NCERT Solutions Class 6 Maths Chapter 6 Perimeter and Area Exercise 6.1

Exercise 6.1 of Class 6 Math Chapter 6 - Perimeter and area- helps you learn how to find the perimeter of different shapes of figures like squares and rectangles. In this exercise, you will learn how to measure the boundary of a shape using simple formulas. These measures are useful, not only for your exams but also in everyday life.

The exercise is made as per the latest NCERT syllabus and helps you to develop your understanding of the key idea of measuring clearly and simply. Find access to the complete NCERT Solutions for class 6 Maths Chapter 6 exercise 6.1 which is available as a PDF. These solutions have been explained in a step by step manner, for better understanding. The NCERT Solutions will help you obtain better marks, and prepare yourself well for the future concepts.

1.0Download NCERT Solutions Class 6 Maths Chapter 6 Perimeter and Area Exercise 6.1: Free PDF

Exercise 6.1 helps you to learn how to measure the perimeter of square and rectangular shapes which helps you develop a solid foundation in measurement and geometry. Download the complete NCERT Solutions for Class 6 Maths Chapter 6 Exercise 6.1 and download the FREE PDF from below:

2.0Key Concepts Covered in Exercise 6.1 Class 6 Maths

• Perimeter formula of rectangle: 2(l + b): Calculating the total boundary length of a rectangle using its length and breadth.
• Perimeter of square: 4a: Determining the perimeter of a square by multiplying the length of one side by four.
• Finding missing sides using perimeter: Using the perimeter formula to determine unknown side lengths of shapes.
• Real-life applications (fencing, rope rounds): Applying perimeter concepts to practical situations such as fencing fields or running around tracks.
• Logical reasoning with track problems: Solving problems involving repeated rounds and position marking along a track.

3.0NCERT Solutions Class 6 Chapter 6 Perimeter and Area: All Exercises

4.0NCERT Class 6 Maths Chapter 6 Perimeter and Area Exercise 6.1: Detailed Solutions

6.1 Perimeter

1. Find the missing terms: a. Perimeter of a rectangle $=14\mathrm{cm}$; breadth $=2\mathrm{cm}$; length = ?. b. Perimeter of a square $=20\mathrm{cm}$; side of a length = ?. c. Perimeter of a rectangle $=12\mathrm{m}$; length $=3\mathrm{m}$; breadth = ?. Sol. a. We know that perimeter of a rectangle $=2(\ell +b)$
   
   Here, the perimeter of the rectangle $=14\mathrm{cm}$ and breadth $\mathrm{b}=2\mathrm{cm},\ell =$ ? Thus $14=2(\ell +2)$ $\Rightarrow 14=2\ell +4$ $\Rightarrow 2\ell =14-4=10$ $\Rightarrow \ell =10÷2=5\mathrm{cm}$ b. Perimeter of a square $=20\mathrm{cm}$; side of a length $=$ ?. We know that the perimeter of the square $=4\times a$ where $\mathrm{a}=$ side of the square
   
   Side $\therefore 20=4\times \mathrm{a}\Rightarrow \mathrm{a}=5\mathrm{cm}$ c. Perimeter of rectangle $=2(\ell +b)$
   
   $\begin{array}{rl}& \Rightarrow 12=2(3+b)\\ & \Rightarrow 12=6+2b\\ & \Rightarrow 12-6=2b\\ & \Rightarrow 2b=6\\ & \Rightarrow b=3\\ & \text{ breadth }=3\mathrm{m}\end{array}$
2. A rectangle having side lengths 5 cm and 3 cm is made using a piece of wire. If the wire is straightened and then bent to form a square, what will be the length of a side of the square? Sol. Perimeter of rectangle $=2(5+3)$ $=2\times 8$ $=16\mathrm{cm}$ Now the wire is straightened and then bent to form a square. $\therefore$ Perimeter of square $=16\mathrm{cm}$ $\Rightarrow 4\mathrm{a}=16\mathrm{cm}$ $\Rightarrow \mathrm{a}=4\mathrm{cm}$, the required length of the side of the square.
   
3. Find the length of the third side of a triangle having a perimeter of 55 cm and having two sides of length 20 cm and 14 cm , respectively. Sol. Let the length of the third side of the triangle be x cm then
   
   Perimeter of triangle $=\mathrm{AB}+\mathrm{BC}+\mathrm{CA}$ $\Rightarrow 55=20+14+x$ $\Rightarrow 55=34+x$ $\Rightarrow \mathrm{x}=55-34$ $\Rightarrow \mathrm{x}=21\mathrm{cm}$
4. What would be the cost of fencing a rectangular park whose length is 150 m and breadth is 120 m , if the fence costs ₹ 40 per metre? Sol. The length of the fence is the perimeter of the rectangular park.
   
   Given that the length of the rectangular park $=150\mathrm{m}$ and breadth $=120\mathrm{m}$ $\therefore$ Perimeter $=2(\ell +\mathrm{b})$ $=2(150+120)$ $=2(270)$ $=540\mathrm{m}$ Now cost of fencing per meter $=$ ₹ 40 Cost of fencing the rectangular park $=₹40\times 540=₹21600$
5. A piece of string measure 36 cm in length. What will be the length of each side if it is used to form: a. A square, b. A triangle with all sides of equal length, and c. A hexagon (a six-sided closed figure) with equal-length sides? Sol. a. Given, a piece of string is 36 cm long $\therefore$ length of each side of the square $=\mathrm{a}$ perimeter $=36$ $\Rightarrow 4\mathrm{a}=36$ $\Rightarrow \mathrm{a}=9\mathrm{cm}$ b. Length of each side of the triangle $=a$ (Given) perimeter $=36$ $\Rightarrow 3\mathrm{a}=36$ $\Rightarrow \mathrm{a}=12\mathrm{cm}$ c. Length of each side of hexagon $=a$ perimeter $=36$ $\Rightarrow 36=6a$ $\Rightarrow \mathrm{a}=6\mathrm{cm}$
6. A farmer has a rectangular field having length 230 m and breadth 160 m . He wants to fence it with 3 rounds of rope as shown. What is the total length of rope needed?
   
   Sol. Perimeter of the rectangular field $=2(\ell +b)$ Here $\ell =230\mathrm{m},\mathrm{b}=160\mathrm{m}$ $\therefore P=2(230+160)$ = 2 (390) $=780\mathrm{m}$ Distance covered by a farmer in one round $=780\mathrm{m}$ $\therefore$ Total length of rope needed $=3\times 780=2340\mathrm{m}$

Matha Pachchi!

7. Find out the total distance Akshi has covered in 5 rounds. Sol. Distance covered by Akshi in 5 rounds $=5\times$ perimeter of PQRS $=5\times 220$ $=1100\mathrm{m}$
8. Find out the total distance Toshi has covered in 7 rounds. Who ran a longer distance? Sol. Distance covered by Toshi in 7 rounds $=7\times 180=1260\mathrm{m}$ $\therefore 1260\mathrm{m}>1100\mathrm{m}$ Hence, Toshi ran the longer distance.
9. Think and mark the positions as directed- a. Mark 'A' at the point where Akshi will be after she runs 250 m . b. Mark 'B' at the point where Akshi will be after she runs 500 m . c. Now, Akshi ran 1000 m . How many full rounds has she finished running around her track? Mark her position as ' $C$ '. d. Mark ' X ' at the point where Toshi will be after she runs 250 m . e. Mark ' $Y$ ' at the point where Toshi will be after she runs 500 m . f. Now, Toshi ran 1000 m . How many full rounds has she finished running around her track? Mark her position as ' Z '. Sol. a. Here, 1.One complete round $=220$ meters. 2.Distance Akshi has run $=250$ meters. 3.Extra distance beyond one round $=250-220=30$ meters. Since Akshi has already completed one full round, she will be 30 meters into her second round. So, after running 30 meters more, she will be on the length side of the track, 30 meters from the starting point. Therefore, mark ' A ' at the point 30 meters along the length of the track from the starting point.
   
   b. Distance per round $=220$ meters Total distance Akshi runs = 500 meters. First, we will find out how many complete rounds she runs: Number of complete rounds $=500÷220=2.27$ (approx.) This means Akshi completes 2 full rounds and then runs an additional distance $=500-(2\times 220)=60\mathrm{m}$ Therefore, Akshi will be 60 meters along the length of the track from the starting point, we can mark point ' B ' at this position on the track.
   
   c. Now, Akshi ran 1000 meters. To find out how many full rounds she completed, we divide the total distance she ran by the perimeter of the track: Number of full rounds $=1000÷220=4.545$ rounds. Akshi has completed 4 full rounds and is partway through her 5th round. To find her position on the track, we calculate the remaining distance after 4 full rounds: Remaining distance $=1000\mathrm{m}-(4\times 220\mathrm{m})$ $=1000\mathrm{m}-880\mathrm{m}$ $=120\mathrm{m}$ Since she has run an additional 120 meters after completing 4 full rounds, her position will be 120 meters from the starting point. If we mark her starting point as ' P ', her position after running 1000 meters can be marked as ' C ', which is 120 meters from ' P ' along the track.
   
   d. Here, 1.Perimeter of the track $=180$ meters 2.Distance Toshi runs $=250$ meters Since 250 meters is more than one complete round ( 180 meters), Toshi will have completed one full round and will have 70 meters left to run ( $250-180=70$ meters). So, Toshi will be 70 meters along the length of the track from the starting point. You can mark ' X ' at this point on the track.
   
   e. Given, that Toshi has run an additional 140 meters after completing 2 rounds, her position will be 140 meters from the starting point. If we mark her starting point as ' A ', her position after running 500 meters can be marked as ' $Y$ '.
   
   f. Here, we need to find out how many full rounds Toshi has completed by dividing the total distance she ran by the perimeter of the track: Number of full rounds $=1000÷180=5.56$ rounds Toshi has finished 5 full rounds. Remaining distance $=1000$ meters - ( $5\times 180$ meters) = 1000 meters - 900 meters $=100$ meters Starting from the initial point, Toshi would be 100 meters into her 6th round. Since the track is 60 meters long and 30 meters wide, she would be somewhere along the length of the track. Let's mark this position as ' Z '.
   

5.0Key Features and Benefits for Class 6 Maths Chapter 6 Exercise 6.1

• This exercise features questions involving finding the perimeter of basic shapes.
• The problems aid students in grasping the concepts of measurement and space using different shapes.
• Regularly solving NCERT solutions improves problem-solving speed and adds confidence for the exams.
• The step-by-step solutions encourage learning the appropriate methods and aids the preparation of different exams like the olympiads.