Free Practice Questions for Mensuration in Maths

We know

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

Here, r = radius of the cone, h = height of the cone

According to Question

$\frac{17600}{7}=\frac{1}{3} \times \frac{22}{7} \times r^2 \times 24$

$r^2=\frac{800 \times 3}{24}$

$\mathrm{r}=\sqrt{100}$

$\mathrm{r} \quad=10 \mathrm{~cm}$

$h 1: h 2=1: 5$

Perimeter of their base

$\frac{2 \pi r_1}{2 \pi r_2}=\frac{5}{3}$

$\frac{r_1}{r_2}=\frac{5}{3}$

Ratio of volume

$\frac{\pi r_1^2 h_1}{\pi r_2^2 h_2}=\frac{5 \times 5 \times 1}{3 \times 3 \times 5}=\frac{5}{9}$

Radius of cone $=21 \mathrm{~cm}=210 \mathrm{~mm}$
and height of cone $=40 \mathrm{~cm}=400 \mathrm{~mm}$
Clearly,
Required number of spheres
$=\frac{\frac{1}{3} \times \pi \times 210 \times 210 \times 400}{\frac{4}{3} \times \pi \times 5 \times 5 \times 5}$
$=\frac{210 \times 210 \times 400}{4 \times 5 \times 5 \times5}$
$=35280$

We have,
$\begin{aligned}
&2 \pi \mathrm{h}(\mathrm{R}-\mathrm{r})=132 \\
&\Rightarrow 2 \times \frac{22}{7} \times 21(\mathrm{R}-\mathrm{r})=132 \\
&\Rightarrow \mathrm{R}-\mathrm{r}=1 \\
&\text { Also, } \mathrm{R}+\mathrm{r}=21 \mathrm{~cm} \text { (given) } \\
&\Rightarrow \mathrm{R}=11 \mathrm{~cm}, \mathrm{r}=10 \mathrm{~cm}
\end{aligned}$
So, TSA of hollow cylinder $=2 \pi \mathrm{h}(\mathrm{R}+\mathrm{r})+2 \pi$
$\begin{aligned}
&\left(R^{2}-r^{2}\right) \\
&=2 \times \frac{22}{7} \times 21 \times 21+2 \times \frac{22}{7} \times 21 \\
&=2 \times \frac{22}{7} \times 21[21+1] \\
&=6 \times 22 \times 22=2904 \mathrm{~cm}^{2}
\end{aligned}$

Area along the length $=2 \times(30 \times 3)=180 \mathrm{~m}^{2}$

Area along the breadth $=2 \times(15 \times 3)=90 \mathrm{~m}^{2}$

Area of 4 corner squares $=4 \times(3)^{2}=36 \mathrm{~m}^{2}$

Area of path $=180+90+36$

$=306 \mathrm{~m}^{2}$

Cost $=306 \times 20 = 6120$ rupees

Alternate method:
Area of path $=($ Perimeter of rectangle $) \times$ breadth
of path $+4 \times(\text { breadth of path })^{2}$
$=[2 \times(30+15)] \times 3+4 \times(3)^{2}$
$=306 \mathrm{~m}^{2}$
Cost $=306 \times 20 = 6120$ rupees

PQRS is a rectangle
$\mathrm{PR}=20$ given
$\mathrm{PQ}: \mathrm{QR}=4: 3$
In $\triangle \mathrm{PQR}$
$16 x^{2}+9 x^{2}=400$
$25 x^{2}=400$
$x^{2}=16$
Area of rectangle $=4 x \times 3x$
$=12 x^{2}$
$=12 \times 16=192$

Area of field $\Rightarrow \quad 18 \times 15=270$

Area of pit $\Rightarrow \quad 7.5 \times 6=45$

Area of remaining field $=270-45=225$

$\mathrm{H}=\frac{7.5 \times 6 \times 0.8}{225}=0.16=16 \mathrm{~cm}$

Volume of copper $=$ Volume of wire

$\frac{4}{3} \pi \times 9 \times 9 \times 9=\pi \times \frac{2}{10} \times \frac{2}{10} \times h$

$\mathrm{h}=243 \mathrm{~m}$

Given
Side of equilateral triangle = 2a
Diagonal of square = 2a
We know that,
Diagonal of square is $\sqrt{2} a^{\prime}$.
$\sqrt{2} a^{\prime}=2 a$
$a^{\prime}=\frac{2 a}{\sqrt{2}}$
$\left(a^{\prime}\right)^2=(\sqrt{2} a)^2=2 a^2$
then,
Area of equilateral triangle $=\frac{\sqrt{3} a^2}{4}=\frac{\sqrt{3}(2 a)^2}{4}=\sqrt{3} a^2$
Area of square $=\left(a^{\prime}\right)^2=(\sqrt{2} a)^2=2 a^2$
Ratio $=\frac{\text { area of an equilateral triangle }}{\text { area of a square }}$
Ratio $=\frac{\sqrt{3} a^2}{2 a^2}=\frac{\sqrt{3}}{2}$

Ratio $=\frac{\text { curved surface area of cylinder }}{\text { total surface area of cylinder }}$
$\frac{2}{3}=\frac{2 \pi r h}{2 \pi r(h+r)}$
$\frac{2}{3}=\frac{h}{h+r}$
$2 h+2 r=3 h$
$2 r=h$
$\frac{r}{h}=\frac{1}{2}$

Given
Diameter $=14 \mathrm{~cm}$ Radius $=\frac{d}{2}=\frac{14}{2}=7 \mathrm{~cm}$
Volume of Cylinder $=2002$
So,
$
\begin{aligned}
& \pi r^2 h=2002 \\
& \frac{22 \times 7 \times 7 \times h}{7}=2002 \\
& h=13 \mathrm{~cm}
\end{aligned}
$
Lateral surface area of cylinder $=2 \pi r(r+h)$
$
=\frac{2 \times 22 \times 7(7+13)}{7}=880 \mathrm{~cm}^2
$

Ratio of sides is 1:2:3
Let side of cuboid are 1x, 2x & 3x
Total Surface area of cuboid = 2(lb+bh+hl)
$1100=2(1 x \times 2 x+2 x \times 3 x+3 x \times 1 x)$
$1100=2\left(11 x^2\right)$
$
x=5 \sqrt{2}
$
Volume of cuboid=lbh
$5 \sqrt{2} \times 10 \sqrt{2} \times 15 \sqrt{2}=1500 \sqrt{2}$

Volume of solid cube $\left(A^3\right)=8 \times$ Volume of small cubes $\left(a^3\right)$
$\Rightarrow 46656=8 \times a^3$
$\Rightarrow$ $a^3=5832$
$\Rightarrow$ $a=18$
Side of the small cube is 18
Side of original cube $=\sqrt[3]{46656}=36$
Ratio $=\frac{\text { Surface area of original cube }}{\text { total surface areas of the smaller } 8 \text { cubes }}$
$\frac{6 A^2}{8 \times 6 a^2}=\frac{6 \times 36 \times 36}{8 \times 6 \times 18 \times 18}=\frac{1}{2}$

Number of spheres $=\frac{\pi \times(10)^{2} \times 25}{\pi \frac{4}{3}(5)^{3}}=15$

Let side of triangle is 2x, 3x and 4x.
$\Rightarrow \quad 2 x+3 x+4 x=18$
$\Rightarrow \quad 9 x=18$
$x=2$
length of three sides,
$4 \mathrm{~cm}, 6 \mathrm{~cm}, 8 \mathrm{~cm}$
$s=\frac{4+6+8}{2}=9$
Area of triangle
$=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{9 \times 5 \times 3 \times 1}$
$=3 \sqrt{15} \mathrm{~cm}^{2}$

ATQ
Volume of sphere = volume of wire
$\Rightarrow$ $\frac{4}{3} \times \pi \times(2)^{3}$ = $\pi \times(r)^{2} \times 100$
$\Rightarrow$ $\frac{4}{3} \times \pi \times 8=$ $\pi \times r^{2} \times 100$
r = 0.33 cm

Radius of the sphere, $r=63 \mathrm{~cm}$

Total surface area of the sphere $=4 \pi r^2$

$=4  \times \frac{22}{7} \times 63 \mathrm{~m} \times 63 \mathrm{~m}$

$=88\times 9 \mathrm{~m} \times 63 \mathrm{~m}$

$=49896 \mathrm{~m}^2$ 

Radius of each sphere $=\frac{\text { Diameter }}{2}=\frac{10}{2} = 5\mathrm{~cm}$

$\text { Number of  balls }=\frac{\text { Volume of cuboid }}{\text { Volume of each sphere ball }}$

$=\frac{\text { length } \times \text { breadth } \times \text { height }}{\frac{4 \pi}{3} \times \text { radius }^3} $

$=\frac{40 \times 20 \times 16}{\frac{4}{3} \times \frac{22}{7} \times\left(5\right)^3}=24.436∼24$

Let internal and external radius of the track be r and R respectively.

Internal circumference of the track $=2 \pi r$

$2 \pi r=440 $

$\Rightarrow 2 \times(22 / 7) \times r=440 $

$\Rightarrow r=70 m$

External circumference of the track $=2 \pi R$

$2 \pi R=506 $

$\Rightarrow 2 \times(22 / 7) \times R=506$

$\Rightarrow R=506 \times(7 / 44)$

$\Rightarrow R=80.5 \mathrm{~m}$

Area of track $=\pi\left(R^2-r^2\right)=(22 / 7) \times\left(80.5^2-70^2\right)=22 / 7 \times 150.5 \times 10.5=4966.5 \mathrm{~m}^2$

Cost of levelling $1 \mathrm{~m}^2=$ Rs. 6

$\therefore$ Cost of levelling $4966.5 \mathrm{~m}^2=4966.5 \times 6=$ Rs. 29,799

Length of the rectangular paper will be height of the cylinder and width of the rectangular paper is circumference of the cylinder

$\therefore 2 \pi \mathrm{r}=10 \mathrm{~cm} \text { and } \mathrm{h}=22 \mathrm{~cm}$

Volume of the cylinder $=\pi r^2 \mathrm{~h} $

$\frac{22}{7} \times \frac{10}{2 \pi} \times \frac{10}{2 \pi} \times 22 \mathrm{~cm}^3 $

$=\frac{22}{7} \times \frac{10}{2} \times \frac{7}{22} \times \frac{10}{2} \times \frac{7}{22} \times 22 \mathrm{~cm}^3$

$=175 \mathrm{~cm}^3$