## The volume of a right circular cone is $9856 \mathrm{~cm}^{3}$.

[question] Question. The volume of a right circular cone is $9856 \mathrm{~cm}^{3}$. If the diameter of the base is $28 \mathrm{~cm}$, find (i) height of the cone (ii) slant height of the cone (iii) curved surface area of the cone $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: (i) Radius of cone $=\left(\frac{28}{2}\right) \mathrm{cm}=14 \mathrm{~cm}$ Let the height of the cone be h. Volume of cone $=9856 \mathrm{~cm}^{3}$ $\Rightarrow \frac{1}{3} \pi r^{...

## A conical pit of top diameter 3.5 m is 12 m deep.

[question] Question. A conical pit of top diameter $3.5 \mathrm{~m}$ is $12 \mathrm{~m}$ deep. What is its capacity in kilolitres? $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [solution] Solution: Radius $(r)$ of pit $=\left(\frac{3.5}{2}\right) \mathrm{m}=1.75 \mathrm{~m}$ Height $(h)$ of pit = Depth of pit $=12 \mathrm{~m}$ Volume of pit $=\frac{1}{3} \pi r^{2} h$ $=\left[\frac{1}{3} \times \frac{22}{7} \times(1.75)^{2} \times 12\right] \mathrm{cm}^{3}$ $=38.5 \mathrm{~m}^{3}$ Thus, ...

## If the volume of a right circular cone of height $9 \mathrm{~cm}$ is $48 \pi \mathrm{cm}^{3}$,

[question] Question. If the volume of a right circular cone of height $9 \mathrm{~cm}$ is $48 \pi \mathrm{cm}^{3}$, find the diameter of its base. [/question] [solution] Solution: Height (h) of cone = 9 cm Let the radius of the cone be r. Volume of cone $=48 \pi \mathrm{cm}^{3}$ $\Rightarrow \frac{1}{3} \pi r^{2} h=48 \pi \mathrm{cm}^{3}$ $\Rightarrow r^{2}=16 \mathrm{~cm}^{2}$ $\Rightarrow r=4 \mathrm{~cm}$ Diameter of base $=2 r=8 \mathrm{~cm}$ [/solution]...

## The height of a cone is 15 cm.

[question] Question. The height of a cone is $15 \mathrm{~cm}$. If its volume is $1570 \mathrm{~cm}^{3}$, find the diameter of its base. [Use $\left.\pi=3.14\right]$ [solution] Solution: Height (h) of cone = 15 cm Let the radius of the cone be r. Volume of cone $=1570 \mathrm{~cm}^{3}$ $\frac{1}{3} \pi r^{2} h=1570 \mathrm{~cm}^{3}$ $\Rightarrow\left(\frac{1}{3} \times 3.14 \times r^{2} \times 15\right) \mathrm{cm}=1570 \mathrm{~cm}^{3}$ $\Rightarrow r^{2}=100 \mathrm{~cm}^{2}$ ⇒ r = 10 cm There...

## Find the volume of the right circular cone with

[question] Question. Find the volume of the right circular cone with (i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cm $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$ [/question] [solution] Solution: (i) Radius (r) of cone = 6 cm Height $(h)$ of cone $=7 \mathrm{~cm}$ $=\left[\frac{1}{3} \times \frac{22}{7} \times(6)^{2} \times 7\right] \mathrm{cm}^{3}$ $=(12 \times 22) \mathrm{cm}^{3}$ $=264 \mathrm{~cm}^{3}$ Therefore, the volume of the cone is $264 \mathrm{~cm}^{3}$. (ii) R...

## A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m.

[question] Question. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. [/question] [solution] Solution: Draw a line BE parallel to AD and draw a perpendicular BF on CD. It can be observed that ABED is a parallelogram. BE = AD = 13 m ED = AB = 10 m EC = 25 − ED = 15 m For $\triangle B E C$, Semi-perimeter, $s=\frac{(13+14+15) \mathrm{m}}{2}=21 \mathrm{~m}$ By Heron’s formula, Area of triangle $=\sqr...

## A floral design on a floor is made up of 16 tiles which are triangular,

[question] Question. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see the given figure). Find the cost of polishing the tiles at the rate of 50p per cm2. [/question] [solution] Solution: It can be observed that Semi-perimeter of each triangular-shaped tile, $s=\frac{(35+28+9) \mathrm{cm}}{2}=36 \mathrm{~cm}$ By Heron’s formula, Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ Area of each tile $=[\sqrt{36(36-35)(36-28)(...

## A kite in the shape of a square with a diagonal 32 cm

[question] Question. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangles of base 8 cm and sides 6 cm each is to be made of three different shades as shown in the given figure. How much paper of each shade has been used in it? [/question] [solution] Solution: We know that Area of square $=\frac{1}{2}$ (diagonal) $^{2}$ Area of the given $k i t e=\frac{1}{2}(32 \mathrm{~cm})^{2}=512 \mathrm{~cm}^{2}$ Area of $1^{\text {st }}$ shade $=$ Area of $2^{\text {nd }}$ shade $...