Chapter 6 exercise answers: Perimeter and Area

Open NCERT p. 132Matches NCERT’s answer

a. Perimeter of a rectangle = 14 cm; breadth = 2 cm

1. Perimeter of a rectangle = 2 × (length + breadth).
2. 14 = 2 × (length + 2), so length + 2 = 7.
3. Length = 7 − 2 = 5 cm.

Answer5 cm

b. Perimeter of a square = 20 cm; side of a length

1. Perimeter of a square = 4 × side.
2. 20 = 4 × side, so side = 20 ÷ 4.

Answer5 cm

c. Perimeter of a rectangle = 12 m; length = 3 m

1. Perimeter of a rectangle = 2 × (length + breadth).
2. 12 = 2 × (3 + breadth), so 3 + breadth = 6.
3. Breadth = 6 − 3 = 3 m.

Answer3 m

Watch this explained “Backwards: a known perimeter fixes a missing side”, 1:58 into Perimeter as the distance all the way round · हिंदी में देखें

Open NCERT p. 132Matches NCERT’s answer

1. The wire's length does not change when it is bent into a new shape.
2. Length of wire = perimeter of the rectangle = 2 × (5 + 3) = 16 cm.
3. This same 16 cm now forms the square's perimeter.
4. Side of square = 16 ÷ 4 = 4 cm.

Answer4 cm

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Open NCERT p. 132Matches NCERT’s answer

1. Perimeter = sum of the three sides.
2. 20 + 14 = 34 cm is the sum of the two known sides.
3. Third side = 55 − 34 = 21 cm.

Answer21 cm

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Open NCERT p. 132Matches NCERT’s answer

1. Perimeter of the park = 2 × (150 + 120) = 2 × 270 = 540 m.
2. Fencing costs ₹40 per metre, so cost = 540 × 40.
3. Cost = ₹21,600.

Answer₹21,600

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Open NCERT p. 132Matches NCERT’s answer

a. A square,

1. The string's 36 cm becomes the perimeter of the square.
2. Side = 36 ÷ 4.

Answer9 cm

b. A triangle with all sides of equal length

1. The same 36 cm now forms an equal-sided triangle.
2. Side = 36 ÷ 3.

Answer12 cm

c. A hexagon (a six sided closed figure) with sides of equal length

1. The same 36 cm now forms an equal-sided hexagon (6 sides).
2. Side = 36 ÷ 6.

Answer6 cm

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Open NCERT p. 132Matches NCERT’s answer

1. Perimeter of the field = 2 × (230 + 160) = 2 × 390 = 780 m.
2. One round of rope needs 780 m, and there are 3 rounds.
3. Total rope = 3 × 780 = 2340 m.

Answer2340 m

Watch this explained “Backwards: a known perimeter fixes a missing side”, 1:58 into Perimeter as the distance all the way round · हिंदी में देखें

Open NCERT p. 133Matches NCERT’s answer

1. Akshi's outer track is 70 m by 40 m.
2. One round = 2 × (70 + 40) = 220 m.
3. 5 rounds = 5 × 220 = 1100 m.

Answer1100 m

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Open NCERT p. 133Matches NCERT’s answer

1. Toshi's inner track is 60 m by 30 m.
2. One round = 2 × (60 + 30) = 180 m.
3. 7 rounds = 7 × 180 = 1260 m.
4. Akshi covered 1100 m (from the earlier part) and Toshi covered 1260 m.
5. 1260 m is more than 1100 m, so Toshi ran the longer distance — even though her track is the shorter loop.

AnswerToshi covered 1260 m, which is more than Akshi's 1100 m, so Toshi ran the longer distance.

Watch this explained “More laps of a shorter loop beats fewer of a longer one”, 3:37 into Perimeter as the distance all the way round · हिंदी में देखें

Open NCERT p. 133Checked by computer

a. Mark ‘A’ at the point where Akshi will be

1. Both girls start at the bottom-right corner of their track and run the way the arrows point: left along the bottom, up the left side, right along the top, then down the right side. The dots on each track are 10 m apart.
2. One round of Akshi's track = 2 × (70 + 40) = 220 m.
3. 250 m = 220 m (1 full round) + 30 m.
4. So A is on the bottom side, 30 m (3 dots) to the left of her starting corner.

AnswerA is 30 m from the start, on the bottom side.

b. Mark ‘B’ at the point where Akshi will be

1. 500 m = 2 × 220 m (2 full rounds) + 60 m.
2. So B is on the bottom side, 60 m (6 dots) to the left of the start, just 10 m before the bottom-left corner.

AnswerB is 60 m from the start, on the bottom side.

c. How many full rounds has she finished running

1. 4 × 220 = 880 and 1000 − 880 = 120, while 5 rounds would need 1100 m.
2. So Akshi has finished 4 full rounds and is 120 m into her 5th round.
3. 120 m = 70 m (bottom) + 40 m (left side) + 10 m, so C is on the top side, 10 m to the right of the top-left corner.

Answer4 full rounds; C is 120 m from the start, on the top side.

d. Mark ‘X’ at the point where Toshi will be

1. One round of Toshi's track = 2 × (60 + 30) = 180 m.
2. 250 m = 180 m (1 full round) + 70 m.
3. 70 m = 60 m (bottom) + 10 m, so X is on the left side, 10 m up from the bottom-left corner.

AnswerX is 70 m from the start, on the left side.

e. Mark ‘Y’ at the point where Toshi will be

1. 500 m = 2 × 180 m (2 full rounds) + 140 m.
2. 140 m = 60 m (bottom) + 30 m (left side) + 50 m, so Y is on the top side, 50 m to the right of the top-left corner (10 m before the top-right corner).

AnswerY is 140 m from the start, on the top side.

f. How many full rounds has she finished running

1. 5 × 180 = 900 and 1000 − 900 = 100, while 6 rounds would need 1080 m.
2. So Toshi has finished 5 full rounds and is 100 m into her 6th round.
3. 100 m = 60 m (bottom) + 30 m (left side) + 10 m, so Z is on the top side, 10 m to the right of the top-left corner.

Answer5 full rounds; Z is 100 m from the start, on the top side.

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Open NCERT p. 138Matches NCERT’s answer

1. Area of a rectangle = length × width.
2. 300 = 25 × width, so width = 300 ÷ 25.

Answer12 m

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Open NCERT p. 138Matches NCERT’s answer

1. Area of the plot = 500 × 200 = 1,00,000 sq m.
2. Rate is ₹8 per hundred sq m, so number of hundreds = 1,00,000 ÷ 100 = 1000.
3. Cost = 1000 × 8 = ₹8000.

Answer₹8000

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Open NCERT p. 138Matches NCERT’s answer

1. Area of the grove = 100 × 50 = 5000 sq m.
2. Each tree needs 25 sq m, so number of trees = 5000 ÷ 25.

Answer200 trees

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Open NCERT p. 138Matches NCERT’s answer

a.

1. Go round the outline from the bottom-left corner: right 3, up 3, right 4, up (not marked), left (not marked), down 1, left 3, down 2, left 2, down 4.
2. The outline closes, so the lengths going right equal the lengths going left: 3 + 4 = top side + 3 + 2, so the top side is 2 m.
3. Likewise the lengths going up equal the lengths going down: 3 + right side = 1 + 2 + 4, so the right side is 4 m.
4. Cut the figure with vertical lines into 4 rectangles: 2 m × 4 m, 1 m × 6 m, 2 m × 3 m and 2 m × 4 m.
5. Areas: 8 + 6 + 6 + 8 = 28 sq m.

Answer28 sq m

b.

1. Along the bottom, 1 + 3 + 1 = 5, which matches the top, so the gap in the middle is 3 m wide and 2 m high.
2. Split the figure into 3 rectangles: a top strip 5 m long and 3 − 2 = 1 m thick, and two legs, each 1 m × 2 m.
3. Areas: 5 + 2 + 2 = 9 sq m.
4. Check another way: the whole 5 m × 3 m rectangle is 15 sq m, the gap is 3 × 2 = 6 sq m, and 15 − 6 = 9 sq m.

Answer9 sq m

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Open NCERT p. 139One way to think about it

1. Using the tangram pieces, we can overlay and compare them
2. Shapes C and E match exactly when placed on top of each other—they have equal area
3. Shape D can be covered exactly by shapes C and E together—D has twice the area of C (or of E)
4. Shape F can be covered exactly by two C's—F has the same area as D
5. Shape G (parallelogram) also covers exactly two C's—G has the same area as D and F
6. Shapes A and B are large triangles—each covers four C's
7. Summary: C = E; D = F = G = 2C; A = B = 4C

In shortC and E have the same area; D, F, and G have the same area (each twice that of C or E); A and B have the same area (each four times that of C)

Watch this explained “Every piece in one unit: the square as sixteen C”, 7:08 into Area as a count of unit squares · हिंदी में देखें

Open NCERT p. 139Checked by computer

1. Placing C and E together exactly covers D, and C and E have equal area.
2. So D's area = area of C + area of E = 2 × area of C.

AnswerShape D is 2 times as big as Shape C. C and E have equal area, and D is made up of exactly a C and an E, so D = C + E = 2C.

Watch this explained “Comparing by covering, with no numbers at all”, 6:24 into Area as a count of unit squares · हिंदी में देखें

Open NCERT p. 139Checked by computer

1. Use the area of Shape C as the unit. Lay C on E: they match exactly, so E has the same area as C.
2. Shape D (the square) is covered exactly by C and E placed together (the hint says so), so D = C + E = 2 × (area of C).
3. Shape F is the medium triangle at the bottom right. Cut it from its square corner to the middle of its long side: each half is exactly the size of C. So C and E together cover F exactly, and F = 2 × (area of C).
4. D and F are both 2 × (area of C), so they have the same area, even though one is a square and the other a triangle.

AnswerShapes D and F have the same area: each is exactly covered by Shapes C and E, so each is 2 times the area of Shape C.

Watch this explained “Every piece in one unit: the square as sixteen C”, 7:08 into Area as a count of unit squares · हिंदी में देखें

Open NCERT p. 139Checked by computer

1. Use the area of Shape C as the unit; Shape E matches C exactly.
2. Shape F (the medium triangle): cut it from its square corner to the middle of its long side and each half is exactly the size of C. So C and E cover F exactly, and F = 2 × (area of C).
3. Shape G (the parallelogram): cut it along its shorter diagonal and each half is again exactly the size of C. So C and E cover G exactly, and G = 2 × (area of C).
4. F and G are both 2 × (area of C), so they have the same area.

AnswerShapes F and G have the same area: each is exactly covered by Shapes C and E, so each is 2 times the area of Shape C.

Watch this explained “Every piece in one unit: the square as sixteen C”, 7:08 into Area as a count of unit squares · हिंदी में देखें

Open NCERT p. 139Checked by computer

1. Shape G's area = 2C (it is covered exactly by C and E, and E = C).
2. Shape A (a large triangle) can be covered exactly by 4 copies of C.
3. So Shape A's area ÷ Shape G's area = 4C ÷ 2C = 2.

AnswerShape A is 2 times as big as Shape G: twice, not four times.

Watch this explained “Every piece in one unit: the square as sixteen C”, 7:08 into Area as a count of unit squares · हिंदी में देखें

Open NCERT p. 139Checked by computer

1. In units of C: A = 4C, B = 4C, C = C, D = 2C, E = C, F = 2C, G = 2C.
2. Adding all seven: 4 + 4 + 1 + 2 + 1 + 2 + 2 = 16.
3. So the whole square's area = 16 × (area of Shape C).

Answer16 × (area of Shape C)

Watch this explained “Every piece in one unit: the square as sixteen C”, 7:08 into Area as a count of unit squares · हिंदी में देखें

Open NCERT p. 139Checked by computer

1. The rectangle uses the exact same 7 pieces as the square, just rearranged.
2. Rearranging pieces with no gaps and no overlaps cannot change the total area.
3. So the rectangle's area is also 16 × (area of Shape C).

Answer16 × (area of Shape C) — the same as the square, since it is made of the same pieces.

Watch this explained “Rearranging changes nothing the count can see”, 7:55 into Area as a count of unit squares · हिंदी में देखें

Open NCERT p. 139Checked by computer

1. Area does not change when the pieces are rearranged, but perimeter can.
2. When pieces are pushed together differently, some edges that used to be on the outside boundary now touch another piece and become hidden, while other, previously hidden, edges become part of the new boundary.
3. For the same area, a rectangle that is not a square always has a bigger perimeter than the square — the square is the most 'compact' shape for a given area.
4. So the square's perimeter and the rectangle's perimeter are different (the rectangle's is larger).

AnswerNo — they are different. Even though the square and the rectangle are made of the same 7 pieces and have the same area, the rectangle has a larger perimeter, because rearranging the pieces changes which edges lie on the boundary.

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Open NCERT p. 144Matches NCERT’s answer

(a)

1. Each grid square is 1 square unit. A right triangle is half of the rectangle formed by its two shorter sides, so its area is ½ × base × height.
2. The left and right sides of the figure are upright, each 6 units long, and 4 units apart.
3. Draw a horizontal line at the lower end of the top slanted edge, and another at the upper end of the bottom slanted edge. This splits the figure into a rectangle and two right triangles.
4. Rectangle: 4 × 5 = 20 square units.
5. Top triangle: ½ × 4 × 1 = 2 square units. Bottom triangle: ½ × 4 × 1 = 2 square units.
6. Total: 20 + 2 + 2 = 24 square units.

Answer24 square units

(b)

1. The left side is upright and 5 units long, the right side is upright and 10 units long, and they are 4 units apart.
2. Draw horizontal lines across from the top and the bottom of the left side. This gives a rectangle with a right triangle above it and another below it.
3. Rectangle: 4 × 5 = 20 square units.
4. Top triangle: ½ × 4 × 3 = 6 square units. Bottom triangle: ½ × 4 × 2 = 4 square units.
5. Total: 20 + 6 + 4 = 30 square units.

Answer30 square units

(c)

1. Draw an upright line from the top corner straight down to the bottom corner (12 units long). The figure splits into a left part and a right triangle.
2. Left part: a rectangle 3 units wide and 8 units tall (the length of the upright left side), with a right triangle above and below it. Rectangle: 3 × 8 = 24; each triangle: ½ × 3 × 2 = 3. Left part = 24 + 3 + 3 = 30 square units.
3. Right part: draw a horizontal line from the right corner across to the upright line. Upper triangle: ½ × 3 × 2 = 3. Lower triangle: ½ × 3 × 10 = 15. Right part = 3 + 15 = 18 square units.
4. Total: 30 + 18 = 48 square units.

Answer48 square units

(d)

1. Draw a horizontal line across the figure at the bottom point of the V-shaped notch (2 units below the top).
2. Below the line is a rectangle 4 units wide and 3 units tall: 4 × 3 = 12 square units.
3. Above the line are two right triangles. Left: ½ × 1 × 2 = 1 square unit. Right: ½ × 3 × 2 = 3 square units.
4. Total: 12 + 1 + 3 = 16 square units.

Answer16 square units

(e)

1. Join the left and right corners with a horizontal line (4 units long). Draw an upright line from the top corner down to it, and another from the bottom corner up to it. The figure is now four right triangles.
2. Top two triangles: ½ × 2 × 2 = 2 each, so 2 + 2 = 4 square units.
3. Bottom-left triangle: ½ × 3 × 4 = 6 square units. Bottom-right triangle: ½ × 1 × 4 = 2 square units.
4. Total: 4 + 6 + 2 = 12 square units.

Answer12 square units

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Open NCERT p. 149Checked by computerAnswers can differ: one example

1. Area of first rectangle = 5 × 10 = 50 sq m.
2. Area of second rectangle = 2 × 7 = 14 sq m.
3. Sum of areas = 50 + 14 = 64 sq m.
4. One rectangle with this area is 8 m × 8 m, since 8 × 8 = 64.

Answer8 m × 8 m (any length and breadth that multiply to 64 sq m works, for example 8 m by 8 m).

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Open NCERT p. 149Matches NCERT’s answer

1. Area = length × width.
2. 1000 = 50 × width, so width = 1000 ÷ 50.

Answer20 m

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Open NCERT p. 149Matches NCERT’s answer

1. Area of the floor = 5 × 4 = 20 sq m.
2. Area of the carpet = 3 × 3 = 9 sq m.
3. Area not carpeted = 20 − 9 = 11 sq m.

Answer11 sq m

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Open NCERT p. 149Matches NCERT’s answer

1. Area of the garden = 15 × 12 = 180 sq m.
2. Area of one flower bed = 2 × 1 = 2 sq m.
3. Area of all four beds = 4 × 2 = 8 sq m.
4. Area left for the lawn = 180 − 8 = 172 sq m.

Answer172 sq m

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Open NCERT p. 149Checked by computerAnswers can differ: one example

1. Many pairs of shapes work. The idea: a long, thin shape has a long boundary, and a compact shape has a short one.
2. Shape A: a strip 1 unit wide and 18 units long. Area = 1 × 18 = 18 square units. Perimeter = 2 × (1 + 18) = 38 units.
3. Shape B: a rectangle 4 units by 5 units. Area = 4 × 5 = 20 square units. Perimeter = 2 × (4 + 5) = 18 units.
4. 38 units is longer than 18 units, so Shape A has the longer perimeter even though its area is smaller.

AnswerOne possible pair (many others work): Shape A is a 1 × 18 rectangle (area 18 square units, perimeter 38 units), and Shape B is a 4 × 5 rectangle (area 20 square units, perimeter 18 units). 38 > 18.

Watch this explained “Drawing to order: 18 units with the longer edge”, 8:35 into Same area, many perimeters: why one does not determine the other · हिंदी में देखें

Open NCERT p. 149One way to think about it

1. Measure your page. Call its length (top to bottom) L cm and its width (left to right) W cm.
2. The border is 1 cm from the top and 1 cm from the bottom, so its length is L − 1 − 1 = L − 2 cm.
3. It is 1.5 cm from the left and 1.5 cm from the right, so its width is W − 1.5 − 1.5 = W − 3 cm.
4. Perimeter of the border = 2 × ((L − 2) + (W − 3)) = 2 × (L + W) − 10 cm. It is always 10 cm less than the perimeter of the page.
5. Example: on a page 28 cm long and 20 cm wide, the border is 26 cm by 17 cm, so its perimeter is 2 × (26 + 17) = 86 cm. (The page's perimeter is 96 cm, and 96 − 10 = 86.)

In shortIt depends on the size of your page. The border's perimeter is 2 × (L + W) − 10 cm, which is 10 cm less than the page's perimeter. For example, a 28 cm × 20 cm page gives a 26 cm × 17 cm border with perimeter 86 cm.

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Open NCERT p. 149One way to think about it

1. Area of the outer rectangle = 12 × 8 = 96 square units, so the inner rectangle needs area 96 ÷ 2 = 48 square units.
2. To avoid touching, the inner rectangle must be shorter than 12 units along the long side and shorter than 8 units along the short side.
3. With whole-number sides, 48 = 1 × 48 = 2 × 24 = 3 × 16 = 4 × 12 = 6 × 8. Only 8 × 6 fits: a 4 × 12 rectangle is as long as the outer one, so it would touch.
4. Draw the 8 × 6 rectangle with its 8-unit side along the 12-unit side. Leave a 2-unit gap on the left and right (12 − 8 = 4 = 2 + 2) and a 1-unit gap at the top and bottom (8 − 6 = 2 = 1 + 1).
5. Other answers exist if the sides need not be whole numbers. For example, 10 × 4.8 and 9.6 × 5 also have area 48 and fit inside.

In shortOne answer: an 8 × 6 rectangle drawn with a 2-unit gap on the left and right and a 1-unit gap at the top and bottom. Its area is 8 × 6 = 48 square units, exactly half of 12 × 8 = 96. Other sizes with area 48 that fit inside, such as 10 × 4.8, also work.

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Open NCERT p. 149Matches NCERT’s answer

1. Let the square's side be s. Folding and cutting gives two rectangles, each s by s/2.
2. Each rectangle's perimeter = 2 × (s + s/2) = 3s. Two rectangles together = 6s.
3. The square's perimeter = 4s.
4. Compare: 6s ÷ 4s = 3/2 = 1½ — so the two rectangles' perimeters together are always 1½ times the square's perimeter. This matches statement (c).
5. Checking the others: each rectangle's area is s × s/2 = half the square's area (not larger, so (a) is false); the square's perimeter 4s is less than 6s (so (b) is false); the two rectangles' areas add up to exactly the square's area, not one third of it (so (d) is false).

AnswerStatement (c) is always true.

Watch this explained “Drawing to order: 18 units with the longer edge”, 8:35 into Same area, many perimeters: why one does not determine the other · हिंदी में देखें