From the question, it is given that, Length of the park $(L) = 700\ m$Breadth of the park $(B) = 300\ m$Then, Area of the park $= length\times breadth$$= 700\times 300$$= 210000\ m^2$Let us assume that $ABCD$ is the one crossroad and $EFGH$ is another crossroad in the park.The length ... Read More

Side of the square sheet$=6\ m$Area of the sheet$=(Side)^2=(6)^2=36\ cm^2$The radius of the circle$=2\ cm$Area of the circle to be cut out$=\pi r^2$$=\frac{22}{7}\times2\times2=\frac{88}{7}\ cm^2$Area of the leftover sheet$=36\ cm^2-\frac{88}{7}\ cm^2$$=\frac{252-88}{7}\ cm^2=\frac{164}{7}\ cm^2$$=23.44\ cm^2$

Circumference of the circle$=31.4\ cm$$2\pi r=31.4$$r=\frac{31.4}{2\times3.14}=5\ cm$​Area of the circle$ = 7\pi r^2=3.14\times5\times5=78.5\ cm^2$Hence, the required radius$=5\ cm$ and area$=78.5\ cm^2$

Diameter of the flower bed$=66\ m$Radius$=\frac{66}{2}=33\ m$ Width of the path$=4\ m$The radius of the flower bed included a path$=33\ m+4\ m=37\ m$Let $r_2=37\ m$Area of the circular path$=\pi\left(r_2^2-r_1^2\right)$$=3.14\left(37^2-33^2\right)$$=3.14\times(37+33)(37-33)$     [$a^2-b^2=(a+b)(a-b)$]$=3.14\times70\times4=879.20\ m^2$Hence, the required area$=879.20\ m^2$Read More

Area of the flower garden$= 314\ m^2$The radius of the circular portion covered by the sprinkler$= 12\ m$Area$= 7\pi r^2=3.14\times12\times12$$=3.14\times144\ m^2=452.16\ m^2$Since $452.16\ m^2$>$314\ m^2$Yes, the sprinkler will water the entire garden.

The radius of the outer circle$= 19\ m$Circumference of the outer circle$= 2\pi r$$=2\times3.14\times19=3.14\times38\ m$$=119.32\ m$The radius of the inner circle$=19\ m-10\ m=9\ m$Circumference$=2\pi r=2\times3.14\times9$$=56.52\ m$Here the required circumferences are $56.52\ m$ and $119.32\ m$

Given:Diameter of the table top$=1.6\ m$.The rate of polishing $= Rs.\ 15/m^2$To do:We have to find the cost of polishing the circular table top.Solution:Radius of the table top$=\frac{1.6}{2}\ m=0.8\ m$.Area of the table top$=\pi (radius)^2$$=3.14\times(0.8)^2$$=3.14\times0.64$$=2.0096\ m^2$ Therefore, Cost of polishing the table top$=Area\ of\ the\ table\ top\times\ Rate\ of\ polishing$ ... Read More

Length of the wire to be bent into circle $=44\ cm$$2\pi r=44$$2\times\frac{22}{7}\times r=44$$r=\frac{44\times7}{2\times22}=7cm$Area of such circle$=\pi r^2$ $=\frac{22}{7}\times7\times7$$=154\ cm^2$Now , the length of the wire is bent into a square.Here perimeter of square$=$ Circumference of line kLength of each side of the square$= \frac{Perimeter}{4}$$=\frac{44}{4}$$=11\ cm$ Area of the square$=(side)^2$$=(11)^2$$=121\ cm^2$Since, $154\ cm^2$>$121\ ... Read More

Radius of the circular sheet$=14\ cm$Area$=\pi r^2=\frac{22}{7}\times14\times14\ cm^2$$=616\ cm^2$Area of 2 small circles$=2\pi r^2$$=2\times\frac{22}{7}\times3.5\times3.5\ cm^2$$=77.0\ cm^2$Area of the rectangle$=l\times b$$=3\times 1\ cm^2$$=3\ cm^2$Area of the remaining sheet after removing the 2 circles and 1 rectangle$=616\ cm^2-(77+3)\ cm^2$$=616\ cm^2-80\ cm^2$$=536\ cm^2$Read More

$(a)$. Radius$(r)=14\ cm$Circumference$=2\pi r=2\times\frac{22}{7}\times14$$=88\ cm$$(b)$. Radius$(r)=28\ mm$Circumference$=2\pi r=2\times\frac{22}{7}\times28$$=176\ mm$$(c)$.Radius$(r)=21\ cm$Circumference$=2\pi r=2\times\frac{22}{7}\times21$$=132\ cm$