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MATHS 2 Marks Questions

Mensuration · Gujarat Board (GSEB) STD 6 (GSEB - English Medium). 22 questions.

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2 Marks Questions

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22 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks
A room is $4\ m$ long and $3\ m \ 50 \ cm$ wide. How many square metre of carpet is needed to cover the floor of the room?
Answer
Length of the room $= 4\ m$
Breadth of the room $= 3\ m\ 50\ cm$
$= 3.50\ m$
$\therefore$ Area of the room $=$ Length $\times$ Breadth
$= 4 \times 3.5\ sq. m$
$= 14.0\ sq\ m$
Hence, $14.0$ square metres of carpet is needed to cover the floor of the room.
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Question 22 Marks
The area of a rectangular garden $50\ m$ long is $300\ sq. m.$ Find the width of the garden.
Answer
Area of the rectangular garden $= 300\ sq\ m$
Length of the rectangular garden $= 50\ m$
$\therefore $ Width of the rectangular garden
$=\frac{\text { Area of the rectangular garden }}{\text { Length of the rectangular garden }}$
$=\frac{300}{50} \mathrm{m}$
$= 6\ m$
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Question 32 Marks
Find the area of the rectangle whose side are $2\ m$ and $70\ cm.$
Answer
$1\ m = 100 \ cm$
$2 \  m = 2 \times 100 \ cm = 200 \ cm$
Area of the rectangle = Length $\times$ Breadth $= 200 \times 70\ sq \ cm = 14,000\ sq \ cm$
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Question 42 Marks
By splitting the figure into rectangle, find their area.(The measures are given in centimetre).
Answer


Area of the figure $= (3 \times 1 + 3 \times 1 + 3 \times 1) sq \ cm$
$= (3 + 3 + 3) sq \ cm$
$= 9 sq \ cm$
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Question 52 Marks
What is the perimeter of the figure? What do you infer from the answer?
Answer


Area of the figure $= (3 \times 3 + 1 \times 2 + 3 \times 3 + 4 \times 2)\ sq \ cm$
$= (9 + 2 + 9 + 8)\ sq \ cm$
$= 28\ sq \ cm$
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Question 62 Marks
Find the area of the figure by counting squares:
Answer
Full-filled squares $= 4$
half-filled squares $= 4$
Area covered by full squares $= 4 \times 1\ sq$ unit $= 4\ sq$ units
Area covered by half squares $= 4 \times \frac{1}{2}\ sq$ unit $= 2\ sq$ units
$\therefore$ Total Area$ = 4\ sq$ units $+\ 2\ sq$ units $= 6\ sq$ units
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Question 72 Marks
Find the area of the figure by counting squares:
Answer
Full-filled squares $= 2$
Half-filled squares $= 4$
$Area covered by full sq$uares$ = 2 \times 1 $ sq units $= 2$ sq units
Area covered by half squares $= 4 \times \frac{1}{2}$ sq units $= 2$ sq units
$\therefore$ Total Area $= 2$ sq units $+\ 2$ sq units $= 4$ sq units
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Question 82 Marks
Find the area of the figure by counting squares:
Answer
Full-filled squares $= 2$
half-filled squares $= 4$
Area covered by full squares $= 2 \times 1$ sq unit =$ 2$ sq units
Area covered by half squares $= 4 \times \frac{1}{2}$ sq unit $= 2$ sq units
$\therefore$ Total Area $= 2$ sq units $+\ 2$ sq units $= 4$ sq units
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Question 92 Marks
Find the side of the square whose perimeter is $20 m.$
Answer
Perimeter of the square
$= 4\ \times $ Length of a side
$\therefore $ Length of one side = $\frac{\text { Perimeter of the square }}{4}$
$=\frac{20}{4} \mathrm{m}$
$= 5 m$
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Question 102 Marks
Find the perimeter of a regular hexagon with each side measuring $8 m.$
Answer
Perimeter of the regular hexagon $= 6\ \times $ length of a side
$= 6 \times (8 m)$
$= 48 m$
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Question 112 Marks
Find the perimeter of a triangle with sides measuring $10 \ cm, 14 \ cm$ and $15 \ cm.$
Answer
Perimeter of the triangle $=$ Sum of the length of its three sides
$= 10 \ cm + 14 \ cm + 15 \ cm$
$= 39 \ cm$
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Question 122 Marks
Find the perimeter of the shape:
An isosceles triangle with equal sides $8 \ cm$ each and third side $6 \ cm$.
Answer
We know in isosceles triangle two sides are equal.
So,
Perimeter $= (2 \times 8) + 6$
$= 16 + 6$
$= 22 \ cm$
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Question 132 Marks
Find the perimeter of the shape:
An equilateral triangle of side $9 \ cm.$
Answer
We know that an equilateral triangle has all sides equal.
Thus, Perimeter of an equilateral triangle $= 3\ \times$ Side of triangle
$= (3 \times 9) \ cm$
$= 27 \ cm$
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Question 142 Marks
Find the perimeter of the shape:
A triangle of sides $3 \ cm, 4 \ cm,$ and $5 \ cm.$
Answer
$i.\ $From the given figure, the side of one slab is $\frac{1}{2}m$ 

Therefore, the side of the square formed by Avneet $=\left(3 \times \frac{1}{2}\right) m=\frac{3}{2} m$
Now,
We know that the perimeter of a square $ = 4 \times$ Side $=4 \times \frac{3}{2}=6 \mathrm{m}$
$ii.\ $We know that,
Perimeter $=$ Sum of all sides

Therefore Perimeter of the Shari's arrangement $= 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1 = 10 m$
$iii.\ $The arrangement which is in the shape of the cross has a greater perimeter $($i.e. $10 m)$
$iv.\ $If we put all $9$ slabs in a line then, the perimeter will be $10 m,$ as shown below.

Thus, from the given arrangements, arrangements with perimeters greater than $10 m$ cannot be determined.
Hence $10$ is the largest possible value if the perimeter.
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Question 152 Marks
A piece of string is $30 \ cm$ long. What will be the length of each side if the string is used to form a square?
Answer
Given:
Perimeter of square $= 30 \ cm$
We know that,
The perimeter of square $= 4\ \times$ Side
$30 = 4\ \times$ Side
Side $ = \frac{30}{4}$
$= 7.5 \ cm$
Hence, the side of a square is $7.5 \ cm$
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Question 162 Marks
The perimeter of a regular hexagon is $18 \ cm.$ How long is its one side$?$
Answer
Given the Perimeter of regular hexagon $= 18 \ cm$
A regular hexagon has $6$ sides, so we can divide the perimeter by $6$ to get the length of one side.
One side of the hexagon $= 18 \ cm \div 6 = 3 \ cm$
Therefore, the length of each side of the regular hexagon is $3 \ cm.$
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Question 172 Marks
Find the distance travelled by Shaina if she takes three rounds of a square park of side $70 m.$
Answer
Given that length of a side $= 70m$
Perimeter of the square park $= 4\ \times$ length of a side $= 4 \times 70 m = 280 m$
Distance covered in one round $= 280 m$
Therefore, distance travelled in three rounds $= 3 \times 280m = 840m$
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Question 182 Marks
A farmer has a rectangular field of length and breadth $240 m$ and $180 m$ respectively. He wants to fence it with $3$ rounds of rope as shown in the figure. What is the total length of rope he must use$?$
Answer
The farmer has to cover three times the perimeter of that field.
Thus, the total length of rope required is thrice its perimeter.
Perimeter of the field $= 2 \times ($length $+$ breadth$)$
$= 2 \times ( 240 m + 180 m)$
$= 2 \times 420 m = 840 m$
Therefore, Total length of rope required $= 3 \times 840 m = 2520 m$
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Question 192 Marks
Find the area in a square metre of a piece of cloth $1\ m\ 25\ cm$ wide and $2\ m$ long.
Answer
Given:
Length of the cloth $= 2\ m$
Breadth of the cloth $= 1\ m\ 25 \ cm = 1\ m + 0. 25\ m = 1.25\ m ($since $25 \ cm = 0.25\ m)$
Therefore, Area of the cloth = length of the cloth $\times$ breadth of the cloth
$= 2\ m \times 1.25\ m = 2.50\ sq\ m$
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Question 202 Marks
The area of a rectangular piece of cardboard is $36\ sq\ cm$ and its length is $9\ cm.$ What is the width of the cardboard$?$
Answer
Given:
Area of the rectangle $= 36\ sq \ cm$
Length $= 9 \ cm$
Width $= ?$
Area of a rectangle = length $\times$ width
So, width = $\frac{\text { Area }}{\text { Length }} = \frac{36}{9} = 4 \ cm$
Therefore, the width of the rectangular cardboard is $4 \ cm.$
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Question 212 Marks
Find the area of a rectangle whose length and breadth are $12 \ cm$ and $4 \ cm$ respectively.
Answer
Given that,
Length of the rectangle $= 12 \ cm$
Breadth of the rectangle $= 4 \ cm$
Therefore, Area of the rectangle $=$ length $\times$ breadth
$= 12 \ cm \times 4 \ cm = 48\ sq \ cm$
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Question 222 Marks
Find the area of the shape shown in the figure.
Answer
This figure is made up of line-segments. Moreover, it is covered by full squares and half squares only.
Here,
Fully-filled squares $= 3$
Half-filled squares $= 3$
If the area of one such square is taken to be as $1$ square unit.
Area covered by full squares$ = 3  \times 1$ sq units $= 3$ sq units.
Area covered by Half-filled squares$ = 3 \times 0.5 = 1.5$
Therefore, Total area $= 3 + 1.5 = 4.5 $ sq units.
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