CBSE Class 6 Mathematics Perimeter and Area Worksheet Set 01

Long Answer Questions

Question 1. Perimeter and area are two different measures of the same figure.
(i) Define the perimeter of a closed plane figure and the area of a closed figure.
(ii) Write the formulas for the perimeter and the area of a rectangle and of a square.
(iii) A rectangular table top is 9 m long and 4 m wide. Find its perimeter and its area, and explain why one answer is written in metres and the other in square metres.
Answer:
(i) The perimeter of a closed plane figure is the distance covered along its boundary when we go around it once. The area of a closed figure is the amount of region enclosed by it.
(ii) Perimeter of a rectangle = 2 × (length + breadth); area of a rectangle = length × breadth.
Perimeter of a square = 4 × side; area of a square = side × side.
(iii) Perimeter = 2 × (9 + 4) = 2 × 13 = 26 m. Area = 9 × 4 = 36 sq m.
The perimeter is a length, measured with a tape along the boundary, so it is in metres. The area tells how many unit squares of side 1 m cover the top (here 36), so it is in square metres. We can never say the area is 36 m or the perimeter is 26 sq m.
In simple words: Perimeter is a length in m; area counts squares in sq m.

Teacher's Note:
1. Write all four formulas neatly.
2. Units are part of the answer.

Question 2. A rectangular park in a colony is 40 m long and 25 m wide. The residents want to fence it all around and then cover the whole ground with grass.
(i) Find the perimeter of the park.
(ii) Find the cost of fencing it at ₹ 45 per metre.
(iii) Find the area of the park and the cost of covering it with grass at ₹ 30 per square metre.
Answer:
(i) Fencing goes along the boundary: perimeter = 2 × (40 + 25) = 2 × 65 = 130 m.
(ii) Cost of fencing = 130 × 45 = ₹ 5850.
(iii) Grass covers the whole ground: area = 40 × 25 = 1000 sq m. Cost of grass = 1000 × 30 = ₹ 30000.
Fencing is paid by length because it lies along the boundary; grass is paid by area because it covers the inside.
In simple words: Fence uses the perimeter (130 m); grass uses the area (1000 sq m).

Teacher's Note:
1. Decide first: boundary or inside?
2. Match the rate unit to the measure.

Question 3. Take a rectangular sheet of paper 8 cm long and 5 cm wide and cut it along one diagonal.
(i) Find the area of the rectangle.
(ii) Place one triangular piece on the other. What do you observe, and what does it tell you about their areas?
(iii) Find the area of each triangle.
(iv) Write the relation between a triangle and the rectangle that just encloses it.
Answer:
(i) Area = 8 × 5 = 40 sq cm.
(ii) The two pieces cover each other exactly. Shapes that cover each other exactly enclose the same region, so the two triangles have equal areas.
(iii) Together the two equal triangles make the whole rectangle, so each is half of it: \( \frac{40}{2} \) = 20 sq cm.
(iv) The area of a triangle is half the area of the rectangle that just encloses it. For a rectangle with base b and height h, the triangle has area \( \frac{1}{2} \) × b × h. This holds for any triangle whose base lies along one side of the rectangle and whose third corner lies on the opposite side.
In simple words: A diagonal cuts a rectangle into two equal halves, so each triangle is half.

Teacher's Note:
1. Do the activity with real paper.
2. This is where ½ × base × height comes from.

Question 4. All sides of a regular polygon are equal in length.
(i) Find the perimeter of an equilateral triangle of side 14 cm.
(ii) Find the perimeter of a square of side 14 cm.
(iii) Find the perimeter of a regular hexagon of side 14 cm.
(iv) Arrange the three perimeters in increasing order and write a general rule for the perimeter of a regular polygon.
Answer:
(i) 3 equal sides: 3 × 14 = 42 cm.
(ii) 4 equal sides: 4 × 14 = 56 cm.
(iii) 6 equal sides: 6 × 14 = 84 cm.
(iv) 42 cm < 56 cm < 84 cm, so the order is equilateral triangle, square, regular hexagon.
Rule: perimeter of a regular polygon = number of sides × length of one side. With the same side length, more sides means a larger perimeter.
In simple words: Same side, more sides: bigger perimeter.

Teacher's Note:
1. Use the same rule three times.
2. Use the < sign when ordering.

Question 5. The floor of a verandah has the shape shown below. All measurements are in metres.
(i) Find the perimeter of the verandah.
(ii) Find its area by splitting the figure into two rectangles.
(iii) Find the area once more by a different splitting and check that both answers agree.
Answer:
(i) Adding every side once: 5 + 3 + 4 + 4 + 9 + 7 = 32 m.
(ii) Draw a horizontal line at the level of the step. Lower rectangle: 9 m × 4 m = 36 sq m. Upper rectangle: 5 m × 3 m = 15 sq m.
Total area = 36 + 15 = 51 sq m.
(iii) Complete the figure into a full 9 m × 7 m rectangle and take away the missing corner piece of 4 m × 3 m.
Full rectangle = 63 sq m; corner piece = 12 sq m; area = 63 − 12 = 51 sq m.
Both methods give 51 sq m. A figure can be split in many ways, but the area is always the same.
In simple words: Split it or complete it and subtract; both give 51 sq m.

Teacher's Note:
1. Mark each split line on the figure.
2. Two methods are a good check.

Teacher's Note:
1. Length + breadth must be 12.
2. List them in order so none is missed.

Question 7. A house is built on a rectangular plot that is 30 ft long and 20 ft wide. In the plan, a bedroom 12 ft wide runs along the whole 20 ft depth on the left. The part that remains on the right is divided by a line across it into a kitchen 12 ft deep and a store room that takes the rest.
(i) Write the dimensions of the bedroom and find its area.
(ii) Find the dimensions of the whole part remaining on the right of the bedroom.
(iii) Find the dimensions and the areas of the kitchen and of the store room.
(iv) Check that the three areas together give the area of the plot.
Answer:
(i) The bedroom is 12 ft × 20 ft; area = 12 × 20 = 240 sq ft.
(ii) Width = 30 − 12 = 18 ft, depth = 20 ft, so the part is 18 ft × 20 ft (area 360 sq ft).
(iii) Kitchen: 18 ft × 12 ft, area = 216 sq ft. Store room depth = 20 − 12 = 8 ft, so it is 18 ft × 8 ft, area = 144 sq ft.
(iv) 240 + 216 + 144 = 600 sq ft, and the plot is 30 × 20 = 600 sq ft. They agree: the parts always add up to the whole.
In simple words: Find missing sides by subtracting; the three rooms add up to 600 sq ft.

Teacher's Note:
1. Draw a rough plan first.
2. Subtract to find each missing side.

Question 8. Two farmers own one rectangular field each. Rehana’s field is 30 m by 20 m and Bhim’s field is 60 m by 10 m.
(i) Find the area of each field.
(ii) Find the perimeter of each field.
(iii) Who will need more fencing wire, and how much more?
(iv) What does this example teach you about area and perimeter?
Answer:
(i) Rehana: 30 × 20 = 600 sq m. Bhim: 60 × 10 = 600 sq m.
(ii) Rehana: 2 × (30 + 20) = 100 m. Bhim: 2 × (60 + 10) = 140 m.
(iii) Bhim needs more wire, 140 − 100 = 40 m more, although both fields have the same area.
(iv) Two figures can have the same area but different perimeters. The field nearer to a square needs a shorter fence; the long thin field needs a longer one.
In simple words: Same land, but the long thin field needs 40 m more fence.

Teacher's Note:
1. Area and perimeter answer different questions.
2. Long and thin means more boundary.

Question 9. A tent is drawn on grid paper as shown. Each small square of the grid is 1 unit by 1 unit.
(i) Name the two simple shapes into which the figure can be split along the dashed line.
(ii) Find the area of the lower part.
(iii) Find the area of the upper part, explaining the rule you use.
(iv) Find the total area of the figure, and explain why its exact perimeter cannot be read off the grid as easily as its area.
Answer:
(i) The dashed line splits the figure into a rectangle at the bottom and a triangle at the top.
(ii) The rectangle is 6 units long and 4 units high: area = 6 × 4 = 24 square units.
(iii) The triangle has base 6 units and height 3 units. Its enclosing rectangle is 6 × 3 = 18 square units, and a triangle covers half of its enclosing rectangle: area = \( \frac{1}{2} \) × 6 × 3 = 9 square units.
(iv) Total area = 24 + 9 = 33 square units.
The area can be read from the grid because every part is made of whole and half unit squares. The perimeter cannot, because the two slanting sides of the triangle do not run along grid lines; a slanting line is longer than a grid side, so it must be measured with a scale or written in diagonal units.
In simple words: Rectangle 24 + triangle 9 = 33 square units; slanting sides are not grid units.

Teacher's Note:
1. The triangle is half its enclosing rectangle.
2. Slanting edges are longer than grid steps.

Question 10. A rectangular sheet of cardboard is divided into four rectangular parts by one straight line drawn across it and one straight line drawn down it. Two of the four areas and two of the measurements are shown below.
(i) Find the height of the upper row of parts.
(ii) Find the width of the right-hand column of parts.
(iii) Find the two missing areas.
(iv) Find the total area of the whole sheet and its perimeter.
Answer:
(i) The upper left part has area 24 sq cm and width 6 cm: height = 24 ÷ 6 = 4 cm.
(ii) The upper right part has area 20 sq cm and the same height 4 cm: width = 20 ÷ 4 = 5 cm.
(iii) Both lower parts have height 3 cm. Lower left = 6 × 3 = 18 sq cm; lower right = 5 × 3 = 15 sq cm.
(iv) Total area = 24 + 20 + 18 + 15 = 77 sq cm. Check: the sheet is 6 + 5 = 11 cm long and 4 + 3 = 7 cm wide, and 11 × 7 = 77 sq cm.
Perimeter = 2 × (11 + 7) = 2 × 18 = 36 cm.
In simple words: Divide areas by known sides to get missing sides, then multiply.

Teacher's Note:
1. Parts in the same row share a height.
2. Parts in the same column share a width.

Case Study Based Questions

Question 11. Meera’s kitchen garden. Meera has made a small kitchen garden behind her house. It is in the shape of a rectangle that is 14 m long and 9 m wide. She grows tomatoes, chillies and coriander in it. Stray goats often enter the garden, so Meera decides to put a wire fence all around it. The shopkeeper charges ₹ 30 for one metre of wire. Meera’s younger brother is curious about something else: he wants to know how much ground the garden actually covers, so that they can plan how many rows of plants will fit inside it.
(i) Find the perimeter of the kitchen garden.
(ii) Find the area of ground covered by the garden.
(iii) Find the cost of putting one round of wire all around the garden.
(iv) Meera decides to put two rounds of wire instead of one. What total length of wire will she need?
Answer:
(i) Perimeter = 2 × (14 + 9) = 2 × 23 = 46 m.
(ii) Area = 14 × 9 = 126 sq m.
(iii) One round is 46 m: cost = 46 × 30 = ₹ 1380.
(iv) Two rounds = 2 × 46 = 92 m.
In simple words: Fence uses the perimeter, ground uses the area.

Teacher's Note:
1. Wire goes round, so use perimeter.
2. Two rounds means double the perimeter.

Question 12. A handkerchief with a lace border. Ritu buys a square cotton handkerchief whose side is 24 cm. She wants to stitch a thin lace all along its four edges so that the handkerchief looks pretty. The shop sells the lace at ₹ 2 for each centimetre. In one corner of the handkerchief there is a printed triangular motif. The base of this triangle is 8 cm long and its height is also 8 cm, and the triangle fits exactly inside a small square of the cloth. Ritu’s friend has a bigger handkerchief of side 30 cm.
(i) Find the length of lace needed for Ritu’s handkerchief.
(ii) Find the area of cloth in Ritu’s handkerchief.
(iii) Find the area of the printed triangular motif.
(iv) How much more lace would be needed for her friend’s handkerchief of side 30 cm?
Answer:
(i) Lace goes along the boundary: 4 × 24 = 96 cm.
(ii) Area = 24 × 24 = 576 sq cm.
(iii) The enclosing square is 8 × 8 = 64 sq cm, and the triangle is half of it: \( \frac{1}{2} \) × 8 × 8 = 32 sq cm.
(iv) Bigger handkerchief: 4 × 30 = 120 cm, so extra lace = 120 − 96 = 24 cm.
In simple words: Lace is the perimeter, cloth is the area, the motif is half its square.

Teacher's Note:
1. The rate ₹ 2 per cm is extra information here.
2. The triangle is half of its enclosing square.

Question 13. The school notice board. The notice board of Class VI is a rectangle 180 cm long and 120 cm wide. Before Children’s Day the students decide to cover the whole board with coloured chart paper. Each sheet of chart paper in the market is a square of side 60 cm, and the sheets can be joined without any overlap. The students also want to fix a thin wooden strip all along the boundary of the board to give it a neat finish. The carpenter charges ₹ 12 for every metre of the wooden strip he fixes.
(i) Find the area of the notice board.
(ii) Find the perimeter of the notice board in centimetres and in metres.
(iii) Find the area of one sheet of chart paper and the number of sheets needed to cover the board.
(iv) Find the cost of fixing the wooden strip all around the board.
Answer:
(i) Area = 180 × 120 = 21600 sq cm.
(ii) Perimeter = 2 × (180 + 120) = 2 × 300 = 600 cm = 6 m.
(iii) One sheet = 60 × 60 = 3600 sq cm; number of sheets = 21600 ÷ 3600 = 6 sheets.
(iv) The strip is 6 m long: cost = 6 × 12 = ₹ 72.
In simple words: Area tells the sheets needed; perimeter tells the strip length.

Teacher's Note:
1. 100 cm = 1 m.
2. Check: 3 sheets along the length × 2 along the width = 6.

Question 14. Tiling the verandah. The verandah outside a classroom is a rectangle 8 m long and 4 m wide. The school has decided to lay square tiles on the whole floor of this verandah. Each tile is a square of side 40 cm, and the tiles are laid without any gap and without breaking any tile. The shop sells each tile for ₹ 18. The headmaster also wants a low railing along the complete boundary of the verandah, so he first asks the mason to measure the distance all around it.
(i) Find the area of the verandah floor.
(ii) Find the distance all around the verandah.
(iii) Find the area of one tile in square metres and the number of tiles needed.
(iv) Find the total cost of the tiles.
Answer:
(i) Area = 8 × 4 = 32 sq m.
(ii) Distance all around = perimeter = 2 × (8 + 4) = 24 m.
(iii) 40 cm = 0.4 m, so one tile = 0.4 × 0.4 = 0.16 sq m. Along the length 800 ÷ 40 = 20 tiles fit and along the width 400 ÷ 40 = 10, so 20 × 10 = 200 tiles (check: 32 ÷ 0.16 = 200).
(iv) Cost = 200 × 18 = ₹ 3600.
In simple words: Count tiles along each side and multiply.

Teacher's Note:
1. Use the same unit for floor and tile.
2. Counting along each side avoids decimals.

Question 15. The school playground. The playground of a school is a rectangle 90 m long and 60 m wide. Every morning the students of Class VI run along the complete boundary of this playground as a warm-up exercise. The physical education teacher tells them that one round means going once around the whole boundary and coming back to the starting point. On Monday each child was asked to run three rounds. On Friday the teacher set a target of 1.5 km of running for each child. The school is also planning to lay fresh grass over the whole playground before the sports day.
(i) Find the perimeter of the playground.
(ii) Find the area of the playground.
(iii) Find the distance a child runs on Monday.
(iv) How many complete rounds must a child run on Friday to meet the target of 1.5 km?
Answer:
(i) Perimeter = 2 × (90 + 60) = 2 × 150 = 300 m.
(ii) Area = 90 × 60 = 5400 sq m.
(iii) Three rounds = 3 × 300 = 900 m.
(iv) 1.5 km = 1500 m, so rounds = 1500 ÷ 300 = 5 rounds.
In simple words: One round is 300 m; use it to count distance and rounds.

Teacher's Note:
1. Change km to m before dividing.
2. One round = perimeter.

Question 16. The Republic Day banner. For Republic Day, a school orders cloth banners. Each banner is a rectangle 5 m long and 3 m wide. A tricolour ribbon is to be stitched all along the four edges of every banner, and the tailor buys this ribbon at ₹ 25 per metre. The cloth itself is bought by area. The school wants four such banners, one for each of the four gates of the school building. The art teacher says that the cloth left over after cutting can be used for small flags.
(i) Find the perimeter of one banner.
(ii) Find the area of cloth in one banner.
(iii) Find the cost of the ribbon needed for one banner.
(iv) Find the total area of cloth needed for all four banners.
Answer:
(i) Perimeter = 2 × (5 + 3) = 16 m.
(ii) Area = 5 × 3 = 15 sq m.
(iii) Ribbon = 16 m: cost = 16 × 25 = ₹ 400.
(iv) Four banners = 4 × 15 = 60 sq m.
In simple words: Ribbon follows the perimeter; cloth follows the area.

Teacher's Note:
1. Edges mean perimeter.
2. Multiply by 4 only in part (iv).

Teacher's Note:
1. Less than half counts as zero.
2. One unit square here is 4 sq cm, not 2.

Question 18. Two fields with the same fence. Rahim and Salma each own one rectangular field, and both fields need exactly the same length of fencing wire. Rahim’s field measures 50 m by 30 m. Salma’s field is a square of side 40 m. Both of them pay the same amount to the fencing contractor, so Rahim believes that the two fields must be equally large. Salma is not so sure, because her field looks wider to her. They decide to settle the argument by actually working out the ground covered by each field.
(i) Find the perimeter of Rahim’s field.
(ii) Find the perimeter of Salma’s field and check Rahim’s statement about the fencing.
(iii) Find the area of each field.
(iv) Who gets more ground for the same fencing, and what rule does this example show?
Answer:
(i) Rahim: 2 × (50 + 30) = 160 m.
(ii) Salma: 4 × 40 = 160 m. The perimeters are equal, so Rahim is right that both need the same fencing.
(iii) Rahim: 50 × 30 = 1500 sq m. Salma: 40 × 40 = 1600 sq m.
(iv) Salma gets 1600 − 1500 = 100 sq m more. Figures with the same perimeter need not have the same area, and among rectangles with the same perimeter the square encloses the most.
In simple words: Same fence, but the square field holds 100 sq m more.

Teacher's Note:
1. Same perimeter, different area.
2. The square is the best shape for a fixed fence.

Question 19. The path around the lawn. A rectangular lawn in a park is 18 m long and 12 m wide. The municipality decides to build a cemented walking path of uniform width 2 m all around the outside of this lawn, so that visitors can walk around the grass without stepping on it. The path touches the lawn along its whole boundary. The cement work on the path costs ₹ 90 for each square metre. The grass of the lawn itself is not disturbed during the work.
(i) Find the area of the lawn.
(ii) Find the length and the breadth of the big rectangle formed by the lawn together with the path.
(iii) Find the area of the path only.
(iv) Find the cost of cementing the path.
Answer:
(i) Lawn = 18 × 12 = 216 sq m.
(ii) The path adds 2 m on both sides in each direction: length = 18 + 2 + 2 = 22 m, breadth = 12 + 2 + 2 = 16 m.
(iii) Big rectangle = 22 × 16 = 352 sq m; path = 352 − 216 = 136 sq m.
(iv) Cost = 136 × 90 = ₹ 12240.
In simple words: Big rectangle minus lawn gives the path.

Teacher's Note:
1. Add the path width on both sides.
2. Path area = outer minus inner.

Question 20. Painting the classroom wall. The front wall of a Class VI room is a rectangle 8 m long and 4 m high. A glass window of size 2 m by 1 m is fitted in the middle of this wall. Before the new session the school decides to paint the whole wall except the window, and the painter charges ₹ 25 for each square metre that he paints. The art teacher also wants a thin decorative strip to be fixed all along the four edges of the wall, and this strip costs ₹ 15 per metre.
(i) Find the area of the wall.
(ii) Find the area of the window and hence the area that has to be painted.
(iii) Find the cost of painting the wall once.
(iv) Find the length of the decorative strip needed and its cost.
Answer:
(i) Wall = 8 × 4 = 32 sq m.
(ii) Window = 2 × 1 = 2 sq m; area to be painted = 32 − 2 = 30 sq m.
(iii) Cost = 30 × 25 = ₹ 750.
(iv) Strip = perimeter = 2 × (8 + 4) = 24 m; cost = 24 × 15 = ₹ 360. Painting is paid by area, the strip by length.
In simple words: Paint uses the area minus the window; the strip uses the perimeter.

Teacher's Note:
1. Leave out the window before multiplying.
2. Two different rates for two different measures.