4 Mark Question

Question 14 Marks

A suitcase with measures $80 cm \times 48 cm \times 24 cm$ is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?
TIPS Firstly, find the total surface area of 1 suitcase. Here, the area of cloth used to cover 1 suitcase will be equal to surface area of 1 suitcase. Use this result to find the length of cloth required for 1 suitcase and then multiply it by 100 to get required length of cloth to cover 100 suitcases.

Answer

Given, length of suitcase $(l)=80 cm$
Breadth of suitcase $(b)=48 cm$
and height of suitcase $(h)=24 cm$
$\therefore$ Total surface area of one suitcase
$
\begin{array}{l}
=2(l b+b h+h l) \\
=2(80 \times 48+48 \times 24+24 \times 80) \\
=2(3840+1152+1920) \\
=2 \times 6912 \\
=13824 cm^2
\end{array}
$
Also, width of tarpaulin $=96 cm$
Since, area of tarpaulin required to cover one suitcase will be equal to the total surface area of suitcase.
So, area of tarpaulin required to cover one suitcase
$
=13824 cm^2
$
$\Rightarrow$ Length $\times$ Width $=13824$
[ $\because$ tarpaulin is in the shape of a rectangle]
$
\begin{array}{l}
\Rightarrow \quad \text { Length } \times 96=13824 \\
\Rightarrow \quad \text { Length }=\frac{13824}{96}=144 cm
\end{array}
$
Thus, length of tarpaulin required to cover
$
\begin{aligned}
100 \text { suitcase } & =100 \times 144=14400 cm \\
& =\frac{14400}{100} m=144 m
\end{aligned}
$
$\left[\because 1 m=100 cm \Rightarrow 1 cm=\frac{1}{100} m\right]$
Hence, 144 m of tarpaulin with width 96 cm is required to cover 100 suitcases.

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Question 24 Marks

Most of the sailboats have two sails, the jib and the mainsail. Assume that the sails are triangles. Find the total area of each sail of the sailboats to the nearest tenth.

Answer

(i) In Ist sailboat, area of sails is calculated below
Area of $\triangle A B D=\frac{1}{2} \times$ Base $\times$ Height
$\begin{array}{l}=\frac{1}{2} \times B D \times A E=\frac{1}{2} \times(20+22) \times 22.3 \\ =\frac{1}{2} \times 42 \times 22.3=468.3 m^2\end{array}$
Area of $\triangle B C D=\frac{1}{2} \times B D \times F C$
$\begin{array}{l}=\frac{1}{2} \times(20+22) \times 16.8 \\ =\frac{1}{2} \times 42 \times 16.8=352.8 m^2\end{array}$
(ii) In Ind sailboat, area of sails is calculated below:
Area of $\triangle A B E=\frac{1}{2} \times$ Base $\times$ Height
$=\frac{1}{2} \times 10.9 \times 19.5=106.3 m^2$
Similarly, area of $\triangle B C D=\frac{1}{2} \times 23.9 \times 8.6=102.8 m^2$
(iii) In IIIrd sailboat, area of sails is calculated below:
Area of $\triangle A B C=\frac{1}{2} \times$ Base $\times$ Height
$=\frac{1}{2} \times 3 \times 8.9=13.35 m^2$
$\begin{array}{l}\text { Now, area of } \triangle D E H=\frac{1}{2} \times 96 \times 168=80.64 m^2 \\ \text { and area of } \triangle E F H=\frac{1}{2} \times 25 \times 124=155 m^2\end{array}$
So, area of sail $D F H=80.64+155=235.64 m^2$

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Question 34 Marks

Word Maze

Find the names of the solids from the given word mare whose areas or volumes are given below by colouring the boxes using the given colour code

S.No	Area/Volume	Colour Code
1	$\frac{1}{2} d_1 \times d_2$	red
2	$l b h$	blue
3	$\pi r^2 h$	yelloe
4	$\pi r^2$	green
5	$\frac{1}{2} b h$	orange
6	$\frac{1}{2}(a+b) \times h$	pink

Answer

Self

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Question 44 Marks

Find the area of the shaded portion in the following figure.

Answer

Area of a trapezium $A B C D=\frac{1}{2} \times(A B+C D) \times A E$
$=\frac{1}{2} \times(120+160) \times 100$
$\begin{array}{l}=\frac{1}{2} \times 280 \times 100 \\ =14000 cm^2\end{array}$
Now, area of a rectangle $=l \times b$
$=40 \times 20=800 cm^2$
Now, area of a rectangle $=l \times b$
$=40 \times 20=800 cm^2$
$\begin{aligned} \text { and area of a circle } & =\pi r^2=\frac{22}{7} \times 7 \times 7 \\ =154 cm^2\end{aligned}$
$\therefore$ Area of shaded portion = Area of trapezium - (Area of rectangle + Area of circle)
$\begin{array}{l}=14000-(800+154) \\ =13046 cm^2\end{array}$
Hence, the area of the shaded portion is $13046 cm^2$.

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Question 54 Marks

A rectangular sheet of a paper is rolled in two different ways to form two different cylinders. Find the volume of cylinders in each case, if the sheet measures $44 cm \times 33 cm$.

Answer

There are two cases
Case I When the rectangular sheet is rolled along its length, then length of the sheet forms the circumference of its base and breadth of sheet becomes the height of cylinder.
Let $r cm$ be the radius of the base and $h cm$ be the height.
Then, $h=33 cm$
and circumference of the base $=44 cm$
$\begin{array}{l}\Rightarrow \quad 2 \pi r=44 \\ \Rightarrow \quad 2 \times \frac{22}{7} \times r=44 \Rightarrow r=\frac{44 \times 7}{2 \times 22} \Rightarrow r=7 cm\end{array}$
$\therefore$ Volume of the cylinder $=\pi r^2 h$
$=\frac{22}{7} \times 7 \times 7 \times 33=5082 cm^3$
Case II When the rectangular sheet is rolled along its breadth, then breadth of the sheet forms the circumference of its base and length of the sheet becomes the height of the cylinder.
Now, $h=44 cm$ and circumference $=33 cm$
$\begin{array}{ll}\Rightarrow & 2 \pi r=33 \Rightarrow 2 \times \frac{22}{7} \times r=33 \\ \Rightarrow & r=\frac{33 \times 7}{2 \times 22} \Rightarrow r=\frac{21}{4} cm\end{array}$
$\therefore$ Volume of the cylinder $=\pi r^2 h$
$\begin{array}{l}=\frac{22}{7} \times \frac{21}{4} \times \frac{21}{4} \times 44 \\ =3811.5 cm^3\end{array}$

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