Question 13 Marks
What will happen to the area of a rectangle if its
Length and breadth are trebled.
AnswerIf the length and breadth of a rectangle are trebled.
Let the initial length and breadth be $l$ and $b$, respectively.
Original area $= l \times b = lb$
Now,
the length and breadth are trebled which means they become three times of their original value.
Therefore New length $= 3l$
New breadth $= 3b$
New area $= 3l \times 3b = 9lb$
Thus, the area of the rectangle will become $9$ times that of its original area.
View full question & answer→Question 23 Marks
Find the perimeter of a rectangle whose area is $500cm^2$ and breadth is $20\ cm.$
AnswerArea $= 500\ cm^2$
Breadth $= 20\ cm$
Area of rectangle $=$ (Length $\times $ Breadth)
Therefore Length $=\frac{\text{Area}}{\text{Breadth}}$
$=\frac{500}{20}=25\text{cm}$
Perimeter of a rectangle $= 2$(Length $+$ Breadth)
$= 2(25 + 20)cm$
$= 2 \times 45\ cm$
$= 90\ cm$
View full question & answer→Question 33 Marks
What will happen to the area of a square if its side is: Tripled.
AnswerLet the original side of the square be s.
Original area $= s \times s = s^2$
If the side of a square is tripled, new side will be equal to $3s.$
New area $= 3s \times 3s = 9s^2$
This means that the area becomes $9$ times that of the original area.
View full question & answer→Question 43 Marks
Using tracing paper and centimeter graph paper to compare the areas of the following pairs of figure:
Answer
Using tracing paper, we traced the figure on a graph paper. This figure contains $8$ complete squares, $11$ more than half squares and $10$ less than half squares. Let us assume that the area of one square is $1cm^2$. If we neglect the less than half squares and consider the area of more than half squares as equal to area of complete square, we get: Area of this shape $= (8 + 11) = 19cm^2$ On comparing the areas of these two shapes, we get that the area of Fig.$(ii)$ is more than that of Fig.$(i).$ View full question & answer→Question 53 Marks
The side of a square field is $65m$. What is the length of the fence required all around it?
AnswerSide of the square field $= 65m$
Length of the fence around the square field = Perimeter of the square field $= 4 \times $ (Side of the square)
Perimeter of the square field $= 4 \times 65 = 260m$
Thus, the length of the fence around the square filed $= 260m$
View full question & answer→Question 63 Marks
How many tiles with dimension $5\ cm$ and $12\ cm$ will be needed to fit a region whose length and breadth are respectively? $70\ cm$ and $36\ cm$
AnswerDimension of the tile $= 5\ cm \times 12\ cm$
Dimension of the region $= 70\ cm \times 36\ cm$
Area of the tile $= 5\ cm \times 12\ cm$
$= 60\ cm^2$
Area of the region $= 70\ cm \times 36\ cm$
$= 2,520\ cm^2$
Number of tiles required to cover the region $=\frac{\text{Area of the region}}{\text{Area of one tile}}$
$=\frac{2520}{42}=42\text{tiles}$
View full question & answer→Question 73 Marks
The dimensions of a photographs are $30\ cm \times 20\ cm$. What length of wooden frame is needed to frame the picture?
AnswerDimensions of the photograph $= 30\ cm \times 20\ cm$
So, the required length of wooden frame $=$ Perimeter of the photograph $= 2$(Length $+$ Breadth)
$= 2(30 + 20)cm = 2 \times 50\ cm = 100\ cm$
View full question & answer→Question 83 Marks
Draw any circle on the graph paper, Count the squares and use them to estimate the area the area of the circular region.
Answer
This circle on the squared paper consists of $21$ complete squares, $15$ more than half squares and $8$ less than half squares.
Let us assume that the area of $1$ square is $1cm^2$.
If we neglect the less than half squares while approximating more than half square as equal to a complete square, we get:
Area of this shape $= (21 + 15) = 36cm^2$ View full question & answer→Question 93 Marks
On a squared paper, draw any irregular closed figure. Find the approximate area by counting the number of squares complete, more than half and exactly half.
AnswerAny irregular figure: This figure consists of $10$ complete squares, $1$ exactly half square, $7$ more than half squares and 6 less than half squares.
If we assume that the area of one complete square is $1\ cm^2$,
Then the area of this shape $= (10 + 1 \times 12 + 7 \times 1) = 17.5\ cm^2$
View full question & answer→Question 103 Marks
A marble tile measures $15\ cm \times 20\ cm$. How many tiles will be required to cover a wall of size $4m \times 6m?$
AnswerDimensions of the tile $= 15\ cm \times 20\ cm$
Dimensions of the wall $= 4m \times 6m = 400\ cm \times 600\ cm$ (Since, $1m = 10\ cm)$
Area of the tile $= 15\ cm \times 2\ cm = 300\ cm^2$
Area of the wall $= 400\ cm \times 600\ cm$
$= 2,40,000\ cm^2$
Number of tiles required to cover the well $=\frac{\text{Area of wall}}{\text{Area of one tile}}$
$=\frac{2400}{300}=800\text{tiles}$
View full question & answer→Question 113 Marks
To fix fence wires in a garden, $70m$ long and $50m$ wide, Arvind bought metal pipes for bosts, he fixed a post every $5$ metres apart. Each post was $2m$ long. What is total length of the pipes he bought for the posts?
AnswerLength of the garden $= 70m$
Breadth of the garden $= 50m$
Perimeter of the garden $= 2$(Length + Breadth) $= 2(70 + 50) = 2 \times 120 = 240m$ On the perimeter of the garden,
it is given that Arvind fixes a post every $5$ metres apart.
So, the number of posts required $=\frac{240}{5}=48$
Since, Length of each post $= 2m$
View full question & answer→Question 123 Marks
How many tiles with dimension $5\ cm$ and $12\ cm$ will be needed to fit a region whose length and breadth are respectively? $100\ cm$ and $144\ cm$
AnswerDimension of the tile $= 5\ cm \times 12\ cm$
Dimension of the region $= 100\ cm \times 144\ cm$
Area of the tile $= 5\ cm \times 12\ cm$
$ = 60\ cm^2$
Area of the region $= 100\ cm \times 144\ cm$
$ = 14,400\ cm^2$
Number of tiles required to cover the region $=\frac{\text{Area of region}}{\text{Area of one tile}}$
$=\frac{14400}{60}=240\text{tiles}$
View full question & answer→Question 133 Marks
A square piece of land has each side equal to $100m$. If $3$ layers of metal wire has to be used to fence it, what is the length of the wire needed?
AnswerSide of the square field $= 100m$
Wire required to fence the square field $=$ Perimeter of the square field $= 4 \times $ (Side of the square field Perimeter)
$= 4 \times 100 = 400m$
This perimeter is the length of wire required to fence one layer.
Therefore, the length of wire required to fence three layers $= 3 \times 400m = 1200m$
View full question & answer→Question 143 Marks
The length of a rectangular fields is $100m$. If its perimeter is $300m$, what is its breadth?
AnswerLength of the rectangular field $= 100m$
Perimeter of the rectangular field $= 300m$
Perimeter of a rectangle $= 2$(Length $+$ Breadth) Applying the above formula,
we get: Breadth of the rectangular field $=\frac{\text{Perimeter}}{2}-\text{Length}$
$=\frac{300}{2}-100$$=150-100$
$=50\text{m}$
View full question & answer→Question 153 Marks
A rectangular piece of lawn is $55m$ wide and $98m$ long. Find the length of the fence around it.
AnswerLength of the lawn $= 98m$
Breadth of the lawn $= 55m$
Length of the fence around the lawn $=$ Perimeter of the lawn $= 2 \times $ (Length $+$ Breadth)
Perimeter of the lawn $= 2(98 + 55)m = 2(153) = 306m$
Thus, the length of the fence around the lawn $= 306m$
View full question & answer→Question 163 Marks
The following figure have been split into rectangles. Find the area. (The measures are given in centimeter)
AnswerThis figure consists of three rectangles $I, II$ and $III.$
Area of rectangle $I =$ (Length $\times $ Breadth)
$= (3 \times 1)$
$= 3\ cm^2$
Similarly, area of rectangle $II =$ (Length $\times $ Breadth)
$= (3 \times 1)$
$= 3\ cm^2$
Area of rectangle $III=$(Length $\times $ Breadth)
$= (3 \times 1)$
$ = 3\ cm^2$
Thus, the total area of this figure $=$ (Area of rectangle $I +$ area of rectangle $II +$ area of rectangle $III)$
$= (3 + 3 + 3)$
$= 9\ cm^2$ View full question & answer→Question 173 Marks
Find the cost of fencing a rectangular park of length $175m$ and breadth $125m$ at the rate of $Rs. 12$ per meter.
AnswerLength of the park $= 175m$
Breadth of the park $= 125m$
Perimeter of the park $= 2$(Length + Breadth)
$= 2(175 + 125)$
$= 2 \times 300$
$= 600m$
Rate of fencing $= Rs. 12$ per meter
Cost of fencing $= Rs. 12 \times 600$
$= Rs. 7,200$
View full question & answer→Question 183 Marks
What will happen to the area of a rectangle if its.
Length is doubled and breadth is same.
AnswerIf the length is doubled and the breadth is same.
Let the initial length and breadth be $l$ and $b$, respectively.
Original area $= l \times b = lb$
Now, length is doubled and breadth remains same.
Therefore New length $= 2l$
New breadth $= b$
New area $= 2l \times b = 2lb$
Thus, the area of the rectangle will become $2$ times that of its original area.
View full question & answer→Question 193 Marks
The total cost of flooring a room at $Rs. 85$ per $m^2$ is $Rs. 5100$. If the length of the room is $8$ metres, then find its breadth.
AnswerAs, the total cost of flooring $= Rs. 5100$ and
The rate of flooring $= Rs. 85$ per $m^2$
So, the area of the floor $=\frac{\text{total cost of the flooring}}{\text{Rate of the flooring}}$
$=\frac{5100}{85}$
$=60\text{m}^{2}$
Also, the length of the room $= 8m$ Now,
the breadth of the room $=\frac{\text{Area of the floor}}{\text{Length of the room}}$
$=\frac{60}{8}$
$=\frac{15}{2}$
$=7.5\text{m}$
So, the breadth of the room is $7.5m.$
View full question & answer→Question 203 Marks
A room is $12.5m$ long and $8m$ wide. A square carpet of side $8m$ is laid on its floor. Find the area of the floor which is not carpeted.
AnswerWe have,
Length of the room $= 12.5m,$
Breadth of the room $= 8m$ and
Side of the square carpet $= 8m$
The area of the room = (Length $\times $ Breadth)
$= 12.5 \times 8$
$= 100m^2$
Also, the area of the square carpet = (Side $\times $ Side)
$ = 8 \times 8 = 64m^2$
Now, the area of the floor which is not carpeted $=$ (Area of the room $-$ Area of the carpet)
$= 100 - 64$
$= 36m^2$
View full question & answer→Question 213 Marks
A wire of length $20m$ is to be folded in the form of a rectangle. How many rectangles can be formed by folding the wire if the sides are positive integers in metres?
AnswerIt is given that a wire of length $20 m$ is to be folded in the form of a rectangle;
Therefore, we have: Perimeter of the rectangle $= 20m$
$\Rightarrow 2$ (Length $+$ Breadth) $= 20m$
$\Rightarrow $ (Length $+$ Breadth) $=\frac{20}{2}=10\text{m}$
Since, length and breadth are positive integers in metres,
therefore, the possible dimensions are: $(1m, 9m), (2m, 8m), (3m, 7m), (4m, 6m)$ and $(5m, 5m)$
Thus, five rectangles can be formed with the given wire.
View full question & answer→Question 223 Marks
What will happen to the area of a square if its side is: Increased by half of it.
AnswerLet the original side of the square be $s.$
Original area $= s \times s = s^2$
If the side of a square is increased by half of it, new side
$=\Big(\text{s}+\frac{1}{2}\text{s}\Big)=\frac{3}{2}\text{s}$
$\text{New area}=\frac{3}{2}\text{s}\times\frac{3}{2}\text{s}=\frac{9}{4}\text{s}$
This means that the area becomes $\frac{9}{4}$ times that of the original area.
View full question & answer→Question 233 Marks
Area of a rectangle of breadth $17\ cm$ is $340\ cm^2$. Find the perimeter of the rectangle.
AnswerArea of the rectangle $= 340\ cm^2$
Breadth of the rectangle $= 17\ cm$
Applying the formula:
Length of a rectangle $=\frac{\text{Area}}{\text{Breadth}}$
We get:
Length of the rectangle $=\frac{340}{17}=20\text{cm}$
Perimeter of rectangle $= 2$(Length $+$ Breadth)
$= 2(20 + 17)$
$= 2 \times 37$
$= 74\ cm$
View full question & answer→Question 243 Marks
What will happen to the area of a rectangle if its.
Length is doubled and breadth is halved.
AnswerIf the Length is doubled and breadth is halved.
Let the initial length and breadth be land b, respectively.
Original area $=l \times b = lb$
Now, length is doubled and breadth is halved.
Therefore New length $= 2l$
New breadth $=\frac{\text{b}}{2}$
New area $=2\text{l}\times\frac{\text{b}}{2}=\text{lb}$
New area is also $lb.$
This means that the areas remain the same.
View full question & answer→Question 253 Marks
The perimeter of a regular pentagon is $100\ cm$. How long is each sides?
AnswerA regular pentagon is a closed polygon having five sides of equal length.
Perimeter of the regular pentagon $= 100cm$
Perimeter of the regular pentagon $= 5$(Side of the regular pentagon)
Therefore, side of the regular pentagon $=\frac{\text{Perimeter}}{2}$
$=\frac{100}{5}$$=20\text{cm}$
View full question & answer→Question 263 Marks
The following figure have been split into rectangles. Find the area. (The measures are given in centimeter)
AnswerThis figure consists of two rectangles $II$ and $IV$ and two squares $I$ and $III$. Area of square $I =$ (Side $\times $ Side)
$= (3 \times 3)$
$= 9\ cm^2$
Similarity, area of rectangle $II =$ (Side $\times $ Side)
$= (2 \times 1)$
$ = 2\ cm^2$
Area of square $III =$ (Side $\times $ Side)
$= (3 \times 3)$
$ = 9\ cm^2$
Area of rectangle $IV =$ (Side $\times $ Side)
$= (2 \times 4)$
$= 8\ cm^2$
Thus, the total area of this figure $=$ (Area of square $I +$ Area of rectangle $II +$ Area of square $III +$ Area of rectangle $IV)$
$= (9 + 2 + 9 + 8)$
$ = 28\ cm^2$
View full question & answer→Question 273 Marks
A rectangle has the area equal to that of a square of side $80\ cm$. If the breadth of the rectangle is $20\ cm$, Find its length.
AnswerSide of the square $= 80\ cm$
Area of square = (Side $\times $ Side)
$= 80 \times 80$
$= 6400\ cm^2$
Given that:
Area of the rectangle = Area of the square $= 6400\ cm^2$
Breadth of the rectangle $= 20\ cm$
Applying the formula:
Length of the rectangle $=\frac{\text{Area}}{\text{Breadth}}$
We get:
Length of the rectangle $=\frac{6400}{20}=320\text{cm}$
View full question & answer→Question 283 Marks
A marble tile measures $10\ cm \times 12\ cm$. How many tiles will be required to cover a wall of size $3m \times 4m?$ Also, find the total cost of the tiles at the rate of $Rs 2$ per tile.
AnswerDimension of the tile $= 10\ cm \times 12\ cm$
Dimension of the wall $= 3m \times 4m$
$= 300\ cm \times 400\ cm$ (Since, $1m = 100\ cm)$
Area of the tile $= 10\ cm \times 12\ cm$
$ = 120\ cm^2$
Area of the wall $= 300\ cm \times 400\ cm$
$ = 1,20,000\ cm^2$
Number of tiles required to cover the wall $=\frac{\text{Area of wall}}{\text{Area of one tile} }$
$=\frac{120000}{120}=1000\text{ tiles}$
Cost of tiles at the rate of $Rs. 2$ per tile $= 2 \times 1,000$
$= Rs. 2,000$
View full question & answer→Question 293 Marks
Shikha runs around a square of side $75m$. Priya runs around a rectangle with length $60m$ and breadth $45m$. Who covers the smaller distance?
AnswerShikha and Priya, while running around the square and rectangular field respectively,
actually cover a distance equal to the perimeters of these fields.
Distance covered by Shikha $=$ Perimeter of the square $= 4 \times 75m = 300m$
Similarly, distance covered by Priya = Perimeter of the rectangle $= 2(60 + 45) = 2 \times 105 = 210m$
Thus, it is evident that the distance covered by Priya is less than that covered by Shikha.
View full question & answer→Question 303 Marks
On a squared paper, draw a triangle Find the approximate area by counting the number of squares complete, more than half and exactly half.
AnswerA triangle: This triangle contains $4$ complete squares,
$6$ more than half squares and $6$ less than half squares.
If we assume that the area of one complete square is $1\ cm^2$,
Then the area of this shape $= (4 + 6 \times 1) = 10\ cm^2$
View full question & answer→Question 313 Marks
One tile of a square plot is $250m$, find the cost of leveling it at the rate of $Rs \ 2$ per square meter.
AnswerSide of the square plot $= 250m$
Area of the square plot $=$ (Side $\times $ Side)
$= 250 \times 250$
$= 62,500m^2$
Rate of leveling the plot $= Rs. 2$ per $m^2$
Cost of leveling the square plot $= Rs. 62,500 \times 2$
$= Rs. 1,25,000$
View full question & answer→Question 323 Marks
Using tracing paper and centimeter graph paper to compare the areas of the following pairs of figure:
Answer
Using tracing paper, we traced the figure on a graph paper.
This figure contains $4$ complete squares, $9$ more than half squares and $9$ less than half squares.
Let us assume that the area of one square is $1cm^2$
If we neglect the less than half squares and consider the area of more than half squares as equal to area of complete square,
we get:
Area of this shape $= (4 + 9) = 13cm^2$ View full question & answer→Question 333 Marks
On a squared paper, draw a rectangle Find the approximate area by counting the number of squares complete, more than half and exactly half.
AnswerA rectangle: This contains $18$ complete squares. If we assume that the area of one complete square is $1cm^2$, Then the area o this rectangle will be $18cm^2$.
View full question & answer→Question 343 Marks
A rectangular piece of land measure $0.7km$ by $0.5km$. Each side is to be fenced with four rows of wires. What length of the wire is needed?
AnswerDimensions of the rectangular land $= 0.7\ km \times 0.5\ km$
Perimeter of the rectangular land $= 2$(Length $+$ Breadth) $= 2 (0.7 + 0.5)km = 2 \times 1.2\ km = 2.4\ km$
This perimeter is equal to one row of wire required to fence the land.
Therefore, length of wire required to fence the land with four rows of wire $= 4 \times 2.4\ km = 9.6\ km$
View full question & answer→Question 353 Marks
Two sides of a triangle are $15\ cm$ and $20\ cm$. The perimeter of the triangles is $50\ cm$. What is the third side?
AnswerGiven: Perimeter $= 50\ cm$
Length of the first side $= 15\ cm$
Length of the second side $= 20\ cm$
We have to find the length of the third side.
Perimeter of a triangle $=$ Sum of all three sides of the triangle
Length of the third side $=$ (Perimeter of the triangle) $-$ (Sum of the length of the other two sides)
$= 50 - (15 + 20) = 50 - 35 = 15\ cm$
View full question & answer→