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Perimeter and Area

CBSE · Class 6 · Mathematics

NCERT Solutions for Perimeter and Area — CBSE Class 6 Mathematics.

42 questions54 flashcards5 concepts

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37 Questions Solved · 5 Sections

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Figure it Out (Perimeter of Rectangles and Squares)

1aFind the missing term: Perimeter of a rectangle = 14 cm; breadth = 2 cm; length = ?Show solution
Given: Perimeter = 14 cm, Breadth (b) = 2 cm

Formula: Perimeter of a rectangle = 2×(l+b)2 \times (l + b)

Working:
14=2×(l+2)14 = 2 \times (l + 2)
142=l+2\frac{14}{2} = l + 2
7=l+27 = l + 2
l=7−2=5 cml = 7 - 2 = 5 \text{ cm}

Answer: Length = 5 cm

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1bFind the missing term: Perimeter of a square = 20 cm; side of length = ?Show solution
Given: Perimeter of square = 20 cm

Formula: Perimeter of a square = 4×side4 \times \text{side}

Working:
20=4×side20 = 4 \times \text{side}
side=204=5 cm\text{side} = \frac{20}{4} = 5 \text{ cm}

Answer: Side = 5 cm

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1cFind the missing term: Perimeter of a rectangle = 12 m; length = 3 m; breadth = ?Show solution
Given: Perimeter = 12 m, Length (l) = 3 m

Formula: Perimeter of a rectangle = 2×(l+b)2 \times (l + b)

Working:
12=2×(3+b)12 = 2 \times (3 + b)
122=3+b\frac{12}{2} = 3 + b
6=3+b6 = 3 + b
b=6−3=3 mb = 6 - 3 = 3 \text{ m}

Answer: Breadth = 3 m

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2A 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?Show solution
Given: Rectangle with length = 5 cm, breadth = 3 cm.

Step 1: Find the length of the wire (perimeter of the rectangle).
Perimeter of rectangle=2×(l+b)=2×(5+3)=2×8=16 cm\text{Perimeter of rectangle} = 2 \times (l + b) = 2 \times (5 + 3) = 2 \times 8 = 16 \text{ cm}

So the wire is 16 cm long.

Step 2: Find the side of the square formed from this wire.

The perimeter of the square = length of wire = 16 cm.
Side of square=Perimeter4=164=4 cm\text{Side of square} = \frac{\text{Perimeter}}{4} = \frac{16}{4} = 4 \text{ cm}

Answer: The length of each side of the square = 4 cm

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3Find 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.Show solution
Given: Perimeter of triangle = 55 cm, Side 1 = 20 cm, Side 2 = 14 cm.

Concept: Perimeter of a triangle = Sum of all three sides.

Working:
Perimeter=Side1+Side2+Side3\text{Perimeter} = \text{Side}_1 + \text{Side}_2 + \text{Side}_3
55=20+14+Side355 = 20 + 14 + \text{Side}_3
55=34+Side355 = 34 + \text{Side}_3
Side3=55−34=21 cm\text{Side}_3 = 55 - 34 = 21 \text{ cm}

Answer: The length of the third side = 21 cm

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4What 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?Show solution
Given: Length = 150 m, Breadth = 120 m, Cost of fencing = ₹40 per metre.

Step 1: Find the perimeter of the rectangular park.
Perimeter=2×(l+b)=2×(150+120)=2×270=540 m\text{Perimeter} = 2 \times (l + b) = 2 \times (150 + 120) = 2 \times 270 = 540 \text{ m}

Step 2: Find the total cost of fencing.
Total cost=Perimeter×Cost per metre=540×40=₹21,600\text{Total cost} = \text{Perimeter} \times \text{Cost per metre} = 540 \times 40 = ₹21{,}600

Answer: The total cost of fencing = ₹21,600

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5aA piece of string is 36 cm long. What will be the length of each side if it is used to form a square?Show solution
Given: Length of string = 36 cm. The string forms a square.

Concept: Perimeter of square = 4×side4 \times \text{side}

Working:
Side=Perimeter4=364=9 cm\text{Side} = \frac{\text{Perimeter}}{4} = \frac{36}{4} = 9 \text{ cm}

Answer: Each side of the square = 9 cm

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5bA piece of string is 36 cm long. What will be the length of each side if it is used to form a triangle with all sides of equal length?Show solution
Given: Length of string = 36 cm. The string forms an equilateral triangle (all sides equal).

Concept: Perimeter of equilateral triangle = 3×side3 \times \text{side}

Working:
Side=Perimeter3=363=12 cm\text{Side} = \frac{\text{Perimeter}}{3} = \frac{36}{3} = 12 \text{ cm}

Answer: Each side of the triangle = 12 cm

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5cA piece of string is 36 cm long. What will be the length of each side if it is used to form a hexagon (a six-sided closed figure) with sides of equal length?Show solution
Given: Length of string = 36 cm. The string forms a regular hexagon (all 6 sides equal).

Concept: Perimeter of regular hexagon = 6×side6 \times \text{side}

Working:
Side=Perimeter6=366=6 cm\text{Side} = \frac{\text{Perimeter}}{6} = \frac{36}{6} = 6 \text{ cm}

Answer: Each side of the hexagon = 6 cm

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6A farmer has a rectangular field having length 230 m and breadth 160 m. He wants to fence it with 3 rounds of rope. What is the total length of rope needed?Show solution
Given: Length = 230 m, Breadth = 160 m, Number of rounds of rope = 3.

Step 1: Find the perimeter of the rectangular field.
Perimeter=2×(l+b)=2×(230+160)=2×390=780 m\text{Perimeter} = 2 \times (l + b) = 2 \times (230 + 160) = 2 \times 390 = 780 \text{ m}

Step 2: Find the total length of rope for 3 rounds.
Total length of rope=3×780=2340 m\text{Total length of rope} = 3 \times 780 = 2340 \text{ m}

Answer: The total length of rope needed = 2340 m

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Figure it Out (Matha Pachchi — Running Track)

1Find out the total distance Akshi has covered in 5 rounds. (Akshi's track: length = 70 m, breadth = 40 m; one round = 220 m)Show solution
Given: One complete round of Akshi's track = 220 m, Number of rounds = 5.

Working:
Total distance=5×220=1100 m\text{Total distance} = 5 \times 220 = 1100 \text{ m}

Answer: Akshi covered a total distance of 1100 m in 5 rounds.

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2Find out the total distance Toshi has covered in 7 rounds. Who ran a longer distance? (Note: Toshi's track dimensions are shown in the figure. Based on the context, Toshi's track has length 60 m and breadth 30 m, giving one round = 180 m — assumption based on standard textbook values.)Show solution
Assumption: Toshi's track has length = 60 m and breadth = 30 m (as given in the standard textbook figure).

Step 1: Find one round of Toshi's track.
One round=2×(60+30)=2×90=180 m\text{One round} = 2 \times (60 + 30) = 2 \times 90 = 180 \text{ m}

Step 2: Find total distance in 7 rounds.
Total distance=7×180=1260 m\text{Total distance} = 7 \times 180 = 1260 \text{ m}

Comparison:
- Akshi in 5 rounds = 1100 m
- Toshi in 7 rounds = 1260 m

Answer: Toshi covered 1260 m, which is more than Akshi's 1100 m. So Toshi ran a longer distance.

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3aMark 'A' at the point where Akshi will be after she ran 250 m.Show solution
Given: One round of Akshi's track = 220 m.

Working:
250÷220=1 complete round with remainder 30 m250 \div 220 = 1 \text{ complete round with remainder } 30 \text{ m}

After 1 complete round, Akshi is back at the starting point. She then runs 30 m more along the track.

Starting from the starting point and moving along the track: the first side (length) = 70 m. Since 30 m < 70 m, Akshi is 30 m along the length side from the starting point.

Answer: Mark 'A' at a point 30 m from the start along the longer side of the track.

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3bMark 'B' at the point where Akshi will be after she ran 500 m.Show solution
Given: One round of Akshi's track = 220 m.

Working:
500÷220=2 complete rounds with remainder 500−440=60 m500 \div 220 = 2 \text{ complete rounds with remainder } 500 - 440 = 60 \text{ m}

After 2 complete rounds, Akshi is at the starting point. She then runs 60 m more.

First side (length) = 70 m. Since 60 m < 70 m, Akshi is 60 m along the length side from the starting point.

Answer: Mark 'B' at a point 60 m from the start along the longer side of the track.

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3cAkshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as 'C'.Show solution
Given: One round of Akshi's track = 220 m, Total distance = 1000 m.

Working:
1000÷220=4 complete rounds with remainder 1000−880=120 m1000 \div 220 = 4 \text{ complete rounds with remainder } 1000 - 880 = 120 \text{ m}

So Akshi has finished 4 full rounds.

Now she runs 120 m more from the starting point:
- First side (length) = 70 m → she completes this side. Remaining = 120−70=50120 - 70 = 50 m.
- Second side (breadth) = 40 m → she completes this side. Remaining = 50−40=1050 - 40 = 10 m.
- Third side (length) = 70 m → she runs 10 m along this side.

Answer: Akshi finishes 4 full rounds. Mark 'C' at a point 10 m along the third side (opposite length side) from the second corner.

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3dMark 'X' at the point where Toshi will be after she ran 250 m.Show solution
Given: One round of Toshi's track = 180 m.

Working:
250÷180=1 complete round with remainder 250−180=70 m250 \div 180 = 1 \text{ complete round with remainder } 250 - 180 = 70 \text{ m}

After 1 complete round, Toshi is at the starting point. She then runs 70 m more.

First side (length) = 60 m → she completes this. Remaining = 70−60=1070 - 60 = 10 m.
Second side (breadth) = 30 m → she runs 10 m along this side.

Answer: Mark 'X' at a point 10 m along the breadth side from the first corner (after the starting length side).

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3eMark 'Y' at the point where Toshi will be after she ran 500 m.Show solution
Given: One round of Toshi's track = 180 m.

Working:
500÷180=2 complete rounds with remainder 500−360=140 m500 \div 180 = 2 \text{ complete rounds with remainder } 500 - 360 = 140 \text{ m}

After 2 complete rounds, Toshi is at the starting point. She then runs 140 m more.

- First side (length) = 60 m → completed. Remaining = 140−60=80140 - 60 = 80 m.
- Second side (breadth) = 30 m → completed. Remaining = 80−30=5080 - 30 = 50 m.
- Third side (length) = 60 m → completed. Remaining = 50−6050 - 60... wait: 50<6050 < 60, so she runs 50 m along the third side.

Answer: Mark 'Y' at a point 50 m along the third side (opposite the starting length side) from the second corner.

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Figure it Out (Area of Rectangles)

1The area of a rectangular garden 25 m long is 300 sq m. What is the width of the garden?Show solution
Given: Length = 25 m, Area = 300 sq m.

Formula: Area of rectangle = l×bl \times b

Working:
300=25×b300 = 25 \times b
b=30025=12 mb = \frac{300}{25} = 12 \text{ m}

Answer: The width of the garden = 12 m

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2What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹8 per hundred sq m?Show solution
Given: Length = 500 m, Width = 200 m, Rate = ₹8 per 100 sq m.

Step 1: Find the area of the plot.
Area=l×b=500×200=1,00,000 sq m\text{Area} = l \times b = 500 \times 200 = 1{,}00{,}000 \text{ sq m}

Step 2: Find the cost.
Cost=1,00,000100×8=1000×8=₹8,000\text{Cost} = \frac{1{,}00{,}000}{100} \times 8 = 1000 \times 8 = ₹8{,}000

Answer: The total cost of tiling = ₹8,000

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3A rectangular coconut grove is 100 m long and 50 m wide. If each coconut tree requires 25 sq m, what is the maximum number of trees that can be planted in this grove?
4By splitting the following figures into rectangles, find their areas (all measures are given in metres). [Figures refer to L-shaped or irregular rectilinear figures in the textbook.]

Figure it Out (Tangram Pieces)

1Explore and figure out how many tangram pieces have the same area.
2How many times bigger is Shape D as compared to Shape C? What is the relationship between Shapes C, D and E?
3Which shape has more area: Shape D or F? Give reasons for your answer.
4Which shape has more area: Shape F or G? Give reasons for your answer.
5What is the area of Shape A as compared to Shape G? Is it twice as big? Four times as big?
6Can you now figure out the area of the big square formed with all seven pieces in terms of the area of Shape C?
7Arrange these 7 pieces to form a rectangle. What will be the area of this rectangle in terms of the area of Shape C now? Give reasons for your answer.
8Are the perimeters of the square and the rectangle formed from these 7 pieces different or the same? Give an explanation for your answer.

Figure it Out (Area and Perimeter — Mixed Problems)

1Give the dimensions of a rectangle whose area is the sum of the areas of these two rectangles having measurements: 5 m × 10 m and 2 m × 7 m.
2The area of a rectangular garden that is 50 m long is 1000 sq m. Find the width of the garden.
3The floor of a room is 5 m long and 4 m wide. A square carpet whose sides are 3 m in length is laid on the floor. Find the area that is not carpeted.
4Four flower beds having sides 2 m long and 1 m wide are dug at the four corners of a garden that is 15 m long and 12 m wide. How much area is now available for laying down a lawn?
5Shape A has an area of 18 square units and Shape B has an area of 20 square units. Shape A has a longer perimeter than Shape B. Draw two such shapes satisfying the given conditions.
6On a page in your book, draw a rectangular border that is 1 cm from the top and bottom and 1.5 cm from the left and right sides. What is the perimeter of the border?
7Draw a rectangle of size 12 units × 8 units. Draw another rectangle inside it, without touching the outer rectangle, that occupies exactly half the area.
8A square piece of paper is folded in half. The square is then cut into two rectangles along the fold. Which of the following statements is always true?
a. The area of each rectangle is larger than the area of the square.
b. The perimeter of the square is greater than the perimeters of both the rectangles added together.
c. The perimeters of both the rectangles added together is always 1½ times the perimeter of the square.
d. The area of the square is always three times as large as the areas of both rectangles added together.

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Frequently Asked Questions

What are the important topics in Perimeter and Area for CBSE Class 6 Mathematics?
Perimeter and Area covers several key topics that are frequently asked in CBSE Class 6 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Perimeter and Area — CBSE Class 6 Mathematics?
Understand the core concepts first, then work through the 42 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
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