Subject Experts Solutions for Chapter: Mensuration, Exercise 2: PRACTICE QUESTIONS (COMPETITION WING)

The side of a square metal sheet is $14 \mathrm{cm}$. A circular sheet of the greatest diameter is cut out from it and removed. Find the area of the remaining metal sheet.
The following steps are involved in solving the above problem. Arrange them in sequential order.
$\left(A\right)$ Area of circular sheet $=\pi r^2=\frac{22}{7}\times 7\times 7=154 {\mathrm{cm}}^2$
Area of square metal sheet $=(\text{ side }{)}^2=(14{)}^2=196 {\mathrm{cm}}^2$
$\left(B\right)$ Since the circular sheet has greatest diameter, diameter of the circle $=$ side of the square $=14 \mathrm{cm}$.
$\left(C\right)$ $\therefore$ Radius of the circular sheet $=\frac{14}{2}=7 \mathrm{cm}$
$\left(D\right)$ $\therefore$ The area of the remaining metal sheet $=$ Area of the square metal sheet $-$ Area of the circular sheet $=196-154=42 {\mathrm{cm}}^2$.

The inner and the outer radii of a circular track are $7 \mathrm{m}$ and $14 \mathrm{m}$, respectively. Find the cost of leveling the track at the rate of $₹100$ per ${\mathrm{m}}^2$.
The following steps are involved in solving the above problem. Arrange them in sequential order
$\left(A\right)$ Area of circular track $=\pi \left(R^2-r^2\right)$ and given $R=14 \mathrm{m}$ and $r=7 \mathrm{m}$
$\left(B\right)$ $\therefore$ Area of circular track $=\frac{22}{7}(196-49)$

$\left(C\right)$ Area of the track $=\frac{22}{7}\times 147=22\times 21=154 {\mathrm{cm}}^2$
$\left(D\right)$ Cost of levelling the path $=462\times 100=₹46,200$

The perimeter of an equilateral triangle is $18 \mathrm{cm}$; find the area of the triangle.
The following steps are involved in solving the above problem. Arrange them in sequential order.
$\left(A\right)$ Area of the equilateral triangle $\left(A\right)$ $=\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}\times (6{)}^2}{4}$
$\left(B\right)$ $A=9\sqrt{3} {\mathrm{cm}}^2$
$\left(C\right)$ Let the side of the equilateral triangle be $a$ and its perimeter $=3a=18 \mathrm{cm}$ (given).
$\left(D\right)$ $a=6 \mathrm{cm}$

The area of the base of a cuboid is $800 {\mathrm{cm}}^2$ and its height is $60 \mathrm{cm}$. Find the cost of the quantity of milk filled in the cuboid at the rate of $₹13$ per litre.

The following steps are involved in solving the above problem. Arrange them in sequential order.
$\left(A\right)$ Since $1000 {\mathrm{cm}}^3=1 \mathrm{l}\Rightarrow$ Capacity of the cuboid $=48 \mathrm{l}$
$\left(B\right)$ $\therefore$ Volume of the cuboid $=800\times 60=48000 {\mathrm{cm}}^3$
$\left(C\right)$ The cost of the milk filled in the cuboid $=48\times 13=₹624$
$\left(D\right)$ Given area of the base $=800 {\mathrm{cm}}^2$ and height $h=60 \mathrm{cm}$
Volume of cuboid $=$ Area of the base $\times$ Height

Select the correct answer from the given options:

Select the correct answer from the given options:

In the figure given below, $ABCD$ is a rectangle. $\mathrm{ED}=16 \mathrm{cm}, \mathrm{BF}=20 \mathrm{cm}, \mathrm{FG}=\mathrm{FC}=8 \mathrm{cm}$, and $\mathrm{FG}$ is perpendicular to $\mathrm{BC}$. If $\mathrm{CD}$ is half of $\mathrm{AD}$ then find the area of the shaded region. (in ${\mathrm{cm}}^2$)

Two cubes of edge $10 \mathrm{m}$each joined together to form a single cuboid. The ratio of the total surface area of the one of the cubes and that of the single cuboid formed is _____.