$(i)$ Area of the rectangle$= l\times b$$=18cm\times\left(6cm+4cm\right)$$=18cm\times10cm=180cm^2$Area of the right $\Delta BCE=\frac{1}{2}\times b\times h$$=\frac{1}{2}\times8\times10=40\ cm^2$Area of the shaded portion$=180cm^2-70cm^2=110cm^2$$(ii)$. Area of the square $PQRS= \left(Side\right)^2$$=\left(20\right)^2=400cm^2$Area of triangle$=\ \frac{1}{2}\times b\times h$$\frac{1}{2}\times10\times10=50cm^2$ Area of triangle$=\ \frac{1}{2}\times b\times h$$=\frac{1}{2}\times10\times20=100cm^2$Area of triangle$=\ \frac{1}{2}\times b\times h$$=\frac{1}{2}\times10\times20=100cm^2$ Area of the three triangles$=50cm^2+100cm^2+100cm^2=250cm^2$Area of the shaded portion$=400cm^2-250cm^2=150cm^2$Read More

$(i)$. Length of the lawn$= 10\ m$Breadth of the lawn$= 5\ m$Area of the lawn$= l\times b$$=10m\times5m=50\ m^2$$(ii)$.  Area of the circular flower bed$= \pi r^2$ $\frac{22}{7}\times2\times2=\frac{88}{7}m^2=12.57\ m^2$ $(iii)$.  Area of the lawn excluding the area of the flower bed$=50m^2-\frac{88}{7}\ m^2$$=\frac{350-88}{7}\ m^2=\frac{262}{7}\ m^2=37.43\ m^2$$(iv)$. Circumference of the flower bed$= 2\pi r^2$$=2\times\frac{22}{7}\times2=12.56\ m$Read More

Length of the road along the length of the field$= 90\ m$Breadth$= 3\ m$Area of the road$= l\times b$$=90m\times3m=270m^2$Similarly, the area of the road parallel to the breadth of the field$=l\times b$$=60\ m\times3\ m=180\ m^2$Area of the common portion$=3\ m\times3\ m=9m^2$ $(i)$. Area of the two roads$=270\ m^2+180\ m^2-9\ m^2$$=450\ m^2-9\ ... Read More

Given: Pragya wrapped a cord around a circular pipe of radius $4\ cm$ $(adjoining figure)$ and cut off the length required of the cord. Then she wrapped it around a square box of side $4\ cm$ $(also shown)$To do: To check whether she have any cord left? Solution:Side of square box $=4\ cm$The ... Read More

Radius of the wheel$=28\ cm$Circumference$=2\pi r=2\times\frac{22}{7}\times28=176\ cm$Number of rotations made by the wheel in going $352\ m$ or $35200\ cm=\frac{35200}{176}=200$Hence, the required number of rotation$=200$

From the question, it is given that,Length of the minute hand of the circular clock$ = 15\ cm$Then, Distance traveled by the tip of a minute hand in 1 hour $=$ circumference of the clock$=2\pi r$$=2\times 3.14\times 15$$=94.2\ cm$

Given: A garden is 90m long and 75m broad. A path 5m wide is to be built outside and around it.To do:We have to find the area of the path.Solution:Area of the garden = $90\times75\ m^2$$=6750\ m^2$Area of the outer rectangle including the park = $(90+5+5)\times(75+5+5)\ m^2$$=100\times85\ m^2$$=8500\ m^2$Therefore, Area of ... Read More

Length of the park$= 125\ m$The breadth of the park$= 65\ m$Area of the park$= l\times b$$=125\ m\times65\ m=8125\ m^2$ Length of the park including path$=125\ m+3\ m+3\ m=131\ m$The breadth of the park including path$= 65\ m+3\ m+3\ m=71\ m$Area of the park including path$131\ m\times71\ m=9301\ m^2$ Area of the ... Read More

Length of the room$= 5.5\ m$Breadth of the room$= 4\ m$Area of the room$=l\times b=5.5\ m\times4\ m=22\ m^2$Width of the verandah$= 2.25\ m$Length of the room including verandah$=5.5\ m+2\times2.25\ m=10\ m$The breadth of the room including verandah$=4\ m+2\times2.25\ m=8.50\ m^2$Area of the room including verandah$=l\times b$$=10\ m\times8.50\ m=85\ m^2$$(i)$. Area ... Read More

Area of the square garden$=\left(side\right)^2$$=30\ m\times30\ m=900\ m^2$Length if the garden excluding the path$=30\ m-2\times1\ m=28\ m$Area of the garden excluding the path$=28\ m\times28\ m=784\ m^2$$(i)$. Area of the path$= 900\ m^2-784\ m^2$$=116\ m^2$$(ii)$. Cost of the planting the remaining portion at the rate of $Rs\ 40\ per\ m^2$$=Rs\ 40\times784=Rs\ ... Read More