# Exercise 11.1 Perimeter-and-Area -NCERT Solutions Class 7

## Chapter 11 Ex.11.1 Question 1

The length and the breadth of a rectangular piece of land are \(500 \rm{}\,m\) and \(300 \,\rm{}m\) respectively. Find

(i) its area

(ii) the cost of the land, if \(1\,\rm{}m^2\) of the land costs \(₹10,000\).

**Solution**

**Video Solution**

**What is given known?**

The length and the breadth of a rectangular piece of land.

**What is the unknown?**

(i) its area

(ii) the cost of the land, if \(1\,\rm{}m^2\) of the land costs \(₹10,000.\)

**Reasoning: **

As the length and breadth of the land are known, the area of the rectangular piece of land can be calculated (\(\rm area = \rm length \times \rm breadth\)). After calculating the area, the cost of land can be calculated by multiplying the total area with the cost of \(1\,\rm{}m^2\) of the land.

**Steps:**

(i) We know that,

\(\begin{align}&\text{Area of rectangle}\\& = {\rm{Length}} \times {\rm{Breadth}} \end{align} \)

\(\begin{align}&= 500 \times 300\\ &= 150000\;\rm{m^2}\end{align}\)

(ii) Cost of \(1\,\rm{}m^2\) of the land \(= \rm{}Rs \,10,000\)

\[\begin{align}\therefore {\text{Cost of 1500 }}{{\rm{m}}^{\rm{2}}}\;{\text{of the land}}\end{align}\]

\[\begin{align}&= 10,000 \times 1500 \\&= {\rm{Rs}}1,500,000,000\end{align}\]

## Chapter 11 Ex.11.1 Question 2

Find the area of a square park whose perimeter is \(320\,\rm{} m.\)

**Solution**

**Video Solution**

**What is known?**

Perimeter of a square park.

**What is the unknown?**

The area of a square park.

**Reasoning: **

In this question, perimeter of square park is given. By using the fact that the perimeter of square is equal to the sum of all the sides, we can find out the sides of square. Once the value of sides are known, the area of square can be easily obtained.

**Steps:**

Given,

Perimeter of square park \(= 320\,\rm{}m\)

Let length of each side of the square be \(a\)

\[\begin{align}\text{Perimeter} &= \text{Sum of all the sides}\\320\,\rm m&= a + a + a + a\\4a &= 320 \, \rm m\\a &= 80\;\rm{m}\end{align}\]

So, \(a = 80\,\rm{}m\)

\[\begin{align}\therefore{\text{Area of square}}&= {\text{side}} \times {\rm{side}}\\ &= 80 \times 80{\rm{ }}\\ &= 6400{\rm{ }}{\rm \,m^2}\end{align}\]

## Chapter 11 Ex.11.1 Question 3

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

**Solution**

**Video Solution**

**What is known?**

Area and length of rectangular plot of land.

**What is unknown?**

The breadth of a rectangular plot and its perimeter.

**Reasoning: **

This question is based on the concept of area and perimeter of rectangle.using the formula of area of rectangle, find the breadth of rectangular plot. Now,the length and breadth of the rectangular plot is known, use the formula of perimeter of rectangle and find the perimeter of plot of land.

**Steps:**

Area of rectangular plot of land \(= 440 \,\rm{m^2}\)

Length of rectangular plot of land \(= 22 \,\rm{}m\)

Let us say that breadth of rectangular plot of land \(= x \,\rm{}m\)

We know that,

\(\begin{align}&\text{Area of rectangle}\\&= {\rm{Length}} \times {\rm{Breadth}} \end{align} \)

\[\begin{align} {\rm{440}} &= {\rm{22}}\,x\\x &= \frac{{440}}{{22}}\\x &= 20\; \rm m\end{align}\]

So, breadth of rectangular plot of land \(= 20 \,\rm{}m\)

Perimeter of a rectangle

\[\begin{align} &= 2\left( {{\rm{Length}} + {\rm{Breadth}}} \right)\\&= 2\left( {22 + 20} \right)\\ &= 84{\,\rm{ m}}\end{align}\]

## Chapter 11 Ex.11.1 Question 4

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

**Solution**

**Video Solution**

**What is known?**

The perimeter of a rectangular sheet and the length.

**What is unknown?**

The area and breadth of rectangular sheet.

**Reasoning: **

Perimeter of the rectangular sheet is \(100 \,\rm{}cm\) and its length is \(35 \,\rm{}cm.\) By using the formula of perimeter of rectangle, find the breadth of rectangular sheet. Once length and breadth of rectangular sheet are known, area of rectangular sheet can be calculated.

**Steps:**

Given,

Perimeter of a rectangular sheet \(= 100 \,\rm{}cm\)

Length of a rectangular sheet \(= 35 \,\rm{}cm\)

Let, the breadth \(= b \,\rm{}cm\)

Perimeter of a rectangular sheet

\[\begin{align} &= {\rm{2}}\left( {{\rm{Length}} + {\rm{Breadth}}} \right)\\{\rm{100}} &= {\rm{2}}\,l + {\rm{2}}\,b\\{\rm{100}} &= {\rm{2}} \times {\rm{35}} + {\rm{2}}\,b \\{\rm{100}} &= {\rm{70}} + {\rm{2}}\,b\\{\rm{2}}\,b &= {\rm{100}} - {\rm{70}}\\b &= \frac{{{\rm{30}}}}{{\rm{2}}}\\b &= {\rm{15 }}\,\rm{}cm\end{align}\]

Area of a rectangular sheet

\[\begin{align} &= {\rm{Length}} \times {\rm{Breadth}}\\&= 35 \times 15{\rm{ }}\\&= 525{\rm{ }}\,\rm{}c{m^2}\end{align}\]

## Chapter 11 Ex.11.1 Question 5

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

**Solution**

**Video Solution**

**What is known?**

The area of a square park is the same as of a rectangular park. Also, the side of the square park is \(60 \,\rm{}m\) and the length of the rectangular park is \(90\,\rm{m}\)

**What is unknown?**

The breadth of the rectangular park.

**Reasoning: **

This question is based on the concept of area of square and rectangle. The side of the square park is \(60\,\rm{m}\) and the length of the rectangular park is \(90\,\rm{m}.\) Use the formula’s related to area of square and rectangle put the given values and find the breadth of rectangular park.

**Steps:**

Given,

Area of a square park \(=\) Area of a rectangular park

Side of the square park \(= 60\,\rm{m}\)

Length of the rectangular park \(= 90\,\rm{m}\)

Let breadth of the rectangular park \(= b \,\rm{}m\)

Since, area of a square park \(=\) area of a rectangular park

So,

\[\begin{align}{\rm{Side}} \times {\rm{Side}} &= {\rm{Length}} \times {\rm{Breadth}}\\60 \times 60 &= 90 \times b\\b &= \frac{{3600}}{{90}}\\b &= 40 \, \rm{ } m\end{align}\]

## Chapter 11 Ex.11.1 Question 6

A wire is in the shape of a rectangle. Its length is \(40 \,\rm{} cm\) and breadth is \(22 \,\rm{}cm.\) 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 ?

**Solution**

**Video Solution**

**What is known?**

A wire is in the shape of a rectangle. Its length is \(40 \,\rm{} cm\) and breadth is \(22 \,\rm{}cm\) and the same wire is rebent in the shape of a square.

**What is unknown?**

The measure of each side of square and which shape (rectangle or square) encloses more area.

**Reasoning: **

The wire is initially in the shape of rectangle and the same wire is rebent in the shape of square. This means the perimeter of both the rectangle and square is the same. By using the equations related to perimeter of square and rectangle, we can find the side of the square. Once side of the square is known, its area can be calculated. Similarly, by using the length and breadth of the rectangle, its area can be obtained. Now, by knowing the area of both the shapes you can easily decide which shape encloses more area.

**Steps:**

Given,

Length of rectangle \(= 40 \,\rm{}cm\)

Breadth \(= 22 \,\rm{}cm\)

Since same length of wire is used to from rectangle and square,

Perimeter of square \(=\) perimeter of rectangle

\[\begin{align}4 \times {\rm{Side}} &= 2\left( {{\rm{Length}} + {\rm{Breadth}}} \right)\\4 \times {\rm{Side}} &= 2\left( {40 + 22} \right){\rm{ }}\\4 \times {\rm{Side}} &= 2 \times 62{\rm{ }}\\4 \times {\rm{Side}} &= 124\\{\rm{Side}} &= \frac{{124}}{4}\\{\rm{Side}} &= 31\;{\rm{cm}}\end{align}\]

\[\begin{align}\text{Area of square}& = {\rm{side}} \times {\rm{side}}\\& = 31 \times 31\\ &= 961\;{\rm c{m^2}}\end{align}\]

\(\begin{align}& \text{Area of rectangle} \\ &= {\rm{Length}} \times {\rm{Breadth}}\end{align} \)

\(\begin{align} & = 40 \times 22\\&= 880{\rm{ }}{\rm\, c{m^2}}\end{align}\)

Therefore, it is clear from the above that the figure square encloses more area than rectangle.

## Chapter 11 Ex.11.1 Question 7

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

**Solution**

**Video Solution**

**What is known?**

The perimeter and the breadth of the rectangle.

**What is unknown?**

Length and area of the rectangle.

**Reasoning: **

Since the perimeter and breadth of rectangle are given, we can find out the length of the rectangle. From the values of length and breadth, area of rectangle can be calculated.

**Steps:**

Given,

Breadth of the rectangle \(= 30 \,\rm{}cm\)

Perimeter of a rectangle \(= 130 \,\rm{}cm.\)

We know that,

\(\begin{align}&\text{Perimeter of rectangle} \\ &= 2 \times \left( {{\rm{Length}} + {\rm{Breadth}}} \right)\end{align} \)

\[\begin{align}130 &= 2 \times \left( {l + 30} \right)\\130 &= 2\,l + 60\\2\,l &= 130-60\\2\,l &= 70\\l &= \frac{{70}}{2}\\l &= 35\;\rm{cm}\end{align}\]

Area of rectangle

\[\begin{align}&=\text{ Length} \times\text{ Breadth} \\ & = 35 \times 30 \\ &= 1050 \,\rm{cm^2}\end{align} \]

## Chapter 11 Ex.11.1 Question 8

A door of length \(2\,\rm{}m\) and breadth \(1\,\rm{}m\) is fitted in a wall. The length of the wall is \(4.5\,\rm{}m\) and the breadth is \(3.6\,\rm{}m\) (See the below Fig). Find the cost of white washing the wall, if the rate of white washing the wall is \(\rm{}Rs\, 20 \,per\, m^2\)

Fig 11.6.

**Solution**

**Video Solution**

**What is known?**

A door of length \(2\,\rm{}m\) and breadth \(1\,\rm{}m\) is fitted in a wall of length \(4.5\,\rm{}m\) and the breadth \(3.6\,\rm{}m\)

**What is unknown?**

The cost of white washing the wall, if the rate of white washing the wall is \(\rm{}Rs\, 20 \,per\, m^2.\)

**Reasoning: **

Since door will not be whitewashed, we will have to subtract area of the door from the area of wall. After, finding area to be whitewashed, multiply the area with rate of white washing \(\rm{}per\, m^2\) to get the cost.

**Steps:**

Given,

Length of wall \(= 4.5\,\rm{}m\)

Breadth of wall \(= 3.6\,\rm{}m\)

\[\begin{align}\text{Area of wall}&= {\rm{Length}} \times {\rm{Breadth}}\\ &= 4.5 \times 3.6{\rm{ }}\\&= 16.2{\rm{ }}{\rm \,m^2}\end{align}\]

Given,

Length of door \(= 2\,\rm{}m\)

Breadth of door \(= 1\,\rm{}m\)

So,

\[\begin{align}\text{Area of door}&= {\rm{Length}} \times {\rm{Breadth}}\\&= 2 \times 1\\&= 2{\rm{ }}{\rm \,m^2}\end{align}\]

\(\begin{align}&\text{Area of wall for white wash}\\ &= {\text{Area of wall}} - {\text{Area of door}} \end{align}\)

\(\begin{align} &= 16.2{\rm{ }}{\rm \,m^2} - 2{\rm{ }}{\rm \,m^2}\\&= 14.2{\rm{ }}{\rm \,m^2}\end{align}\)

The rate of white washing the wall

\(= \rm{Rs }\,20\,{\rm{ per\, }}{\rm m^2} \)

Therefore,

\(\begin{align}{\text{The cost of white washing}}\;1618{\rm{ }}{\,\rm m^2} \end{align} \)

\[\begin{align} &= 14.2 \times 20\\ &= {\rm{Rs\, }}284\end{align}\]