3 Marks Question

Question 13 Marks

A road roller takes $750$ complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is $84 cm$ and length is $1 m.$

Answer

Diameter of the road roller $= 84 cm$
$\therefore$ Radius $(r)$ of the road roller $ = \frac{{84}}{2}cm = 42 cm$
Length $(h)$ of the road roller $= 1m = 100 cm$
$\therefore$ Lateral surface area of the road roller $ = 2\pi rh$
$ = 2 \times \frac{{22}}{7} \times 42 \times 100$
$= 26400\ cm^2$
$\therefore$ Area of the road covered in $1$ complete revolution $= 26400\ cm^2$
$\therefore$ Area of the road covered in $750$ complete revolutions
$= 26400\ cm^2$ $\times$ $750\ cm^2$
$= 19800000\ cm^2$
$ = \frac{{19800000}}{{100 \times 100}}{m^2}$
$= 1980\ m^2$

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Question 23 Marks

The lateral surface area of a hollow cylinder is $4224\ cm^2$. It is cut along its height and formed a rectangular sheet of width $33 \ cm.$ Find the perimeter of rectangular sheet?

Answer

Lateral surface area of the hollow cylinder $= 4224\ cm^2$
$\therefore$ Area of the rectangular sheet $= 4224\ cm^2$
$\therefore$ Length $\times 33 = 4224$
$\therefore$ Length $ = \frac{{4224}}{{33}}$
$\therefore$ Length $= 128 cm$
$\therefore$ Perimeter of the rectangular sheet
$= 2(Length + Breadth)$
$= 2(128 + 33) cm$
$= 2(161) cm$
$= 322 cm$
Hence, the perimeter of the rectangular sheet is $322 \ cm.$

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Question 33 Marks

Rukhsar painted the outside of the cabinet of measure $1 m \times 2 m \times 1.5 m$. How much surface area did she cover if she painted all except the bottom of the cabinet.

Answer

$l = 2 m$
$b = 1 m$
$h = 1.5 m$
Required area
$= 2 (l \times b + b \times h + h \times l) – l \times b$
$= 2 (2 \times 1 + 1 \times 1.5 + 1.5 \times 2) m^2– (2 \times 1) m^2$
$= 13\ m^2– 2\ m^2$
$= 11\ m^2$
Hence, she covered $11\ m^2$ of surface area.

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Question 43 Marks

Mohan wants to buy a trapezium-shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is $10500 m^2$ and the perpendicular distance between the two parallel sides is $100 \ m$, find the length of the side along the river.

Answer

Let the length of the side along the river be $2 x m.$
Then, the length of the side along the road is $x m.$
Area of the field $= 10500$ square metres
$\therefore \frac{{(2x + x) \times 100}}{2} = 10500$
$\therefore 150x = 10500$
$\therefore x = \frac{{10500}}{{150}}$
$\therefore x = 70$
$\therefore 2x = 2 \times 70 = 140 m.$
Hence, the length of the side along the river is $140 m.$

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Question 53 Marks

Length of the fence of a trapezium-shaped field $A B C D$ is $120 m$ . If $B C=48 m, C D=17 m$ and $A D=40 m$, find the area of this field. Side $A B$ is perpendicular to the parallel sides $A D$ and $B C$.

Answer

Fence of the trapezium shaped field $ABCE = 120 m$
$\therefore AB + BC + CD + DA = 120$
$ \therefore AB + 48 + 17 + 40 = 120$
$ \therefore AB + 105 = 120$
$ \therefore AB = 120 – 105$
$ \therefore AB = 15 m$
$\therefore$ Area of the field $ = \frac{{(BC + AD) \times AB}}{2}$
$ = \frac{{(48 + 40) \times 15}}{2}$
$= 660\ m^2$.

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Question 63 Marks

The area of trapezium is $34\ cm^62$ and the length of one of the parallel sides is $10 \ cm$ and its height is $4 \ cm$. Find the length of the another parallel side.

Answer

Area of trapezium
$ = \frac{1}{2}h(a + b)$
$\therefore 34 = \frac{1}{2} \times 4(10 + b)$
$\therefore 34 = 2 \times (10 + b)$
$\therefore 10 + b = \frac{{34}}{2}$
$\therefore 10 + b = 17$
$\therefore b = 17 – 10$
$\therefore b = 7 cm$
Hence, the length of another parallel side is $7 cm.$

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Question 73 Marks

Diagram of the adjacent picture frame has outer dimensions $= 24 cm \times 28 cm$ and inner dimensions $16 cm \times 20 cm$. Find the area of each section of the frame, if the width of each section is the same

Answer

Area of the right section of the frame
$ = \frac{{(28 + 20) \times \frac{1}{2}(24 - 16)}}{2}c{m^2}$
$ = \frac{{48 \times 4}}{2}c{m^2}$
$= 96\ cm^2$
Area of the left section of the frame $= 96\ cm^2$
Area of the upper section of the frame
$ = \frac{{(24 + 16) \times \frac{1}{2}(28 - 20)}}{2}c{m^2}$
$ = \frac{{40 \times 4}}{2}c{m^2}$
$= 80\ cm^2$
Area of the lower section of the frame
$= 80\ cm^2$

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Question 83 Marks

The internal measures of a cuboidal room are $12 m \times 8 m \times 4 m$. Find the total cost of whitewashing all four walls of a room, if the cost of white washing is ₹$5$ per $m^2$. What will be the cost of white washing if the ceiling of the room is also whitewashed?

Answer

Let the length of the room $= l = 12 m$
Width of the room $= b = 8 m$
Height of the room $= h = 4 m$
Area of the four walls of the room = Perimeter of the base $\times$ Height of the room
$= 2 (l + b) \times h = 2 (12 + 8) \times 4$
$= 2 \times 20 \times 4 = 160\ m^2 $
Cost of white washing per $m^2= ₹5$
Hence, the total cost of white washing four walls of the room $= ₹ (160 \times 5) = ₹800$
Area of ceiling $= 12 \times 8 = 96\ m^2$
Cost of white washing the ceiling $= ₹(96 \times 5) = ₹480$
So, the total cost of white washing $= ₹(800 + 480) = ₹1280$

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Question 93 Marks

An aquarium is in the form of a cuboid whose external measures are $80 cm \times 30 cm \times 40 cm$. The base, side faces and back face are to be covered with a coloured paper. Find the area of the paper needed.

Answer

The length of the aquarium $= l = 80 cm$
Width of the aquarium $= b = 30 cm$
Height of the aquarium $= h = 40 cm$
Area of the base$ = l × b = 80 × 30 = 2400\ cm^2$
Area of the side face $= b × h = 30 × 40 = 1200\ cm^2$
Area of the back face $= l × h = 80 × 40 = 3200\ cm^2$
Required area = Area of the base $+$ area of the back face $+ (2 \times$ area of a side face)
$= 2400 + 3200 + (2 \times 1200) = 8000\ cm^2$

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