Free Practice Questions for Mensuration in Maths

Given that,

Volume of hemisphere = Volume of Cylinder 

$\frac{2}{3} \pi r^3=\pi r^2 h $

$\frac{2}{3} r=h $

$\text { So, } \frac{r}{h}=\frac{3}{2}$

Volume of Sphere $=\frac{4}{3} \pi r^3$, where $\mathrm{r}=21$

Volume of Sphere $=\frac{4}{3} \times \frac{22}{7} \times 21\times 21\times 21=88 \times 441$

Volume of Cone $=\frac{1}{3} \pi R^2 h$, where $\mathrm{R}=7$ and $\mathrm{h}=21$

Volume of Cone $=\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 21=22 \times 49$

So, Difference between the volumes of them

$=88 \times 441-22 \times 49=38808-1078=37730$

Difference between two perpendicular sides = $(84-13)=71$

Pythagorean Triplet used $13,84$ and 85 

$\frac{1}{2} \times 13\times 84=546 \mathrm{~cm}^2$
So, Perimeter $=13+84+85=182 \mathrm{~cm}$
Basic Method:
Let the perpendicular sides are a and $b$, then $a-b=71$
And $\frac{1}{2} \times a \times b=546$
$a b=1092$
$(a+b)^2=(a-b)^2+4 a b$
$(a+b)^2=71^2+4 \times 1092$
$(a+b)^2=5041+4368$
$(a+b)^2=9409$
$(a+b)=97$
So, $a=\frac{97+71}{2}=\frac{168}{2}=84$
And $b=\frac{97-71}{2}=\frac{26}{2}=13$
Hypotenuse $=\sqrt{84^2+13^2}=85$
Perimeter $=13+84+85=182$

Volume of frustum $=$ Volume of Cylinder

 $\frac{1}{3} \pi\left(R_1^2+R_2^2+R_1 R_2\right) H=\pi r^2 h$
Where $R_1=15, R_2=21, H=42$ and $r=18$

$\frac{1}{3}\left(15^2+21^2+15 \times 21\right) \times 42=18^2 \times h $

$\frac{1}{3}(225+441+315) \times 42=324 \times h$

$h=\frac{327 \times 42}{324}=42.38 \mathrm{~cm}$

Total change $=\pm A \pm B+\frac{\pm A \times \pm B}{100}$
Total Change in Area $=10-12+\frac{10 \times(-12)}{100}=-2-1.2=-3.2 \%$

Side of bigger square $=a+2 \times 5.5=a+11$
Area of path = Area of bigger square - Area of smaller square

$130. 75=(a+11)^2-a^2$
$130.75=a^2+121+22 a-a^2$
$22 a=130.75-121$
$22 a=9.75$
$a=\frac{975}{2200}$
Perimeter of field $=4 \times \frac{975}{2200}=\frac{39}{22}$
Cost of fencing the field $=\frac{39}{22} \times 110=$₹ 195

Total surface area of a cylinder $=2 \pi r(h+r)$
$
\begin{aligned}
&3.52=2 \times \frac{22}{7} \times 0.35(h+0.35) \\
&3.52=2.2(h+0.35) \\
&h+0.35=\frac{3.52}{2.2}=1.6 \\
&h=1.6-0.35 \\
&h=1.25=\frac{125}{100}=\frac{5}{4} \mathrm{~m}
\end{aligned}
$

Area of square $=$ Side $^2=x^2$
Area of triangle $=\frac{1}{2} \times$ Base $\times$ Height $=\frac{1}{2} \times x \times h=\frac{x h}{2}$
According to question
$
\begin{aligned}
&x^2=2 \times \frac{x h}{2} \\
&h=\frac{x^2}{x}=x \text { unit }
\end{aligned}
$

Area of rhombus $=\frac{1}{2} \times d_1 \times d_2$
$5880=\frac{1}{2} \times 70 \times d_2$
$d_2=\frac{5880}{35}=168$
Side of rhombus $=\sqrt{\frac{d_1^2}{4}+\frac{d_2^2}{4}}=\sqrt{\frac{(168)^2}{4}+\frac{(70)^2}{4}}=\sqrt{\frac{168 \times 168}{4}+\frac{70 \times 70}{4}}$
Side of rhombus $=\sqrt{84 \times 84+35 \times 35}=\sqrt{7056+1225}=\sqrt{8281}=91 \mathrm{~m}$

$120 \times 60 \times 48=200 \times a^3$, where $a$ is the side of cube
$
\begin{aligned}
&a^3=\frac{7200 \times 48}{200}=36 \times 6 \times 8 \\
&a=12 \mathrm{~cm}
\end{aligned}
$
Lateral surface area of cube $=4 a^2=4 \times 12 \times 12=4 \times 144=576 \mathrm{~cm}^2$

In a pyramid,
Number of faces $=\mathrm{n}+1$
Number of vertices $=\mathrm{n}+1$
Number of edges $=2 \mathrm{n}$
$\mathrm{F}$ (faces $)=4+1=5$
$\mathrm{V}($ vertices $)=4+1=5$
$\mathrm{E}$ (edges) $=2 \times 4=8$
Hence, $(3 V+2 F-E)=(3 \times 5+2 \times 5-8)=25-8=17$

Circumference of circle $=2 \pi r$
Circumference of circle $=2 \times \frac{22}{7} \times 14=88 \mathrm{~cm}$
According to question
Circumference of circle $=$ Perimeter of square
$
88=4 a
$
$
a=22 \mathrm{~cm}
$
Hence, The side of the square is $22 \mathrm{~cm}$.

Volume of cuboid $=l b h$
So,
Volume of cuboid $=$ Total water displacement
$
50 \times 45 \times h=90 \times 1
$
$
\mathrm{h}=\frac{90}{50 \times 45}=\frac{1}{25} \mathrm{~m}=4 \mathrm{~cm}
$

Volume of cuboid $=\mathrm{a} \times \mathrm{b} \times \mathrm{c}$
Total Surface Area of a cuboid $=2(a b \times b c \times c a)$
When we check the option 3
$
\frac{1}{a \times b \times c}=\left(\frac{2}{2(a b \times b c \times c a)}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)
$
$
\begin{aligned}
\frac{1}{a \times b \times c} &=\left(\frac{1}{(a b \times b c \times c a)}\right)\left(\frac{a b+b c+c a}{a \times b \times c}\right) \\
\frac{1}{a \times b \times c} &=\frac{1}{a \times b \times c}
\end{aligned}
$
So, the option 3 is true

Let $r$ be the circle of radius.
Given, Perimeter of rectangle and circle are same
So, $2(1+b)=2 \pi r$
$
\begin{aligned}
2(15+7) &=2 \times \frac{22}{7} \times r \\
r &=\frac{44 \times 7}{44} \\
r &=7 \mathrm{~cm}
\end{aligned}
$
Now, area of circle $=\pi r^2$
$
\begin{aligned}
&=\frac{22}{7} \times 7 \times 7 \\
&=154 \mathrm{~cm}^2
\end{aligned}
$

Given 

$r=3 \mathrm{~cm}$
$\text { Volume of spherical ball }=\frac{4}{3} \pi (3)^3$
$=36 \pi$
If ball immersed in water the volume of vertical cylinder is increased by volume of ball
$36 \pi=\pi (r)^2 h$                                              ($r=5$)
$36 \pi=\pi (5)^2 h$
$h=\frac{36}{25}$
$h =1.44$

$
\begin{aligned}
&l+b+h=18 \mathrm{~cm} \\
&\text { diagonal } \sqrt{l^2+b^2+h^2}=13 \mathrm{~cm}
\end{aligned}
$
Square of both side
$
\begin{aligned}
&l^2+b^2+h^2=169 \\
&l+b+h=18 \mathrm{~cm} \quad \text { (Square of both side) } \\
&l^2+b^2+h^2+2 l b+2 b h+2 h l=324 \\
&169+2(\mathrm{~b}+\mathrm{bh}+\mathrm{hl})=324 \\
&2(\mathrm{lb}+\mathrm{bh}+\mathrm{hl})=324-169 \\
&\text { total surface area of the cuboid }=155 \mathrm{~cm}^2
\end{aligned}
$

Perimeter of circle $=$ perimeter of square
$2 \pi r=4 a$
$2 \times \frac{22}{7} \times 21=4 \times a$
$a=33 \mathrm{~m}$
side of the square park $=33 \mathrm{~m}$

Side of park $=14 \mathrm{~m}$ and $8 \mathrm{~m}$
Side of park with path $=16 \mathrm{~m}$ and $10 \mathrm{~m}$
Area of path = area of big rectangular park - area of small rectangular park
$
\begin{aligned}
&=16 \times 10-14 \times 8 \\
&=160-112 \\
&=48 \mathrm{~m}^2
\end{aligned}
$

Let $7 \mathrm{x}$ and $5 \mathrm{x}$ be the length and breadth of rectangle

Perimeter of rectangle $=\frac{2880}{15}=192 \mathrm{~m}$

Perimeter of rectangle $=2(1+b)$

$\Rightarrow$ $192 =2(7 \mathrm{x}+5 \mathrm{x})$

$\Rightarrow$ $ 12 \mathrm{x} =96 $

$\Rightarrow$  $ \mathrm{x} =8 $

So, length of rectangle $=7 \times 8=56 \mathrm{~m}$

breadth of rectangle $=5 \times 8=40 \mathrm{~m}$