Chapter 11: Perimeter and Area

The length and the breadth of a rectangular piece of land are $500\text{ m}$ and $300\text{ m}$ respectively. Find

(i)             its area

(ii)          the cost of the land, if $1{\text{ m}}^2$ of the land costs $Rs10,000.$

Given:

Length of the rectangular piece of land $=500\text{ m}$ 

Breadth of the rectangular piece of land $=300\text{ m}$ we Know that,

(i)             Area of the rectangular piece of land

$=\text{Length}\times \text{Breadth}$

$=500\times 300$

$=1,50,000{\text{ m}}^2$

(ii)          Since, the cost of $1{\text{ m}}^2\text{ land}=Rs10,000$

Therefore, the cost of $1,50,000{\text{ m}}^2$ land

$=10,000\times 1,50,000$

$=Rs1,50,00,00,000$

Find the area of a square park whose perimeter is $320\text{ m}$.

It is given in the question that,

We know that,
Perimeter of square park $=4\times$ Length of the side of park

Perimeter of square park $=320\text{ m}$

$\Rightarrow 4\times \text{side}=320$

$\Rightarrow \text{ side}=\frac{320}{4}=80\text{ m}$

Now, Area of square park $\begin{array}{l}=\text{side}\times \text{side}\\ ={\left(\text{side}\right)}^2\\ =80\times 80=6400{\text{ m}}^2\end{array}$

Thus, the area of square park is $6400{\text{ m}}^2.$

Find the breadth of a rectangular plot of land, if its area is $440{\text{ m}}^2$ and the length is $22\text{ m}$. Also find its perimeter.

Length of the rectangular park $=22\text{ m}$

Area of rectangular park $=440{\text{ m}}^2$

We know that,

$\begin{array}{l}\Rightarrow \text{ Area of rectangle}\text{= length}\times \text{breadth}\\ =440{\text{ m}}^2\end{array}$

$\Rightarrow 22\times \text{breadth}=440$

$\Rightarrow \text{ breadth}=\frac{440}{22}=20\text{ m}$

Now, Perimeter of rectangular park $=2\left(\text{length}+\text{breadth}\right)$

$\begin{array}{l}=2\left(22+20\right)\\ =2\times 42=84\text{ m}\end{array}$

Thus, the perimeter of rectangular park is $84$ m.

The perimeter of a rectangular sheet is $100\text{ cm}$. If the length is $35\text{ cm}$, find its breadth. Also find the area.

Length of the rectangular sheet $=35$ cm

Perimeter of the rectangular sheet $=100\text{ cm}$

We know that,

 Perimeter of rectangle $=2(\text{length}\times \text{breadth)}$

$\Rightarrow 2\left(\text{length}+\text{breadth}\right)=100\text{ cm}$

$\Rightarrow 2\left(35+\text{breadth}\right)=100$

$\Rightarrow 35+\text{breadth}=\frac{100}{2}$

$\Rightarrow 35+\text{breadth}=50$

$\Rightarrow \text{ breadth}=50-35$

$\Rightarrow \text{ breadth}=15\text{ cm}$

Now, Area of rectangular sheet

$=\text{length}\times \text{breadth}$

$=35\times 15=525{\text{ cm}}^2$

Thus, breadth and area of rectangular sheet are $15\text{ cm}$ and $525{\text{ cm}}^2$ respectively.

The area of a square park is the same as of a rectangular park. If the side of the square park is $60\text{ m}$ and the length of the rectangular park is $90\text{ m,}$ find the breadth of the rectangular park.

It is given in the question that, side of the square park $=60\text{ m}$

Length of the rectangular park $=90\text{ m}$

According to the question,

Area of square park $=$ Area of rectangular park

We know that,

Area of the square $=$ $\text{side}\times \text{side}$ $={\left(\text{side}\right)}^2$ 

Area of the rectangle $=\text{length}\times \text{breadth}$

So, according to question

$\Rightarrow \text{ side}\times \text{side}=\text{length}\times \text{breadth}$

$\Rightarrow 60\times 60=90\times \text{breadth}$

$\Rightarrow \text{ breadth}=\frac{60\times 60}{90}=40\text{ m}$

Thus, the breadth of the rectangular park is $40\text{ m}\text{.}$

A wire is in the shape of a rectangle. Its length is $40\text{ cm}$ and breadth is $22\text{ cm}$. If the same wire is rebent in the shape of a square, what will be the measure of each side. Also find which shape encloses more area?

Length of the rectangle shape wire $=40$ cm

Breadth of the rectangle shape wire $=22$ cm

According to the question,

Perimeter of square $=$ Perimeter of rectangle

$\Rightarrow 4\times \text{side}=2\left(\text{length}+\text{breadth}\right)$

$\Rightarrow 4\times \text{side}=2\left(40+22\right)$

$\Rightarrow 4\times \text{side}=2\times 62$

$\Rightarrow \text{ side}=\frac{2\times 62}{4}=31\text{ cm}$

Thus, the side of the square is $31$ cm.

Now, Area of rectangle $=$ length $\times$ breadth

$=40\times 22=880{\text{ cm}}^2$

and Area of square $=$ side $\times$ side $=31\times 31=961{\text{ cm}}^2$

Therefore, on comparing, the area of square is greater than that of rectangle.

That is, area of square $>$ area of rectangle.

The perimeter of a rectangle is $130\text{ cm}$. If the breadth of the rectangle is $30\text{ cm}$, find its length. Also find the area of the rectangle.

It is given in the question that, Breadth of rectangle $=30$ cm

Perimeter of rectangle $=130\text{ cm}$

We know that,

Perimeter of rectangle $=2$ (Length $+$ Breadth)

$\Rightarrow 2\left(\text{length}+\text{breadth}\right)=130\text{ cm}$

$\Rightarrow 2\left(\text{length}+30\right)=130$

$\Rightarrow \text{ length}+30=\frac{130}{2}$

$\Rightarrow \text{ length}+30=65$

$\Rightarrow \text{ length}=65–30=35\text{ cm}$

Therefore, length of rectangle $=35$ cm

Now, area of rectangle $=$ length $\times$ breadth

$=35\times 30=1050{\text{ cm}}^2$

Thus, the area of rectangle is $1050{\text{ cm}}^2.$

A door of length $2\text{ m}$ and breadth $1\text{ m}$ is fitted in a wall. The length of the wall is $4.5\text{ m}$ and the breadth is $3.6\text{ m}$ (Fig. 11.6). Find the cost of white washing the wall, if the rate of white washing the wall is $Rs20{\text{ per m}}^2.$

        Fig. 11.6

Given:

Length and breadth of the door are $2\text{ m}$  and $1\text{ m}$  respectively.

$\therefore$ Area of rectangular door $=$ length $\times$ breadth

$=2\text{ m}\times 1\text{ m}=2{\text{ m}}^2$

It is given in the question that, Length and breadth of the wall are $4.5$ m and $3.6$ m respectively.

$\therefore$ Area of wall including door $=$ length $\times$ breadth  

$=4.5\text{ m}\times 3.6\text{ m}=16.2{\text{ m}}^2$

Now, Area of wall excluding door

$=$ Area of wall including door $-$ Area of door

$=16.2-2=14.2{\text{ m}}^2$

Since, the rate of white washing $1{\text{ m}}^2$ of the wall $=Rs20$

Therefore, the rate of white washing $14.2{\text{ m}}^2$ of the wall

$=20\times 14.2$

$=Rs284$

Thus, the cost of white washing the wall excluding the door is $Rs284$.

 

 

Find the area of each of the following parallelograms:

a)  

 

b) 

 

c)  

 

 

d) 

 

 

e)  

 

 

 

 

 

We know that the area of parallelogram $=$ base $\times$ height

 

a)   Here base $=7$ cm and height $=4\text{ cm}$

$\therefore$ Area of parallelogram $=7\times 4=28{\text{ cm}}^2$

b)  Here base $=5$ cm and height $=3\text{ cm}$

$\therefore$ Area of parallelogram $=5\times 3=15{\text{ cm}}^2$

c)   Here base $=2.5$ cm and height $=3.5\text{ cm}$

$\therefore$ Area of parallelogram $=2.5\times 3.5=8.75{\text{ cm}}^2$

d)  Here base $=5$ cm and height $=4.8\text{ cm}$

$\therefore$ Area of parallelogram $=5\times 4.8=24{\text{ cm}}^2$

e)   Here base $=2$ cm and height $=4.4\text{ cm}$

$\therefore$ Area of parallelogram $=2\times 4.4=8.8{\text{ cm}}^2$

Find the area of each of the following triangles:

 

 

 

 

 

 

 

We know that the area of triangle
$=\frac{1}{2}\times$ base $\times$ height

Here, base $=4\text{ cm}$ and height $=3\text{ cm}$

$\therefore$ Area of triangle $=\frac{1}{2}\times 4\times 3=6{\text{ cm}}^2$

Here, base $=5\text{ cm}$ and height $=3.2\text{ cm}$

$\therefore$ Area of triangle $=\frac{1}{2}\times 5\times 3.2=8{\text{ cm}}^2$

Here, base $=3\text{ cm}$ and height $=4\text{ cm}$

$\therefore$ Area of triangle $=\frac{1}{2}\times 3\times 4=6{\text{ cm}}^2$

Here, base $=3\text{ cm}$ and height $=2\text{ cm}$

$\therefore$ Area of triangle $=\frac{1}{2}\times 3\times 2=3{\text{ cm}}^2$

 

 

Find the missing values:

S.No.	Base	Height	Area of the Parallelogram
a.	$20$ cm		$246{\text{ cm}}^2$
b.		$15\text{ cm}$	$154.5{\text{ cm}}^2$
c.		$8.4\text{ cm}$	$48.72{\text{ cm}}^2$
d.	$15.6$ cm		$16.38{\text{ cm}}^2$

 

We know that the area of parallelogram
$=$ base $\times$ height

a)   Here, base $=20\text{ cm and area}=246{\text{ cm}}^2$

$\because$ Area of parallelogram $=$ base $\times$ height

$\Rightarrow 246=20\times \text{height}$

$\Rightarrow \text{ height}=\frac{246}{20}=12.3\text{ cm}$

b)  Here, height $=15\text{ cm and area}=154.5{\text{ cm}}^2$

$\because$ Area of parallelogram $=$ base $\times$ height

$\Rightarrow 154.5=\text{base}\times 15$

$\Rightarrow \text{ base}=\frac{154.5}{15}=10.3\text{ cm}$

c)   Here, height $=8.4\text{ cm and area}=48.72{\text{ cm}}^2$

$\because$ Area of parallelogram $=$ base $\times$ height

$\Rightarrow 48.72=\text{base}\times 8.4$

$\Rightarrow \text{ base}=\frac{48.72}{8.4}=5.8\text{ cm}$

d)  Here, base $=15.6\text{ cm and area}=16.38{\text{ cm}}^2$

$\therefore$ Area of parallelogram $=$ base $\times$ height

$\Rightarrow 16.38=15.6\times \text{height}$

$\Rightarrow \text{ height}=\frac{16.38}{15.6}=1.05\text{ cm}$

Thus, the missing values are:

S.No.	Base	Height	Area of the Parallelogram
a.	$20$ cm	$12.3\text{ cm}$	$246{\text{ cm}}^2$
b.	$10.3\text{ cm}$	$15\text{ cm}$	$154.5{\text{ cm}}^2$
c.	$5.8\text{ cm}$	$8.4\text{ cm}$	$48.72{\text{ cm}}^2$
d.	$15.6$ cm	$1.05\text{ cm}$	$16.38{\text{ cm}}^2$

 

Find the missing values:

 

We know that the area of triangle
$=\frac{1}{2}\times \text{base}\times \text{height}$

In first row, base $=15\text{ cm}$ and area $=87{\text{ cm}}^2$

$\therefore 87=\frac{1}{2}\times 15\times \text{height}$

$\Rightarrow \text{ height}=\frac{87\times 2}{15}=11.6\text{ cm}$

In second row, height $=31.4$ mm and area $=1256{\text{ mm}}^2$

$\therefore 1256=\frac{1}{2}\times \text{base}\times 31.4$

$\Rightarrow \text{ base}=\frac{1256\times 2}{31.4}=80\text{ mm}$

In third row, base $=22$ cm and area $=170.5{\text{ cm}}^2$

$\therefore 170.5=\frac{1}{2}\times 22\times \text{height}$

$\Rightarrow \text{ height}=\frac{170.5\times 2}{22}=15.5\text{ cm}$

Thus, the missing values are:

 

PQRS is a parallelogram (Fig 11.23). QM is the height from Q to SR and QN is the height from Q to PS. If SR $=12$ cm and QM $=7.6$ cm. Find:

               Fig. 11.23

the area of the parallegram PQRS

QN, if PS $=8$ cm

Given: SR $=12$ cm, QM $=7.6$ cm, PS $=8$ cm.

Area of parallelogram $=$ base $\times$ height

$=12\times 7.6=91.2{\text{ cm}}^2$

Area of parallelogram $=$ base $\times$ height

$\Rightarrow 91.2=8\times \text{QN}$

$\Rightarrow \text{ QN}=\frac{91.2}{8}=11.4\text{ cm}$

DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (Fig 11.24). If the area of the parallelogram is $1470{\text{ cm}}^2$, $\text{AB}=35$ cm and AD $=49$ cm, find the length of BM and DL.

            Fig 11.24

Given: Area of parallelogram ABCD $=1470{\text{ cm}}^2$

Base (AB) $=35$ cm and base (AD) $=49$ cm

Since Area of parallelogram $=$ base $\times$ height

$\begin{array}{l}\Rightarrow 1470=\text{AB}\times \text{DL}\\ \Rightarrow 1470=35\times \text{DL}\end{array}$

$\Rightarrow \text{ DL}=\frac{1470}{35}$

$\Rightarrow \text{ DL}=42\text{ cm}$

Again, Area of parallelogram $=$ base $\times$ height

$\Rightarrow 1470=49\times \text{BM}$

$\Rightarrow \text{ BM}=\frac{1470}{49}$

$\Rightarrow \text{ BM}=30\text{ cm}$

Thus, the lengths of DL and BM are $42$ cm and $30$ cm respectively.

$\Delta \text{ABC}$ is right angled at A (Fig 11.25). AD is perpendicular to BC. If AB $=5$ cm, BC $=13$ cm and AC $=12$ cm, Find the area of $\Delta \text{ABC}$. Also find the length of AD.

               Fig. 11.25

In right angled triangle BAC, $\text{AB}=5\text{ cm}$ and $\text{AC}=12\text{ cm}$

Area of triangle $=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times \text{AB}\times \text{AC}$

$=\frac{1}{2}\times 5\times 12=30{\text{ cm}}^2$

Now, in $\Delta \text{ABC}$,

Area of triangle $\text{ABC}=\frac{1}{2}\times \text{BC}\times \text{AD}$

$\Rightarrow 30=\frac{1}{2}\times 13\times \text{AD}$

$\Rightarrow \text{ AD}=\frac{30\times 2}{13}=\frac{60}{13}\text{ cm}$

$\Delta \text{ABC}$ is isosceles with AB $=$ AC $=7.5$ cm and BC $=9$ cm (Fig 11.26). The height AD from A to BC, is $6\text{ cm}$. Find the area of $\Delta \text{ABC}$. What will be the height from C to AB i.e., CE?

              Fig. 11.26

In $\Delta \text{ABC}$, AD $=6$ cm and BC $=9$ cm

Area of triangle $=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times \text{BC}\times \text{AD}$

$=\frac{1}{2}\times 9\times 6=27{\text{ cm}}^2$

Again, Area of triangle $=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times \text{AB}\times \text{CE}$

$\Rightarrow 27=\frac{1}{2}\times 7.5\times \text{CE}$

$\Rightarrow \text{ CE}=\frac{27\times 2}{7.5}$

$\Rightarrow \text{ CE}=7.2\text{ cm}$

Thus, height from C to AB i.e., CE is $7.2\text{ cm}$.

 

 

Find the circumference of the circles with the following radius: $\left(\text{Take }\pi =\frac{22}{7}\right)$

a)   $14\text{ cm}$ 

b)  $28\text{ cm}$

c)   $21\text{ cm}$

 

 

a)   Radius of the circle $=14\text{ cm}$ 
We know that,

Circumference of the circle $=2\pi r$

$\begin{array}{l}=2\times \frac{22}{7}\times 14\\ =88\text{ cm}\end{array}$

b)  Radius of the circle $=28\text{ cm}$ 

We know that,

Circumference of the circle $=2\pi r$
$\begin{array}{l}=2\times \frac{22}{7}\times 28\\ =176\text{ cm}\end{array}$

 

c)   Radius of the circle $=21\text{ cm}$ 

We know that,

Circumference of the circle $=2\pi r$
$\begin{array}{l}=2\times \frac{22}{7}\times 21\\ =132\text{ cm}\end{array}$

Find the area of the following circles, given that:

a)   radius $=14\text{ mm}$ $\left(\text{Take }\pi =\frac{22}{7}\right)$ 

b)  diameter $=49\text{ m}$

c)   radius $=5\text{ cm}$

a)   It is given that,

Radius of the circle $=14\text{ mm}$ 

We know that,

Area of circle $\begin{array}{l}=\pi r^2\\ =\frac{22}{7}\times 14\times 14\end{array}$
                       $\begin{array}{l}=22\times 2\times 14\\ =616{\text{ mm}}^2\end{array}$

b)  It is given that,

Diameter of the circle $=49\text{ m}$

We know that,

$\therefore \text{ Radius}=\frac{49}{2}=24.5\text{ m}$ 
$\begin{array}{l}\therefore \text{ Area of circle}=\pi r^2\\ =\frac{22}{7}\times 24.5\times 24.5\end{array}$ 
$\begin{array}{l}=22\times 3.5\times 24.5\\ =1886.5{\text{ m}}^2\end{array}$

c)   Radius of the circle $=5\text{ cm}$

We know that

Area of circle $\begin{array}{l}=\pi r^2\\ =\frac{22}{7}\times 5\times 5\end{array}$ 
                      $=\frac{550}{7}{\text{cm}}^2$

                      $=78.57{\text{cm}}^2$

If the circumference of a circular sheet is $154$ m, find its radius. Also find the area of the sheet. $\left(\text{Take }\pi =\frac{22}{7}\right)$

Circumference of the circular sheet $=154\text{ m}$ 

We know that,

Circumference of the circular sheet $=$ $2\pi r$

$\therefore 2\pi r=154\text{ m}$

$\Rightarrow r=\frac{154}{2\pi }$

$\Rightarrow r=\frac{154\times 7}{2\times 22}=24.5\text{ m}$

Now, area of circular sheet

Area of the circle $=\pi r^2$ 
$=\pi r^2=\frac{22}{7}\times 24.5\times 24.5$ 
$=22\times 3.5\times 24.5=1886.5{\text{ m}}^2$

Hence, the radius and area of circular sheet are
$24.5$ m and $1886.5{\text{ m}}^2$ respectively.

A gardener wants to fence a circular garden of diameter $21$ m. Find the length of the rope he needs to purchase, if he makes $2$ rounds of fence. Also find the costs of the rope, if it costs $Rs4$ per meter. $\left(\text{Take }\pi =\frac{22}{7}\right)$

Diameter of the circular garden $=21\text{ m}$

$\therefore$ Radius of the circular garden $=\frac{21}{2}\text{ m}$

Now, circumference of circular garden $=2\pi r$ 

After putting the values we get,

$=2\times \frac{22}{7}\times \frac{21}{2}$ 

$=22\times 3=66\text{ m}$

Now, the length of rope required for fencing the garden will be equal to the two times the circumference of the garden, because the fencing is done in 2 rounds.

$=2\times 2\pi r$

$=2\times 66=132\text{ m}$

Since, the cost of $1$ meter rope $=Rs4$

Hence, cost of $132$ meter rope $=4\times 132=Rs528$

From a circular sheet of radius $4$ cm, a circle of radius $3$ cm is removed. Find the area of the remaining sheet. (Take $\pi =3.14$ )

It is given in the question that,

Outer radius of circular sheet $(R)=4\text{ cm}$ 

Inner radius of circular sheet $(r)=3\text{ cm}$ 

 

Area of remaining sheet

$=$ Area of circular sheet $-$ Area of removed circle

$=\pi {\text{R}}^2-\pi r^2=\pi ({\text{R}}^2-r^2)$

$=\pi (4^2-3^2)=\pi (16-9)$

$=3.14\times 7=21.98{\text{ cm}}^2$

Hence, the area of remaining sheet is $21.98{\text{ cm}}^2.$

Saima wants to put a lace on the edge of a circular table cover of diameter $1.5\text{ m}$. Find the length of the lace required and also find its cost if one meter of the lace costs $Rs15.$ (Take $\pi =3.14$ )

Diameter of the circular table cover $=1.5\text{ m}$

$\therefore$ Radius of the circular table cover $=\frac{1.5}{2}\text{ m}$

Circumference of circular table cover

$=2\pi r$

$=2\times 3.14\times \frac{1.5}{2}$

$=4.71\text{ m}$

Therefore, the length of required lace is $4.71\text{ m}$.

Now, the cost of $1$ m lace $=Rs15$

Then the cost of $4.71$ m lace

$=15\times 4.71$

$=Rs70.65$

Therefore, the cost of $4.71$ m lace is $Rs70.65$.

Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Diameter of the semicircle $=10\text{ cm}$

$\therefore$ Radius of the semicircle $=\frac{10}{2}=5\text{ cm}$

We know that perimeter of a complete circle $=2\pi r,$ where r is the radius of the circle. So, the perimeter of a semi-circle $=\pi r$

Perimeter of the figure

$=$ Circumference of semi-circle $+$ diameter

$=\pi r+\text{D}$

$\begin{array}{l}=\frac{22}{7}\times 5+10\\ =\frac{110}{7}+10\end{array}$

$\begin{array}{l}=\frac{110+70}{7}\\ =\frac{180}{7}\\ =25.71\text{ cm}\end{array}$

Hence, the perimeter of the given figure is $25.71$ cm.

Find the cost of polishing a circular table-top of diameter $1.6$ m, if the rate of polishing is $Rs15{\text{/m}}^2.$ (Take $\pi =3.14$ )

Diameter of the circular table top $=1.6\text{ m}$

$\therefore$ Radius of the circular table top $=\frac{1.6}{2}=0.8\text{ m}$

Area of circular table top
$=\pi r^2$

$=3.14\times 0.8\times 0.8$

$=2.0096{\text{ m}}^2$

As per the question,

Cost of polishing $1{\text{ m}}^2$ of the table top $=Rs15$

Cost of polishing $2.0096{\text{ m}}^2$ of table top
$=15\times 2.0096$ 
$=Rs30.14$ (approx.)

Hence, the cost of polishing a circular table top is $Rs30.14$ (approx.)

Shazli took a wire of length $44$ cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? $\left(\text{Take }\pi =\frac{22}{7}\right)$

Total length of the wire $=44\text{ cm}$

This wire has been bent into the shape of a circle.

$\therefore$ The circumference of the circle $=2\pi r=44\text{ cm}$

$\Rightarrow 2\times \frac{22}{7}\times r=44$

$\Rightarrow r=\frac{44\times 7}{2\times 22}=7\text{ cm}$

Now, area of the circle
$=\pi r^2$

$=\frac{22}{7}\times 7\times 7=154{\text{ cm}}^2$

Now the wire is rebent into a square.

Then perimeter of square $=44\text{ cm}$

$\Rightarrow 4\times \text{side}=44$

$\Rightarrow \text{ side}=\frac{44}{4}=11\text{ cm}$

Now area of square $={(\text{side)}}^{\text{2}}$
$\begin{array}{l}=11\times 11\\ =121{\text{ cm}}^2\end{array}$

Therefore, the area of circle is greater than that of the square, so the circle encloses more area.

From a circular card sheet of radius $14$ cm, two circles of radius $3.5$ cm and a rectangle of length $3\text{ cm}$ and breadth 1cm are removed. (as shown in the adjoining figure). Find the area of the remaining sheet. $\left(\text{Take }\pi =\frac{22}{7}\right)$

Radius of circular sheet (R) $=14$ cm

Area of the circle $=\pi {\text{R}}^2$

$\therefore$ Area of bigger circle $=\frac{22}{7}\times 14\times 14$

$=22\times 2\times 14$

$=44\times 14$

$=616{\text{ cm}}^2$

Radius of smaller circle $\text{(}r\text{)}=3.5\text{ cm}$

Area of the circle $=\pi r^2$

$\therefore$ Area of 2 small circles $=2\times \frac{22}{7}\times 3.5\times 3.5$

$=44\times 0.5\times 3.5$

$=22\times 3.5$

$=77{\text{ cm}}^2$

Length of rectangle $\text{(}l\text{)}=3$ cm and breadth of rectangle $\text{(}b\text{)}=1\text{ cm}$

Area of rectangle $=$ Length $\times$ Breadth

$\therefore$ Area of rectangle $=3\times 1$

$=3{\text{ cm}}^2$

According to question,

Area of remaining sheet $=$ Area of circular sheet– (Area of two smaller circles $+$ Area of rectangle)

$=\pi {\text{R}}^2-\left[2(\pi r^2)+(l\times b)\right]$

$=616–77–3$

$=616-80$

$=536{\text{ cm}}^2$

Hence, the area of remaining sheet is $536{\text{ cm}}^2$.

A circle of radius $2$ cm is cut out from a square piece of an aluminium sheet of side $6$ cm. What is the area of the left over aluminium sheet? (Take $\pi =3.14$ )

Radius of circle $=2$ cm

As we know,

Area of the circle $=\pi r^2$

$\therefore$ Area $=3.14\times 2\times 2$

$=3.14\times 4$

$=12.56{\text{ cm}}^2$

Side of square $=6$ cm

Area of square $={\left(\text{side}\right)}^2$

$\therefore$ Area of square shaped sheet $={\left(6\right)}^2$

and

According to question,

Area of aluminium sheet left $=$ Total area of aluminium sheet $-$ Area of circle

$=36-12.56$

$=23.44{\text{ cm}}^2$

Hence, the area of aluminium sheet left is $23.44{\text{ cm}}^2.$

The circumference of a circle is $31.4$ cm. Find the radius and the area of the circle? (Take $\pi =3.14$ )

Given,

The circumference of the circle $=31.4$ cm

According to question,

$\Rightarrow 2\pi r=31.4$

$\Rightarrow 2\times 3.14\times r=31.4$

$\Rightarrow r=\frac{31.4}{2\times 3.14}=5\text{ cm}$

Area of the circle $\begin{array}{l}=\pi r^2=3.14\times 5\times 5\\ =78.5{\text{ cm}}^2\end{array}$

Hence, the radius and the area of the circle are $5$ cm and $78.5{\text{ cm}}^2$ respectively.

A circular flower bed is surrounded by a path $4$ m wide. The diameter of the flower bed is $66$ m. What is the area of this path? ( $\pi =3.14$ )

Diameter of the circular flower bed $=66\text{ m}$

$\therefore$ Radius of circular flower bed $(r)=\frac{66}{2}=33\text{ m}$

$\therefore$ Radius of circular flower bed with $4\text{ m}$ wide path $\left(\text{R}\right)=33+4=37\text{ m}$

According to the question,

Area of the circular path $=$ Area of bigger circle $-$ Area of smaller circle

$=\pi {\text{R}}^2-\pi r^2=\pi ({\text{R}}^2-r^2)$

$=\pi \left[{(37)}^2-{(33)}^2\right]$

$=3.14\left[\left(37+33\right)\left(37-33\right)\right]$          $\left[\because a^2-b^2=(a+b)(a-b)]\right]$

$=3.14\times 70\times 4$

$=879.20{\text{ m}}^2$

Hence, the area of the circular path is $879.20{\text{ m}}^2.$

A circular flower garden has an area of $314{\text{ m}}^2$. A sprinkler at the centre of the garden can cover an area that has a radius of $12$ m. Will the sprinkler water the entire garden? (Take $\pi =3.14$ )

Radius of the circular sprinkler $=12$ m

Area of the circular sprinkler $=\pi r^2$

$=3.14\times 12\times 12$

$=3.14\times 144$

$=452.16{\text{ m}}^2$

Area of the circular flower garden $=314{\text{ m}}^2$

Since, the area of sprinkler is greater than the area of the circular flower garden, hence the sprinkler will water the entire garden.

Find the circumference of the inner and the outer circles, shown in the adjoining figure?
(Take $\pi =3.14$ )

Radius of outer circle $=19\text{ m}$

We know that,

Circumference of the circle $=2\pi r$

$=2\times 3.14\times 19$

$=38\times 3.14$

$=119.32\text{ m}$

Radius of inner circle $=19–10=9\text{ m}$ 

As we know that,

Circumference of the circle $=2\pi r$ 

$=2\times 3.14\times 9$

$=18\times 3.14$

$=56.52$

Hence,

Circumference of outer circle $=119.32\text{ m}$ 

Circumference of inner circle $=56.52\text{ m}$ 

How many times a wheel of radius $28$ cm must rotate to go $352$ m? $\left(\text{Take }\pi =\frac{22}{7}\right)$

Let us consider that the wheel rotates $n$ times of its circumference.

Radius of wheel $=28$ cm

Total distance covered $=352\text{ m}=35200\text{ cm}$

Circumference of the wheel $=2\pi r$ 

$\therefore$ Distance covered by wheel $=n\times$ circumference of wheel

$\Rightarrow 35200=n\times 2\pi r$

$\Rightarrow 35200=n\times 2\times \frac{22}{7}\times 28$

$\Rightarrow n=\frac{35200\times 7}{2\times 22\times 28}$

$\Rightarrow n=200\text{ revolutions}$

Therefore, wheel must rotate $200$ times to cover a distance of $352\text{ m}$.

The minute hand of a circular clock is $15$ cm long. How far does the tip of the minute hand move in $1$ hour? (Take $\pi =3.14$ )

In $1$ hour, minute hand completes one round of the clock which means it makes a circle.

We have to find out that how far will the tip of minute hand move in 1 hour.

For this we have to find out the distance travelled by the tip of minute hand.

$\therefore$ Distance travelled by the minute hand in 1 hour $=$ Circumference of the clock

We know that,

Circumference of the circle $=2\pi r$ 

Radius of the circle $(r)=15\text{ cm}$

Circumference of circular clock $=2\pi r$

$=2\times 3.14\times 15$

$=94.2\text{ cm}$

Therefore, the tip of the minute hand moves $94.2\text{ cm}$ in $1$ hour.

 

 

 

A garden is $90$ m long and $75$ m broad. A path $5\text{ m}$ wide is to be built outside and around it. Find the area of the path. Also find the area of the garden in hectare.

Length of garden $=90\text{ m}$ 

Breadth of garden $=75\text{ m}$ 

We know that,

Area of rectangle $=$ Length $\times$ Breadth

$\therefore$ Area of garden $=90\times 75$

$=6750{\text{ m}}^2$

Outer length of rectangular garden with path $=90+5+5=100\text{ m}$

Outer breadth of rectangular garden with path $=75+5+5=85\text{ m}$

Outer area of the rectangular garden $=$ outer length $\times$ outer breadth $=100\times 85=8,500{\text{ m}}^2$

Now, Area of path $=$ Outer area of the rectangular garden $-$ Inner area of the rectangular garden

$=8,500-6,750$

$=1,750{\text{ m}}^2$

Since, $1{\text{ m}}^2=\frac{1}{10000}\text{ hectares}$

Therefore, $6,750{\text{ m}}^2=\frac{6750}{10000}=0.675\text{ hectares}$ 

A $3\text{ m}$ wide path runs outside and around a rectangular park of length $125\text{ m}$ and breadth $65\text{ m}$. Find the area of the path.

Length of rectangular park $=125\text{ m},$

Breadth of rectangular park $=65\text{ m}$ 

Area of rectangle $=$ Length $\times$ Breadth

$\therefore$ Area of park $=125\times 65$

$=8125{\text{ m}}^2$

Width of the path $=3\text{ m}$

Length of rectangular park with path $=125+3+3=131\text{ m}$

Breadth of rectangular park with path $=65+3+3=71\text{ m}$

Area of rectangle $=$ Length $\times$ Breadth

$\therefore$ Area of garden including path $=131\times 71$

$=9301{\text{ m}}^2$

$\therefore$ Area of path
$=$ Area of park with path $-$ Area of park without path

$=\left(\text{AB}\times \text{AD}\right)-\left(\text{EF}\times \text{EH}\right)$

$=9301-8125$

$=1,176{\text{ m}}^2$

Thus, area of path around the park is $1,176{\text{ m}}^2$.

A picture is painted on a cardboard $8$ cm long and $5$ cm wide such that there is a margin of $1.5$ cm along each of its sides. Find the total area of the margin.

Length of painted cardboard $=8$ cm 

Breadth of painted cardboard $=5$ cm

We know that,

Area of rectangle $=$ Length $\times$ Breadth

$\therefore$ Area of cardboard including margin $=8\times 5$

$=40{\text{ cm}}^2$

Since, there is a margin of $1.5$ cm long from each of its side.

Therefore, reduced length $=8-\left(1.5+1.5\right)=8-3=5\text{ cm}$

And reduced breadth $=5-\left(1.5+1.5\right)=5-3=2\text{ cm}$

$\therefore$ Area of cardboard not including margin $=5\times 2$

$=10{\text{ cm}}^2$

Area of margin
$=$ Area of cardboard (ABCD) $-$ Area of cardboard (EFGH)

$=\left(\text{AB}\times \text{AD}\right)-\left(\text{EF}\times \text{EH}\right)$

$=40-10$

$=30{\text{ cm}}^2$

Thus, the total area of margin is $30{\text{ cm}}^2$.

A verandah of width $2.25$ m is constructed all along outside a room which is $5.5$ m long and $4$ m wide. Find:

(i)             the area of the verandah.

(ii)          the cost of cementing the floor of the verandah at the rate of $Rs200{\text{ per m}}^2.$

(i)             The length of room $=5.5$ m and width of the room $=4$ m

We know that,

Area of rectangle $=$ Length $\times$ Breadth

$\therefore$ Area of room $=5.5\times 4$

$=22{\text{ m}}^2$

The length of room with verandah $=5.5+2.25+2.25=10\text{ m}$

The width of room with verandah $=4+2.25+2.25=8.5\text{ m}$

$\therefore$ Area of room including verandah $=10\times 8.5$

$=85{\text{ m}}^2$

Area of verandah

$=$ Area of room with verandah $-$ Area of room without verandah

$=\text{Area of ABCD}-\text{Area of EFGH}$

$=\left(\text{AB}\times \text{AD}\right)-\left(\text{EF}\times \text{EH}\right)$

$=85-22$

$=63{\text{ m}}^2$

(ii)          The cost of cementing $1{\text{ m}}^2$ of the floor of verandah $=Rs200$

The cost of cementing $63{\text{ m}}^2$ of the floor of verandah $=200\times 63=Rs12,600$

A path $1$ m wide is built along the border and inside a square garden of side $30$ m. Find:

(i)             the area of the path

(ii)          the cost of planting grass in the remaining portion of the garden at the rate of $Rs40{\text{ per m}}^2.$

Side of the square garden $=30$ m and
We know that,

Area of square $={\left(\text{side}\right)}^2$

$\therefore$ Area of square garden $={\left(30\right)}^2$

$=900{\text{ m}}^2$

Width of the path along the border $=1\text{ m}$ 

Side of square garden without path $=30-\left(1+1\right)=30-2=28\text{ m}$

$\therefore$ Area of garden not including path $={\left(28\right)}^2$

$=784{\text{ m}}^2$

Now Area of path
$=$ Area of ABCD $-$ Area of EFGH

$=900-784$

$=116{\text{ m}}^2$

(ii)   Area of remaining portion $=28\times 28=784{\text{ m}}^2$

The cost of planting grass in $1{\text{ m}}^2$ of the garden $=\text{ Rs}40$ 

The cost of planting grass in $784{\text{ m}}^2$ of the garden $=Rs40\times 784=Rs31,360$

Two cross roads, each of width $10$ m, cut at right angles through the centre of a rectangular park of length $700$ m and breadth $300$ m and parallel to its sides. Find the area of the roads. Also find the area of the park excluding cross roads. Give the answer in hectares.

Length of park $=700\text{ m}$ 

Breadth of park $=300\text{ m}$ 

We know that,

Area of rectangle $=$ Length $\times$ Breadth

$\therefore$ Area of park $=700\times 300$

$=210000{\text{ m}}^2$

Here, PQ $=10$ m and PS $=300$ m, EH $=10$ m and EF $=700$ m

And KL $=10$ m and KN $=10$ m

 

Area of roads
$=$ Area of PQRS $+$ Area of EFGH $-$ Area of KLMN

[ $\because$ KLMN is taken twice, which is to be subtracted]

$=\text{PS}\times \text{PQ}+\text{EF}\times \text{EH}-\text{KL}\times \text{KN}$

$=\left(300\times 10\right)+\left(700\times 10\right)-\left(10\times 10\right)$

$=3000+7000-100$

$=9,900{\text{ m}}^2$

Area of road in hectares, $1{\text{ m}}^2=\frac{1}{10000}\text{ hectares}$

$9,900{\text{ m}}^2=\frac{9900}{10000}=0.99\text{ hectares}$

Now, Area of park excluding cross roads

$=\text{Area of park}-\text{Area of road}$

$=2,10,000-9,900$

$=2,00,100{\text{ m}}^2$

$=\frac{200100}{10000}\text{ hectares}=20.01\text{ hectares}$

Through a rectangular field of length $90$ m and breadth $60$ m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is $3$ m, find

(i)             the area covered by the roads.

(ii)          the cost of constructing the roads at the rate of $Rs110{\text{ per m}}^2.$

(i)             Here, $\text{PQ}=3\text{ m and PS}=60\text{ m},$ $\text{EH}=3\text{ m}$ and

$\text{EF}=90\text{ m and KL}=3\text{ m and KN}=3\text{ m}$

 

Area of roads
$=$ Area of PQRS $+$ Area of EFGH $-$ Area of KLMN

[ $\because$ KLMN is taken twice, which is to be subtracted]

$=\text{PS}\times \text{PQ}+\text{EF}\times \text{EH}-\text{KL}\times \text{KN}$

$=\left(60\times 3\right)+\left(90\times 3\right)-\left(3\times 3\right)$

$=180+270-9$

$=441{\text{ m}}^2$

(ii)          The cost of  constructing $1{\text{ m}}^2$ of the roads $=Rs110$

The cost of  constructing $441{\text{ m}}^2$ of the roads $=Rs110\times 441=Rs48,510$

Therefore, the cost of constructing the roads $=Rs48,510$

Pragya wrapped a cord around a circular pipe of radius $4$ cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side $4$ cm (also shown). Did she have any cord left? ( $\pi =3.14$ )

Here, cord wrapped around the circular pipe is equal to the circumference of that pipe.

We know that circumference of the pipe or circle $=2\pi r$ 

Radius of pipe $=4$ cm

Putting the values, we get:

$=2\times 3.14\times 4=25.12\text{ cm}$

Again, wrapping cord around a square is equal to the perimeter of the square $=4\times \text{side}$

$=4\times 4=16\text{ cm}$

Remaining cord
$=$ Cord wrapped on pipe $-$ Cord wrapped on square

$=25.12-16$

$=9.12\text{ cm}$

Thus, she has left $9.12$ cm cord.

The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find:

(i)             the area of the whole land

(ii)          the area of the flower bed

(iii)      the area of the lawn excluding the area of the flower bed

(iv)       the circumference of the flower bed.

Length of rectangular lawn $=10$ m,

breadth of the rectangular lawn $=5$ m

And radius of the circular flower bed $=2$ m

(i)             Area of the whole land $=$ length $\times$ breadth
$=10\times 5=50{\text{ m}}^2$

(ii)          Area of flower bed $=\pi r^2$ 
$=3.14\times 2\times 2=12.56{\text{ m}}^2$

(iii)      Area of lawn excluding the area of the flower bed
$=$ area of lawn $-$ area of flower bed
$=50-12.56$
$=37.44{\text{ m}}^2$

(iv)       The circumference of the flower bed $=2\pi r$ 
$=2\times 3.14\times 2=12.56\text{ m}$

In the following figures, find the area of the shaded portions:

(i)            

 

(ii)           

 

 

 

 

 

 

(i)             Here, $\text{AB}=18\text{ cm},\text{ BC}=10\text{ cm},$ $\text{AF}=6\text{ cm},$ $\text{AE}=10\text{ cm and BE}=8\text{ cm}$ 

Area of shaded portion (EFDC)

$=$ Area of rectangle ABCD $-$ (Area of $\Delta \text{FAE}+$ area of $\Delta \text{EBC}$ )

$=\left(\text{AB}\times \text{BC}\right)-\left(\frac{1}{2}\times \text{AE}\times \text{AF}+\frac{1}{2}\times \text{BE}\times \text{BC}\right)$

$=\left(18\times 10\right)-\left(\frac{1}{2}\times 10\times 6+\frac{1}{2}\times 8\times 10\right)$

$=180–\left(30+40\right)$

$=180-70$

$=110{\text{ cm}}^2$

(ii)          Here, SR $=$ SU $+$ UR $=$ $10+ 10= 20$ cm,
QR $=20$ cm

PQ $=$ SR $=20$ cm, PT $=$ PS $-$ TS $=$ $20- 10$ cm

TS $=10$ cm, SU $=10$ cm, QR $=20$ cm and
UR $=10$ cm

Area of shaded region (QTU)

$=$ Area of square PQRS $-$ Area of $\Delta \text{QPT}$ 
$-$ Area of $\Delta \text{TSU}$ $-$ Area of $\Delta \text{UQR}$

 

$=\left(\text{SR}\times \text{QR}\right)-\frac{1}{2}\times \text{PQ}\times \text{PT}-\frac{1}{2}\times \text{ST}\times \text{SU}-\frac{1}{2}\times \text{QR}\times \text{UR}$

$=20\times 20-\frac{1}{2}\times 20\times 10-\frac{1}{2}\times 10\times 10-\frac{1}{2}\times 20\times 10$

$=400-100-50-100$

$=150{\text{ cm}}^2$

Find the area of the quadrilateral ABCD. Here, $\text{AC}=22\text{ cm,}$ $\text{BM}=3\text{ cm,}$ $\text{DN}=3\text{ cm,}$ and $\text{BM}\perp \text{AC,}$ $\text{DN}\perp \text{AC}$

 

Here, $\text{AC}=22\text{ cm},\text{ BM}=\text{3 cm},\text{ DN}=3\text{ cm}$

Area of quadrilateral ABCD $=\text{Area of }\Delta \text{ABC}+\text{Area of }\Delta \text{ADC}$

$=\frac{1}{2}\times \text{AC}\times \text{BM}+\frac{1}{2}\times \text{AC}\times \text{DN}$

$=\frac{1}{2}\times 22\times 3+\frac{1}{2}\times 22\times 3$

$=3\times 11+3\times 11$

$=33+33$

$=66{\text{ cm}}^2$

Thus, the area of quadrilateral ABCD is $66{\text{ cm}}^2$.