Chapter 6: Mensuration

Ans: Area of one page of the magazine $=15 cm \times 24 cm =360$ $cm ^{2}$

Than,area of half page of the magazine $=\dfrac{360}{2} cm ^{2}=180 cm ^{2}$

Cost of advertising for $10 cm ^{2}=$ Rs.300

Cost of advertising for $1 cm ^{2}=\dfrac{Rs.300}{10}$

Cost of advertising for $180 cm ^{2}$=$\text { Rs. } \dfrac{300}{10} \times 180$

$=\text { Rs. } 30 \times 180=\text { Rs. } 5400$

Ans: Given perimeter of the square garden $=48 m =4 \times$ side

$\text { side }=\dfrac{48}{4} =12 m$

Now,area of square garden$=Side\times Side=12\times 12=144 m ^{2}$

Also,given area of small flower bed $=18 m ^{2}$

Than,area of the garden not covers with flower bed$=144-18=126 m ^{2}$

Now,the fractional part of the garden covered by the flower bed $=\dfrac{\text { Area of flower bed }}{\text { Area of the square garden }}=\dfrac{18}{144}=\dfrac{1}{8}$

In addition,the ratio of the area covered by the flower bed and the remaining area $=\dfrac{18}{126}=\dfrac{1}{7} \quad$ i.e. $1: 7$

Ans: Given,side of the square = $15 cm$ ,length of the rectangle=$18 cm$

According to the question

Also given,perimeter of the square $=$ perimeter of the rectangle=$4 \times \text { sides }=2 \times(\text { length }+\text { breadth })$

$4 \times 15  =2 \times(18  + breadth )$

$60 =2 \times(18 + breadth )$

$breadth =30-18 =12 cm$

Now,area of the rectangular $=18 \times 12=216 cm ^{2}$

Area of square =${15}^{2}cm ^{2}=225cm ^{2}$

Ans:Given,square of side= $2 cm$

Length of one piece $=$ Perimeter of the square $=4 \times 2 cm =8 cm$

Let the number of pieces = $n$

According to the question

$n \times(\text { side })^{2}=28$

$n \times(2)^{2}=28$

$n=\dfrac{28}{4}=7$

The original length of the wire $=\text { Number of pieces } \times \text { length of one piece }$

$=7 \times 8=56 cm$

Ans: 

Total area of the park $=(150 \times 100)+(270 \times 180)$

$=15000+48600=63600 m ^{2}$

Now one packet of fertilizer used for area $=300 m ^{2}$

Than, number of total packets$=\dfrac{63600}{300}$ $=212$

Ans: Given, area of a rectangular field $=1600 m ^{2}$

And length $=80 m$

Now, area $=length \times breadth$

$80 \times breadth =1600$

$breadth =\dfrac{1600}{80}$

$=20 m$

Now, perimeter of rectangular field $=2(length+breadth)$

$=2(80+20)=200 m$

Ans: Given,the area of the square on a chess board $=4cm^{2}$

There are 64 squares on a chess board.

So,area of the chess board $=64 \times 4 sq cm =256 cm ^{2}$

Area of the squares left unoccupied in the beginning$=32 \times$ area of  a square$=32 \times 4 =128 cm ^{2}$

Ans: Area of the squares occupied by chessmen$=32 \times$ area of one square$=32 \times 4=128 cm ^{2}$

Ans: Perimeter of the rectangle $=36 cm$

$2 \times(\text { length }+\text { breadth })=36 cm$

$(\text { length }+\text { breadth })=\dfrac{36}{2}=18 cm$

Than,the possible dimension of the rectangle and their area as follows:

Ans: Area of a rectangle =$length\times  breadth$=$36cm^{2}$

Than,the possible dimension of the rectangle and their area as follows:

Ans: Given,area of each small square =$1{cm}^{2}$

And there are 11 small squares.

So, area =$11\times 1=11{cm}^{2}$

Perimeter=sum of all sides

Perimeter=18cm

Ans: Given,area of each small square = $1cm^{2}$

And there are 13 small squares.

So, area =$13\times 1=13cm^{2}$

Perimeter=sum of all sides

Perimeter=28cm

Ans: Given,area of each small square = $1cm^{2}$

And there are 13 small squares.

So, area =$13\times 1=13cm^{2}$

Perimeter=sum of all sides

Perimeter=28cm

Ans: Given,area of the figure $=96 cm ^{2}$

Total no. of small square in the given figure $=24$

So, area of each small square $=\dfrac{96}{24}$$=4 cm ^{2}$

Now, side of each small square $=2 cm$

Perimeter of the figure $=34 \times 2$$=68 cm$