumference of the second circle }(C_2)& =2\pi (r_2)\\ & =2\pi (3x)\\ C_2& =6\pi x\\ \text{The ratio of their circumferences}& =C_1:C_2\\ & =\frac{C1}{C2}\\ & =\frac{4\pi x}{6\pi x}\\ & =\frac{4}{6}\\ & =\frac{2}{3}\\ & =2:3\end{array}$$

Answer The ratio of their circumferences $=2:3$

$$\begin{array}{rl}\text{Distance from the Earth to the Moon (radius)}& =3,85,000\text{km}\\ \text{Circumference}& =2\pi r\\ & =2\times \frac{22}{7}\times 385000\\ \\ & =2\times \frac{22}{{\cancel{7}}^1}\times {\cancel{385000}}^{55000}\\ \\ & =2\times 22\times 55000\\ & =44\times 55000\\ & =24,20,000\text{km}\end{array}$$

Answer The moon travels $24,20,000km$ in one month.

$$\begin{array}{rl}\text{Diameter of the wheel}& =35\text{cm}\\ \text{Number of revolutions}& =1000\\ \text{Distance covered in 1 revolutions}& =\text{Circumference}\\ \text{Circumference}& =\pi d\\ \\ & =\frac{22}{{\cancel{7}}^1}\times {\cancel{35}}^{5}cm\\ \\ & =22\times 5cm\\ \text{Distance covered in 1 revolutions}& =110\text{cm}\\ \\ \text{Distance covered in 1000 revolutions}& =1000\times \text{Circumference}\\ & =1000\times 110\\ & =110000\text{cm}\\ \\ \text{Convert cm to km}& ⟹1cm=\frac{1}{100000}km\\ \\ 110000cm& =\frac{110000}{100000}km\\ \\ & =1.1km\end{array}$$

Answer Distance covered by the wheel in 1000 revolutions $=1.1km$

$$\begin{array}{rl}\text{Radius of the wheel}& =0.70m\\ \text{Distance covered in 1 revolution}& =\text{Circumference}\\ \text{Circumference}& =2\pi r\\ & =2\times \frac{22}{{\cancel{7}}^1}\times {\cancel{0.70}}^{0.10}m\\ & =2\times 22\times 0.10m\\ & =44\times 0.10m\\ & =4.40m\\ \text{Circumference}& =4.4m\\ \\ \text{Total distance}& =22km\\ \\ \text{Convert km to m}& ⟹1km=1000m\\ & =22\times 1000\\ \text{Total distance}& =22000m\\ \\ \text{Number of revolutions}& =\frac{\text{Total distance}}{\text{Circumference}}\\ \\ \text{Revolutions}& =\frac{22000}{4.4}\\ \\ & =\frac{22000}{4.4}\times \frac{10}{10}\\ \\ & =\frac{{\cancel{220000}}^{20000}}{{\cancel{44}}^4}\\ \\ & =\frac{20000}{4}\\ \\ & =5000\end{array}$$

Answer Number of revolutions covered in 22 km $=5000$

$$\begin{array}{rl}Radius& =42\text{cm}\\ \text{Perimeter of the square}& =\text{Circumference of the Circle}\\ 4\times Side& =2\pi r\\ \\ 4\times Side& =2\times \frac{22}{{\cancel{7}}^1}\times {\cancel{42}}^6\\ \\ 4\times Side& =44\times 6\\ \\ Side& =\frac{{\cancel{44}}^{11}\times 6}{{\cancel{4}}^1}\\ \\ Side& =11\times 6\\ \\ Side& =66\text{cm}\end{array}$$

Answer Sides of the square $=66cm$

$$\begin{array}{rl}\text{Diameter of the park}& =140\text{m}\\ \text{Radius}& =\frac{d}{2}\\ \text{Radius of the Inner circle }(r_1)& =70\text{m}\\ \text{Width of the path}& =7\text{m}\\ \text{Radius of the Outer cicle }(r_2)& =70\text{m}+7\text{m}\\ (r_2)& =77m\\ \\ \text{Circumference of the inner circle}& =2\pi (r_1)\\ \\ & =2\times \frac{22}{{\cancel{7}}^1}\times {\cancel{70}}^{10}m\\ \\ & =44\times 10m\\ & =440m\\ \text{Circumference of the outer circle}& =2\pi (r_2)\\ \\ & =2\times \frac{22}{{\cancel{7}}^1}\times {\cancel{77}}^{11}m\\ \\ & =44\times 11m\\ & =484m\\ \text{Total length to be fenced}& =440m+484m\\ & =924m\\ \text{Cost of fencing per meter}& =\text{₹ }7\\ \text{Cost}& =924\times 7\\ & =\text{₹ }6468\end{array}$$

Answer The total cost of fencing $=\text{₹ }6468$

$$\begin{array}{rl}\text{Diameter}& =280\text{m}\\ \text{Distance covered in 1 revolution}& =\text{Circumference of the circle}\\ \text{Circumference of the circle}& =\pi d\\ \\ & =\frac{22}{{\cancel{7}}^1}\times {\cancel{280}}^{40}\\ \\ & =22\times 40\\ & =880m\\ \text{Distance covered in 10 revolution}& =10\times \text{Circumference}\\ & =10\times 880\\ & =8800\text{m}\\ \text{Convert }& m\text{ to }km\\ \\ 1m& =\frac{1}{1000}km\\ \\ 8000m& =\frac{8800}{1000}\text{km}\\ \\ & =8.8\text{km}\\ \end{array}$$

Answer Distance covered by the athlete $=8.8km$

$$\begin{array}{rl}\text{Side}& =11cm\\ \text{Circumference of circle}& =\text{Perimeter of rhombus}\\ 2\pi r& =4\times \text{side}\\ \\ 2\times \frac{22}{7}\times r& =4\times 11cm\\ \\ \frac{44}{7}\times r& =44cm\\ \\ r& ={\cancel{44}}^{1}cm\times \frac{7}{{\cancel{44}}^1}\\ \\ r& =7\text{cm}\\ \text{Diameter}& =2\times r\\ & =2\times 7cm\\ \text{Diameter}& =14cm\end{array}$$

Answer The diameter of the circular piece $=14cm$

$$\begin{array}{rl}\text{Let inner radius }& =r_1\\ \text{ outer radius }& =r_2\\ \\ \text{Inner circumference }(C_1)& =352m\\ 2\pi r_1& =352m\\ 2\times \frac{22}{7}\times r_1& =352\\ r_1& =\frac{352\times 7}{2\times 22}\\ r_1& =\frac{{\cancel{352}}^{16}\times 7}{2\times {\cancel{22}}^1}\\ r_1& =\frac{{\cancel{16}}^8\times 7}{{\cancel{2}}^1\times 1}\\ r_1& =8\times 7\\ r_1& =56m\\ \\ \text{Outer circumference}(C_2)& =396m\\ 2\pi r_2& =396m\\ 2\times \frac{22}{7}\times r_2& =396\\ r_2& =\frac{396\times 7}{2\times 22}\\ r_2& =\frac{{\cancel{396}}^{18}\times 7}{2\times {\cancel{22}}^1}\\ r_2& =\frac{{\cancel{18}}^9\times 7}{{\cancel{2}}^1\times 1}\\ r_2& =9\times 7\\ r_2& =63m\\ \text{Width of the track}& =\text{outer radius}-\text{inner radius}\\ & =r_2-r_1\\ & =63m-56m\\ & =7m\end{array}$$

Answer Width of the track $=7m$

$$\begin{array}{rl}\text{Radius (r)}& =63\text{cm}\\ \text{Length of rope}& =\text{Circumference}\\ \text{Circumference}& =2\pi r\\ & =2\times \frac{22}{{\cancel{7}}^1}\times {\cancel{63}}^{9}cm\\ & =2\times 22\times 9cm\\ & =44\times 9cm\\ & =396\text{cm}\end{array}$$

Answer Least length of rope required $=396cm$