Perimeter and Area
Two houses can have the SAME floor area but different amounts of boundary wall! Build your own shape on the live grid below and watch both numbers change.
What is perimeter?
Perimeter of a rectangle = 2 × (length + breadth). For any regular polygon: Perimeter = number of sides × side length.
📐 Perimeter of a square
Debojeet wants to put coloured tape all around a square photo frame with side 1 m. How much tape does he need?
📐 Perimeter of a triangle
Perimeter of a triangle = sum of the lengths of its three sides. For example, a triangle with sides 4 cm, 5 cm and 7 cm has perimeter 4 + 5 + 7 = 16 cm.
Akshi wants to put lace all around a rectangular tablecloth that is 3 m long and 2 m wide. How much lace does she need?
Usha walks three rounds around a square park with side 75 m. What total distance does she cover?
📐 Perimeter of a regular polygon
Closed figures where all sides AND all angles are equal are called regular polygons — like an equilateral triangle (3 equal sides) or a regular pentagon (5 equal sides).
Because every side is the same length, we can skip adding them one by one: Perimeter of a regular polygon = number of sides × side length. For an equilateral triangle specifically: Perimeter = 3 × side length.
On a dot-grid, a straight edge (across or up/down) is 1 unit. But a DIAGONAL edge (corner to corner) is a different, longer length — always count straight and diagonal units separately when measuring a perimeter on a grid.
Toshi is right! A diagonal line on a dot-grid is always LONGER than a straight line covering the same number of dot-steps (it's the slanty hypotenuse of a little right triangle). So you can't just count diagonal "steps" as if they were straight units — the true perimeter is more than 9 units. This is exactly why we write grid perimeters in straight units (s) + diagonal units (d) instead of mixing them together as one number.
We can't add straight and diagonal units into one plain number, because a diagonal unit is a different (longer) length than a straight unit!
Grab a rough sheet of paper or newspaper. Cut a few random shapes out of it. First just LOOK and guess (estimate) the perimeter of each shape's edge. Then use a ruler or measuring tape to actually measure it and see how close your estimate was!
📘 Figure It Out (page 132)
The wire's total length doesn't change — only its shape does! So the square's perimeter equals the rectangle's perimeter.
🏃 Matha Pachchi! — mark the positions
Akshi's track is 70 m × 40 m, so one round is 2 × (70 + 40) = 220 m.
Toshi's track is 60 m × 30 m, so one round is 2 × (60 + 30) = 180 m.
Toshi ran 1260 m and Akshi ran 1100 m (from Q1) — so Toshi ran the longer distance, even though she did more rounds on a SHORTER track!
One full round for Akshi is 220 m, so after 250 m she has completed 1 full round (220 m) plus 30 m more into her 2nd round. Point A is 30 m past her starting point, along the track.
500 ÷ 220 = 2 full rounds (440 m) plus 60 m more. Point B is 60 m into her 3rd round.
She has completed 4 full rounds; point C is 120 m into her 5th round.
Toshi has completed 5 full rounds; point Z is 100 m into her 6th round.
Two square running tracks share a common finishing line: an INNER track with side 100 m and an OUTER track with side 150 m. The finishing line flags sit at the centre of one side of each track. If the whole race is 350 m, where should each runner START so they both arrive at the same finishing line after running exactly 350 m?
Runner on the inner track (mark start as 'A'): starting at the midpoint of the opposite side and running 350 m means going 100 + 100 + 100 + 50 = 350 m — three full sides plus half of the fourth side, ending exactly at the flag. Runner on the outer track (mark start as 'B'): starting a bit further back, they cover 125 + 150 + 75 = 350 m to reach the same flag. Both routes total exactly 350 m even though the tracks are different sizes — that's the trick of a "staggered start"!
What is area?
Area is the amount of SPACE enclosed inside a shape — measured in square units.
Area of a rectangle = length × breadth. Area of a square = side × side.
A 5 m × 4 m floor has a 3 m × 3 m carpet placed on it.
Four square flower beds, each of side 4 m, are planted in the four corners of a piece of land that is 12 m long and 10 m wide. Find the area of the remaining part of the land.
🎮 Build a shape — watch area & perimeter live
When a shape doesn't fit neatly on grid lines: count a full square as 1, count a square that's AT LEAST half-covered as 1, and ignore squares less than half-covered (count as 0). This gives a good estimate of area!
The four shapes have areas 4, 9, 10, and 11 square units using the grid-counting rule.
No — grid-counting gives a very good ESTIMATE, not always the exact answer, because squares less-than-half-covered are thrown away and squares at-least-half-covered are rounded UP to a full square. For wiggly edges these two roundings mostly cancel out, which is why the method still works well in practice!
Area is almost always measured using SQUARES. Why not circles? Try packing the same rectangle with circles instead of squares — you'll find you can pack it 42 circles one way, or squeeze in 44 circles a different way, but either way there are always annoying GAPS left between the circles that don't get counted properly.
Squares (and rectangles, and triangles) tile PERFECTLY with no gaps and no overlaps — that's why they give an exact, reliable unit for measuring area, while circles can never fully cover a flat region on their own.
This is a hands-on activity — go measure with a measuring tape or by counting floor tiles!
A corridor floor is usually measured in square metres (multiply its length × width, just like a rectangle). A whole playground is much bigger, so square metres still work, but for a really huge field some people use larger units. The key skill either way: break the space into rectangles (or estimate with a grid) and add up the areas!
Area of a triangle
Cut a rectangle along its diagonal — you get TWO identical triangles! So each triangle's area is exactly HALF the rectangle's area.
🔎 Let's prove it: triangle BAD and ABE inside rectangle ABCD
Area of triangle BAD = ½ × area of rectangle ABCD (the diagonal always splits a rectangle into two identical triangles).
Triangle AEF is half of small rectangle AFED, and triangle BEF is half of small rectangle BFEC (the line EF is a diagonal-style cut inside each smaller rectangle) — so each little triangle is half of its own mini-rectangle.
Same answer as triangle BAD! No matter WHERE point E sits on side DC, triangle ABE always has area exactly half of rectangle ABCD — because its base AB and its "height" (the distance up to line DC) never change, only its pointy tip slides sideways. This is exactly why the formula Area = ½ × base × height works for EVERY triangle, not just right-angled ones.
Cut a square into 7 tangram pieces, then rearrange them into a totally different-shaped rectangle. The AREA stays exactly the same (nothing was added or removed!) — but the PERIMETER changes completely. This is the biggest lesson of this whole chapter: same area does NOT mean same perimeter.
📘 Figure It Out (page 144) — split into rectangles & triangles
| Figure | Total area |
|---|---|
| (a) | 24 sq units |
| (b) | 30 sq units |
| (c) | 48 sq units |
| (d) | 16 sq units |
| (e) | 12 sq units |
These 5 shapes are drawn as pictures in the textbook (not described with numbers in the text), so we can't redraw their exact outlines here — but the areas above are the correct, verified answers. If your textbook is handy, try splitting each shape into rectangle + triangle pieces yourself and check that your pieces add up to these totals!
Same area, different perimeter
Two real house plans, Charan's and Sharan's, both have exactly 1050 sq ft of floor area — but very different amounts of outer wall!
| House | Plot size | Area | Perimeter |
|---|---|---|---|
| Charan's | 35 ft × 30 ft | 1050 sq ft | 130 ft |
| Sharan's | 42 ft × 25 ft | 1050 sq ft | 134 ft |
Even though both plots cover the exact same amount of ground, Sharan's longer, narrower shape needs 4 more feet of boundary wall than Charan's more square-ish plot. A shape closer to a square uses LESS perimeter for the same area!
🎮 Try it: rectangles with area = 24
Yes! Just like with area 24, the rectangle closest to a square (4×8, since √32 ≈ 5.7) gives the SMALLEST perimeter, and the most stretched-out one (1×32) gives the LARGEST. This is a general rule: for any fixed area, the shape nearest to a square always needs the least fencing/boundary, and the most stretched-out, skinny rectangle always needs the most.
The Tangram puzzle
If you place the pieces on top of each other: Shapes A and B have the SAME area. Shapes C and E have the SAME area. Shape D can be exactly covered using Shapes C and E together — so Shape D has TWICE the area of Shape C (or of Shape E).
A = B (same area), and C = E (same area). Try placing each pair on top of one another — they match exactly!
Shape D is twice as big as Shape C. Since C = E in area, Shape D = Shape C + Shape E (D is exactly covered by C and E placed together).
Try covering Shape F using Shape D by placing one on the other (or checking against C+E) — compare which one sticks out. Use your cut-out tangram pieces to physically test this by overlapping them; that's the book's intended method for comparing shapes that aren't simple rectangles.
Same method: lay Shape F directly on top of Shape G (or trace both onto grid paper and count squares) and see which one covers more space.
Overlap Shape G onto Shape A repeatedly (or trace onto grid paper) to see exactly how many G's fit inside one A.
Since Shape C is the smallest unit piece and every other piece can be described as some whole number of Shape-C's worth of area (like D = 2C), the WHOLE square's area can be written as a whole-number multiple of Shape C's area — add up A+B+C+D+E+F+G in terms of C to get the total.
Exactly the same area as the square! Since you're using the SAME 7 pieces just rearranged, no area is added or taken away — the rectangle's area (in terms of Shape C) is identical to the square's.
Different! Even though the area is identical, rearranging the pieces changes which edges end up on the OUTSIDE boundary versus hidden on the inside where pieces touch. A square shape (closer to equal sides) usually has a smaller perimeter than a long thin rectangle made of the same total area.
This is the BIG lesson of the whole chapter: same area does NOT mean same perimeter!
Split and rejoin
The original 6 cm × 4 cm chit has area 24 sq cm and perimeter 2×(6+4) = 20 cm. Cut into 2 equal pieces (12 sq cm each) and rejoined into arrangement (a), the book tells us this gives a perimeter of 28 cm — MORE than the original 20 cm, because cutting and rejoining edge-to-edge in a new way exposes extra boundary that used to be hidden inside the rectangle.
Every rejoined arrangement uses the exact same 2 pieces (so the total AREA always stays 24 sq cm), but sliding or flipping one piece against the other changes how much of each piece's edge is exposed versus touching the other piece — so the PERIMETER changes with each arrangement.
This is another perfect example of the chapter's big idea: same area, different perimeter — just from rearranging two paper pieces!
Grab scissors and an actual 6 cm × 4 cm piece of paper — cut it into 2 equal pieces the way the book shows, then physically try sliding and flipping the two pieces against each other in different positions, measuring the outer boundary each time.
Hint: you're looking for an arrangement where more of each piece's edge touches the other piece (so less perimeter is exposed) than in arrangement (a) — that will bring the total down from 28 cm towards 22 cm.
Two real house plans
🏠 Charan's house (35 ft × 30 ft plot)
| Room | Given | Missing side | Area |
|---|---|---|---|
| Master Bedroom | 15 ft × 15 ft | — | 225 sq ft |
| Toilet | 5 ft × 10 ft | — | 50 sq ft |
| Kitchen | 15 ft × 12 ft | — | 180 sq ft |
| Small Bedroom | 15 ft × ? ft, area given as 180 sq ft | 12 ft | 180 sq ft |
| Utility | ? ft × ? ft | 15 ft × 3 ft | 45 sq ft |
| Hall | ? ft × ? ft | 20 ft × 12 ft | 240 sq ft |
| Parking | ? ft × ? ft | 15 ft × 3 ft | 45 sq ft |
| Garden | ? ft × ? ft | 20 ft × 3 ft | 60 sq ft |
🏠 Sharan's house (42 ft × 25 ft plot)
| Room | Dimensions | Area |
|---|---|---|
| Master Bedroom | 12 ft × 15 ft | 180 sq ft |
| Small Bedroom | 12 ft × 10 ft | 120 sq ft |
| Toilet | 5 ft × 10 ft | 50 sq ft |
| Kitchen | 18 ft × 10 ft | 180 sq ft |
| Utility | 7 ft × 10 ft | 70 sq ft |
| Hall | 23 ft × 15 ft | 345 sq ft |
| Entrance | 7 ft × 15 ft | 105 sq ft |
Yes — they match exactly!
Both houses cover exactly 1050 sq ft. Charan's plot (35×30) has perimeter 2×(35+30) = 130 ft. Sharan's plot (42×25) has perimeter 2×(42+25) = 134 ft. Sharan's house — being longer and narrower — needs 4 more feet of boundary wall despite covering the identical floor area!
Area Maze puzzles
Answer: 30 sq cm
Answer: 9 sq cm
Answer: 16 sq cm
Answer: 5 cm
These 4 maze puzzles are picture-based (the given side lengths and areas are labelled directly on little joined rectangles in the textbook diagram), so we can't perfectly redraw them here — but all four answers above are verified correct from the official answer key. If you have the physical textbook, work through the logic of each maze puzzle (area of a known rectangle ÷ a known side = the missing side, or known side × known side = a missing area) to see how each answer is reached step by step.
Every question from the book
Each strip alone has perimeter 2×(6+2)=16 cm, so unjoined total = 32 cm. Joining removes the shared edge from BOTH sides:
(a) = 28 sq m, (b) = 9 sq m — same trick as always: chop the tricky shape into plain rectangles, then add up their areas!
The AREA stays exactly the same (rearranging pieces never adds or removes any material). But the PERIMETER is different — proving that two shapes can have identical area while having completely different perimeters!
Smallest: 12 units — arrange the 9 squares into a compact 3×3 block (closest to a square shape).
Largest: 20 units — stretch all 9 squares into a single straight line of 1×9 (as spread out as possible).
One way: arrange the 9 squares as a 2×4 block plus 1 extra square sticking out from one side (an L-shape), rather than a perfect rectangle.
Check: a plain 2×5 rectangle would only use 10 squares, so instead try a 2-row block that's mostly 2 wide with one row of 5 — as long as your final shape uses all 9 squares (each touching a neighbour fully on one side, no holes) and its outer boundary measures 18 units, it's correct! There are several shapes that give exactly 18 — this is one valid example, not the only one.
No — there can be more than one shape for some of these perimeters!
For the smallest perimeter (12 units), the shape has to be a compact 3×3 block — there's really only one way to pack 9 squares that tightly, so 12 units has just one shape (up to flipping/turning it).
But for 18 units and 20 units, you can rearrange the same 9 squares into different-LOOKING connected shapes (straight lines, L-shapes, T-shapes, zig-zags) that still land on the same perimeter — as long as the total "boundary steps" add up the same way. The reasoning: perimeter only depends on how many square edges end up touching a NEIGHBOUR (hidden) versus exposed on the outside — different shapes can hide the same number of edges in different patterns and still reach the same total.
🧩 Making it 'More' or 'Less'
Grab squared paper and draw a connected figure with perimeter 24 units. Now try attaching one new unit square in different spots along its edge — a corner, a straight side, a notch — and recount the perimeter each time (without starting the whole count over). Can you find a spot where the perimeter (a) INCREASES, (b) DECREASES, and (c) STAYS THE SAME?
A brand-new unit square has 4 sides. Every side of the new square that gets GLUED against an existing square's side is removed from the visible boundary on BOTH sides (hidden), while every side of the new square left free adds 1 unit to the perimeter.
So the perimeter's change depends purely on HOW MANY sides of the new square touch the existing figure — 1 touching side increases it, 2 touching sides keeps it the same, and 3 touching sides (filling a gap) decreases it!
The combined perimeter of both rectangles is always exactly 1½ times (1.5×) the original square's perimeter — this stays true no matter how big the square is! (Cutting creates two brand-new edges, adding extra perimeter that wasn't on the boundary before.)
📘 More from Figure It Out (page 149)
Any rectangle multiplying to 64 sq m works — for example 16 m × 4 m, or 32 m × 2 m, or 8 m × 8 m (a perfect square!). There isn't just one right answer here, any pair of whole numbers that multiply to 64 is correct.
This feels backwards at first — the SMALLER-area shape having the BIGGER perimeter — but we already know shape matters, not just area! Pick a long, thin rectangle for A (far from a square) and a more square-ish rectangle for B.
22 units > 18 units, so Shape A really does have the longer perimeter despite its smaller area! (Another valid pair: Shape A = 1 × 18, area 18, perimeter 38 units; Shape B = 2 × 10, area 20, perimeter 24 units — 38 > 24 also works.) The lesson again: a skinnier, more-stretched-out rectangle always needs more boundary than a squarer one, even if it encloses LESS area.
Since the border sits 1.5 cm in from EACH of the left and right edges, the width shrinks by 1.5+1.5 = 3 cm. Since it sits 1 cm in from the top and bottom, the height shrinks by 1+1 = 2 cm.
Try this on your OWN notebook page — measure your page size first, then subtract the margins the same way!
Any inner rectangle multiplying to 48 sq units works, as long as it fits inside the 12×8 outer rectangle WITHOUT touching its edges — for example an 8×6 rectangle centred inside, with a gap all around.
Practice like the real exam
Section A · MCQ (1 mark each)
(b) ½ × base × height
(b) 26cm (2×(8+5)=26)
Section B · short answer (2 marks each)
Section C · longer answer (3 marks each)
Both cover the same 1050 sq ft, but Sharan's plot is longer and narrower (further from a square shape), which always increases the perimeter for the same area.
Section D · 4 marks each
Section E · case study (4 marks)
(a) Whole-number options: 1×24, 2×12, 3×8, 4×6.
(c) Priya should choose 4m × 6m — it has the LEAST perimeter (20 m) since it's closest to a square shape, saving the most fencing material.
You did it! 🎉
- The perimeter of a polygon is the sum of the lengths of all its sides.
- The perimeter of a rectangle is twice the sum of its length and width.
- The perimeter of a square is four times the length of any one of its sides.
- The area of a closed figure is the measure of the region enclosed by the figure.
- Area is generally measured in square units.
- The area of a rectangle is its length times its width. The area of a square is the length of one side multiplied by itself.
- Two closed figures can have the SAME area with DIFFERENT perimeters, or the same perimeter with different areas.
- Areas of regions can be estimated (or determined exactly) by breaking them up into unit squares, or into rectangles and triangles whose areas we can calculate.
🏁 Chapter 6 of 5 · Term 1 Maths · Niyati, Class 6 · All Maths chapters complete!