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)
$r(2r+23)=259$
$2r^2+23r-259=0$
$(\mathrm{r}-7)(\mathrm{r}-\frac{37}{2})=0$
$\mathrm{r}=7\mathrm{cm}$
$\mathrm{h}=23+7=30\mathrm{cm}$

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, $2A+4.48=2\times 75.46+4.48=150.92+4.48=155.40{\mathrm{m}}^2$

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
$\begin{array}{rl}& l\times b=540\\ & 30\times b=540\\ & b=\frac{540}{30}=18\end{array}$
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
$\begin{array}{rl}& \pi r^{2}h=12\times \frac{4}{3}\times \pi \times r^3\\ & 20\times 20\times h=12\times \frac{4}{3}\times 10\times 10\times 10\\ & \mathrm{h}=\frac{16000}{400}=40\mathrm{cm}\end{array}$

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)
$\begin{array}{rl}& \Rightarrow r^2=l^2-h^2\\ & \Rightarrow r^2={30}^2-{24}^2\\ & \Rightarrow r=18\mathrm{m}\end{array}$
Curved surface area of cone $=\pi \mathrm{rl}$
$\begin{array}{rl}& =\frac{22}{7}\times 18\times 30\\ & =1697.143{\mathrm{m}}^2\end{array}$

Volume of big cube $=n\times$ Volume of small cube, (where $\mathrm{n}$ is number of small cubes formed)
$\begin{array}{rl}& 8\times 8\times 8=n\times 4\times 4\times 4\\ & n=\frac{512}{64}=8\end{array}$
Total surface area of cube $=6a^2$
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 $3x$.
Perimeter of rectangular field $=2$ (Length $+$ Breadth $)$

$400=2(x+3x)$

$2\times 4x=400$

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

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 $$\begin{gathered} d_1 \\ \mathrm{\&} \\ {d}_2 \end{gathered}$$ are diagonals.
$\begin{array}{rl}& 240=\frac{1}{2}\times 30\times d_2\\ & d_2=16\mathrm{cm}\end{array}$
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 $)=5+2=7$

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

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

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, $DE\Vert BC$

$\begin{array}{rlrl}& \therefore & \frac{AD}{DB}& =\frac{AE}{EC}\\ \Rightarrow & & \frac{x}{x-2}& =\frac{x+2}{x-1}\\ \Rightarrow & & x^2-x& =x^2-4\\ & \therefore & x& =4\end{array}$

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

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

∴ $PQ=3AQ\Rightarrow \frac{PQ}{AQ}=\frac{3}{1}$
∴ $\frac{PQ}{AQ}=\frac{PR}{BR}=\frac{3}{1}$
∴ $\frac{PR}{2}=\frac{3}{1}\Rightarrow PR=6$ सेमी
∴ $PB=PR-BR=6-2=4$ सेमी

$ABCD$ एक समलम्ब है, जिसमें $AB\Vert CD$ यदि विकर्ण $AC$ खींचा जाए, तो

$\mathrm{△}ADC$ में, $PR=\frac{7}{2}=3.5$ सेमी
$\mathrm{△}ABC$ में, $RQ=\frac{4}{2}=2$ सेमी
$\therefore PQ=PR+RQ=3.5+2=5.5$ सेमी