Free Surface Areas and Volumes 01 Practice Test - 9th Grade 

Given,
Radius $\left(r\right)$ = 7 $\text{cm}$
Slant height $\left(l\right)$ = 10 $\text{cm}$

Area of the aluminium sheet required
=Total surface area of the cone
= $\pi r(l+r)$
= $\frac{22}{7}\times 7(10+7)$
= 374 ${\text{cm}}^2$

Hence, the area of the aluminium sheet required is 374 ${\text{cm}}^2.$

Radius of the sphere
$=\frac{diameter}{2}=\frac{14}{2}=7m$
 
Area available to the motorcyclist = Surface area of the sphere
$=4\pi r^2$
$=4\times \frac{22}{7}\times 7^2$
$=616m^2$

Hence, the available area to the motorcyclist is 616 $m^2$.

$1\text{litre =}1000{\text{cm}}^3$
$40\text{litres =}40000{\text{cm}}^3$

Volume of water in the container
$=40\text{litres =}40000{\text{cm}}^3$

Volume of each ice cube
$=(\text{Side}{)}^3$
$=5^3$
$=125{\text{cm}}^3$

$\therefore$ Number of ice cubes that can be fitted

$=\frac{\text{Volume of water container}}{\text{Volume of each cube}}$

$=\frac{40000{\text{cm}}^3}{125{\text{cm}}^3}$

$=320\text{ice cubes}$

Given,
Total surface area of the cube $=54{\text{cm}}^2$

We know that,
Cube has 6 number of faces.
Each face of the Rubik's cube contains 9 tiles.
Each Rubik's cube contains 9 red tiles.

$\therefore \text{Area of each face}$
$=\frac{\text{Total surface area of the cube}}{\text{Total number of faces of the cube}}=\frac{54}{6}=9{\text{cm}}^2$

Now, $\text{area of each tile}$
$=\frac{\text{Area of each face}}{\text{Total number of tiles in each face}}=\frac{9}{9}=1{\text{cm}}^2$

Hence, the area occupied by every single red tile of the cube is $1{\text{cm}}^2$.

We know that,
volume of a cylinder $=\pi r^{2}h$

Number of glasses of juice that are sold
$=\frac{\text{volume of the vessel}}{\text{volume of each glass}}=\frac{\pi \times 10\times 10\times 35.2}{\pi \times 4\times 4\times 10}=22$

$\therefore$ The amount earned by the shopkeeper
$=\text{Rate of each glass}\times \text{Total number of glasses}$
$=10\times 22=₹220/-$

Given,
Radius of the cone
 $r=7cm$
Slant height of the cone
$l=14cm$

From the pythagoras theorem
$l^2=h^2$ + $r^2$,

height of the cone
$h=\sqrt{{14}^2-7^2}$
$h=\sqrt{147}=12.1243cm$

Volume of a Cone
$=\frac{1}{3}\times \pi \times r^{2}h$

$=\frac{1}{3}\times \frac{22}{7}\times 7^2\times 12.1243$

$=622.381c{m}^3$

We know that,
volume of cone $=\frac{1}{3}\pi r^{2}h$
volume of hemisphere $=\frac{2}{3}\pi r^3$

Total volume
    = Volume of the cone + Volume of              the hemisphere

    $=\frac{1}{3}\pi r^{2}h+\frac{2}{3}\pi r^3=\frac{1}{3}\pi r^2(h+2r)=\frac{1}{3}\times \frac{22}{7}\times {3.5}^2\times (5+7)=\frac{1}{3}\times \frac{22}{7}\times {3.5}^2\times 12=22\times 0.5\times 3.5\times 4=154{\text{cm}}^3$

$\therefore$ The volume of the solid formed is
    $154{\text{cm}}^3$      

Cuboids can be formed by stacking rectangles as shown in the figure.

A   square can be defined as a rectangle in which the adjacent sides are equal. Thus, stacking squares also yields cuboids. In fact, we obtain a special type of cuboid in this case, called the cube.

If right angle triangle is rotated along any of it's perpendicular side then we will get a right circular cone. See the image below