Free Practice Questions for Mensuration in Maths

Given $(\mathrm{h}-\mathrm{r})=23 \mathrm{~cm}$
$
h=23+r
$
The total surface area of the cylinder is $=1628 \mathrm{~cm}^2$
$2 \pi \mathrm{r}(\mathrm{r}+\mathrm{h})=1628$
$2 \times \frac{22}{7} \times \mathrm{r}(\mathrm{r}+\mathrm{h})=1628$
$\mathrm{r}(\mathrm{r}+\mathrm{h})=259$
$
\mathrm{r}(\mathrm{r}+23+\mathrm{r})=259 ....... $from (1)
$r(2 r+23)=259$
$2 r^2+23 r-259=0$
$(\mathrm{r}-7)\left(\mathrm{r}-\frac{37}{2}\right)=0$
$\mathrm{r}=7 \mathrm{~cm}$
$\mathrm{~h}=23+7=30 \mathrm{~cm}$

Distance travelled by road roller in 1 revolution $=2 \pi \mathrm{rh}$
Where $r=\frac{84}{2}=42 \mathrm{~cm}$ and $\mathrm{h}=1 \mathrm{~m}$
So,
Distance travelled by road roller in 750 revolutions
$
=750 \times 2 \times \frac{22}{7} \times \frac{42}{100} \times 1=1980 \mathrm{~m}^2
$

$
r=\frac{9.8}{2}=4.9 \mathrm{~m}
$
Area of circular garden, $\mathrm{A}=\pi r^2=\frac{22}{7} \times 4.9 \times 4.9=75.46 \mathrm{~m}^2$
Hence, $2 A+4.48=2 \times 75.46+4.48=150.92+4.48=155.40 \mathrm{~m}^2$

As $24^2+45^2=51^2$, So it is a right-angled triangle.
Area of triangle $=\frac{1}{2} \times 24 \times 45=540 \mathrm{~cm}^2$
According to question
Area of rectangle $=$ Area of triangle
$
\begin{aligned}
&l \times b=540 \\
&30 \times b=540 \\
&b=\frac{540}{30}=18
\end{aligned}
$
Perimeter of rectangle $=2(l+b)=2(30+18)=2 \times 48=96 \mathrm{~cm}$

Linear symmetry :  when we draw a line segment exactly in the middle of a pattern or drawing, if one part is identical to the other then the pattern is said to be linearly symmetrical.

For example : Equilateral triangle, Rectangle, Rhombus and Kite have linear symmetry.

Rotational symmetry : If a figure is rotated around a centre point and it still appears exactly as it did before the rotation, it is said to have rotational symmetry.

For example : Equilateral triangle, Rhombus and Rectangle have rotational symmetry.

Hence, Kite has linear symmetry but no rotational symmetry.

Total volume of solid which is melted to form a new solid cube $=$ volume of new

solid cube
So

$6^3+8^3+10^3=a^3$

$\Rightarrow a^3=1728 $

$\Rightarrow a^3=12^3 $

$\Rightarrow a =12$ cm

Hcf of 15 and 12 is 3
Now, area of a tile is $=3 \times 3=9 \mathrm{~m}$
Area of floor of room $=15 \times 12=180 \mathrm{~m}$
Required numbers of tiles $=\frac{180}{9}=20$

Let $\mathrm{h}$ be the height of the water
According to question
$
\begin{aligned}
&\pi r^2 h=12 \times \frac{4}{3} \times \pi \times r^3 \\
&20 \times 20 \times h=12 \times \frac{4}{3} \times 10 \times 10 \times 10 \\
&\mathrm{~h}=\frac{16000}{400}=40 \mathrm{~cm}
\end{aligned}
$

The area of the floor of a cubical room is $=192 \mathrm{~m}^2$
$\begin{array}{ll}\text { side }^2 & =192 \mathrm{~m}^2 \\ \text { side } & =\sqrt{192}=8 \sqrt{3} \mathrm{~m}\end{array}$
The length of the longest rod that can be kept in that room is $=\sqrt{3} \times$ side
$
=\sqrt{3} \times 8 \sqrt{3}=24 \mathrm{~m}
$

We know that, 

Slant Height, l = √(r²+h²)
$
\begin{aligned}
&\Rightarrow r^2=l^2-h^2 \\
&\Rightarrow r^2=30^2-24^2 \\
&\Rightarrow r=18 \mathrm{~m}
\end{aligned}
$
Curved surface area of cone $=\pi \mathrm{rl}$
$
\begin{aligned}
&=\frac{22}{7} \times 18 \times 30 \\
&=1697.143 \mathrm{~m}^2
\end{aligned}
$

Volume of big cube $=n \times$ Volume of small cube, (where $\mathrm{n}$ is number of small cubes formed)
$
\begin{aligned}
&8 \times 8 \times 8=n \times 4 \times 4 \times 4 \\
&n=\frac{512}{64}=8
\end{aligned}
$
Total surface area of cube $=6 a^2$
Total surface area of big cube $=6 \times 8 \times 8=384 \mathrm{~cm}^2$
Total surface area of small cubes $=8 \times 6 \times 4 \times 4=768 \mathrm{~cm}^2$
Difference $=768-384=384 \mathrm{~cm}^2$

Let the breadth of rectangular field is $x$ then the length is $3 x$.
Perimeter of rectangular field $=2$ (Length $+$ Breadth $)$

$400=2(x+3 x) $

$2 \times 4 x=400 $

$x=\frac{400} {8}=50$

Breadth $=50 \mathrm{~m}$, Length $=3 \times 50=150 \mathrm{~m}$
Hence, $(2 \times$ length $+3 \times$ breadth $)=(2 \times 150+3 \times 50)=300+150=450 \mathrm{~m}$

Area of rhombus $=\frac{1}{2} \times d_1 \times d_2$, where $d_1 \\\ \& \\\ d_2$ are diagonals.
$
\begin{aligned}
&240=\frac{1}{2} \times 30 \times d_2 \\
&d_2=16 \mathrm{~cm}
\end{aligned}
$
Side of rhombus $=\frac{1}{2} \sqrt{\left(d_1\right)^2+\left(d_2\right)^2}$
                               $=\frac{1}{2} \sqrt{(16)^2+(30)^2}=\frac{1}{2} \sqrt{256+900}=\frac{1}{2} \sqrt{1156}=\frac{1}{2} \times 34=17 \mathrm{~cm}$
Perimeter of rhombus $=4 \times$ Side $=4 \times 17=68 \mathrm{~cm}$

In a prism,

Number of faces $=\mathrm{n}+2$

Number of vertices $=2 \mathrm{n}$

Number of edges $=3 \mathrm{n}$

$\mathrm{F}($ faces $)=5+2=7$

$\begin{aligned}&\mathrm{V} \text { (vertices) }=2 \times 5=10 \\&\mathrm{E}(\text { edges })=3 \times 5=15\end{aligned}$

So,

Option (1), $F+E-V=7+15-10=22-10=12$

Option (2), $V+E-F=10+15-7=25-7=18$

Option (3), $2 V+F-E=2 \times 10+7-15=27-15=12$

Hence, Statement (3) is not true.

Linear symmetry :  When we draw a line segment exactly in the middle of a pattern or drawing, if one part is identical to the other then the pattern is said to be linearly symmetrical.

Isosceles triangle and kite have 1 line of symmetry.

Rhombus and rectangle have 2 lines of symmetry.

Scalene triangle and parallelogram have no line of symmetry.

Kite has 1 line of symmetry while rhombus has 2 lines of symmetry.

Hence, Kite and rhombus do not have the same number of lines of symmetry.

दिया है, $D E \| B C$

$\begin{aligned} & \therefore & \frac{A D}{D B} &=\frac{A E}{E C} \\ \Rightarrow & & \frac{x}{x-2} &=\frac{x+2}{x-1} \\ \Rightarrow & & x^{2}-x &=x^{2}-4 \\ & \therefore & x &=4 \end{aligned}$

समरूप त्रिभुजों के परिमापों में अनुपात $=$ संगत भुजाओं में अनुपात<>

समरूप त्रिभुजों के क्षेत्रफलों में अनुपात $=$ संगत ऊँचाइयों के वर्गों में अनुपात
$
=2^{2}: 3^{2}=4: 9
$

∴ $P Q=3 A Q \Rightarrow \frac{P Q}{A Q}=\frac{3}{1}$
∴ $\frac{P Q}{A Q}=\frac{P R}{B R}=\frac{3}{1}$
∴ $\frac{P R}{2}=\frac{3}{1} \Rightarrow P R=6$ सेमी
∴ $P B=P R-B R=6-2=4$ सेमी

$A B C D$ एक समलम्ब है, जिसमें $A B \| C D$ यदि विकर्ण $A C$ खींचा जाए, तो

$\triangle A D C$ में, $P R=\frac{7}{2}=3.5$ सेमी
$\triangle A B C$ में, $R Q=\frac{4}{2}=2$ सेमी
$\therefore P Q=P R+R Q=3.5+2=5.5$ सेमी