Mensuration - Class 6 Maths - Extra Questions

Find the length of a rectangle, which has:
$$area\, =\,120\,\, cm^2$$ and $$breadth\, = \,8\,cm$$

Let the length of the rectangle $$=l$$ cm

Area of the rectangle $$ = l\times b = 120{ cm }^{ 2 } $$
$$=> 8l = 120$$
$$ => l = 15  cm$$


Find the perimeter (in metres) of a rectangle of length 15 m and its breadth is 8 m.

$$Perimeter of Rectangle = 2(Length + Breadth)$$
$$ \Rightarrow Perimeter = 2 (15+8) $$
$$ \Rightarrow Perimeter = 46 \  m $$


The perimeter of the rectangle whose length $$= 25$$ cm breadth $$= 15$$ cm is_______cm

Perimeter of  the rectangle is sum of all sides
$$25 + 15 + 25 + 15 = 80 cm$$
Perimeter$$=80 cm$$


The perimeter(in m) of a rectangular plot whose length is $$75$$ m and breadth is $$50$$ m is

Perimeter of a rectangle $$=$$ sum of its sides
                                     
                                          $$=2 (l+b)$$

                                          $$=2(50+75)$$

                                          $$=250\ m$$


The perimeter of a square field whose side is $$4$$ m is______m

perimeter of square$$=4 \times$$ side
                                 $$=4\times4$$
                                 $$=16 m$$


The length of rectangle is 10 cm.If the perimeter is 34 cm , find the breadth (cm).

The length of the rectangle is 10 cm and perimeter is 34 cm.
We know for a rectangle
perimeter = 2 ( length + breadth ) 
Hence, $$ 2 (10 + \textrm{breadth} .) = 34 \\ \Rightarrow \textrm{breadth} = 17 - 10 = 7 \, cm.$$


How many metres of carpet $$75cm$$ wide will be required to cover the floor of a room which is $$20$$ metres long and $$12$$ metres broad?

Area of the carpet
$$=$$ area of the floor of the room
$$=20\times 12$$ sq.m.
$$=240$$ sq.m.
Width of the carpet $$=75cm=\cfrac { 3 }{ 4 } m$$
$$\therefore $$ Length of the carpet $$=\cfrac{Area}{Width}$$
$$=\dfrac{240}{ \cfrac { 3 }{ 4 }} $$
$$=320$$ metres


Madhu has divided the rectangle given into $$2$$ triangles. The red triangle is half of the big rectangle. Area of the big rectangle is $$20$$ square cm. So the area of the red triangle is _________ square cm.
695909_0664a9a4b11d4e7091116e0c226f1aee.png

the area of the red triangle is 10 square cm.
Since the red triangle is half of the big rectangle therefore area of red triangle = 10 sq cm



Fill in the blanks with proper method:
Perimeter of the rectangle of length 3 m and breadth 2 m = ____ m

Perimeter of a rectangle = 2(l + b)
Given, Length of rectangle = 3 m 
Breadth of a rectangle = 2 m
So, Perimeter of rectangle = 2 (3 + 2)
= 2 (5)
= 10 m


Find the length of each side of a square plot having perimeter $$172\ m$$.

Perimeter of a square having side $$a$$ is equal to $$4a$$.

perimeter of square $$= 4a$$

Thus,

$$4a=172$$

$$a=\dfrac{172}{4}$$

$$a=43$$m

Hence side of the square is $$43$$m



The area of rectangle  is 540 cm and its length is 36 cm. Find its width and perimeter.

$$Area = length \times width$$
$$540 = 36 \times width$$
$$width = \dfrac{{540}}{{36}}$$
$$width = 15\,cm$$
$$perimeter = 2 \times \left( {l + b} \right)$$
$$ = 2 \times \left( {36 + 15} \right)$$
$$ = 102cm$$
We get, $$perimeter = 102cm.$$


The adjacent sides of a rectangle are $$(4x^{2}-3y)$$ and $$5xy$$. Find its area.

Given that,
Length $$(l)= 4x^2 - 3y $$
Breadth $$(b) = 5xy $$

Area of rectangle $$= l \times b$$
                              $$= (4x^2 - 3y)\times 5xy$$
                              $$= 20x^3 y - 15xy^2 $$

Hence, the area is $$20x^3 y - 15xy^2 $$


Find the missing measurement (in $$cm$$) in the figure.

1186315_92121263e00a40dfb748e0893f75528f.png

Perimeter of the square $$= 28$$ cm

$$\implies 4 \times Side = 28$$

$$\implies Side = \dfrac{28}{4} $$

$$\implies Side = 7 $$

Missing measurement $$= 7 \ cm $$


Find the area of a square one of whose diagonals is $$5$$ cm.

Let $$a$$ be the side of the square.

Then by the problem, we've

Diagonal $$(d)=5$$

$$\Rightarrow a\sqrt{2}=5$$

$$\Rightarrow a=\dfrac{5}{\sqrt{2}}$$.

Then area of the square will be $$a^2=\dfrac{25}{2}$$ cm.


Area of a square is $$144\,cm^2$$. Find its sides.

Given, Area o the square MNOP is $$144 cm^2$$
We know that the area of a square is $$A=side\times side$$
So,
$$side\times side=144$$
$$side^2=144$$
$$side=\sqrt{144}$$
$$side=12\ cm$$



Calculate the area of the rectangle whose length $$20\ m$$ and breadth $$11\ m$$.

Length $$l=20m$$
Breadth $$b=11m$$
Area of rectangle 
$$=l\times b \\ =20\times 11\\ =220m^2$$


Find the length of a rectangle, which has:
$$area = 768\,m^2$$ and $$breadth = 24\,m$$

Area of Rectangle = Length $$\times$$ Breadth
$$\Rightarrow$$ 768 = Length $$\times$$ 24     
$$\Rightarrow$$ Length=32 m


The length of a hall is $$18\ m$$ and its width is $$13.5\ m$$. Find the least number of square tiles, each of side $$25\ cm$$, required to cover the floor of the hall, without leaving any of the margin.

Area of the hall  $$ = l\times b = 18 \times 135 = 2430\ { m }^{ 2 } $$

Area of each square tile $$ = { a }^{ 2 } = { 25 }^{ 2 } = 625  sq cm = 0.625  sq  m $$

Number of tiles required $$ = \dfrac {Area\ of\  the\  hall} {Area\  of\  each\  square\  tile}  = \dfrac {2430} { 0.625} = 3888 $$


The length of a rectangle is 16 cm and its perimeter is equal to perimeter of a square with side 12.5 cm. Find the area of the rectangle.

Perimeter of the square $$ = 4a = 4 \times 12.5 = 50  cm $$

Let the breadth of the rectangle $$=b$$ cm.
Perimeter of the rectangle $$ = 2(l + b) = 50 cm$$

$$  => 16 + b = 25 $$
$$=> b= 9 \  cm $$

Area of the rectangle $$ = l\times b = 16 \times 9 = 144{ cm }^{ 2 } $$



Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is $$800\displaystyle \  { m }^{ 2 }$$ ? If so, find its length and breadth.

Let the breadth be $$x \ $$ and the length will be $$2x \ m$$ .
The area = $$2{ x }^{ 2 } \ sq m$$ .

$$2{ x }^{ 2 }=800\\\Rightarrow { x }^{ 2 }=400\\\Rightarrow x =20$$

So, the breadth is $$20 \ m$$ and the length $$= 40 \ m$$ 


The area of a square park is the same as a rectangular park. If the side of the square park is $$16\ m$$ and the length of the rectangular park is $$90\ m$$, find the breadth of the rectangular park.


1345295_1329195_ans_3a9c752e5edb4273b0770a185b640496.jpg



On a square handkerchief, nine circular designs each of radius $$7$$ cm are made. Find the area of the remaining portion of the handkerchief. $$\bigg[\text{Use}\, \pi=\dfrac{22}{7}\bigg]$$
766174_127c0ea773bc4825a65053d9130454f1.png

Area of the remaining portion $$=$$ Area of square $$-$$ Area of $$9$$ circles.

Here, the handkerchief is made up of contributing $$9$$ circles.

So, side of the square $$=3 \times $$ Diameter of circle

Here, the radius of a circle is $$7\:cm$$

$$\therefore$$ Diameter of circle is $$14\:cm$$

$$\Rightarrow $$side of a square $$=3\times 14=42\:cm$$

$$\therefore$$ Area of the square $$=42\times 42=1764\:cm^2$$

Area of one circle
 $$=\pi r^2\\=\dfrac{22}{7}\times 7^2\\=154\:cm^2$$

Hence area of $$9$$ circles
 $$=9\times 154\\=1386\:cm^2$$

$$\therefore$$ Area of the remaining portion $$=$$ Area of square $$-$$ Area of $$9$$ circles.

                                                      $$=1764-1386$$

                                                      $$=378\:cm^2$$

Hence, area of the remaining portion is $$378\:cm^2$$


A square room of side $$12\ m$$ is paved with square tiles of side $$50\ cm$$ then find the number of tiles required to space the floor of the room?

Let the number of square tiles required be $$x$$
so,
Area of room = Area of 'x' tiles
Area of room$$=side\times side=12\times12=144m^2=1440000cm^2$$
Area of 1 tile$$=50\times50=2500cm^2$$
Area of 'x' tile$$=2500x cm^2$$
Number of tiles required$$=\dfrac{1440000}{2500}=576$$
So, 576 tiles are required


The perimeter of a square field is $$40$$ m more than another square field. Three times the area of the smaller field is $$50$$ $$m^2$$ more than the area of the larger field. Find the sides of both the fields.

Let the side of small square be $$x \ m.$$
so , the area of small square will be $$x^2\ m^2$$
Thus the side of big square will be $$x+40$$ $$ m.$$
so , the area of big square will be $$(x+40)^2m^2$$
According to question,
$$3x^2=(x+40)^2+50$$
$$3x^2=x^2+1600+80x+50$$
$$2x^2=1650+80x$$
$$x^2=825+40x$$
$$x^2-40x-825$$
by quadratic formula,
 $$x=$$ $$\dfrac { 40\pm \sqrt { { (40) }^{ 2 }-4(1)(-825) }  }{ 2(1) } $$
$$x=$$ $$\dfrac { 40\pm \sqrt { 1600+3300 }  }{ 2(1) } $$
$$x=$$ $$\dfrac { 40\pm \sqrt { 4900 }  }{ 2 } $$
$$x=$$ $$\dfrac { 40\pm \sqrt { 70^2 }  }{ 2 } $$
$$x=$$ $$\dfrac { 40\pm 70  }{ 2 } $$
$$x=$$ $$20\pm 35 $$
Length can not be negative so,
$$x=$$ $$20+35\ m $$
$$x=$$ $$55\ m$$
Thus length of side of a larger square will be $$95\ m$$


The length of a rectangular filed is 50 m and its breadth is 35 m. Calculate its area.

Given: Length $$l=50m$$, Breadth $$b=35m$$
Area of rectangle $$=l\times b=50\times 35$$
                               $$=1750m^2$$


If the side of a rectangle are $$7\ cm$$ and $$5\ cm$$, then find its area.

$$\begin{array}{l} We\, \, know\, \, that, \\ Rectangle\, \, area=l\times b \\ Given\, \, that\, \, l=7cm,b=5cm \\ Therefore, \\ Area=7\times 5c{ m^{ 2 } }=35c{ m^{ 2 } } \\ \end{array}$$


The area of a rectangle if its length and breadth are $$8cm,12cm$$ respectively.

The length of the rectangle is $$8cm.$$
The breadth of the rectangle is $$12cm.$$

The area of the rectangle $$=$$ $$8\times 12$$
                                            $$=96cm^2$$


The following figures are drawn on a squared paper. Count the number of squares enclosed by each figure and find its area, taking the area of each square as $$1 cm^{2}$$.
1675225_2c923070e2a446c6b131faaae7f53d12.PNG

In the given figure, there is 18 complete square.
So, area=$$18\ cm^2$$


The following figures are drawn on a squared paper. Count the number of squares enclosed by each figure and find its area, taking the area of each square as $$1 cm^{2}$$.
1675226_f7928a79956b4a1c88a24376a4ec65a5.PNG

There are 6 complete and 4 half squares in the given shape.
Since Area of one square $$= 1\ cm^2$$
Therefore, Area of this shape $$= (6 + 4 × 1/2) = 8\ cm^2$$


On a squared paper, draw the following. Find the approximate area of each by counting the number of squares complete, more than half and exactly half.
A 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 $$= (4 + 6  ××  1) = 10\ cm^2$$

1674250_1675232_ans_de0baf65f39c46e8a67c413c05038dbf.PNG


The following figures are drawn on a squared paper. Count the number of squares enclosed by each figure and find its area, taking the area of each square as $$1 cm^{2}$$.
1675227_ebf9b4a978724d18b377e78f2bf47d31.JPG

The given shape has $$20$$ complete and $$8$$ half squares.

It is given that area of one square $$ = 1\ cm^2$$

So the area of the given shape $$= 20 + 8 × 12 = 24\ cm^2 $$


Find the perimeters of the squares whose sides are given below:
$$115.5$$cm.

If $$s$$ is the length of the side of the square then its perimeter is given as $$4s$$

Given side $$S=115.5cm$$
Perimeter $$=4\times 115.5$$
                  $$=462cm$$


The following figures are drawn on a squared paper. Count the number of squares enclosed by each figure and find its area, taking the area of each square as $$1 cm^{2}$$.
1675228_71f68ab2185d45eb90c3f35c2652176f.JPG

The given shape has $$13$$ complete, $$8$$ more than half and $$7$$ less than half squares.

It is given that area of one square $$= 1\ cm^2$$

So the area of the given shape $$= 13 + 8 × 1 = 21\ cm^2 $$


On a squared paper, draw the following. Find the approximate area of each by counting the number of squares complete, more than half and exactly half.
A rectangle.

 This contains 18 complete squares.
If we assume that the area of one complete square is $$1\ cm^2$$
then the area of this rectangle will be $$18\ cm^2$$.

1674255_1675231_ans_2bdac69148264d5fb770ca2310c910ef.PNG


Find the area of a rectangle, whose
Length $$=4.5 cm$$, breadth $$=2 cm$$.

Given,
Length$$=4.5\ cm$$
Breadth$$=2\ cm$$

Area$$=length\times breadth$$
        $$=4.5\times 2\ cm^2$$
        $$=9\ cm^2$$

Hence, the required area is $$9\ cm^2$$.


Use tracing paper and centimeter graph paper to compare the areas of the following pairs of figures:
1675237_0265c4f2e1884a8183ed9e1d5ac65f2f.JPG

With the help of tracing paper trace both the figures on a graph

Figure $$(I)$$ has $$4$$ complete, $$9$$ more than half and $$9$$ less than half squares.

Assume that area of one square $$= 1\ cm^2$$

By neglecting less than half squares, we get

Area of the shape $$= 4 + 9 = 13\ cm^2$$

Figure $$(II)$$ has $$8$$ complete, $$11$$ more than half and 10 less than half squares.

Assume that area of one square $$= 1\ cm^2$$

By neglecting less than half squares, we get

Area of the shape $$= 8 + 11 = 19\ cm^2$$

By comparing the areas of both the shapes, we know that the figure $$(II)$$ has area greater than that of figure $$(I)$$.
1680938_1675237_ans_1893d4df48014cb2866169b5928b3546.png


On a squared paper, draw any irregular closed figure. Find the approximate area of each by counting the number of squares complete, more than half, and exactly half.

The figure is shown above 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 approximate area of this shape $$= (10 +  1 ×121 ×12  + 7  ×  1) $$
                                                                     $$= 17.5\ cm^2$$

1674241_1675233_ans_5fbdafb706b647d29aaf57eefd1efc82.png


Find the area of a rectangle whose length is $$5.5\ m$$ and breadth is $$3.4\ m$$.   

Given the length and breadth of a rectangle are $$5.5$$ m and $$3.4$$ m respectively.
Then area of the rectangle $$=5.5\times 3.4=18.7$$ m$$^2$$.


How many square metres of glass will be required for a window, which has $$12$$ panes, each pane measuring $$25\ cm$$ by $$16\ cm$$


1683238_1676760_ans_ec53faacb3424423ba78eab8d0f84d03.PNG


The perimeter of an isosceles triangle is $$50\ cm$$. If one of the two equal sides is $$18\ cm$$, find the third side.

Length of equal sides = 18cm

Let the length of the third side $$= x\ cm$$
                    Perimeter = sum of all sides
                       
                    $$\therefore 50 = 18+18+x$$
                        $$50 = 36+x$$
                        $$x= 50-36$$
                           $$=14cm$$
$$\textbf{Thus, the length of the third side is 14 cm}$$


Find the area and perimeter of each of the following figures, if the area of each small square is $$1\ sq\ cm$$.
1790577_d692afd7746e4f41b4dad79740b69071.png

For figure (i) there are $$11$$ small square and area of each small square is $$1\ sq\ cm$$.
Area of the given figure $$= (11\times 1) cm^{2} = 11\ cm^{2}$$
Perimeter of the given figure $$= (18 \times 1) cm = 18\ cm$$

For figure (ii) there are $$13$$ and area of each small square is $$1\ sq\ cm$$.
Area of the given figure $$= (13\times 1) cm^{2} = 13\ cm^{2}$$
Perimeter of the given figure $$= (28 \times 1) cm = 28\ cm$$

For figure (iii) there are $$13$$ and area of each small square is $$1\ sq\ cm$$.
Area of the given figure $$= (13\times 1) cm^{2} = 13\ cm^{2}$$
Perimeter of the given figure $$= (28 \times 1) cm = 28\ cm$$.


Divide the park shown in Fig. into two rectangles. Find the total area of this park. If one packet of fertilizer is used for $$300\ sqm$$, how many packets of fertilizer are required for the whole park?
1790506_92c03f4d128949d698364f7bdd10b0f5.png

Area of the rectangular block $$ABCD = length \times breadth$$
$$= (150\times 100)m^{2}$$
$$= 15000\ m^{2}$$
Area of the rectangular block $$ADFE = length \times breadth$$
$$= (270\times 180) m^{2}$$
$$= 48600\ m^{2}$$
$$\therefore$$ Total area of the park $$= 15000\ m^{2} + 48600\ m^{2} = 63600\ m^{2}$$
Number of the packets of the fertilizer used for $$300\ m^{2} = 1$$
$$\therefore$$ Number of the packets of fertilizer used for $$= 636000\ m^{2} = \dfrac {63600}{300} = 212$$.


The cross-section of a tunnel perpendicular to its length is a trapezium $$ABCD$$ as shown in the following figure; also given that :
$$AM=BN; AB=7\ m; CD=5\ m$$. The height of the tunnel is $$2.4m$$.The tunnel is $$40\ m$$ long 
Calculate:
The cost of painting the internal surface of the tunnel (excluding the floor) at the rate of Rs. $$5$$ per $$m^2$$ (sq. metre).
1842438_7b73a3478dd346cb98af64129cadae03.png

Given, $$AB=7m,\ CD=5\ m$$ and $$AM=BN$$
The height is $$2.4\ m$$ and its length is $$40m$$
$$AM=BN\\=\dfrac{7-5}{2}\\=\dfrac{2}{2}\\=1m$$
$$\therefore $$ In $$\triangle ADM$$,
$$AD^2=AM^2+DM^2$$    [Using pythagoras theorem]
$$=1^2+(2.4)^2$$
$$=1+5.76$$
$$=6.76$$
$$=(2.6)^2$$
$$\Rightarrow AD=2.6\ m$$
Perimeter of the cross section tunnel $$=(7+2.6+2.6+5)\ m$$
                                                           $$=17.2\ m$$
Length $$=40\ m$$
$$\therefore$$ Internal surface area of the tunnel (except floor)
$$=(17.2 \times 40-40 \times 7)m^2$$
$$=(688-280)m^2$$
$$=408m^2$$
Rate of painting $$=Rs 5\ per m^2$$
Hence, total cost of painting $$=Rs \ 5 \times 408$$
                                             $$=Rs \ 2040$$


The side of the square of area $$9y^{2}$$ is $$3y$$, if true enter 1, else 0

Given,
Area of square $$=9y^2$$

We know that,
Area of a square $$=(side)^{2}$$
 So,               $$ 9y^{2} = (side)^{2}$$       (Given)
Taking square root on both sides, we get,
   $$\Rightarrow \sqrt{9y^2}=side$$
        $$ \Rightarrow  side=3y$$
Hence, side of the square is $$3y$$


Class 6 Maths Extra Questions