CBSE Class 6 Mathematics Perimeter and Area Assignment Set 01

Short Answer Questions

Question 1. What is meant by the perimeter of a closed plane figure? How do we find the perimeter of a polygon? Write one situation from daily life where we need to find a perimeter.
Answer:
The perimeter of a closed plane figure is the distance covered along its boundary when we go around it once.
For a polygon (a closed figure made of line segments), the perimeter is the sum of the lengths of all its sides.
Daily life: the length of wire needed to fence a field, or of lace to go all around a tablecloth, is a perimeter.
In simple words: Perimeter is the length of the boundary; for a polygon, add all its sides.

Teacher's Note:
1. Answer all three parts.
2. Any example where something goes 'all around' works.

Question 2. Find the perimeter of each of the following.
(i) A rectangular chart paper that is 15 cm long and 9 cm wide.
(ii) A square tile whose side is 11 cm.
Answer:
Perimeter of a rectangle = 2 × (length + breadth); perimeter of a square = 4 × side.
(i) 2 × (15 + 9) = 2 × 24 = 48 cm.
(ii) 4 × 11 = 44 cm.
In simple words: Rectangle: double (length + breadth); square: 4 times the side.

Teacher's Note:
1. Write the formula before using it.
2. Perimeter is a length, so the unit is cm.

Question 3. The perimeter of a rectangle is 28 cm and its breadth is 4 cm.
(i) Find its length.
(ii) A square has the same perimeter as this rectangle. Find the side of the square.
Answer:
(i) 2 × (length + 4) = 28, so length + 4 = 14 and length = 10 cm.
(ii) Side of the square = 28 ÷ 4 = 7 cm.
In simple words: Halve the perimeter and subtract the breadth; for the square, divide by 4.

Teacher's Note:
1. Work backwards from the formula.
2. Check: 2 × (10 + 4) = 28.

Question 4. The perimeter of a triangular park is 50 cm on a drawing, and two of its sides measure 16 cm and 19 cm. Write the rule for the perimeter of a triangle and use it to find the third side.
Answer:
Rule: perimeter of a triangle = sum of the lengths of its three sides.
Sum of the two known sides = 16 + 19 = 35 cm.
Third side = 50 − 35 = 15 cm.
In simple words: Take away the two known sides from the perimeter.

Teacher's Note:
1. Check: 16 + 19 + 15 = 50.
2. Write the rule first, as asked.

Question 5. Find the area of each figure and explain why the answer carries a “square” unit.
(i) A rectangular notebook cover 13 cm by 6 cm.
(ii) A square sticker of side 9 cm.
Answer:
(i) Area = 13 × 6 = 78 sq cm.
(ii) Area = 9 × 9 = 81 sq cm.
Area is found by counting how many unit squares fit inside the figure, so it is always written in square units such as sq cm or sq m.
In simple words: Area counts unit squares, so its unit is a square unit.

Teacher's Note:
1. Multiply the two sides for area.
2. Never write an area in plain cm.

Question 6. Differentiate between perimeter and area. Give two points of difference and explain them using a rectangle of length 5 cm and breadth 3 cm.
Answer:
1. Perimeter is the distance all around the boundary; area is the amount of region enclosed.
2. Perimeter is measured in units of length (cm, m); area in square units (sq cm, sq m).
For the rectangle: perimeter = 2 × (5 + 3) = 16 cm, area = 5 × 3 = 15 sq cm.
In simple words: Perimeter is the boundary in cm; area is the inside in sq cm.

Teacher's Note:
1. Give the meaning and the unit as the two differences.
2. Use the given rectangle to show both.

Question 7. A rectangular garden is 24 m long and its area is 312 sq m.
(i) Find the width of the garden.
(ii) Find how much wire is needed to go once around the garden.
Answer:
(i) Width = area ÷ length = 312 ÷ 24 = 13 m.
(ii) Wire = perimeter = 2 × (24 + 13) = 2 × 37 = 74 m.
In simple words: Divide the area by the length to get the width, then find the perimeter.

Teacher's Note:
1. Part (ii) needs the answer of part (i).
2. Check: 24 × 13 = 312.

Question 8. A regular polygon has all its sides equal.
(i) Find the perimeter of a regular pentagon of side 8 cm.
(ii) Find the perimeter of an equilateral triangle of side 8 cm.
Also write a general rule for the perimeter of any regular polygon.
Answer:
(i) A regular pentagon has 5 equal sides: 5 × 8 = 40 cm.
(ii) An equilateral triangle has 3 equal sides: 3 × 8 = 24 cm.
General rule: perimeter of a regular polygon = number of sides × length of one side.
In simple words: Multiply the side by the number of sides.

Teacher's Note:
1. Pentagon 5, triangle 3.
2. The same rule works for every regular polygon.

Question 9. A square and a rectangle each have a perimeter of 40 cm. The rectangle is 14 cm long.
(i) Find the breadth of the rectangle and the side of the square.
(ii) Which of the two covers more area, and by how much?
Answer:
(i) Length + breadth = 40 ÷ 2 = 20 cm, so breadth = 20 − 14 = 6 cm. Side of the square = 40 ÷ 4 = 10 cm.
(ii) Rectangle: 14 × 6 = 84 sq cm. Square: 10 × 10 = 100 sq cm.
The square covers more area, by 100 − 84 = 16 sq cm, though both have the same boundary length.
In simple words: Same perimeter, but the square holds 16 sq cm more.

Teacher's Note:
1. Same perimeter does not mean same area.
2. A square encloses the most area for a given perimeter.

Question 10. A kitchen garden is in the shape of a rectangle 18 m long and 12 m wide. Find the cost of fencing it all around at the rate of ₹ 35 per metre. Show the steps clearly.
Answer:
Fencing goes all around, so we need the perimeter.
Perimeter = 2 × (18 + 12) = 2 × 30 = 60 m.
Cost = 60 × 35 = ₹ 2100.
In simple words: Find the perimeter and multiply by the rate per metre.

Teacher's Note:
1. Fencing is charged per metre of boundary.
2. Show the formula, the working and the unit.

Question 11. The floor of a school hall is 12 m long and 9 m wide. Tiles are to be laid on the whole floor at the rate of ₹ 60 per square metre. Find the total cost of tiling.
Answer:
Tiling covers the floor, so we need the area.
Area = 12 × 9 = 108 sq m.
Cost = 108 × 60 = ₹ 6480.
In simple words: Find the area and multiply by the rate per square metre.

Teacher's Note:
1. 'Per square metre' tells you to use area.
2. Compare with Q10, which used perimeter.

Question 12. Nita jogs along the boundary of a rectangular park that is 75 m long and 45 m wide.
(i) How far does she run in 4 complete rounds?
(ii) How many complete rounds must she run to cover 1200 m?
Answer:
One round = perimeter = 2 × (75 + 45) = 240 m.
(i) 4 rounds = 4 × 240 = 960 m.
(ii) Rounds = 1200 ÷ 240 = 5 complete rounds.
In simple words: One round is 240 m; multiply or divide by it.

Teacher's Note:
1. One round equals the perimeter.
2. Check: 5 × 240 = 1200.

Question 13. Inside the same rectangle of length 12 units and breadth 5 units, two different triangles are drawn. Each triangle has one complete side of the rectangle as its base, and its third corner lies on the opposite side of the rectangle.
(i) Find the area of each triangle.
(ii) The two triangles look quite different. Explain why their areas are still equal.
Answer:
(i) Area of the rectangle = 12 × 5 = 60 square units. Each triangle has a full side of the rectangle as base and its third corner on the opposite side, so each is half of the rectangle: 60 ÷ 2 = 30 square units.
(ii) Each triangle covers exactly half of the same rectangle. Moving the third corner along the opposite side does not change the base or the height, so the area stays 30 square units though the shape looks different.
In simple words: Both triangles are half of the same rectangle, so both are 30 square units.

Teacher's Note:
1. Same base and same height give the same area.
2. Shape can change while area stays the same.

Question 14. In a set of tangram pieces, the middle-sized triangle D can be covered exactly by the two small triangles C and E, and C and E are of the same size.
(i) How many times the area of C is the area of D?
(ii) If the area of C is 4 square units, write the areas of D and E.
Also state how two paper shapes can be compared in area without measuring them.
Answer:
(i) D is covered exactly by the two equal pieces C and E, so area of D = 2 times the area of C.
(ii) E is the same size as C, so E = 4 square units, and D = 2 × 4 = 8 square units.
Two paper shapes can be compared by placing one on the other, or by cutting one and covering the other with its pieces. If they cover each other exactly, their areas are equal.
In simple words: D equals two small triangles, so D = 8 and E = 4 square units.

Teacher's Note:
1. Covering exactly means equal area.
2. Try it with real tangram pieces.

Question 15. A cut-out of a leaf has a curved, irregular boundary.
(i) Describe how you would estimate its perimeter using a piece of thread and a scale.
(ii) Describe how you would estimate its area using squared paper, and write the rule followed for a square that is exactly half covered.
Answer:
(i) Lay a thread carefully along the whole curved boundary, starting and ending at the same point. Mark the thread there, straighten it and measure it with a scale. That length is the perimeter.
(ii) Trace the leaf on squared paper with each small square = 1 square unit. Count each full square as 1, a square more than half covered as 1, and ignore a square less than half covered.
A square that is exactly half covered is counted as \( \frac{1}{2} \) square unit. Add all these to get the estimated area.
In simple words: Thread for the boundary; counting squares for the area.

Teacher's Note:
1. A scale alone cannot measure a curved edge.
2. Learn the four counting rules.

Question 16. Two shapes are drawn on a square dot grid. The boundary of shape P is made of 8 straight unit sides and 2 diagonal sides, while the boundary of shape Q is made of 4 straight unit sides and 6 diagonal sides.
(i) Write the perimeter of each shape in the short form used in the chapter.
(ii) Each boundary has 10 sides in all. Which shape has the longer boundary? Give a reason.
Answer:
(i) Perimeter of P = 8s + 2d units; perimeter of Q = 4s + 6d units, where s is a straight unit and d a diagonal unit.
(ii) Shape Q has the longer boundary. A diagonal of a unit square is longer than its side, so d is greater than s. Q has 4 more diagonal sides and 4 fewer straight sides than P.
In simple words: Diagonals are longer, and Q has more diagonals, so Q's boundary is longer.

Teacher's Note:
1. Do not simply add s and d together.
2. Compare by swapping 4 straight sides for 4 diagonal ones.

Question 17. A garden is 16 m long and 11 m wide. Square flower beds of side 3 m are dug at each of its four corners. How much area is left for the lawn? Show each step.
Answer:
Area of the whole garden = 16 × 11 = 176 sq m.
Area of one flower bed = 3 × 3 = 9 sq m.
Area of four flower beds = 4 × 9 = 36 sq m.
Area left for the lawn = 176 − 36 = 140 sq m.
In simple words: Take away the four beds from the whole garden.

Teacher's Note:
1. Remember there are four beds.
2. Whole minus parts gives the remainder.

Question 18. Write all the rectangles with whole-number sides whose area is 18 sq cm. Which of them has the greatest perimeter and which has the least? What rule do you notice?
Answer:
The rectangles are: 1 cm × 18 cm (perimeter 38 cm), 2 cm × 9 cm (perimeter 22 cm), 3 cm × 6 cm (perimeter 18 cm).
Greatest perimeter: 1 cm × 18 cm. Least perimeter: 3 cm × 6 cm.
Rule: for a fixed area, the long thin rectangle has the greatest perimeter and the one closest to a square has the least.
In simple words: Same area: thin means long boundary, nearly square means short boundary.

Teacher's Note:
1. List the factor pairs of 18.
2. 6 × 3 is the same rectangle as 3 × 6.

Question 19. A rectangular plot of land is 400 m long and 250 m wide. It is to be levelled at the rate of ₹ 12 per hundred square metres. Find the total cost of levelling, showing each step of your working.
Answer:
Area of the plot = 400 × 250 = 100000 sq m.
The rate is for every hundred square metres: 100000 ÷ 100 = 1000 hundreds.
Cost of levelling = 1000 × 12 = ₹ 12000.
In simple words: Find the area, count the hundreds and multiply by ₹ 12.

Teacher's Note:
1. Read the rate carefully: per hundred sq m.
2. Count zeros with care.

Question 20. A square of side 4 cm is cut away from one corner of a rectangular sheet of paper measuring 12 cm by 9 cm.
(i) Find the area of the remaining piece.
(ii) Find the perimeter of the remaining piece and compare it with the perimeter of the whole sheet.
Answer:
(i) Whole sheet = 12 × 9 = 108 sq cm; square cut away = 4 × 4 = 16 sq cm; remaining = 108 − 16 = 92 sq cm.
(ii) The remaining piece has sides 12 cm, 5 cm, 4 cm, 4 cm, 8 cm and 9 cm, so its perimeter = 42 cm. The whole sheet has perimeter 2 × (12 + 9) = 42 cm.
The perimeters are equal: the 4 cm lost from the length and the 4 cm lost from the breadth are replaced by the two new 4 cm inner edges.
In simple words: The area drops by 16 sq cm, but the perimeter stays 42 cm.

Teacher's Note:
1. Draw the sheet and mark all six sides.
2. Cutting a corner square keeps the perimeter the same.