NCERT Solutions for Class 6 Maths Chapter 06 Perimeter and Area

Question 1. Find the missing terms:
(a) Perimeter of a rectangle = 14 cm; breadth = 2 cm; length = ?.
(b) Perimeter of a square = 20 cm; side of a length = ?.
(c) Perimeter of a rectangle = 12 m; length = 3 m; breadth = ?.

Exam Tip: Use the perimeter formula for rectangles (2 × (length + breadth)) and squares (4 × side) to solve for missing dimensions.

Question 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?
Answer: The perimeter of the rectangle is 2 × (5 + 3) = 16 cm. When this wire is straightened out and then bent into a square shape, the perimeter of the square is also 16 cm. Since all sides of a square are equal, the side length equals 16 ÷ 4 = 4 cm.
In simple words: The wire's total length (the perimeter) stays the same. Divide this by 4 to get each side of the square.

Exam Tip: Remember that the length of the wire does not change when it is reshaped - only the form changes, not the total length.

Question 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.
Answer: The perimeter of a triangle is the sum of all three sides. We know the perimeter is 55 cm and two sides measure 20 cm and 14 cm. Therefore, the third side equals 55 - (20 + 14) = 55 - 34 = 21 cm.
In simple words: Add the two known sides, then subtract from the total to find the missing side.

Exam Tip: Always check: the sum of any two sides must be greater than the third side (triangle inequality).

Question 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?
Answer: First, find the perimeter of the park: 2 × (150 + 120) = 2 × 270 = 540 m. Next, multiply the perimeter by the cost per metre: 540 × 40 = Rs. 21,600.
In simple words: Calculate the distance around the park, then multiply by the cost per metre.

Exam Tip: Fencing covers the entire boundary (perimeter) of the park, not the inside area.

Question 5. A piece of string is 36 cm long. 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 sides of equal length?
Answer:
(a) For a square: The perimeter is 36 cm. Since a square has 4 equal sides, each side = 36 ÷ 4 = 9 cm.
(b) For a triangle with all sides equal: The perimeter is 36 cm. Since a triangle has 3 equal sides, each side = 36 ÷ 3 = 12 cm.
(c) For a hexagon with all sides equal: The perimeter is 36 cm. Since a hexagon has 6 equal sides, each side = 36 ÷ 6 = 6 cm.
In simple words: Divide the total string length by the number of sides to get the length of each side.

Exam Tip: For any regular polygon (all sides equal), side length = perimeter ÷ number of sides.

Question 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?
Answer: The perimeter of the field is 2 × (230 + 160) = 2 × 390 = 780 m. For 3 rounds of rope, multiply the perimeter by 3: 780 × 3 = 2340 m.
In simple words: Find how far around the field once, then multiply by 3 for the three rounds.

Exam Tip: When an object is enclosed multiple times, multiply the perimeter by the number of rounds.

Question. Each track is a rectangle. Akshi's track has length 70 m and breadth 40 m. Running one complete round on this track would cover 220 m, i.e., 2 × (70 + 40) m = 220 m. This is the distance covered by Akshi in one round.
Answer: This is a worked example showing how to calculate the perimeter of a rectangular track. The perimeter formula for any rectangle is 2 × (length + breadth). For Akshi's track with length 70 m and breadth 40 m, the calculation is 2 × (70 + 40) = 2 × 110 = 220 m. This 220 m represents the distance covered in one complete circuit around the track.
In simple words: To find the distance around a rectangular track, add the length and breadth, then double the result.

Exam Tip: When solving problems about running laps or circuits, the perimeter tells you the distance covered in one complete lap.

Question. Matha Pakhchi (Birds Problem):
Akshi and Toshi start running along the rectangular tracks as shown in the figure. Akshi runs along the outer track and completes 5 rounds. Toshi runs along the inner track and completes 7 rounds. Now, they are wondering who ran more. Find out who ran the longer distance.
Answer: Akshi's track is rectangular with length 70 m and breadth 40 m. The perimeter is 2 × (70 + 40) = 220 m. Akshi ran 5 rounds, so the total distance is 5 × 220 = 1100 m. Toshi's track has length 60 m and breadth 30 m. The perimeter is 2 × (60 + 30) = 180 m. Toshi ran 7 rounds, so the total distance is 7 × 180 = 1260 m. Comparing the distances: Akshi covered 1100 m in 5 rounds while Toshi covered 1260 m in 7 rounds. Therefore, Toshi ran a longer distance than Akshi.
In simple words: Calculate how far each person ran by multiplying the perimeter of their track by the number of rounds. Compare the totals to find who ran more.

Exam Tip: Always find the total distance first by multiplying perimeter × number of rounds before comparing distances.

Question. (a) 'A' is marked on picture where Akshi will be after she ran 250 m.
Answer: Since Akshi's track perimeter is 220 m, running 250 m means she completes more than one full round. One full round is 220 m, and she has 250 - 220 = 30 m remaining. Starting from her beginning point and moving 30 m along the 70 m length, point 'A' is marked on the track at 30 m from the starting position.
In simple words: Subtract one complete lap (220 m) from 250 m. The remaining distance (30 m) shows where on the track she will be.

Exam Tip: When a runner completes more than one lap, use modulo arithmetic: find the remainder after dividing the total distance by the perimeter.

Question. (b) 'B' is marked on picture where Akshi will be after she ran 500 m.
Answer: Akshi's track perimeter is 220 m. Running 500 m: 500 ÷ 220 = 2 full rounds with a remainder. 500 - (2 × 220) = 500 - 440 = 60 m remaining. Starting from her beginning position and moving 60 m, point 'B' is marked on the track at the 60 m position along the track's perimeter.
In simple words: Find how many full laps fit into 500 m, then use the leftover distance to locate point 'B' on the track.

Exam Tip: Always find the remainder distance using the perimeter to determine the exact position on the track.

Question. (c) Now, Akshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as 'C'.
Answer: Akshi's track perimeter is 220 m. Dividing 1000 m by 220 m: 1000 ÷ 220 = 4 complete full rounds with a remainder. 1000 - (4 × 220) = 1000 - 880 = 120 m remaining. She has finished exactly 4 full rounds. The remaining 120 m places point 'C' at the 120 m position along her track's perimeter.
In simple words: Divide the total distance by the perimeter to find the number of complete rounds. The remainder shows where she ends up on the track.

Exam Tip: The quotient gives complete laps, and the remainder gives the position within the current lap.

Question. (d) 'X' is marked at the point where Toshi will be after she ran 250 m.
Answer: Toshi's track perimeter is 180 m. Running 250 m: 250 ÷ 180 = 1 full round with a remainder. 250 - 180 = 70 m remaining. Starting from her beginning position and moving 70 m along her track, point 'X' is marked at the 70 m position on Toshi's track perimeter.
In simple words: Subtract one lap from the total distance. Use the leftover amount to find the position on the track.

Exam Tip: Apply the same method for both runners - find remainder after dividing by their respective track perimeters.

Question. (e) 'Y' is marked at the point where Toshi will be after she ran 500 m.
Answer: Toshi's track perimeter is 180 m. Running 500 m: 500 ÷ 180 gives 2 complete full rounds with a remainder. 500 - (2 × 180) = 500 - 360 = 140 m remaining. Point 'Y' is marked at the 140 m position along Toshi's track perimeter after completing 2 full rounds.
In simple words: Find how many full laps fit into 500 m, then locate point 'Y' using the leftover distance.

Exam Tip: Always calculate complete laps first, then use the remainder to pinpoint the exact location on the track.

Question. (f) Now, Toshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as 'Z'.
Answer: Toshi's track perimeter is 180 m. Dividing 1000 m by 180 m: 1000 ÷ 180 = 5 complete full rounds with a remainder. 1000 - (5 × 180) = 1000 - 900 = 100 m remaining. She has finished exactly 5 full rounds. Point 'Z' is marked at the 100 m position along her track perimeter.
In simple words: Divide the total distance by the track perimeter to find the number of complete rounds. The remainder indicates her final position on the track.

Exam Tip: The quotient represents finished laps while the remainder pinpoints the current position within the track.

Question. Deep Dive: In races, usually there is a common finish line for all the runners. Here are two square running tracks with the inner track of 100 m each side and outer track of 150 m each side. The common finishing line for both runners is shown by the flags in the figure which are in the center of one of the sides of the tracks. If the total race is of 350 m, then we have to find out where the starting positions of the two runners should be on these two tracks so that they both have a common finishing line after they run for 350 m. Mark the starting points of the runner on the inner track as 'A' and the runner on the outer track as 'B'.
Answer: The inner track has a perimeter of 4 × 100 = 400 m. The outer track has a perimeter of 4 × 150 = 600 m. For a 350 m race: Runner A on the inner track travels 350 m out of 400 m perimeter. So A starts 400 - 350 = 50 m before the common finishing line (measured backward along the track). Runner B on the outer track travels 350 m out of 600 m perimeter. So B starts 600 - 350 = 250 m before the common finishing line (measured backward along the track). Both runners will reach the common finishing line after running exactly 350 m.
In simple words: Subtract the race distance from each track's perimeter to find where each runner must start so they both finish at the same line.

Exam Tip: To ensure a fair race with different track sizes, calculate how far back each runner must start from the common finishing line.

Question. Estimate and Verify: Take a rough sheet of paper or a sheet of newspaper. Make a few random shapes by cutting the paper in different ways. Estimate the total length of the boundaries of each shape then use a scale or measuring tape to measure and verify the perimeter for each shape.

Question 1. Find the missing terms:
(a) Perimeter of a rectangle = 14 cm; breadth = 2 cm; length = ?.
(b) Perimeter of a square = 20 cm; side of a length = ?.
(c) Perimeter of a rectangle = 12 m; length = 3 m; breadth = ?.
Answer:
(a) Using the perimeter formula for a rectangle: P = 2 × (length + breadth), we get 14 = 2 × (length + 2). Dividing by 2 gives 7 = length + 2, so length = 5 cm.
(b) Using the perimeter formula for a square: P = 4 × side, we get 20 = 4 × side. Therefore, side = 5 cm.
(c) Using the perimeter formula for a rectangle: P = 2 × (length + breadth), we get 12 = 2 × (3 + breadth). Dividing by 2 gives 6 = 3 + breadth, so breadth = 3 m.
In simple words: Use the formulas to work backward from the known perimeter and one measurement to find the missing measurement.

Exam Tip: Always set up the perimeter equation carefully and solve step-by-step to avoid arithmetic errors.

Question 2. The perimeter of the rectangle = 2 × (5 + 3) = 16 cm. This wire is then used to make a square. So, perimeter of square = 16 cm. Side of the square = 16 ÷ 4 = 4 cm.
Answer: When a rectangular wire is reshaped into a square, the total length stays the same. The rectangular wire has perimeter 2 × (5 + 3) = 16 cm. When formed into a square, this same 16 cm length becomes the perimeter of the square. Since all four sides of a square are equal, each side measures 16 ÷ 4 = 4 cm.
In simple words: The wire length does not change when reshaped. Just divide the total length by 4 to find each side of the square.

Exam Tip: Remember that perimeter is the total distance around a shape - it stays the same when you reshape an object.

Question 3. Perimeter of triangle = sum of all sides = 55 cm. ⇒ 40 + 14 + third side = 55. ⇒ Third side = 55 - (20 + 14) = 55 - 34. Thus, the third side = 21 cm.
Answer: For any triangle, the perimeter equals the sum of all three sides. Given that the perimeter is 55 cm and two sides measure 20 cm and 14 cm, the third side is found by subtracting: third side = 55 - 20 - 14 = 55 - 34 = 21 cm.
In simple words: Add the two known sides, then subtract from the total perimeter to get the missing side.

Exam Tip: Always verify that the triangle inequality holds: the sum of any two sides must be greater than the third side.

Question 4. Perimeter of the park = 2 × (150 + 120) = 2 × 270 = 540 m. ⇒ Cost of fencing = 540 × ₹40 = ₹21,600.
Answer: To find the cost of fencing, first calculate the perimeter: 2 × (150 + 120) = 540 m. Then multiply the perimeter by the cost per metre: 540 × 40 = Rs. 21,600.
In simple words: Calculate how far around the park, then multiply by the cost per metre.

Exam Tip: Fencing covers only the boundary (perimeter), not the interior area of the park.

Question 5. The length of each side, if it is used to form:
(a) A square
Perimeter of square = 36 cm. Side of square = 36 ÷ 4 = 9 cm.
(b) A triangle with all sides of equal length
Perimeter of triangle = 36 cm. Side of triangle = 36 ÷ 3 = 12 cm.
(c) A hexagon (a six sided closed figure) with sides of equal length
Perimeter of hexagon = 36 cm. Side of hexagon = 36 ÷ 6 = 6 cm.
Answer:
(a) A square with perimeter 36 cm has side length 36 ÷ 4 = 9 cm.
(b) A triangle with perimeter 36 cm and all equal sides has side length 36 ÷ 3 = 12 cm.
(c) A hexagon with perimeter 36 cm and all equal sides has side length 36 ÷ 6 = 6 cm.
In simple words: Divide the total length by the number of sides to find the length of each side.

Exam Tip: For regular polygons (all sides equal), use the formula: side length = perimeter ÷ number of sides.

Question 6. Perimeter of the field = 2 × (230 + 160) = 2 × 390 = 780 m. Total length of rope needed for 3 rounds = 780 × 3 = 2340 m.
Answer: The perimeter of the field is 2 × (230 + 160) = 780 m. Since the field is fenced with 3 rounds of rope, the total length needed is 780 × 3 = 2340 m.
In simple words: Find the distance around the field once, then multiply by 3 to account for all three rounds.

Exam Tip: When an object is enclosed with multiple rounds, multiply the perimeter by the number of rounds required.

Page 133

Each track is a rectangle. Akshi's track has length 70 m and breadth 40 m. Running one complete round on this track would cover 220 m, i.e., 2 × (70 + 40) m = 220 m. This is the distance covered by Akshi in one round.

Question. Figure it Out

Question 1. Find out the total distance Akshi has covered in 5 rounds.
Answer: Akshi's track perimeter is 2 × (70 + 40) = 220 m. In 5 rounds, the total distance covered is 5 × 220 = 1100 m.
In simple words: Multiply the distance around the track by the number of rounds.

Exam Tip: Total distance = number of rounds × perimeter of track.

Question 2. Find out the total distance Toshi has covered in 7 rounds. Who ran a longer distance?
Answer: Toshi's track has length 60 m and breadth 30 m. The perimeter is 2 × (60 + 30) = 180 m. In 7 rounds, the total distance is 7 × 180 = 1260 m. Comparing the two: Akshi covered 1100 m in 5 rounds, while Toshi covered 1260 m in 7 rounds. Therefore, Toshi ran a longer distance.
In simple words: Calculate each person's total distance by multiplying their track perimeter by the number of rounds, then compare.

Exam Tip: Always compare the final totals - the number of rounds and perimeter size both matter for determining who ran more.

Question 3. Think and mark the positions as directed

Question. (a) Mark 'A' at the point where Akshi will be after she ran 250 m.
Answer: Akshi's track perimeter is 220 m. Running 250 m: 250 - 220 = 30 m remaining after one complete lap. Point 'A' is marked 30 m along the track perimeter from her starting position.
In simple words: Subtract one full lap from the total distance. The leftover amount shows where she is on the track.

Exam Tip: Find the remainder when dividing total distance by perimeter to locate position on the track.

Question. (b) Mark 'B' at the point where Akshi will be after she ran 500 m.
Answer: Akshi's track perimeter is 220 m. Running 500 m: 500 ÷ 220 = 2 complete laps with 500 - 440 = 60 m remaining. Point 'B' is marked 60 m along the track perimeter after completing 2 full laps.
In simple words: Find how many complete laps fit into the total distance, then use the leftover distance to mark the position.

Exam Tip: Use division to find complete laps, then use the remainder for the exact location.

Question. (c) Now, Akshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as 'C'.
Answer: Akshi's track perimeter is 220 m. Dividing: 1000 ÷ 220 = 4 complete full rounds with 1000 - 880 = 120 m remaining. Akshi has completed 4 full rounds. Point 'C' is marked 120 m along the track perimeter.
In simple words: Divide to find complete laps. The remainder shows where she is positioned on the track.

Exam Tip: The quotient gives the number of complete laps; the remainder gives the position within the next lap.

Question. (d) Mark 'X' at the point where Toshi will be after she ran 250 m.
Answer: Toshi's track perimeter is 180 m. Running 250 m: 250 - 180 = 70 m remaining after one complete lap. Point 'X' is marked 70 m along Toshi's track perimeter from her starting position.
In simple words: Subtract one full lap from the total distance to find where she ends up on her track.

Exam Tip: Always account for complete laps before determining the remaining position on the track.

Question. (e) Mark 'Y' at the point where Toshi will be after she ran 500 m.
Answer: Toshi's track perimeter is 180 m. Running 500 m: 500 ÷ 180 = 2 complete laps with 500 - 360 = 140 m remaining. Point 'Y' is marked 140 m along Toshi's track perimeter after completing 2 full laps.
In simple words: Find how many full laps fit into 500 m, then use the extra distance to find the position on the track.

Exam Tip: Calculate laps and remainder separately to get the exact position on the track.

Question. (f) Now, Toshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as 'Z'.
Answer: Toshi's track perimeter is 180 m. Dividing: 1000 ÷ 180 = 5 complete full rounds with 1000 - 900 = 100 m remaining. Toshi has completed 5 full rounds. Point 'Z' is marked 100 m along Toshi's track perimeter.
In simple words: Divide the total distance by the track perimeter to find complete rounds. The remainder shows final position on the track.

Exam Tip: The quotient = complete laps; remainder = position within the current lap.

Page 134

Question. Deep Dive: In races, usually there is a common finish line for all the runners. Here are two square running tracks with the inner track of 100 m each side and outer track of 150 m each side. The common finishing line for both runners is shown by the flags in the figure which are in the center of one of the sides of the tracks. If the total race is of 350 m, then we have to find out where the starting positions of the two runners should be on these two tracks so that they both have a common finishing line after they run for 350 m. Mark the starting points of the runner on the inner track as 'A' and the runner on the outer track as 'B'.
Answer: Inner track: perimeter = 4 × 100 = 400 m. Outer track: perimeter = 4 × 150 = 600 m. For a 350 m race to end at the common finishing line: Runner A (inner track) must start 400 - 350 = 50 m before the finishing line. Runner B (outer track) must start 600 - 350 = 250 m before the finishing line. Both runners will reach the common finishing line exactly after running 350 m.
In simple words: Subtract the race distance from each track's perimeter. The result tells you where to start so both runners finish together.

Exam Tip: For fair races with different track sizes, work backward from the common finish line using each track's perimeter.

Page 134

Question. Estimate and Verify: Take a rough sheet of paper or a sheet of newspaper. Make a few random shapes by cutting the paper in different ways. Estimate the total length of the boundaries of each shape then use a scale or measuring tape to measure and verify the perimeter for each shape.

Question. Figure it Out: For estimating the total length of the boundaries of each shape, we can estimate the perimeter visually. Since the shapes are irregular and torn in different ways, visually estimate the boundary lengths by counting how many straight segments or curves each shape has. Break the boundary into smaller, more manageable parts.
Answer: To estimate perimeter of irregular shapes, break the boundary into smaller, more manageable straight segments. Count straight segments or approximate curves. For verification, use a measuring scale or tape along the actual boundary to measure each part, then add all measurements for the total perimeter. Compare your estimates to actual measurements to check accuracy.
In simple words: To find the perimeter of odd-shaped objects, measure each edge carefully, then add them all up.

Exam Tip: Breaking irregular boundaries into smaller parts makes both estimation and actual measurement more accurate.

Page 134

Question. Akshi says that the perimeter of this triangle shape is 9 units. Toshi says it can't be 9 units and the perimeter will be more than 9 units. What do you think?
Answer: Akshi's statement is incorrect. When we count only the straight horizontal and vertical grid segments, we might get 9 units. However, Toshi is right. The diagonal line segments are longer than straight grid segments. Since the triangle's boundary contains diagonal lines that are longer than the equivalent straight-line distances, the actual perimeter is more than 9 units. This demonstrates that diagonal paths are always longer than the sum of their horizontal and vertical components.
In simple words: Straight lines across a grid are shorter than diagonal lines that go the same direction. The triangle has diagonal sides, so its perimeter must be larger than 9 units.

Exam Tip: Always remember that diagonal distances on a grid are longer than the horizontal and vertical distances that cover the same space.

Page 135

Question. Write the perimeters of the figures below in terms of straight and diagonal units.
Answer:
Figure 1: 8s + 2d units
Figure 2: 4s + 6d units
Figure 3: 12s + 6d units
Figure 4: 18s + 6d units
In simple words: Count the straight grid segments as "s" and the diagonal segments as "d", then add them together for each figure's perimeter.

Exam Tip: Carefully distinguish between straight-line segments and diagonal segments when calculating perimeters on grid paper.

Page 136

Question. Find various objects from your surroundings that have regular shapes and find their perimeters. Also, generalise your understanding for the perimeter of other regular polygons.
Answer:
Book (Rectangular): Length = 20 cm and Width = 15 cm, therefore Perimeter = 2 × (20 + 15) = 70 cm
Tiffin (Rectangular): Length = 18 cm and Width = 12 cm, therefore Perimeter = 2 × (18 + 12) = 60 cm
Desk (Rectangular): Length = 120 cm and Width = 60 cm, therefore Perimeter = 2 × (120 + 60) = 360 cm
Boxes (Cuboidal): Side = 15 cm, Perimeter = 4 × 15 = 60 cm
Generalising Perimeter for Regular Polygons: A regular polygon is a shape where all sides and angles are equal. Perimeter Formula for any regular polygon P = n × side length where n is the number of sides and side length is the length of one side.
In simple words: To find the perimeter of a regular polygon with all equal sides, just multiply the side length by how many sides it has.

Exam Tip: Identify whether a shape is regular (all sides equal) or irregular, then use the appropriate method to calculate perimeter.

Page 136

Question. Find out the length of the boundary (i.e., the perimeter) of each of the other arrangements below.
Answer:
(b) Perimeter = 8 cm + 6 cm + 2 cm + 4 cm + 6 cm + 2 cm = 28 cm
(c) Perimeter = 2 cm + 6 cm + 2 cm + 2 cm + 6 cm + 2 cm + 2 cm + 6 cm = 28 cm
(d) Perimeter = 2 cm + 3 cm + 2 cm + 6 cm + 2 cm + 3 cm + 2 cm + 6 cm = 26 cm
In simple words: Add up all the sides of the shape, including both the outer edges and any inner edges that form part of the boundary.

Exam Tip: When shapes have inner cutouts or steps, make sure to count every edge that forms the outer and inner boundaries.

Page 136

Question. Arrange the two pieces to form a figure with a perimeter of 22 cm.
Answer: Perimeter = 2 cm + 1 cm + 2 cm + 6 cm + 2 cm + 1 cm + 2 cm + 6 cm = 22 cm. When the two pieces are arranged to overlap or connect in a specific way, their combined perimeter equals 22 cm.
In simple words: Arrange the pieces carefully so that when you add up all the outer edges, the total is exactly 22 cm.

Exam Tip: When combining shapes, internal boundaries (where pieces touch) are not counted in the final perimeter.

Page 138

Question 1. The area of a rectangular garden 25 m long is 300 sq m. What is the width of the garden?
Answer: Using the area formula for a rectangle: Area = Length × Width. Substituting the known values: 300 = 25 × Width. Solving for Width: Width = 300 ÷ 25 = 12 m. Therefore, the width of the garden is 12 meters.
In simple words: Divide the area by the length to find the width.

Exam Tip: Always rearrange the area formula correctly: if Area = Length × Width, then Width = Area ÷ Length.

Question 2. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of Rs. 8 per hundred sq m?
Answer: First, calculate the area: Area = 500 × 200 = 100,000 sq m. The cost is Rs. 8 per 100 sq m, so the total cost is (100,000 ÷ 100) × 8 = 1,000 × 8 = Rs. 8,000.
In simple words: Find the area, then divide by 100 and multiply by the rate to get the total cost.

Exam Tip: Pay attention to the rate - is it per sq m, per 100 sq m, or per 1000 sq m? This changes how you calculate the total cost.

Question 3. A 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?
Answer: The area of the grove is: Area = Length × Width = 100 × 50 = 5,000 sq m. Each tree needs 25 sq m, so the maximum number of trees is: 5,000 ÷ 25 = 200 trees.
In simple words: Find the total area, then divide by the space each tree needs.

Exam Tip: Always ensure the units match - if area is in sq m and space per tree is in sq m, you can divide directly.

Question 4. By splitting the following figures into rectangles, find their areas (all measures are given in metres):
Answer:
For figure (a): Area of four rectangles = (2 × 1) + (5 × 2) + (7 × 1) + (3 × 3) = 2 + 10 + 7 + 9 = 28 sq units.
For figure (b): Area of three rectangles = (5 × 2) + (3 × 2) + (3 × 2) = 10 + 6 + 6 = 22 sq units.
In simple words: Divide the complex shape into simple rectangles, find the area of each, then add them together.

Exam Tip: When breaking irregular shapes into rectangles, make sure each rectangle is non-overlapping and all rectangles together cover the entire original shape.

Page 139

Question. Figure it Out: Cut out the tangram pieces given at the end of your textbook.

Question 1. Explore and figure out how many pieces have the same area.
Answer: Shapes A and B have the same area. Similarly, Shapes C and E have the same area. When you physically overlap the tangram pieces or measure them on a grid, you can see that A equals B (same area), and C equals E (same area).
In simple words: Some tangram pieces are the exact same size even if they look different.

Exam Tip: Overlaying shapes is a practical way to compare areas without calculation.

Question 2. How many times bigger is Shape D as compared to Shape C? What is the relationship between Shapes C, D and E?
Answer: Shape D is twice the area of Shape C (or Shape E, since C and E are the same area). The relationship is: Shape D = 2 × Shape C, and also Shape D = 2 × Shape E. This means that two copies of Shape C placed together form Shape D.
In simple words: Shape D is exactly twice as large as Shape C or Shape E. You can place two C's together to make one D.

Exam Tip: Relationships between areas can be expressed as multiplication - one shape can be a multiple of another.

Question 3. Which shape has more area: Shape D or F? Give reasons for your answer.
Answer: Shapes F and D are equal in area. Both shapes occupy the same amount of space when compared using the tangram grid.
In simple words: Shape F and Shape D take up the same amount of space.

Exam Tip: Two shapes can have equal areas even if they have completely different appearances and perimeters.

Question 4. Which shape has more area: Shape F or G? Give reasons for your answer.
Answer: Both F and G have the same area. Shape F is a triangle and Shape G is a different triangle, but when measured or compared on the tangram grid, they occupy equal space.
In simple words: Shape F and Shape G are equal in size even though they might look different.

Exam Tip: Always verify area relationships by measurement or overlay, not just by visual appearance.

Question 5. What is the area of Shape A as compared to Shape G? Is it twice as big? Four times as big?
Answer: Shape A is four times as large as Shape G. This relationship shows that A = 4 × G in terms of area.
In simple words: Shape A is 4 times bigger than Shape G. You would need 4 copies of Shape G to cover Shape A.

Exam Tip: Use grid squares or overlays to verify area multiplication relationships precisely.

Question 6. Can you figure out the area of the big square formed with all seven pieces in terms of the area of Shape C?
Answer: Since Shape D = 2 × C and other relationships exist between the shapes, when all seven pieces are arranged to form the big square, the total area is 16 times the area of Shape C. This is because: A = B = 4C, G = F = D = 2C, and E = C. Hence, the total area is 16C.
In simple words: You can arrange the seven pieces to form a big square that is 16 times bigger than Shape C alone.

Exam Tip: When multiple shapes combine, add their individual areas to get the total - use known relationships to make calculations easier.

Question 7. Arrange 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.
Answer: The total area of the rectangle would still be 16 times the area of Shape C, as rearranging the pieces does not change the total area. The configuration changes, but the total area remains the same because area depends on the total number and size of pieces, not their arrangement.
In simple words: Whether the pieces form a square or a rectangle, the total area stays the same at 16C. Rearranging pieces doesn't create or lose area.

Exam Tip: Area is conserved when shapes are rearranged - the total always remains constant regardless of the new configuration.

Question 8. Are the perimeters of the square and the rectangle formed from these 7 pieces different or the same? Give an explanation for your answer.
Answer: The perimeters of the square and the rectangle are different. Even though both shapes use the same seven pieces and have the same total area, their perimeters differ. This is because perimeter depends on the shape's geometry - how the boundary is arranged. When pieces are rearranged from a square to a rectangle, the outer boundary changes length even though the interior content remains the same.
In simple words: The distance around a square and a rectangle made from the same pieces can be different. Area stays the same when you rearrange, but perimeter can change.

Exam Tip: Area and perimeter are independent - you can have the same area with different perimeters, depending on the shape.

Page 140

Question. Look at the figures below and guess which of them has a larger area.
Answer: Shape (b) has larger area. Even though the shapes have irregular and jagged boundaries, Shape (b) encloses more space than Shape (a).
In simple words: The red spiky shape covers more space than the green bumpy shape.

Exam Tip: For irregular shapes, visual estimation can be a starting point, but always verify with actual measurement or calculation.

Question. Find the area of the following figures.
Answer:
Area of figure 1 = 4 square units.
Area of figure 2 = 9 square units.
Area of figure 3 = 10 square units.
Area of figure 4 = 11 square units.
In simple words: Count the full and partial squares each figure covers on the grid to find the total area.

Exam Tip: When counting grid squares for irregular shapes, count full squares first, then estimate partial squares as half or whole units.

Page 141

Question. Try using different shapes (triangle and rectangle) to fill the given space (without overlaps and gaps) and find out the merits associated with using a square shape to find the area rather than another shape. List out the points that make a square the best shape to use to measure area.
Answer: When trying to fill an area with different shapes, squares are the most efficient because: (1) Squares fit together perfectly with no gaps or overlaps. (2) A square grid creates uniform units that are easy to count. (3) Triangles and other shapes leave partial spaces and make counting difficult. (4) Squares align neatly in rows and columns. (5) Calculating area using squares is straightforward - just count the squares. Therefore, a square is the best shape to measure area because it allows complete coverage without waste and makes counting simple and accurate.
In simple words: Squares fit together better than triangles or other shapes. It's easier to count squares than to deal with leftover spaces from other shapes.

Exam Tip: This explains why area is always measured in "square units" - squares are the most practical shape for dividing and measuring space.

Question 1. Find the area (in square metres) of the floor outside of the corridor.
Answer: Area = 1200 sq m.
In simple words: The floor space outside the corridor measures 1200 square metres.

Exam Tip: When finding areas outside a corridor, subtract the corridor area from the total building or ground area.

Question 2. Find the area (in square metres) occupied by your school playground.
Answer: Area = 9000 sq m.
In simple words: The school playground covers 9000 square metres.

Exam Tip: School playgrounds are typically rectangular - measure length and width, then multiply to get area.

Page 141

Question. Let's Explore! On a squared grid paper (1 square = 1 square unit), make as many rectangles as you can whose lengths and widths are a whole number of units such that the area of the rectangle is 24 square units.
Answer: Rectangles with area 24 square units:
1 × 24 = 24 square units
2 × 12 = 24 square units
3 × 8 = 24 square units
4 × 6 = 24 square units
6 × 4 = 24 square units (same as 4 × 6, just rotated)
8 × 3 = 24 square units (same as 3 × 8, just rotated)
12 × 2 = 24 square units (same as 2 × 12, just rotated)
24 × 1 = 24 square units (same as 1 × 24, just rotated)
In simple words: You can make many different-shaped rectangles that all have the same 24 square unit area - tall thin ones and short wide ones.

Exam Tip: The factors of a number determine how many different rectangles you can make with that area. For 24, the factor pairs are: (1,24), (2,12), (3,8), and (4,6).

Question. (a) Which rectangle has the greatest perimeter?
Answer: First rectangle has the greatest perimeter 50 units.
In simple words: The longest, thinnest rectangle (1 × 24) has the largest distance around it.

Exam Tip: For rectangles with the same area, the most elongated shape (closest to a line) has the greatest perimeter.

Question. (b) Which rectangle has the least perimeter?
Answer: Last rectangle has the least perimeter 20 units.
In simple words: The most square-like rectangle (4 × 6, which is closest to a square) has the smallest distance around it.

Exam Tip: For a fixed area, the closest rectangle is to a square, the smaller its perimeter. A true square would have the minimum perimeter.

Question. (c) If you take a rectangle of area 32 sq cm, what will your answers be? Given any area, is it possible to predict the shape of the rectangle with the greatest perimeter as well as the least perimeter? Give examples and reasons for your answer.
Answer: For a rectangle of area 32 sq cm, the rectangle with greatest perimeter would be 1 × 32 (perimeter = 66 cm) and the rectangle with least perimeter would be 4 × 8 (perimeter = 24 cm). For any given area, we can always predict: (1) Greatest perimeter: the most elongated rectangle, which is 1 × area (or as close to 1 × area as integer factors allow). (2) Least perimeter: the rectangle closest to a square (when length and width are as close to each other as possible). For example, with area 24, the 4 × 6 rectangle (closest to square 5 × 4.8) gives least perimeter. With area 32, the 4 × 8 rectangle is close to square dimensions and gives least perimeter. As the shape becomes more square-like, the perimeter decreases for the same area. This is because a square minimizes the boundary length needed to enclose a given area.
In simple words: For any area, make the thinnest possible rectangle to get the biggest perimeter. Make a shape as close to a square as possible to get the smallest perimeter.

Exam Tip: This is a key relationship: for a fixed area, perimeter is minimized when the shape is a square, and maximized when the shape is elongated.

Page 142

Question. Draw a rectangle on a piece of paper and draw one of its diagonals. Cut the rectangle along that diagonal and get two triangles.
Answer: After cutting the rectangle along its diagonal, you obtain two triangles. When you examine these two triangles closely, you notice that they overlap exactly on top of each other. This demonstrates that both triangles have the same area.
In simple words: A diagonal line cuts a rectangle into two pieces that are exactly the same size.

Exam Tip: This practical demonstration shows that the area of a triangle is half the area of the rectangle it fits in.

Question. Check! Whether the two triangles overlap each other exactly. Do they have the same area?
Answer: Yes, the two triangles overlap each other exactly. They have the same area as shown in figure. When two triangles form a rectangle by combining along their matching sides, each triangle's area is half the rectangle's area. Therefore, both triangles have identical areas.
In simple words: The two triangles from a cut rectangle match perfectly. Each triangle is exactly half the size of the rectangle.

Exam Tip: This shows why the formula for triangle area is: Area of triangle = (1/2) × base × height.

Question. Try this with more rectangles having different dimensions. You can check this for a square as well.
Answer: When you repeat this procedure with rectangles of various dimensions, the result holds true every time: the diagonal always divides the rectangle into two equal triangles. This property also works for squares, which are special cases of rectangles. In every case, each triangle has area = (1/2) × rectangle area.
In simple words: No matter what size rectangle you draw, its diagonal always creates two equal triangles.

Exam Tip: This is a universal property - it works for all rectangles and squares, regardless of their dimensions.

Question. Now, see the figures below. Is the area of the blue rectangle more or less than the area of the yellow triangle? Or is it the same? Why?
Answer: The area of the blue rectangle and the yellow triangle are equal. Both shapes have the same area. The blue rectangle and the yellow triangle are related by a geometric property: a triangle with the same base and height as a rectangle has half the area of that rectangle. In this case, the triangle appears to have the same area as the rectangle because the triangle's base and height create a total area that equals the rectangle's area. In simple terms, the area of the rectangle equals 2 × the area of the triangle, or the triangle and rectangle shown have equal areas due to their specific dimensions.
In simple words: The blue rectangle and yellow triangle cover the same amount of space, even though they look different.

Exam Tip: Always check if shapes share the same base and height - this determines their area relationship.

Question. Can you see some relationship between the blue rectangle and the yellow triangle and their areas? Write the relationship here.
Answer: The relationship is: Area of Rectangle = 2 × Area of Triangle. This means that a rectangle encloses twice the area of a triangle that has the same base and height. Conversely, a triangle with a given base and height has exactly half the area of a rectangle with the same base and height.
In simple words: A rectangle is always twice as big as a triangle with the same base and height.

Exam Tip: This is why the triangle area formula is half the rectangle formula: Area of triangle = (1/2) × base × height.

Page 143

Question. Use your understanding from previous grades to calculate the area of any closed figure using grid paper and
(1) Find the area of blue triangle BAD.
(2) Find the area of red triangle ABE.
Area of rectangle ABCD =
Conclusion _____________________
Answer:
(1) The area of blue triangle BAD = (1/2) × 5 × 4 sq units = 10 sq units
(2) The area of red triangle ABE = (1/2) × 5 × 4 sq units = 10 sq units
Area of rectangle ABCD = 5 × 4 sq units = 20 sq units
Conclusion: On the same base and same height, the area of a triangle is half the area of the rectangle.
In simple words: When a triangle and a rectangle share the same base and height, the triangle's area is always half the rectangle's area.

Exam Tip: This relationship is fundamental - remember it well as it appears in many geometry problems.

Page 144

Question. Figure it Out: Find the areas of the figures below by dividing them into rectangles and triangles.
Answer:
Area of figure a: 24 sq units.
Area of figure b: 29 sq units.
Area of figure c: 40 sq units.
Area of figure d: 16 sq units.
Area of figure e: 12 sq units.
In simple words: Break each shape into simple rectangles and triangles, find the area of each part, then add them all up.

Exam Tip: When a figure is made up of multiple shapes, break it down into manageable pieces and use the appropriate formula for each piece.

Page 145

Question. Using 9 unit squares, solve the following.
(1) What is the smallest perimeter possible?
(2) What is the largest perimeter possible?
(3) Make a figure with a perimeter of 18 units.
(4) Can you make other shaped figures for each of the above three perimeters, or is there only one shape with that perimeter? What is your reasoning?
Answer:
(1) The smallest perimeter is 12 units, achieved by arranging the 9 squares into a 3 × 3 square grid.
(2) The largest perimeter is 20 units, achieved by arranging the squares in a long line (1 × 9 rectangle) or another elongated shape.
(3) A figure with a perimeter of 18 units can be made using various arrangements, such as specific L-shapes or T-shapes with the 9 squares.
(4) Yes, you can make other shaped figures for each perimeter. For 20 units and 18 units, multiple different shapes are possible. However, for 12 units, the 3 × 3 square is the most efficient and yields the smallest perimeter. The reasoning is: for a fixed area, different shapes can have different perimeters. More elongated or jagged shapes create larger perimeters, while more compact, square-like shapes create smaller perimeters. Multiple shapes can share the same perimeter value by using different arrangements.
In simple words: You can arrange the 9 squares in different ways to get different perimeters. The square arrangement (3 × 3) gives the smallest perimeter. Stretched-out shapes give bigger perimeters.

Exam Tip: For a fixed area, compact shapes (closer to squares) have smaller perimeters, while elongated or irregular shapes have larger perimeters. Multiple shapes can have the same perimeter but different areas.

Question. Let's do something tricky now! We have a figure below having perimeter 24 units. Without calculating all over again, observe, think and find out what will be the change in the perimeter if a new square is attached as shown on the right. Experiment placing this new square at different places and think what the change in perimeter will be. Can you place the square so that the perimeter: a) increases; b) decreases; c) stays the same?
Answer: c) stays the same.
In simple words: When you add a new square to the figure, the perimeter may change depending on where you place it. But if you position the square carefully on one of the sides, it does not change at all.

Exam Tip: Always think about how adding a shape affects the outline - sometimes a part of the new shape replaces part of the old outline, so the total distance stays equal.

Question. Below is the house plan of Charan. It is in a rectangular plot. Look at the plan. What do you notice?
Answer: The plan shows that all the rooms and areas inside the house fit together to make one big rectangle. Each room has its own measurements and area. Some measurements are given, and you need to find the missing ones using the fact that all the rooms together form the complete house plot.
In simple words: All the different rooms added together equal the total house area. You can use the given numbers to work out the missing numbers.

Exam Tip: When working with house plans, always check that the lengths on opposite sides match - they should add up to the same total.

Question. Some of the measurements are given. a. Find the missing measurements. b. Find out the area of his house.
Answer: a. The missing measurements are as follows:
Utility: 15 ft × 3 ft, Area = 45 sq ft
Small Bedroom: 15 ft × 12 ft, Area = 180 sq ft
Hall: Area = 265 sq ft
Garden: 20 ft × 3 ft, Area = 60 sq ft
Parking: 15 ft × 3 ft, Area = 45 sq ft
b. The area of his house = 30 × 35 sq ft = 1050 sq ft
In simple words: Work out each missing measurement by looking at how the rooms fit together. Once you have all the pieces, add them up or multiply the outside length and width to get the total house area.

Exam Tip: Always verify your answer by adding all individual room areas - they should match the total house area you calculated.

Page 147

Question. Now, find out the missing dimensions and area of Sharan's home. Below is the plan: Some of the measurements are given. a. Find the missing measurements. b. Find out the area of his house. What are the dimensions of all the different rooms in Sharan's house? Compare the areas and perimeters of Sharan's house and Charan's house.
Answer: The area of Sharan's house and Charan's house are the same and equal to 1050 sq ft. The perimeter of Sharan's house (134 ft) is greater than Charan's house (130 ft).
a. The missing measurements are as follows:
Master Bedroom: 12 ft × 15 ft, Area = 180 sq ft
Toilet: 5 ft × 10 ft, Area = 50 sq ft
Kitchen: 18 ft × 10 ft, Area = 180 sq ft
Small Bedroom: 12 ft × 10 ft, Area = 120 sq ft
Hall: 13 ft × 15 ft, Area = 345 sq ft
b. The area of his house = 25 × 42 sq ft = 1050 sq ft
In simple words: Even though both houses have the same total area, Sharan's house has a different shape, which makes its perimeter longer. Both shapes show how the same area can have different outlines.

Exam Tip: Remember that two shapes can have the same area but different perimeters - the shape matters, not just the size.

Page 148

Area Maze Puzzles

Question. In each figure, find the missing value of either the length of a side or the area of a region. a. [Figure with regions: 13 sq cm, 26 sq cm, 15 sq cm, ? sq cm] b. [Figure with regions: 3 cm, 10 sq cm, ? sq cm, 2 cm, 10 sq cm]
Answer: a. The missing area = 30 sq cm
b. The missing area = 9 sq cm
c. The missing area = 20 sq cm
d. The missing side = 5 cm
In simple words: To find a missing area, add up all the other areas and subtract from the total. To find a missing side length, work backwards from the area of that rectangle.

Exam Tip: Always use the fact that areas of parts must add up to the whole area - this helps you spot which numbers go together.

Figure it Out

Question 1. Give 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.
Answer: The sum of areas of two rectangles = 5 m × 10 m + 2 m × 7 m = (50 + 14) sq m = 64 sq m. The dimensions of the required rectangle are a and b such that a × b = 64 sq m.
In simple words: Find how much space both rectangles cover together. Then draw any rectangle that has this total area - there are many possible answers.

Exam Tip: The question asks for "a rectangle" not "the rectangle" - any rectangle with area 64 sq m works, such as 8 × 8 or 4 × 16.

Question 2. The area of a rectangular garden that is 50 m long is 1000 sq m. Find the width of the garden.
Answer: Area of a rectangular garden = length × width
Width = Area / Length = 1000 / 50 = 20 m
In simple words: Use the area formula rearranged: if you know the area and length, divide to get the width.

Exam Tip: Always arrange the formula correctly before substituting numbers - write it out first, then put in the values.

Question 3. The 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.
Answer: Total area of room = length × width = 5 m × 4 m = 20 sq m. Area of one carpet = side × side = 3 m × 3 m = 9 sq m. Number of carpet required = 20/9 = 2.22 (approx.). So, the area of room that is not carpeted = area of room - area of two carpet = 20 sq m - 18 sq m = 2 sq m.
In simple words: Find how much the whole floor covers. Find how much one carpet covers. Subtract to see what space is left uncovered.

Exam Tip: Remember that "area not carpeted" means the empty space left - subtract the carpet area from the room area, not how many carpets fit.

Question 4. Four 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?
Answer: Area of garden = 15 m × 12 m = 180 sq m. Area of four flower beds = 4 × (2 m × 1 m) = 4 × 2 sq m = 8 sq m. So, the available area for laying down a lawn = (180 - 8) sq m = 172 sq m
In simple words: Find the whole garden area. Find how much space all the flower beds take up. Subtract the flower beds from the garden to see what is left for the lawn.

Exam Tip: Sketch the corners to make sure you understand which parts are being dug up - this prevents mistakes.

Question 5. Shape 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.
Answer: Shape A: a rectangle with dimensions 9 × 2 units. Area = 18 square units, Perimeter = 22 units. Shape B: a rectangle with dimensions 5 × 4 units. Area = 20 square units, Perimeter = 18 units. This shows that a shape can have a smaller area but a longer perimeter than another shape.
In simple words: Area and perimeter are different things. A thin, long shape can have more outline (perimeter) than a fat, square shape, even if it covers less space (area).

Exam Tip: Always calculate both area and perimeter - do not mix them up or assume one determines the other.

Question 6. On 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?
Answer: Perimeter = 2 × (18 + 12) cm = 2 × 30 cm = 60 cm
In simple words: The border is a rectangle whose sides are shorter than the page. Add the two shorter sides, add the two longer sides, and add them all together.

Exam Tip: Use perimeter = 2 × (length + width) rather than adding all four sides one by one - it is faster and less error-prone.

Question 7. Draw a rectangle of size 12 units × 8 units. Draw another rectangle inside it, without touching the outer rectangle that occupies exactly half the area.
Answer: A rectangle of size 12 units × 8 units has area = 96 square units. To occupy exactly half, the inner rectangle should have area = 48 square units. An example is a rectangle of dimensions 8 units × 6 units, which fits inside without touching the outer boundary.
In simple words: The outer rectangle covers 96 square units. You need to draw an inner rectangle that covers only 48 square units - exactly half. Make sure it does not touch the edges.

Exam Tip: Check your inner rectangle area by multiplying length × width - it must equal exactly half of 96, which is 48.

Question 8. A square piece of paper is folded in half. The square is then cut into two rectangles along the fold. Regardless of the size of the square, one of the following statements is always true. Which statement is true here?
(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 1 1/2 times the perimeter of the square.
(d) The area of the square is always three times as large as the areas of both the rectangles added together.
Answer: (c) The perimeters of both the rectangles added together is always 1 1/2 times the perimeter of the square. True
In simple words: When you cut a square in half, you create a fold line inside. This fold line becomes part of the edges of both new rectangles. So the two rectangles together have more outline than the original square had.

Exam Tip: Work through one example with actual numbers to verify - let the square be 10 × 10, then check each statement to see which one is always true.