NCERT Solutions for Class 7 Maths Chapter 11 - Perimeter And Area

Exercise – 11.1

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

Find: 

i. Its area.

Ans: 

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

Breadth of a rectangular piece of land ${\text{ = }}$${\text{300 m}}$

Area of given rectangular piece of land ${\text{ = }}$Length${\text{x}}$ Breadth 

${\text{ = }}$ ${\text{500 x 300}}$

$\therefore $ Area of given rectangular piece of land ${\text{ = }}$ ${\text{1,50,000 }}{{\text{m}}^2}$ 

ii. The cost of the land, if ${\text{1 }}{{\text{m}}^2}$ of the land costs ₹ ${\text{10,000}}$

Ans: 

As the cost of ${\text{1 }}{{\text{m}}^2}$land ${\text{ = }}$  ₹ ${\text{10,000}}$

$\therefore $ The cost of ${\text{1,50,000 }}{{\text{m}}^2}$land${\text{ = }}$${\text{10,000 x 1,50,000}}$ 

$\therefore $ The cost of ${\text{1,50,000 }}{{\text{m}}^2}$land ${\text{ = }}$₹${\text{1,50,00,00,000}}$ 

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

Ans: 

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

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

$\therefore {\text{ side  =  }}\dfrac{{{\text{320}}}}{{\text{4}}}{\text{  =  80 m}}$ 

Now, 

As the side of given square park is ${\text{80 m}}$ 

Area of square park ${\text{ = }}$${\text{side x side}}$ 

${\text{ =  80 x 80}}$ 

$\therefore $ Area of square park ${\text{ =  6400 }}{{\text{m}}^2}$ 

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

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

Ans:

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

$ \Rightarrow {\text{ length x breadth  =  440 }}{{\text{m}}^2}$ 

As the length of the rectangular park is given ${\text{22 m}}$

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

$\therefore {\text{breadth  =  }}\dfrac{{{\text{440}}}}{{{\text{22}}}}{\text{  =  20 m}}$ 

Now, 

Perimeter of rectangular park ${\text{ =  2(length x breadth)}}$ 

${\text{ =  2(22  +  20)  =  84 m}}$ 

Thus, the perimeter of the rectangular park is ${\text{84 m}}$.

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

Ans:

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

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

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

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

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

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

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

Now,

Area of rectangular sheet 

${\text{ =  length x breadth}} $

${\text{ =  35 x 15}} $  

${\text{ =  525 c}}{{\text{m}}^2} $ 

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

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

Ans:

The side of the given square park${\text{ =  60 m}}$ 

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

As per given,

Area of square park = Area of rectangular park

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

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

$\Rightarrow {\text{ breadth  =  }}\dfrac{{{\text{60 x 60}}}}{{90}} $  

$\Rightarrow {\text{ breadth  =  40 m}} $  

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

6. A wire is in the shape of a rectangle. Its length is ${\text{40 cm}}$and breadth is ${\text{22 cm}}$. 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? 

Ans:

According to the given, 

Perimeter of square $ = $ Perimeter of rectangle 

$\Rightarrow {\text{ 4 x side  =  2(length x breadth)}} $  

$\Rightarrow {\text{ 4 x side  =  2(40  +  22)}} $  

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

$\therefore {\text{ side  =  }}\dfrac{{2{\text{ x 62}}}}{4} $  

$\therefore {\text{ side  =  31 cm}} $ 

Thus, 

The side of the square is ${\text{31 cm}}$ 

Now, 

Area of rectangle 

${\text{ =  length x breadth }} $  

${\text{ =  40 x 22}} $  

${\text{ =  880 c}}{{\text{m}}^2} $

Area of square

${\text{ =  side x side}} $

$= 31{\text{ x 31}} $  

${\text{ =  961 c}}{{\text{m}}^2} $  

Therefore, in comparison, the area of a square is greater than that of a rectangle.

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

Ans:

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

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

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

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

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

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

$\Rightarrow {\text{ length  =  35 cm}} $  

Now,

Area of rectangle ${\text{ =  length x breadth  =  35 x 30  =  1050 c}}{{\text{m}}^2}$

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

8. A door of length${\text{2 m}}$ and breadth${\text{1 m}}$is fitted in a wall. The length of the wall is ${\text{4}}{\text{.5 m}}$ and the breadth is ${\text{3}}{\text{.6 m}}$ . Find the cost of white washing the wall, if the rate of white washing the wall is  ₹${\text{20 per }}{{\text{m}}^2}$ .

Ans:

Area of rectangular door $ = {\text{ length x breadth  =  2 x 1  =  2 }}{{\text{m}}^2}{\text{ }}$ 

Area of wall including door ${\text{ =  length x breadth  =  4}}{\text{.5 x 3}}{\text{.6  =  16}}{\text{.2 }}{{\text{m}}^2}$ 

Now, 

Area of wall excluding door $ = $ Area of wall including door${\text{ - }}$ Area of door 

${\text{ =  16}}{\text{.2  -  2  =  14}}{\text{.2 }}{{\text{m}}^2}$

$\because $ the rate of white washing of ${\text{1 }}{{\text{m}}^2}$ of wall $ = $ ₹$20$

$\therefore $the rate of white washing ${\text{14}}{\text{.2 }}{{\text{m}}^2}$ of wall ${\text{ =  20 x 14}}{\text{.2  =  284}}$  

Thus, the cost of white washing the wall excluding the door is ₹$284$ 

Exercise – 11.2

1. Find the area of each of the following parallelograms:

a.

Ans: 

As the area of parallelogram ${\text{ =  base x height}}$ 

Given base ${\text{ =  7 cm}}$  and height ${\text{ =  4 cm}}$ 

$\therefore $ Area of parallelogram ${\text{ =  7 x 4  =  28 c}}{{\text{m}}^2}$ 

b.

Ans:

Given base $ = {\text{ 5 cm}}$and height${\text{ =  3 cm}}$ 

$\therefore $ Area of parallelogram ${\text{ =  5 x 3  =  15 c}}{{\text{m}}^2}$ 

c.

Ans: 

Given base ${\text{ =  2}}{\text{.5 cm}}$ and height ${\text{ =  3}}{\text{.5 cm}}$

$\therefore $ Area of parallelogram ${\text{ =  2}}{\text{.5 x 3}}{\text{.5  =  8}}{\text{.75 c}}{{\text{m}}^2}{\text{ }}$ 

d.

Ans: 

Given base ${\text{ =  5 cm}}$and height ${\text{ =  4}}{\text{.8 cm}}$ 

$\therefore $ Area of parallelogram ${\text{ =  5 x 4}}{\text{.8  =  24 c}}{{\text{m}}^2}$ 

e.

Ans: 

Given base ${\text{ =  2 cm}}$ and height ${\text{ =  4}}{\text{.4 cm}}$ 

$\therefore $ Area of parallelogram ${\text{ =  2 x 4}}{\text{.4  =  8}}{\text{.8 c}}{{\text{m}}^2}$  

2. Find the area of each of the following triangles:

Ans:

As the area of triangle ${\text{ =  }}\dfrac{1}{2}{\text{ x base x height}}$ 

a. Given, base ${\text{ =  4 cm}}$ and height ${\text{ =  3 cm}}$ 

$\therefore $Area of triangle ${\text{ =  }}\dfrac{1}{2}{\text{ x 4 x 3  =  6 c}}{{\text{m}}^2}$ 

b. Given, base ${\text{ =  5 cm}}$ and height $ = {\text{ 3}}{\text{.2 cm}}$

$\therefore $ Area of triangle ${\text{ =  }}\dfrac{1}{2}{\text{ x 5 x 3}}{\text{.2  =  8 c}}{{\text{m}}^2}$ 

c. Given, base ${\text{ =  3 cm}}$ and height ${\text{ =  4 cm}}$ 

$\therefore $ Area of triangle ${\text{ =  }}\dfrac{1}{2}{\text{ x 3 x 4  =  6 c}}{{\text{m}}^2}$ 

d. Given, base ${\text{ =  3 cm}}$ and height ${\text{ =  2 cm}}$ 

$\therefore $ Area of triangle $ = {\text{ }}\dfrac{1}{2}{\text{ x  3 x 2  =  3 c}}{{\text{m}}^2}$ 

3. Find the missing values:

Ans:

As we know that, area of parallelogram ${\text{ =  base x height}}$

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

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

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

$\Rightarrow {\text{ height  =  12}}{\text{.3 cm}} $ 

b. Here, height $ = {\text{ 15 cm}}$ and area ${\text{ =  154}}{\text{.5 c}}{{\text{m}}^2}$ 

$\Rightarrow {\text{ base x 15  =  154}}{\text{.5}} $

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

$\Rightarrow {\text{ base  =  10}}{\text{.3 cm}} $ 

c. Here, height ${\text{ =  8}}{\text{.4 cm}}$ and area ${\text{ =  48}}{\text{.72 c}}{{\text{m}}^2}$ 

$\Rightarrow {\text{ base x 8}}{\text{.4  =  48}}{\text{.72}} $

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

$\Rightarrow {\text{ base  =  5}}{\text{.8 cm}} $   

d. Here, base ${\text{ =  15}}{\text{.6 cm}}$ and area $ = {\text{ 16}}{\text{.38 c}}{{\text{m}}^2}$ 

$\Rightarrow {\text{ 15}}{\text{.6 x height  =  16}}{\text{.38}} $

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

$\Rightarrow {\text{ height  =  1}}{\text{.05 cm}} $ 

Hence, the missing values are:

4. Find the missing values:

Ans:

As the area of triangle ${\text{ =  }}\dfrac{1}{2}{\text{ x base x height}}$ 

a. Given, base ${\text{ =  15 cm}}$ and area ${\text{ =  87 c}}{{\text{m}}^2}$ 

$\Rightarrow {\text{ }}\dfrac{1}{2}{\text{ x 15 x height  =  87}} $

$\Rightarrow {\text{ height  =  }}\dfrac{{87{\text{ x 2}}}}{{15}} $

$\Rightarrow {\text{ height  =  11}}{\text{.6 cm}} $ 

b. Given, height ${\text{ =  31}}{\text{.4 mm}}$ and area ${\text{ =  1256 m}}{{\text{m}}^2}$ 

$\Rightarrow {\text{ }}\dfrac{1}{2}{\text{ x base x 31}}{\text{.44  =  1256}} $

$\Rightarrow {\text{ base  =  }}\dfrac{{1256{\text{ x 2}}}}{{31.44}} $

$\Rightarrow {\text{ base  =  80 mm}} $ 

c. Given, base ${\text{ =  22 cm}}$ and area $ = {\text{ 170}}{\text{.5 c}}{{\text{m}}^2}$ 

$\Rightarrow {\text{ }}\dfrac{1}{2}{\text{ x 22 x height  =  170}}{\text{.5}} $

$\Rightarrow {\text{ height  =  }}\dfrac{{170.5{\text{ x 2}}}}{{22}} $

$\Rightarrow {\text{ height  =  15}}{\text{.5 cm}} $

Hence, the missing values are:

5. PQRS is a parallelogram. QM is the height from Q to SR and QN is the height from Q to PS. If SR ${\text{ =  12 cm}}$ and QM ${\text{ =  7}}{\text{.6 cm}}$.

Find:  a. the area of the parallelogram PQRS 

Ans:

In given parallelogram PQRS, 

SR ${\text{ =  12 cm}}$, QM $ = {\text{ 7}}{\text{.6 cm}}$ , PS ${\text{ =  8 cm}}$ 

Area of parallelogram ${\text{ =  base x height  =  12 x 7}}{\text{.6  =  91}}{\text{.2 c}}{{\text{m}}^2}{\text{ }}$  

b. QN, if PS ${\text{ =  8 cm}}$

Ans: Area of parallelogram ${\text{ =  base x height}}$ 

$\Rightarrow {\text{ 91}}{\text{.2  =  8 x QN}} $

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

$\Rightarrow {\text{ QN  =  11}}{\text{.4 cm}} $ 

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

Ans:

In the given parallelogram ABCD,

Area of parallelogram ${\text{ =  1470 c}}{{\text{m}}^2}$ 

Base (AB) ${\text{ =  35 cm}}$ 

Base (AD) $ = {\text{ 49 cm}}$ 

$\because $ area of parallelogram $ = {\text{ base x height}}$ 

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

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

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

Also,

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

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

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

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

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

Ans:

In right angled triangle BAC,

AB $ = {\text{ 5 cm}}$ and AC $ = {\text{ 12 cm}}$ 

We know that, area of triangle ${\text{ =  }}\dfrac{1}{2}{\text{x base x height}}$ 

$\Rightarrow $ Area of triangle ${\text{ =  }}\dfrac{1}{2}{\text{ x AB x AC  =  }}\dfrac{1}{2}{\text{ x 5 x 12  =  30 c}}{{\text{m}}^2}$ 

Now, in triangle ABC,

Area of triangle ABC $ = {\text{ }}\dfrac{1}{2}{\text{ x BC x AD}}$

$\Rightarrow {\text{ 30  =  }}\dfrac{1}{2}{\text{ x 13 x AD}} $

$\Rightarrow {\text{ AD  =  }}\dfrac{{30{\text{ x 2}}}}{{13}} $

$\Rightarrow {\text{ AD  =  }}\dfrac{{60}}{{13}}{\text{ cm}} $ 

8. ∆ ABC is isosceles with ${\text{AB  =  AC  =  7}}{\text{.5 cm}}$and BC $ = {\text{ 9 cm}}$. The height AD from A to BC, is  ${\text{6 cm}}$. Find the area of ∆ ABC. What will be the height from C to AB i.e., CE?

Ans:

in given triangle ABC,

AD $ = {\text{ 6 cm}}$ and BC ${\text{ =  9 cm}}$ 

We know that, area of triangle $ = {\text{ }}\dfrac{1}{2}{\text{ x base x height}}$ 

$ \Rightarrow $  area of triangle $ = {\text{ }}\dfrac{1}{2}{\text{ x BC x AD  =  }}\dfrac{1}{2}{\text{ x 9 x 6  =  27 c}}{{\text{m}}^2}$ 

Also,

area of triangle $ = {\text{ }}\dfrac{1}{2}{\text{ x base x height}}$

$\Rightarrow {\text{ 27  =  }}\dfrac{1}{2}{\text{ x AB x CE}} $

$\Rightarrow {\text{ 27  =  }}\dfrac{1}{2}{\text{ x 7}}{\text{.5 x CE}} $

$\Rightarrow {\text{ CE  =  }}\dfrac{{27{\text{ x 2}}}}{{7.5}} $

$\Rightarrow {\text{ CE  =  7}}{\text{.2 cm}} $ 

Thus, the height from C to AB i.e., CE is  ${\text{7}}{\text{.2 cm}}$.

Exercise – 11.3

1. Find the circumference of the circles with the following radius: (take  $\pi = \dfrac{{{\text{22}}}}{{\text{7}}}$ )

a. 14 cm

Ans:

We know that, circumference of circle  =  2 $\pi r$ 

$\therefore $ Circumference of the given circle ${\text{ =  2 x }}\dfrac{{22}}{7}{\text{ x 14  =  88 cm}}$ 

b. 28 mm

Ans:

We know that, circumference of circle =  2 $\pi r$

$\therefore $ Circumference of the given circle ${\text{ =  2 x }}\dfrac{{22}}{7}{\text{ x 28  =  176 mm}}$ 

c. 21 cm

Ans:

We know that, circumference of circle 2 $\pi r$

$\therefore $ Circumference of the given circle ${\text{ =  2 x }}\dfrac{{22}}{7}{\text{ x 21  =  132 cm}}$ 

2. Find the area of the following circles, given that: (take  $\pi = \dfrac{{{\text{22}}}}{{\text{7}}}$ )

a. radius $ = {\text{ 14 mm}}$ 

Ans:

We know that, area of circle $ =  \pi {{\text{r}}^{\text{2}}}$ 

Given, radius $ = {\text{ 14 mm}}$ 

$\therefore $  area of circle ${\text{ =  }}\dfrac{{22}}{7}{\text{ x 14 x 14  =  616 m}}{{\text{m}}^2}$ 

b. diameter ${\text{ =  49 m}}$

Ans:

We know that, area of circle $ =  \pi {{\text{r}}^{\text{2}}}$

Given, diameter ${\text{ =  49 m}}$ 

$\because {\text{ radius  =  }}\dfrac{{{\text{diameter}}}}{2}{\text{  =  }}\dfrac{{49}}{2}{\text{ m}}$ 

$\therefore $  area of circle ${\text{ =  }}\dfrac{{22}}{7}{\text{ x }}\dfrac{{49}}{2}{\text{ x }}\dfrac{{49}}{2}{\text{  =  1886}}{\text{.5 }}{{\text{m}}^2}$

c. radius ${\text{ =  5 cm}}$

Ans:

We know that, area of circle $ =  \pi {{\text{r}}^{\text{2}}}$

Given, radius ${\text{ =  5 cm}}$ 

$\therefore $  area of circle ${\text{ =  }}\dfrac{{22}}{7}{\text{ x 5 x 5  =  }}\dfrac{{550}}{7}{\text{ c}}{{\text{m}}^2}$

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

Ans:

Given, circumference of the circular sheet ${\text{ =  154 m}}$

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

$\Rightarrow {\text{ 2 x }}\dfrac{{22}}{7}{\text{ x r  =  154}} $

$\Rightarrow {\text{ r  =  }}\dfrac{{154{\text{ x 7}}}}{{2{\text{ x 22}}}} $

$\Rightarrow {\text{ r  =  24}}{\text{.5 m}} $

Now, area of circular sheet $ =  \pi {{\text{r}}^{\text{2}}}{\text{  =  }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 24}}{\text{.5 x 24}}{\text{.5  =  1886}}{\text{.5 }}{{\text{m}}^{\text{2}}}$

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

4. A gardener wants to fence a circular garden of diameter $21{\text{ m}}$ . 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  ₹$4$ per meter. (take  $\pi = \dfrac{{{\text{22}}}}{{\text{7}}}$ )

Ans:

Given, diameter of the circular garden $ = {\text{ 21 m}}$ 

$\therefore $ radius of circular garden $ = {\text{ }}\dfrac{{21}}{2}{\text{ m}}$ 

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

$\therefore $ circumference of the circular garden $ = {\text{ 2 x }}\dfrac{{22}}{7}{\text{ x }}\dfrac{{21}}{2}{\text{  =  66 m}}$ 

The gardener makes 2 rounds of fence,

So, the total length of the rope of fencing = 2 $\pi r  =  2 x 66  =  132 m$

$\because $  the cost of $1{\text{ m}}$ rope $ = $ ₹ $4$ 

$\therefore $  the cost of $132{\text{ m}}$ rope $ = $ $4{\text{ x 132  = }}$ ₹$528$ 

5. From a circular sheet of radius $4{\text{ cm}}$ , a circle of radius ${\text{3 cm}}$ is removed. Find the area of the remaining sheet. (Take $\pi {\text{  =  3}}{\text{.14}}$) 

Ans:

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

Radius of removed circle(r) $ = {\text{ 3 cm}}$

$\therefore $ area of remaining sheet $ = $ area of circular sheet $ - $ area of removed circle

$\therefore $ area of remaining sheet $=\pi R^{2}-\pi r^{2}$ 

$\therefore $ area of remaining sheet ${\text{ =  (3}}{\text{.14 x 4 x 4)   -  (3}}{\text{.14 x 3 x 3)}}$ 

$\therefore $ area of remaining sheet ${\text{ =  3}}{\text{.14 x (16  -  9)  =  3}}{\text{.14 x 7  =  21}}{\text{.98 c}}{{\text{m}}^2}$

Thus, the area of the remaining sheet is $21.98{\text{ c}}{{\text{m}}^2}$ .

6. Saima wants to put a lace on the edge of a circular table cover of diameter ${\text{1}}{\text{.5 m}}$. Find the length of the lace required and also find its cost if one meter of the lace costs ` ₹$15$. (Take $\pi {\text{  =  3}}{\text{.14}}$)

Ans:

Given, diameter of the circular table cover $ = {\text{ 1}}{\text{.5 m}}$ 

$\therefore $ radius of the circular table $ = {\text{ }}\dfrac{{1.5}}{2}{\text{ m}}$

Now, circumference of the circular table ${\text{ =  2}}\pi {\text{r  =  2 x 3}}{\text{.14 x }}\dfrac{{1.5}}{2}{\text{  =  4}}{\text{.71 m}}$ 

$\therefore $ the length of required lace is $4.71{\text{ m}}$ 

As, the cost of $1{\text{ m}}$ lace = ₹$15$

The cost of $4.71{\text{ m}}$lace $ = {\text{ 4 x 4}}{\text{.71  =  }}$ ₹$70.65$ 

Hence, the cost of $4.71{\text{ m}}$is ₹$70.65$.

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

Ans:

Given, diameter $ = {\text{ 10 cm}}$ 

$\therefore {\text{ radius  =  }}\dfrac{{{\text{diameter}}}}{2}{\text{  =  }}\dfrac{{10}}{2}{\text{  =  5 cm}}$ 

As per the question,

Perimeter of the figure $ = $ circumference of semicircle $ + $ diameter

$\therefore $ perimeter of the figure  $\pi r$  +  $d  =  \dfrac{{22}}{7}{\text{ x 5  +  10 }}$ 

$\therefore $ perimeter of the figure ${\text{ =  }}\dfrac{{110}}{7}{\text{  +  10  =  }}\dfrac{{110 + 70}}{7}{\text{  =  }}\dfrac{{180}}{7}{\text{  =  25}}{\text{.71 cm}}$ 

Hence, the perimeter of the given figure is $25.71{\text{ cm}}$ 

8. Find the cost of polishing a circular table-top of diameter $1.6{\text{ m}}$ , if the rate of polishing is ₹$15{\text{/}}{{\text{m}}^2}$ . (Take $\pi {\text{  =  3}}{\text{.14}}$)

Ans:

Given, diameter of the circular table top $ = {\text{ 1}}{\text{.6 m}}$ 

$\therefore $ radius of the circular table top $ = {\text{ }}\dfrac{{1.6}}{2}{\text{  =  0}}{\text{.8 m}}$ 

Now, area of the circular table top $ =  \pi {{\text{r}}^{\text{2}}}{\text{  =  3}}{\text{.14 x 0}}{\text{.8 x 0}}{\text{.8  =  2}}{\text{.0096 }}{{\text{m}}^2}$ 

The cost of $1{\text{ }}{{\text{m}}^2}$ of polishing $ = $ ₹$15$ 

$\therefore $ the cost of $2.0096{\text{ }}{{\text{m}}^2}$ of polishing ${\text{ =  15 x 2}}{\text{.0096  =  30}}{\text{.14(approx)}}$ 

Hence the cost of polishing a circular table of $2.0096{\text{ }}{{\text{m}}^2}$is ₹$30.14$ (approx).

9. Shazli took a wire of length $44{\text{ cm}}$  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? (take  $\pi = \dfrac{{{\text{22}}}}{{\text{7}}}$ )

Ans:

Given, total length of the wire $ = {\text{ 44 cm}}$ 

$\therefore $ circumference of the circle =2 $\pi r$ 

$\Rightarrow { 2\pi r  =  44} $

$\Rightarrow {\text{ 2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x r  =  44}} $

$\Rightarrow {\text{ r  =  }}\dfrac{{{\text{44 x 7}}}}{{{\text{2 x 22}}}} $

$\Rightarrow {\text{r =  7 cm}} $ 

Now, area of circle $ =  \pi {{\text{r}}^{\text{2}}}{\text{  =  }}\dfrac{{22}}{7}{\text{ x 7 x 7  =  154 c}}{{\text{m}}^2}$ 

Now, the wire is converted into square

$\therefore $ the perimeter of square $ = {\text{ 44 cm}}$ 

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

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

So, the area of square $ = $ side ${\text{x}}$ side $ = $ $11{\text{ x 11  =  121 c}}{{\text{m}}^2}$ 

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

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

Ans:

Given, radius of circular sheet(R) $ = {\text{ 14 cm}}$ 

Radius of smaller circle(r) $ = {\text{ 3}}{\text{.5 cm}}$

Length of rectangle (l) $ = {\text{ 3 cm}}$

Breadth of the rectangle (b) $ = {\text{ 1 cm}}$

As per given,

Area of remaining sheet $ = $ area of circular sheet $ - $ (area of two smaller circle $ + $ area of rectangle)

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

$= {\text{ [}}\dfrac{{22}}{7}{\text{ x 14 x 14]  -  [(2 x }}\dfrac{{22}}{7}{\text{ x 3}}{\text{.5 x 3}}{\text{.5)  +  (3 x 1)]}} $

${\text{ =  [22 x 14 x 2]  -  [(44 x 0}}{\text{.5 x 3}}{\text{.5)  +  3]}} $

${\text{ =  616  -  80}} $

${\text{ =  536 c}}{{\text{m}}^2} $ 

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

11. A circle of radius ${\text{2 cm}}$  is cut out from a square piece of an aluminium sheet of side $6{\text{ cm}}$. What is the area of the leftover aluminium sheet? (Take $\pi {\text{  =  3}}{\text{.14}}$)

Ans:

Given, radius of circle $ = {\text{ 2 cm}}$ 

Side of aluminium square sheet $ = {\text{ 6 cm}}$

As per given,

Area of aluminium sheet left $ = $ total area of square sheet $ - $ area of circle

$=$ side x side $-\pi r^{2}$

${\text{ =  (6 x 6)  -  (3}}{\text{.14 x 2 x 2)}} $

${\text{ =  36  -  12}}{\text{.56  =  23}}{\text{.44 c}}{{\text{m}}^2} $ 

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

12. The circumference of a circle is${\text{31}}{\text{.4 cm}}$. Find the radius and the area of the circle. (Take $\pi {\text{  =  3}}{\text{.14}}$)

Ans:

Given, the circumference of the circle $ = {\text{ 31}}{\text{.4 cm}}$ 

$\Rightarrow 2\pi r  =  31{\text{.4}} $

$\Rightarrow {\text{ 2 x 3}}{\text{.14 x r  =  31}}{\text{.4}} $

$\Rightarrow {\text{ r  =  }}\dfrac{{31.4}}{{2{\text{ x 3}}{\text{.14}}}} $

$\Rightarrow {\text{ r  =  5 cm}} $ 

Now, area of circle $=\pi r^{2}=3.14 \times 5 \times 5$  $=78.5 \mathrm{~cm}^{2}$

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

13. A circular flower bed is surrounded by a path $4{\text{ m}}$  wide. The diameter of the flower bed is ${\text{66 m}}$. What is the area of this path? (Take $\pi {\text{  =  3}}{\text{.14}}$)

Ans:

Given, diameter of the circular flower bed $ = {\text{ 66 m}}$ 

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

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

As per given,

Area of path $ = $ area of bigger circle $ - $ area of smaller circle

$=\pi \mathrm{R}^{\mathrm{2}}-\pi r^{2}=\pi\left({R}^{2}-r^{2}\right)$ 

\[{\text{ =  [(37}}{{\text{)}}^2}{\text{  -  (33}}{{\text{)}}^2}{\text{]}}\]

$= {\text{ 3}}{\text{.14[(37  +  33)(37  -  33)]}} $

${\text{ =  3}}{\text{.14 x 7 x 4  =  879}}{\text{.20 }}{{\text{m}}^2} $ 

$[\because {\text{ }}{{\text{a}}^2} - {\text{ }}{{\text{b}}^2}{\text{  =  (a  +  b)(a  -  b)]}}$ 

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

14. A circular flower garden has an area of $314{\text{ }}{{\text{m}}^2}$ . A sprinkler at the centre of the garden can cover an area that has a radius of $12{\text{ m}}$. Will the sprinkler water the entire garden?  (Take $\pi {\text{  =  3}}{\text{.14}}$)

Ans:

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

Circular area by the sprinkler $ = {\text{ 3}}{\text{.14 x 12 x 12  =  3}}{\text{.14 x 144  =  452}}{\text{.16 }}{{\text{m}}^2}$ 

Given, area of the circular flower garden $ = {\text{ 314 }}{{\text{m}}^2}$ 

As the area of the circular flower garden is smaller than the area by sprinkler, hence sprinkler will water the entire garden. 

15. Find the circumference of the inner and the outer circles, shown in the adjoining figure.  (Take $\pi {\text{  =  3}}{\text{.14}}$)

Ans:

given, radius of outer circle(r) $ = {\text{ 19 m}}$ 

$\therefore $ circumference of outer circle =  2 $\pi r$ $ = {\text{ 2 x 3}}{\text{.14 x 19  =  119}}{\text{.32 m}}$ 

Now, radius of the inner circle(r`) $ = {\text{ 19  -  10  =  9 m}}$ 

$\therefore $ circumference of inner circle =  2 $\pi r$ $ = {\text{ 2 x 3}}{\text{.14 x 9  =  56}}{\text{.52 m}}$

Hence, the circumference of inner and outer circles are ${\text{56}}{\text{.52 m and 119}}{\text{.32 m respectively}}$ 

16. How many times a wheel of radius $28{\text{ cm}}$ must rotate to go $352{\text{ m}}$? (take  $\pi = \dfrac{{{\text{22}}}}{{\text{7}}}$ )

Ans:

Let the wheel rotate n times of its circumference.

Given, radius of the wheel $ = {\text{ 28 cm}}$ 

And total distance $ = {\text{ 352 m  =  35200 cm}}$ 

$\therefore $ distance covered by wheel $ = {\text{ n x circumference of wheel}}$

$\Rightarrow$ 35200  =  n x $2\pi r $

$\Rightarrow {\text{ 35200  =  n x 2 x }}\dfrac{{22}}{7}{\text{ x 28}} $

$\Rightarrow {\text{ n  =  }}\dfrac{{35200{\text{ x 7}}}}{{2{\text{ x 22 x 28}}}} $

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

Thus wheel must rotate $200$times to go $352{\text{ m}}$ .

17. The minute hand of a circular clock is ${\text{15 cm}}$ long. How far does the tip of the minute hand move in 1 hour? (Take $\pi {\text{  =  3}}{\text{.14}}$)

Ans:

In 1 hour, a minute hand completes one round means makes a circle.

Given, radius of the circle(r) $ = {\text{ 15 cm}}$ 

Circumference of the circular clock =  2 $\pi r$  =  $2 x 3{\text{.14 x 15  =  94}}{\text{.2 cm}}$ 

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

Exercise – 11.4

1. A garden is $90{\text{ m}}$ long and $75{\text{ m}}$ 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 hectares. 

Ans:

Given, length of the rectangular garden $ = {\text{ 90 m}}$

Breadth of the rectangular garden $ = {\text{ 75 m}}$ 

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

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

Outer area of rectangular path $ = {\text{ length x breadth  =  100 x 85  =  8500 }}{{\text{m}}^2}{\text{ }}$ 

Inner area of garden without rectangular path $ = {\text{ length x breadth  =  90 x 75  =  6750 }}{{\text{m}}^2}{\text{ }}$

Now,

Area of path $ = $ Outer area of rectangular path $ - $ Inner area of garden without rectangular path

$\therefore $ area of path ${\text{ =  8500  -  6750  =  1750 }}{{\text{m}}^2}$ 

$\because {\text{ 1 }}{{\text{m}}^2}{\text{  =  }}\dfrac{1}{{10000}}{\text{ hectares}} $

$\therefore {\text{ 6750 }}{{\text{m}}^2}{\text{  =  }}\dfrac{{6750}}{{10000}}{\text{  =  0}}{\text{.675 hectares}} $ 

2. 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. 

Ans:

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

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

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

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

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

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

$\therefore $ area of path $ = $ ${\text{(AB x AD)  -  (EF x EH)  =  (131 x 71)  -  (125 x 65)}}$ 

$\therefore $ area of path $ = $ ${\text{9301  -  8125  =  1176 }}{{\text{m}}^2}$

Hence, the area of path around the park is $1176{\text{ }}{{\text{m}}^2}$ 

3. A picture is painted on a cardboard $8{\text{ cm}}$ long and ${\text{5 cm}}$wide such that there is a margin of ${\text{1}}{\text{.5 cm}}$along each of its sides. Find the total area of the margin. 

Ans:

Given, length of painted cardboard $ = {\text{ 8 cm}}$ 

Breadth of painted cardboard  $ = {\text{ 5 cm}}$

$\because $ there is margin of $1.5{\text{ cm}}$ from each of its side

$\therefore $ new length $ = {\text{ 8  -  (1}}{\text{.5  =  +  1}}{\text{.5)  =  8  -  3  =  5 cm}}$ 

$\therefore $ new breadth $ = {\text{ 5  -  (1}}{\text{.5  =  +  1}}{\text{.5)  =  5  -  3  =  2 cm}}$ 

$\therefore $area of margin $ = $ area of cardboard(ABCD) $ - $ area of cardboard(EFGH)

$\therefore $area of margin $ = $ ${\text{(AB x AD)  -  (EF x EH)  =  (8 x 5)  -  (5 x 2)}}$

$\therefore $area of margin $ = $ ${\text{40  -  10  =  30 c}}{{\text{m}}^2}$

Hence, the area of margin is ${\text{30 c}}{{\text{m}}^2}$ 

4. A verandah of width $2.25{\text{ m}}$is constructed all along outside a room which is $5{\text{ m}}$long and ${\text{4 m}}$wide. 

Find: i. the area of the verandah. 

Ans:

Given, the length of the room $ = {\text{ 5}}{\text{.5 m}}$ 

Width of the room $ = {\text{ 4 m}}$

The length of the room with verandah $ = {\text{ 5}}{\text{.5  +  2}}{\text{.25  +  2}}{\text{.25  =  10 m}}$

The width of room with verandah $ = {\text{ 4  +  2}}{\text{.25  +  2}}{\text{.25  =  8}}{\text{.5 m}}$

$\therefore $area of verandah $ = $ area of room with verandah(ABCD) $ - $ area of room without verandah(EFGH)

$\therefore $area of margin $ = $ ${\text{(AB x AD)  -  (EF x EH)  =  (10 x 8}}{\text{.5)  -  (5}}{\text{.5 x 4)}}$

$\therefore $area of margin $ = $ ${\text{85  -  22  =  63 }}{{\text{m}}^2}$

ii. the cost of cementing the floor of the verandah at the rate of  ₹$200{\text{ per }}{{\text{m}}^2}$.

Ans:

$\because $ the cost of cementing $1{\text{ }}{{\text{m}}^2}$ the floor of verandah $ = $ ₹$200$ 

$\therefore $ the cost of cementing $63{\text{ }}{{\text{m}}^2}$ floor of verandah $ = {\text{ 200 x 63  =  }}$ ₹$12,600$ 

5. A path $1{\text{ m}}$ wide is built along the border and inside a square garden of side ${\text{30 m}}$. 

Find: 

i. The area of the path. 

Ans:

Given, side of the garden $ = {\text{ 30 m}}$ 

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

Side of the square garden without path $ = {\text{ 30  -  (1  +  1)  =  3  -  2  =  28 m}}$ 

Now, 

$\therefore $area of path $ = $ area of ABCD $ - $ area of EFGH

$\therefore $area of margin $ = $ ${\text{(AB x AD)  -  (EF x EH)  =  (30 x 30)  -  (28 x 28)}}$

$\therefore $area of margin $ = $ ${\text{900  -  784  =  116 }}{{\text{m}}^2}$

ii. the cost of planting grass in the remaining portion of the garden at the rate of ₹${\text{40 per }}{{\text{m}}^2}$. 

Ans:

Area of remaining portion $ = {\text{ 28 x 28  =  784 }}{{\text{m}}^2}$ 

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

$\therefore $ the cost of cementing ${\text{784 }}{{\text{m}}^2}$ floor of verandah $ = {\text{ 40 x 784  =  }}$ ₹$31,360$ 

6. Two cross roads, each of width $10{\text{ m}}$, cut at right angles through the centre of a rectangular park of length $700{\text{ m}}$ and breadth $300{\text{ m}}$ 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. 

Ans:

Given, 

PQ $ = {\text{ 10 m}}$ and PS $ = {\text{ 300 m}}$

EH $ = {\text{ 10 m}}$ and EF $ = {\text{ 700 m}}$

KL $ = {\text{ 10 m}}$ and KN $ = {\text{ 10 m}}$

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

($\because$ KLMN is taken twice, which is to be subtracted)

$\therefore $area of roads $ = {\text{ PS x PQ  +  EF x EH  -  KL x KN}}$ 

$\therefore $area of roads $ = {\text{ (300 x 10)  +  (700 x 10)  -  (10 x 10)}}$

$\therefore $area of roads $ = {\text{ 3000  +  7000  -  100  =  9,900 }}{{\text{m}}^2}$

$\because {\text{ 1 }}{{\text{m}}^2}{\text{  =  }}\dfrac{1}{{10000}}{\text{ hectares}} $

$\therefore {\text{ 9,900 }}{{\text{m}}^2}{\text{  =  }}\dfrac{{9900}}{{10000}}{\text{  =  0}}{\text{.99 hectares}} $ 

Now,

Area of park excluding cross roads $ = $ area of park $ - $ area of road

$\therefore $ Area of park excluding cross roads      

$ = $${\text{(AB x AD)}} - 9900{\text{  =  (700 x 300)  -  9900  =  210000  -  9900  =  200100 }}{{\text{m}}^2}$ 

$\because {\text{ 1 }}{{\text{m}}^2}{\text{  =  }}\dfrac{1}{{10000}}{\text{ hectares}} $

$\therefore {\text{ 2,10,000 }}{{\text{m}}^2}{\text{  =  }}\dfrac{{210000}}{{10000}}{\text{  =  20}}{\text{.01 hectares}} $ 

7. Through a rectangular field of length $90{\text{ m}}$ and breadth $60{\text{ m}}$ , 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 ${\text{3 m}}$, 

Find: 

i. the area covered by the roads. 

Ans:

Given,

PQ $ = {\text{ 3 m}}$ and PS $ = {\text{ 60 m}}$

EH $ = {\text{ 3 m}}$ and EF $ = {\text{ 90 m}}$

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

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

($\because$ KLMN is taken twice, which is to be subtracted)

$\therefore $area of roads $ = {\text{ PS x PQ  +  EF x EH  -  KL x KN}}$ 

$\therefore $area of roads $ = {\text{ (60 x 3)  +  (90 x 3)  -  (3 x 3)}}$

$\therefore $area of roads $ = {\text{ 180  +  270  -  9  =  441 }}{{\text{m}}^2}$

ii. the cost of constructing the roads at the rate of  ₹${\text{110 per }}{{\text{m}}^2}$.

Ans:

$\because $ the cost of constructing $1{\text{ }}{{\text{m}}^2}$ roads $ = $ ₹$110$ 

$\therefore $ the cost of constructing ${\text{441 }}{{\text{m}}^2}$ roads $ = {\text{ 110 x 441  =  }}$ ₹$48,510$ 

Hence, the cost of constructing the roads is ₹$48,510$ 

8. Pragya wrapped a cord around a circular pipe of radius $4{\text{ cm}}$ (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side $4{\text{ cm}}$ (also shown). Did she have any cord left? (Take $\pi {\text{  =  3}}{\text{.14}}$)

Ans:

Given, radius of pipe $ = {\text{ 4 cm}}$ 

Wrapping cord around circular pipe =  2 $\pi r$  =  $2 x 3{\text{.14 x 4  =  25}}{\text{.12 cm}}$ 

Again,

Wrapping cord around a square $ = {\text{ 4 x side  =  4 x 4  =  16 cm}}$ 

Remaining cord $ = $ cord wrapped on wire $ - $ cord wrapped on square

$\therefore $ remaining cord $ = $ $25.12{\text{  -  16  =  9}}{\text{.12 cm}}$ 

Thus, she has left $9.12{\text{ cm}}$ cord.

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

Find: 

i. the area of the whole land.

Ans:

Given, length of rectangular lawn $ = {\text{ 10 m}}$ 

Breadth of the rectangular lawn $ = {\text{ 5 m}}$ 

Radius of the circular flower bed $ = {\text{ 2 m}}$ 

Area of the whole land $ = {\text{ length x breadth  =  10 x 5  =  50 }}{{\text{m}}^2}$ 

ii. the area of the flower bed.

Ans:

Area of flower bed =  2 $\pi r$  =  $3{\text{.14 x 2 x 2  =  12}}{\text{.56 }}{{\text{m}}^2}$

iii. the area of the lawn excluding the area of the flower bed.

Ans:

Area of lawn excluding the area of the flower bed $ = $ area of lawn $ - $ area of flower bed

$\therefore $ Area of lawn excluding the area of the flower bed $ = $ $50$ $ - $ $12.56{\text{  =  37}}{\text{.44 }}{{\text{m}}^2}$ 

iv. the circumference of the flower bed.

Ans: 

The circumference of the flower bed =  2 $\pi r$  =  2 x 3{\text{.14 x 2  =  12}}{\text{.56 m}}$ 

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

Ans:

i. Here,

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

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

Area of shaded portion $ = $ $({\text{AB x BC)  -  [(}}\dfrac{1}{2}{\text{ x AE x AF)  +  (}}\dfrac{1}{2}{\text{ x BE x BC)]}}$

Area of shaded portion $ = $ $(18{\text{ x 10)  -  [(}}\dfrac{1}{2}{\text{ x 10 x 6)  +  (}}\dfrac{1}{2}{\text{ x 8 x 10)]}}$

Area of shaded portion $ = $ $(18{\text{0)  -  [(30)  +  (40)]  =  180  -  70  =  110 c}}{{\text{m}}^2}$

ii. Here,

SR $ = $ SU $ + $ UR $ = {\text{ 10  +  10  =  20 cm}}$ 

QR $ = {\text{ 20 cm}}$ , PQ $ = $ SR $ = $ $20{\text{ cm}}$ 

PT $ = $ PS $ - $ TS $ = {\text{ 20  -  10 cm}}$ 

TS $ = {\text{ 10 cm}}$ , SU $ = {\text{ 10 cm}}$ , QR $ = {\text{ 20 cm}}$ and UR $ = {\text{ 10 cm}}$ 

Area of shaded region $ = $ Area of square PQRS $ - $ Area of $\vartriangle {\text{QPT}}$ $ - $ Area of $\vartriangle {\text{TSU}}$ $ - $ Area of $\vartriangle {\text{UQR}}$

Area of shaded region $ = $ (SR x QR) $ - $ ($\dfrac{1}{2}{\text{ x PQ x PT}}$) $ - $ ($\dfrac{1}{2}{\text{ x ST x SU}}$) $ - $ ($\dfrac{1}{2}{\text{ x UR x QR}}$) 

Area of shaded region $ = $ (${\text{20 x 20}}$) $ - $ ($\dfrac{1}{2}{\text{ x 20 x 10}}$) $ - $ ($\dfrac{1}{2}{\text{ x 10 x 10}}$) $ - $ ($\dfrac{1}{2}{\text{ x 20 x 10}}$) 

Area of shaded region $ = $ ($400$) $ - $ ($100$) $ - $ ($50$) $ - $ ($100$) $ = {\text{ 150 c}}{{\text{m}}^2}$ 

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

Ans:

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

area of quadrilateral ABCDF $ = $ area of $\vartriangle {\text{ABC}}$ $ + $ area of $\vartriangle {\text{ADC}}$

Area of  quadrilateral ABCDF $ = $ ($\dfrac{1}{2}{\text{ x AC x BM}}$) $ + $ ($\dfrac{1}{2}{\text{ x AC x DN}}$)

Area of  quadrilateral ABCDF $ = $ ($\dfrac{1}{2}{\text{ x 22 x 3}}$) $ + $ ($\dfrac{1}{2}{\text{ x 22 x 3}}$)

Area of  quadrilateral ABCDF $ = $ (${\text{11 x 3}}$) $ + $ (${\text{11 x 3}}$) $ = {\text{ 33  +  33  =  66 c}}{{\text{m}}^2}$ 

Hence, the area of quadrilateral ABCDF is $66{\text{ c}}{{\text{m}}^2}$ 

NCERT Solutions for Class 7 Maths Chapter 11 – Free PDF Download

Class 7 Maths Chapter 11 Includes:

Facts

• The total length of the boundary of a closed rectilinear figure is called its perimeter.

• The surface enclosed by a two-dimensional figure is called its area.

• The area of a parallelogram = Base x Height (Altitude).

• The area of a triangle = ½ x Base x Altitude corresponding to the base.

• Circumference of a circle = 2πr.

• Area of a circle = πr2.

• 1 cm2 = 100 mm2

• 1m2 = 10000 cm2

• 1 hectare = 10000 m2

• The perimeter of a regular polygon = Length of a side x Number of sides.

Rectilinear Figure

A rectilinear figure is a plane figure enclosed by line segments and is said to be simple if no two sides of it intersect at a point other than the vertex.

Triangle

A plane figure that is enclosed by three line segments is called a triangle. 

Quadrilateral

A quadrilateral is a plane figure that is enclosed by of four line segments. 

Parallelogram

A parallelogram is a plane figure enclosed by four line segments. The opposite sides in a parallelogram are parallel and equal in length.

We also know that the sum of lengths of all sides of a rectilinear figure is called its perimeter and the magnitude of a plane region is called its area. 

Example: The side of a square is 40 cm and the breadth of a rectangle is 20cm. If their areas are equal, then find the perimeter of the rectangle.

Solution.

Side of the square = 40 cm.

Area of the square = (Side)2 = (40 cm)2

= 40 cm x 40 cm = 1600 cm2

Since, Area of the rectangle = Area of the square

Area of the rectangle = 1600 cm2

(l x b) = 1600 => l x 20 = 1600

l = 1600 = 80 cm

20

Therefore, length ( l ) = 80 cm

Perimeter = 2 ( l + b ) = 2 ( 80 + 20 ) cm

= 2 x 100 cm = 200 cm

By referring to the examples given above, you can now solve Exercise 11.1 in NCERT solutions.

Area of a Triangle

The area of a triangle is ½ x Base x Altitude corresponding to the base.

Example: Find the area of a triangle when the base (b) is 4cm and height (h) = 3cm.

Solution:

We know, Area of a triangle = ½ x b x h

= (½ x 4 x 3) cm2 = 6 cm2

Area of a parallelogram

The area of a parallelogram is base x height.

Example: Find the area of a parallelogram when the base (b) is 7cm and height (h) = 4cm.

Solution:

We know, Area of a parallelogram = b x h

= (7 x 4) cm2

= 28 cm2

Circumference of a Circle

The perimeter of a circle is called the circumference of the circle. To establish the formula, let us go through the following activity.

Draw four circles of different radii and find their circumference by using a string. In each case, find the ratio of the circumference to the diameter (i.e., circumference + diameter). This ratio is always more than three times its diameter and its constant. In general, we denote this constant ratio by π (pronounced as pi).

The value of π is 3.14 to two decimal places, its approximate value is equal to 22/7. 

If ‘C’ represents the circumference of the circle, and ‘d’ represents the diameter, then,

C/d = π => C = πd

d = 2r (where r is the radius of the circle)

C = 2πr

Area of a Circle

The space occupied by a circle on a flat plane that has length and breadth is called the area of the surface. The formula to find the area of the circle is πr2. So the area of a circle is one complete cycle of the radius of the circle. 

Let us understand an example that will help you with the exercise 11.3 in NCERT Solutions of Chapter 11.

Example:

Find the area of a circle whose circumference is 88 cm.

Solution

Here, the circumference is given as 88cm.

Let the radius of the circle be ‘r’.

2πr = 88

=> 2 x 22/7 x r = 88

=> r = 88 x 7/2 x 22 = 14 cm

Now, area of the circle = πr2 = 22/7 x 14 x 14 cm2 = 22 x 2 x 14 cm2

= 616 cm2

Thus, the area of the circle is 616 cm2. 

Conversion of Units

We have:

Note: Since,   1m = 100 cm ∴ 1 m2 = 10000 cm2

            1 km = 1000 m ∴ 1 km2 = 10,00,000 m2

            1 hectare (ha) = 100 m x 100m = 10000 m2

Applications

Now, we will apply the methods of finding areas of squares, rectangles, circle, etc. to find the areas of some pathways or borders and to find the cost of making them.

Let us see an example.

Example: A rectangular park is 40m long and 25 m wide. A path 2.5m wide is constructed outside the park. Find the area of the path.

Solution: Let ABCD be the rectangular park of sides 40m and 25 m, and the region between the outer and inner rectangles represents the path 2.5 m wide.

Now, PQ = (40 + 2.5 + 2.5) m = 45m

PS = (25 + 2.5 + 2.5) m = 30m

∴   Area of rectangle ABCD = l x b = 40 m x 25 m

                                    = 1000 m2

So,  Area of the path = (Area of rectangle PQRS ) – ( Area of rectangle ABCD )

= 1350 m2 – 1000 m2 

= 350 m2

Benefits of using Vedantu for Class 7 Maths Chapter 11 - Perimeter and Area

Key Features of NCERT Solutions, These solutions are designed to help students achieve proficiency in their studies. They are crafted by experienced educators who excel in teaching Class 7 Maths. Some of the features include:

• Comprehensive explanations for each exercise and questions, promoting a deeper understanding of the subject.

• Clear and structured presentation for easy comprehension.

• Accurate answers aligned with the curriculum, boosting students' confidence in their knowledge.

• Visual aids like diagrams and illustrations to simplify complex concepts.

• Additional tips and insights to enhance students' performance.

• Chapter summaries for quick revision.

• Online accessibility and downloadable resources for flexible study and revision.

Conclusion

The NCERT Solutions for Class 7 Maths Chapter 11 - Perimeter and Area, provided by Vedantu, is a valuable tool for Class 7 students. It helps introduce Maths concepts in an accessible manner. The provided solutions and explanations simplify complex ideas, making it easier for Class 7 students to understand the material. By using Vedantu's resources, Students can develop a deeper understanding of NCERT concepts. These solutions are a helpful aid for grade 7 students, empowering them to excel in their studies and develop a genuine appreciation for Perimeter and Area.