Mensuration

A hemispherical and a conical hole are scooped out of a solid wooden cylinder. The height of the solid cylinder is $7 \; cm$, radius of each of hemisphere, cone and cylinder is $3 \; cm$. The height of the cone is $3 \; cm$. Find the volume of the remaining solid. Give your answer correct to the nearest whole number. Take $\pi = 3.142$

Height of cylinder $= H = 7 \; cm$

Radius of cylinder $= R = 3 \; cm$

Radius of hemisphere $= R_S = 3 \; cm$

Radius of cone $= r = 3 \; cm$

Height of cone $= h = 3 \; cm$

Volume of cylinder $= V = \pi R^2 H$

Volume of hemisphere $= V_S = \dfrac{2}{3} \pi R_S^3$

Volume of cone $= V_C = \dfrac{1}{3} \pi r^2 h$

Volume of remaining solid $= V - \left(V_S + V_C\right)$

$= \pi R^2 H - \left(\dfrac{2}{3} \pi R_S^3 + \dfrac{1}{3} \pi r^2 h\right)$

$= \pi \left[3^2 \times 7 - \left(\dfrac{2}{3} \times 3^3 + \dfrac{1}{3} \times 3^2 \times 3\right)\right]$

$= \pi \left[63 - \left(18 + 9\right)\right] = 3.142 \times 36 = 113.112$

$\therefore \;$ Volume of the remaining solid $= 113 \; cm^3$ $\;\;\;$ [correct to the nearest whole number]

Mensuration

A solid metallic sphere of radius $12 \; cm$ is melted and recast into a solid cylinder of height $36 \; cm$. Find the radius and the curved surface area of the cylinder. Take $\pi = 3.14$

Radius of sphere $= R = 12 \; cm$

Height of cylinder $= h = 36 \; cm$

Let radius of cylinder $= r \; cm$

Volume of sphere $= V_S = \dfrac{4}{3} \pi R^3$

Volume of cylinder $= V_C = \pi r^2 h$

Since the sphere is melted and recast into a solid cylinder,

Volume of sphere $= $ Volume of cylinder

i.e. $\;$ $\dfrac{4}{3} \pi R^3 = \pi r^2 h$

i.e. $\;$ $r^2 = \dfrac{4 R^3}{3 h} = \dfrac{4 \times 12^3}{3 \times 36} = 64$

$\implies$ $r = \pm 8$

$\therefore \;$ Radius of cylinder $= r = 8 \; cm$ $\;$ [negative value is discarded as radius cannot be negative]

Curved surface area of the cylinder $= CSA = 2 \pi r h$

i.e. $\;$ $CSA = 2 \times 3.14 \times 8 \times 36 = 1808.64 \; cm^2$

Mensuration

A circus tent in the form of a cylinder is surmounted by a cone. The height of the tent is $13 \; m$ and the height of the cylinder is $8 \; m$. If the diameter of its base is $24 \; m$, calculate the total surface area of the tent to the nearest square meter. Take $\pi = \dfrac{22}{7}$

Height of tent $= 13 \; m$ (given)

Height of cylinder $= H = 8 \; m$ (given)

$\therefore \;$ Height of cone $= h = 13 - 8 = 5 \; m$

Diameter of base of tent $= 24 \; m$ (given)

$\therefore \;$ Radius of cylinder $= $ Radius of cone $= R = 12 \; m$

Slant height of cone $= \ell = \sqrt{h^2 + R^2} = \sqrt{5^2 + 12^2} = \sqrt{169} = 13 \; m$

Surface area of the conical portion of the tent $= s = \pi R \left(\ell + R\right)$

i.e. $\;$ $s = \pi \times 12 \left(13 + 12\right) = \pi \times 12 \times 25 = 300 \pi \; m^2$

Surface area of the cylindrical portion of the tent $= S = 2 \pi R \left(H + R\right)$

i.e. $\;$ $S = 2 \times \pi \times 12 \left(8 + 12\right) = \pi \times 24 \times 20 = 480 \pi \; m^2$

$\therefore \;$ Total surface area of the tent $= A = s + S$

i.e. $\;$ $A = 300 \pi + 480 \pi = 780 \times \dfrac{22}{7} = 2451.4 \; m^2$

i.e. $\;$ Total surface area of the tent $= 2451 \; m^2$ (to the nearest square meter)

Mensuration

The surface area of a solid metallic sphere is $2464 \; cm^2$. It is melted and recast into solid right circular cones of radius $3.5 \; cm$ and height $7 \; cm$. Calculate the radius of the sphere and the number of cones recast. Take $\pi = \dfrac{22}{7}$.

Let the radius of the sphere $= R$

Surface area of sphere $= 4 \pi R^2 = 2464$

i.e. $\;$ $R^2 = \dfrac{2464}{4 \pi} = \dfrac{2464 \times 7}{4 \times 22} = 196$ $\implies$ $R = \sqrt{196} = 14 \; cm$

$\therefore \;$ Radius of sphere $= R = 14 \; cm$

Radius of cone $= r = 3.5 \; cm$

Height of cone $= h = 7 \; cm$

Let the number of cones recast $= n$

Since the sphere is melted and recast into $n$ number of cones,

Volume of sphere $= n \times$ volume of cone

i.e. $\;$ $\dfrac{4}{3} \pi R^3 = n \times \dfrac{1}{3} \pi r^2 h$

i.e. $\;$ $n = \dfrac{4 R^3}{r^2 h} = \dfrac{4 \times 14^3}{3.5^2 \times 7} = 128$

$\therefore \;$ Number of cones recast $= 128$

Mensuration

The internal and external diameters of a hollow hemispherical vessel are $14 \; cm$ and $21 \; cm$ respectively. Find the total surface area of the vessel. Take $\pi = \dfrac{22}{7}$

Given: $\;$ Internal diameter of the vessel $= 14 \; cm$;

External diameter of the vessel $= 21 \; cm$

$\therefore \;$ Internal radius of the hemispherical vessel $= r = 7 \; cm$

External radius of the hemispherical vessel $= R = 10.5 \; cm$

Total surface area of the hemispherical vessel (T.S.A) $= $ External surface area $+$ Internal surface area $+$ Area of the ring

$\begin{aligned} i.e. \; T.S.A & = 2 \pi R^2 + 2 \pi r^2 + \pi \left(R^2 - r^2\right) \\\\ & = 2 \pi \left(R^2 + r^2\right) + \pi \left(R^2 - r^2\right) \\\\ & = 2 \pi \left(10.5^2 + 7^2\right) + \pi \left(10.5^2 - 7^2\right) \\\\ & = 318.5 \pi + 61.25 \pi \\\\ & = 379.75 \pi = 379.75 \times \dfrac{22}{7} = 1193.5 \; cm^2 \end{aligned}$

Mensuration

The surface area of a solid metallic sphere is $5024 \; cm^2$. It is melted and recast into solid cones of radius $5 \; cm$ and height $10 \; cm$. Calculate the radius of the sphere and the number of cones recast. Take $\pi = 3.14$

Let radius of sphere $= R$

Given: Surface area of sphere $= 5024 \; cm^2$

i.e. $\;$ $4 \pi R^2 = 5024$

i.e. $\;$ $R^2 = \dfrac{5024}{4 \pi} = \dfrac{1256}{3.14} = 400$

$\implies$ $R = \sqrt{400} = 20$ $\;\;\;$ [negative value discarded since radius of sphere cannot be negative]

$\therefore \;$ Radius of sphere $= R = 20 \; cm$

Volume of sphere $= V_S = \dfrac{4}{3} \pi R^3$

Given: Radius of cone $= r = 5 \; cm$; $\;$ Height of cone $= h = 10 \; cm$

Volume of cone $= V_C = \dfrac{1}{3} \pi r^2 h$

Let the number of cones made $= n$

$\because \;$ The sphere is melted and recast into a number of cones, we have

$V_S = n \times V_C$

$\implies$ $n = \dfrac{V_S}{V_C}$

i.e. $\;$ $n = \dfrac{\dfrac{4}{3}\pi R^3}{\dfrac{1}{3}\pi r^2 h}$

i.e. $\;$ $n = \dfrac{4 R^3}{r^2 h} = \dfrac{4 \times 20^3}{5^2 \times 10} = 128$

i.e. $\;$ Number of cones made $= n = 128$

Mensuration

A solid cylinder of radius $7 \; cm$ and height $14 \; cm$ is melted and recast into solid spheres of radius $3.5 \; cm$. Find the number of spheres formed.

Radius of cylinder $= r = 7 \; cm$

Height of cylinder $= h = 14 \; cm$

Volume of cylinder $= V_c = \pi r^2 h = \pi \times 7^2 \times 14 \; cm^3$

Radius of sphere $= R = 3.5 \; cm$

Volume of sphere $= V_s = \dfrac{4}{3} \pi R^3 = \dfrac{4}{3} \times \pi \times 3.5^3 \; cm^3$

Let the number of spheres formed $= n$

$\because \;$ the cylinder is melted and recast into a number of spheres,

$\text{Volume of cylinder} = n \times \text{Volume of sphere}$

$\begin{aligned} \therefore \; n & = \dfrac{\text{Volume of cylinder}}{\text{Volume of sphere}} \\\\ & = \dfrac{\pi \times 7^2 \times 14}{\dfrac{4}{3} \times \pi \times 3.5^3} \\\\ & = \dfrac{3 \times 7^2 \times 14}{4 \times 3.5^3} = 12 \end{aligned}$

$\therefore \;$ Number of spheres formed $= 12$

Mensuration

A conical vessel of radius $6 \; cm$ and height $8 \; cm$ is completely filled with water. A sphere is lowered into the water and its size is such that when it touches the sides, it is just immersed. Find the fraction of water that overflows.

Figure shows the cross-sectional view of the conical vessel (triangle ABC) and the sphere (circle at center $C_1$).

Since the sphere touches the sides of the conical vessel when it is just immersed, the sides of the cone are tangents to the circle.

In the figure,

radius of the conical vessel $= CM = MB = r = 6 \; cm$

height of the conical vessel $= AM = h = 8 \; cm$

slant height of the conical vessel $= AC = \ell = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10 \; cm$

radius of the sphere $= C_1 K = R \; cm$

In triangles $AKC_1$ and $AMC$,

$\angle AKC_1 = \angle AMC = 90^\circ$

[angle between the tangent to a circle and the radius of the circle at the point of contact]

$\angle A = \angle A$ $\;\;\;$ [common angle]

$\therefore \;$ By AA postulate, $\;$ $\triangle AKC_1 \sim \triangle AMC$

$\therefore \;$ $\dfrac{MC}{KC_1} = \dfrac{AC}{AC_1}$ $\;\;\;$ [corresponding parts of similar triangles are in proportion]

i.e. $\;$ $\dfrac{r}{R} = \dfrac{\ell}{8 - R}$

i.e. $\;$ $\dfrac{6}{R} = \dfrac{10}{8 - R}$

i.e. $\;$ $48 - 6R = 10R$

i.e. $\;$ $16 R = 48$ $\implies$ $R = 3 \; cm$

Now,

Volume of sphere $= V_S = \dfrac{4}{3} \pi R^3 = \dfrac{4}{3} \times \pi \times 3^3 = 36 \pi \; cm^3$

Volume of conical vessel $= V_C = \dfrac{1}{3} \pi r^2 h = \dfrac{1}{3} \times \pi \times 6^2 \times 8 = 96 \pi \; cm^3$

$\therefore \;$ Fraction of water that overflows

$= \dfrac{V_C - V_S}{V_C} = \dfrac{96 \pi - 36 \pi}{96 \pi} = \dfrac{60}{96} = \dfrac{5}{8}$ of the total volume of water

Mensuration

A cylindrical vessel of radius $4 \; cm$ contains water. A solid sphere of radius $3 \; cm$ is lowered into water until it is completely immersed. Find the rise in the water level in the vessel.

Let the initial level of water in the cylindrical vessel $= h \; cm$

Radius of the cylindrical vessel $= r = 4 \; cm$

Initial volume of water $= \pi r^2 h = \pi \times 4^2 \times h = 16 \pi h \; cm^3$

Let the radius of the sphere $= R = 3 \; cm$

Volume of sphere $= \dfrac{4}{3} \pi R^3 = \dfrac{4}{3} \times \pi \times 3^3 = 36 \pi \; cm^3$

Let the new height of water $= H \; cm$

New volume of water $= \pi r^2 H = 16 \pi H \; cm^3$

$\therefore \;$ We have,

$16 \pi h + 36 \pi = 16 \pi H$

i.e. $\;$ $16 \pi \left(H - h\right) = 36 \pi$

i.e. $\;$ $H - h = \dfrac{36}{16} = 2.25 \; cm$

i.e. $\;$ Rise in water level $= 2.25 \; cm$

Mensuration

A solid metallic cone of slant height $13 \; cm$ and radius $5 \; cm$ is melted and recast into solid spheres each of radius $1 \; cm$. Find the number of spheres recast.

Slant height of cone $= \ell = 13 \; cm$

Radius of cone $= r = 5 \; cm$

Let height of cone $= h \; cm$

$h = \sqrt{\ell^2 - r^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \; cm$

Volume of cone $= V_1 = \dfrac{1}{3}\pi r^2 h$

Let number of spheres made $= n$

Radius of each sphere $= R = 1 \; cm$

Volume of each sphere $= V_2 = \dfrac{4}{3} \pi R^3$

$\because \;$ The cone is melted and recast into $n$ spheres, we have,

$V_1 = n \times V_2$

i.e. $\;$ $\dfrac{1}{3} \pi r^2 h = n \times \dfrac{4}{3} \pi R^3$

i.e. $\;$ $n = \dfrac{r^2 h}{4 R^3}$

i.e. $\;$ $n = \dfrac{5^2 \times 12}{4 \times 1^3} = 75$

$\therefore \;$ Number of spheres made $= 75$

Mensuration

The total surface area of a hollow metal cylinder, open at both ends, of external radius $8 \; cm$ and height $10 \; cm$ is $338 \pi \; cm^2$. Taking $r$ to be the internal radius, write down an equation in $r$ and use it to find the thickness of the metal in the cylinder.

External radius of the hollow cylinder $= R = 8 \; cm$

Internal radius of the hollow cylinder $= r \; cm$

Height of the hollow cylinder $= h = 10 \; cm$

Total surface area of the hollow cylinder $= A = 2 \pi R h + 2 \pi r h + 2 \pi \left(R^2 - r^2\right)$

i.e. $\;$ $2 \pi \left(Rh + rh + R^2 - r^2\right) = 338 \pi$

i.e. $\;$ $Rh + rh + R^2 - r^2 = 169$

i.e. $\;$ $8 \times 10 + 10 r + 8^2 - r^2 = 169$

i.e. $\;$ $10 r - r^2 = 25$

i.e. $\;$ $r^2 - 10r + 25 = 0$

i.e. $\;$ $\left(r - 5\right)^2 = 0$ $\implies$ $r = 5 \; cm$

$\therefore \;$ Thickness of the metal in the cylinder $= R - r = 8 - 5 = 3 \; cm$

Mensuration

A spherical cannon ball of diameter $28 \; cm$ is melted and cast into a right circular cone with base diameter $35 \; cm$. Find the height of the cone, correct to one decimal place.

Diameter of spherical cannon ball $= 28 \; cm$

$\therefore \;$ Radius of spherical cannon ball $= r = 14 \; cm$

Base diameter of the right circular cone $= 35 \; cm$

$\therefore \;$ Base radius of the right circular cone $= R = 17.5 \; cm$

Let the height of he cone be $= H$

$\because \;$ The spherical cannon ball is melted and recast into a right circular cone,

Volume of sphere $=$ Volume of right circular cone

i.e. $\;$ $\dfrac{4}{3} \pi r^3 = \dfrac{1}{3} \pi R^2 H$

$\implies$ $H = \dfrac{4 r^3}{R^2} = \dfrac{4 \times 14^3}{17.5^2} = 35.84$

$\therefore \;$ Height of cone $= 35.8 \; cm$ $\;$ (correct to one decimal place)

Mensuration

On a square handkerchief, nine circular designs each of radius 7 cm are made. Find the area of the remaining portion of the handkerchief.

Radius of each circular design $=r=7 \ cm$

$\therefore$ Diameter of each circular design $= 14 \ cm$

$\therefore$ Length of side of the square handkerchief $=s= 14 \times 3 = 42 \ cm$


$\therefore$ Area of the square handkerchief $=s^2 = 42^2 = 1764 \ cm^2$ $\cdots$ (1)

Area of each circular design $=\pi r^2 = \dfrac{22}{7} \times 7^2 = 154 \ cm^2$

$\therefore$ Area of circular designs $= 9 \times 154 = 1386 \ cm^2$ $\cdots$ (2)

$\therefore$ Required area [from equations (1) and (2)] $=1764-1386=378 \ cm^2$

Mensuration

A solid metallic cone of slant height 13 cm and radius 5 cm is melted and recast into solid spheres each of radius 1 cm. Find the number of spheres recast.

Slant height of cone $= \ell = 13 \ cm$

Radius of cone $=r=5 \ cm$

Let height of cone $= h$

Now, $\ell^2 = h^2 + r^2$

$\implies$ $h = \sqrt{\ell^2 - r^2} = \sqrt{13^2 - 5^2} = \sqrt{144} = 12 \ cm$

$\therefore$ Volume of cone $= \dfrac{1}{3} \pi r^2 h = \dfrac{1}{3} \times \pi \times 5^2 \times 12 = 100 \pi \ cm^3$

Radius of sphere $=R=1 \ cm$

$\therefore$ Volume of sphere $= \dfrac{4}{3} \pi R^3 = \dfrac{4}{3} \times \pi \times 1^3 = \dfrac{4}{3} \pi \ cm^3$

Let number of spheres $=N$

Since the cone is recast into a number of small spheres,

volume of cone $= N \times$ volume of 1 sphere

i.e. $100 \pi = N \times \dfrac{4}{3} \pi$

$\implies$ $N = \dfrac{100 \times 3}{4} = 75$

$\therefore$ Number of spheres recast = 75

Mensuration

The total surface area of a hollow metal cylinder, open at both ends, of external radius 8 cm and height 10 cm, is $338 \pi \ cm^2$. Taking r to be the internal radius, write down an equation in r and use it to determine the thickness of the metal in the cylinder.

For the hollow cylinder,

outer radius $=R=8 \ cm$

inner radius $=r$

height $=h=10 \ cm$

total surface area (TSA) $=338 \pi \ cm^2$

$\begin{aligned} \text{TSA} & = \text{Lateral surface area} + \text{Base area} \\ & = 2 \pi h \left(R+r\right) + 2 \pi \left(R^2 - r^2\right) \end{aligned}$

$\begin{aligned} \text{i.e. } & 2 \times \pi \times 10 \times \left(8+r\right) + 2 \times \pi \left(8^2 - r^2\right) = 338 \pi \\ \text{i.e. } & 20 \left(8+r\right) + 2 \left(8+r\right) \left(8-r\right) = 338 \\ \text{i.e. } & \left(8+r\right) \left(10+8-r\right) = 169 \\ \text{i.e. } & \left(8+r\right) \left(18-r\right) = 169 \\ \text{i.e. } & 144-8r+18r-r^2=169 \\ \text{i.e. } & r^2 - 10r+25 = 0 \\ \text{i.e. } & \left(r-5\right)^2 = 0 \\ \implies & r = 5 \end{aligned}$

$\therefore$ Inner radius $= 5 \ cm$

$\therefore$ Thickness of metal in the cylinder $= R-r=8-5=3 \ cm$

Mensuration

An association offered to contribute 50% of the cost for making 100 tents. The lower part of each tent is of the form of a cylinder of diameter 4.2 m and height 4 m with the conical upper part of same diameter but of height 2.8 m. If the canvas to be used costs ₹ 100 per square meter, find the amount the association pays.

For the conical portion of the tent:

Diameter $= 4.2 \ m$

$\therefore$ Radius $= r = 2.1 \ m$

Height $=h=2.8 \ m$

$\begin{aligned} \therefore \text{Slant height} = \ell & = \sqrt{h^2+r^2} \\\\ & = \sqrt{2.8^2 + 2.1^2} \\\\ & = \sqrt{12.25} = 3.5 \ m \end{aligned}$

$\therefore$ Curved surface area of the conical portion

$= \pi r \ell = \dfrac{22}{7} \times 2.1 \times 3.5 = 23.1 \ m^2$ $\cdots$ (1)

For the cylindrical portion of the tent:

Diameter $= 4.2 \ m$

$\therefore$ Radius $= R = 2.1 \ m$

Height $=H = 4 \ m$

$\therefore$ Curved surface area of the cylindrical portion $=2\pi R H = 2 \times \dfrac{22}{7} \times 2.1 \times 4 = 52.8 \ m^2$ $\cdots$ (2)

For the entire tent:

$\therefore$ From equations (1) and (2),

curved surface area of the entire tent $=23.1+52.8 = 75.9 \ m^2$

Cost of canvas per square meter $=$ ₹ 100

$\therefore$ Cost of canvas for 1 tent $= 75.9 \times 100 =$ ₹ 7,590

$\therefore$ Cost of canvas for 100 tents $=7590 \times 100 = $ ₹ 7,59,000

$\therefore$ Amount paid by the association $= 50\%$ of ₹ 7,59,000 $= \dfrac{759000}{2}=$ ₹ 3,79,500

Mensuration

PQRS is a square lawn with side $PQ=42 \ m$. Two circular flower beds are there on the sides PS and QR with center at O, the intersection of its diagonals. Find the total area of the two flower beds (shaded portion).

From the figure, $PQ=QR=RS=SP=42 \ m$

$\therefore$ Area of the square lawn $= \left(\text{side}\right)^2 = 42^2 = 1764 \ m^2$ $\cdots$ (1)

$\begin{aligned} \text{Length of diagonal of the square lawn}& =SQ=PR & = & \sqrt{PQ^2 + QR^2} \\ & & & \\ & & = & \sqrt{42^2 + 42^2} = 42 \sqrt{2} \ m \end{aligned}$

$\therefore$ Diameter of the circular flower bed $=SQ=PR=42\sqrt{2} \ m$

$\therefore$ Radius of the circular flower bed $=r=21 \sqrt{2} \ m$

$\therefore$ Area of the circular flower bed $= \pi r^2 = \dfrac{22}{7} \times \left(21\sqrt{2}\right)^2 = 2772 \ m^2$ $\cdots$ (2)

$\therefore$ Area of the circular flower bed NOT covered by the square lawn

$= 2772-1764 = 1008 \ m^2$ [from equations (1) and (2)]

$\therefore$ Area of the shaded portion $= \dfrac{1008}{2} = 504 \ m^2$

Mensuration

The cost of fencing a circular field at the rate of ₹ 24 per meter is ₹ 5,280. The field is to be ploughed at the rate of ₹ 0.50 per square meter. Find the cost of ploughing the field.

Let radius of the circular field $=r$

Cost of fencing the circular field at the rate of ₹ 24 per meter = ₹ 5,280

$\implies$ Circumference of the circular field $= \dfrac{5280}{24} = 220 \ m$

i.e. $2 \pi r = 220$

$\implies$ $r = \dfrac{220}{2\pi} = \dfrac{110 \times 7}{22} = 35 \ m$

$\therefore$ Area of the circular field $= \pi r^2 = \dfrac{22}{7} \times 35^2 = 3850 \ m^2$

Cost of ploughing the field per square meter = ₹ 0.50

$\therefore$ Cost of ploughing the entire field $= 3850 \times 0.50 =$ ₹ 1,925