Perimeter and Area — Detailed Solutions
Complete chapter explanations with step-by-step textbook solutions, HTML Canvas diagrams, area–perimeter reasoning, composite figures, tangram logic, house plans and higher-order SK Tuitions practice.
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Canvas Visual Learning Lab
These diagrams are drawn directly with the HTML <canvas> element and JavaScript. They illustrate boundary, area, tracks, tangrams, decomposition and area–perimeter comparisons.
Perimeter
Boundary length, tracks, regular polygons and split–rejoin problems
The perimeter is the total distance travelled along the boundary of a closed figure when we go around it once.
For a polygon, add the lengths of all its sides.
The canvas shows how the boundary lengths combine for a rectangle, square and triangle.
A rectangle has two sides of length 12 cm and two sides of breadth 8 cm.
Answer: 40 cm.
All four sides of a square are equal.
4 m of tape is required.
Add the three side lengths:
Answer: 16 cm.
Because the lace goes all the way around, we need the perimeter.
Answer: 10 m.
First find the distance in one round.
Answer: 900 m.
Use P=2(l+b).
Answer: 5 cm.
For a square, P=4s.
Answer: 5 cm.
Use P=2(l+b).
Answer: 3 m.
The wire length does not change, so the rectangle’s perimeter becomes the square’s perimeter.
Answer: 4 cm.
The three sides together must total 55 cm.
Answer: 21 cm.
First find the length of fencing required.
Answer: ₹21,600.
The full string becomes the perimeter in every case.
- Square: 4 equal sides, so 36÷4=9 cm.
- Equilateral triangle: 3 equal sides, so 36÷3=12 cm.
- Regular hexagon: 6 equal sides, so 36÷6=6 cm.
One round is one perimeter.
Answer: 2340 m.
The canvas marks the starting points and the positions after 250 m, 500 m and 1000 m.
One round equals the perimeter of the outer rectangle.
Answer: 1100 m.
One inner round is:
Akshi ran 1100 m, while Toshi ran 1260 m.
Toshi runs 160 m farther.
Akshi’s track perimeter is 220 m and she starts at the bottom-right corner, moving left along the bottom side.
After 250 m:
She completes 1 full round and then moves 30 m left along the bottom side. Mark this point A.
After 500 m:
She completes 2 full rounds and then moves 60 m left along the 70 m bottom side. So she is 10 m from the bottom-left corner. Mark B.
After 1000 m:
She has completed 4 full rounds. The remaining 120 m takes her 70 m along the bottom, 40 m up the left side and 10 m along the top. Mark C 10 m to the right of the top-left corner.
Toshi’s track perimeter is 180 m and she starts at the bottom-right corner.
After 250 m:
After one full round, she travels 60 m along the bottom and then 10 m up the left side. Mark X there.
After 500 m:
After two full rounds, she travels 60 m left, 30 m up and 50 m right along the top. Mark Y 10 m before the top-right corner.
After 1000 m:
She completes 5 full rounds. The extra 100 m is 60 m left + 30 m up + 10 m right. Mark Z 10 m to the right of the top-left corner.
The inner and outer runners need different starting positions even though they share the same finish line.
Inner track: one side is 100 m. Starting at the bottom-right corner gives:
So mark A at the inner bottom-right corner.
Outer track: one side is 150 m and the finish is at the midpoint of the bottom side. The source solution breaks the route into:
This places B on the top side, 25 m to the left of the top-right corner. From there the runner travels 125 m to the top-left corner, 150 m down the left side and 75 m to the finish.
This is a hands-on activity, so there is no single numerical answer.
- Cut a closed shape from paper.
- Estimate its boundary length.
- Measure each straight section with a ruler, or trace the boundary with thread and measure the thread.
- Add the measured boundary lengths.
- Compare the measured perimeter with the estimate.
The boundary contains 6 straight unit segments and 3 diagonal unit segments.
A diagonal across a unit square is longer than one straight unit, so 3 diagonal units are longer than 3 straight units.
Therefore the perimeter is more than 9 straight units. Toshi is correct.
Four dot-grid outlines are redrawn on Canvas and labelled with their perimeter expressions.
Count every horizontal/vertical unit segment as s and every diagonal unit segment as d.
- First figure: 8s + 2d
- Second figure: 4s + 6d
- Third figure: 12s + 6d
- Fourth figure: 18s + 6d
Both are regular polygons: all sides of each figure are equal, and all angles within each figure are equal. Their numbers of sides are different, but the same idea of equal side length lets us write a compact perimeter formula.
If a regular polygon has n equal sides and each side has length s, adding s repeatedly n times is the same as multiplication.
Examples: equilateral triangle = 3s, square = 4s, regular pentagon = 5s, regular hexagon = 6s.
Two 6 cm × 2 cm rectangles have the same total area, but their perimeter changes with the length of the shared edge.
Each separate 6 cm × 2 cm rectangle has perimeter 16 cm, so before joining them the combined boundary is 32 cm.
Whenever two pieces share an edge of length x, that edge disappears from the outside boundary twice:
- b: shared edge = 2 cm → P=32−4=28 cm.
- c: shared edge = 2 cm → P=32−4=28 cm.
- d: shared edge = 3 cm → P=32−6=26 cm.
We want:
Therefore join the two 6 cm × 2 cm rectangles along 5 cm of their long sides, leaving a 1 cm offset. The outside perimeter becomes 22 cm.
Area
Square units, composite regions, tangrams and fixed-area rectangles
The area of a closed figure is the amount of region enclosed by it.
Area is measured in square units, such as cm² or m².
Boundary length and enclosed region are different measurements; the canvas highlights each separately.
Area of the whole floor:
Area of the carpet:
Uncovered area:
Answer: 11 m².
Whole plot:
One flower bed:
Remaining area:
Answer: 56 m².
For a rectangle, A=l×b. Therefore:
Answer: 12 m.
First find the total area.
Number of groups of 100 m²:
Answer: ₹8,000.
Total area of the grove:
Each tree requires 25 m².
Answer: 200 trees.
The two textbook outlines are redrawn and split into simpler regions.
One convenient split is into horizontal strips:
- bottom strip: 3×3 = 9 m²
- next strip: 7×1 = 7 m²
- next strip: 5×2 = 10 m²
- top strip: 2×1 = 2 m²
Answer: 28 m².
Think of it as a 5 m × 3 m outer rectangle with a 3 m × 2 m rectangular opening removed.
Answer: 9 m².
The seven tangram pieces are drawn as one square and labelled A–G.
Using the hint and by matching/covering pieces:
- A and B have the same area.
- C and E have the same area.
- D, F and G have the same area.
C and E have equal area. D can be covered exactly by C and E together.
So D is twice C, and also twice E.
D and F have the same area. Each is equivalent to two small-triangle units such as C+E.
F and G have equal area. Their shapes are different, but rearrangement/covering shows they occupy the same amount of region.
If C is one small-area unit, then G is 2 such units while A is 4 such units.
A is twice as large as G, not four times.
Let Area(C)=1 unit. Then:
- A=4, B=4
- C=1, E=1
- D=2, F=2, G=2
Area of the big square = 16 × Area(C).
Rearranging pieces changes only their positions, not their individual areas. No piece is added, removed or overlapped.
The area remains exactly the same.
No. The total area stays the same because the same seven pieces are used, but the outside boundary can change when the pieces are rearranged. Therefore equal area does not force equal perimeter.
A visual guess is possible, but it is not reliable enough for measurement. The chapter’s method is to trace each shape on a square grid and estimate the area by counting full and partial squares.
The four dot-grid figures are shown with their source-provided areas.
Using the grid-square conventions—full square = 1, more than half ≈ 1, exactly half = 1/2, less than half ignored—the source solutions give:
- 4 square units
- 9 square units
- 10 square units
- 11 square units
This is an estimation activity, so the exact counted value can vary slightly with how the circle is placed on the grid.
- Trace a circle whose diameter spans 3 grid units.
- Count every full square inside it as 1 square unit.
- Ignore pieces smaller than half a square.
- Count pieces larger than half as 1 square unit and exact halves as 1/2.
The estimate should be close to 7 square units; a slightly different grid estimate is possible because the chapter is demonstrating approximation rather than an exact circle-area formula.
Squares tile a surface without gaps or overlaps. Equal squares line up in rows and columns, so counting and multiplication are straightforward.
Circles leave gaps when packed together. The chapter even shows that the same rectangle can appear to contain different numbers of circles under different packings, so circles do not provide a consistent unit for covering the region completely.
These are measurement activities and therefore depend on your actual school.
- Measure the required length(s) and width(s) in metres.
- Break irregular regions into rectangles if needed.
- Calculate each rectangular area using length×breadth.
- Add or subtract the pieces appropriately.
- Write the final answer in m².
List the factor pairs of 24:
| Dimensions | Perimeter |
|---|---|
| 1×24 | 2(1+24)=50 |
| 2×12 | 28 |
| 3×8 | 22 |
| 4×6 | 20 |
Greatest perimeter: 1×24 → 50 units.
Least perimeter: 4×6 → 20 units.
The rectangle whose side lengths are closest to each other has the smaller perimeter.
Factor pairs:
| Dimensions | Perimeter |
|---|---|
| 1×32 | 66 cm |
| 2×16 | 36 cm |
| 4×8 | 24 cm |
Greatest: 1×32. Least: 4×8.
Within the chapter’s whole-number-side setting, the most stretched factor pair gives the greatest perimeter, while the closest factor pair gives the least.
Area of a Triangle & Extended Applications
Triangle reasoning, grid figures, unit-square perimeters, house plans and area mazes
The canvas animates the rectangle-to-two-triangles idea and a triangle with the same base and height.
Yes. The two pieces overlap exactly when one is turned over, so they are congruent and have equal area.
Together they make the full rectangle. Therefore each one has half the rectangle’s area.
The diagram is designed so that the triangle has the same height as the rectangle and twice its base. Therefore:
So the areas are equal, even though the shapes look different.
The rectangle is 5 units wide and 4 units high.
Triangle BAD is exactly half of the rectangle:
Triangle ABE has the same base AB and the same perpendicular height 4 units; its apex E lies on the opposite side of the rectangle.
Both triangles have area 10 square units.
A triangle with base b and perpendicular height h occupies half the area of a rectangle (or parallelogram) with the same base and height.
The five grid figures from the exercise are reconstructed on Canvas with decomposition guides.
Enclose the slanted quadrilateral in a 4×7 rectangle.
The bounding rectangle has area 28. The sloping top removes a triangle of area 2 and the sloping bottom removes another triangle of area 2.
Answer: 24 square units.
Use a 4×10 bounding rectangle, area 40. The upper slant cuts off a triangle of base 4 and height 3, area 6. The lower slant cuts off a triangle of base 4 and height 2, area 4.
Answer: 30 square units.
Split the figure through the central vertical line.
On the left, use a 3×8 rectangle (24) plus two right triangles, each of area 3. This gives 30.
On the right is a triangle with vertical base 12 and horizontal height 3:
Answer: 48 square units.
Imagine a 4×5 rectangle, area 20. The V-shaped notch at the top is a triangle with base 4 and height 2.
Answer: 16 square units.
Draw a horizontal line through the two side vertices. This makes two triangles.
Top triangle: base 4, height 2 → area 4.
Bottom triangle: base 4, height 4 → area 8.
Answer: 12 square units.
All shapes use exactly 9 unit squares; only the arrangement changes the outside boundary.
The most compact arrangement is a 3×3 square.
Smallest perimeter: 12 units.
Keep the squares in one 1×9 strip. Each new square shares only one side with the previous square, producing the longest connected boundary allowed by this arrangement.
Largest perimeter: 20 units.
One example is an L-like chain of 8 squares with the ninth square placed inside the corner so it shares two sides. The canvas shows one valid arrangement.
A useful counting rule is:
where n is the number of squares and E is the number of shared edges. For n=9 and P=18, we need E=9 shared edges.
Yes for many perimeters such as 18 and 20: different connected arrangements can have the same number of exposed edges.
According to the supplied solution, the minimum perimeter 12 is the exception here: the compact 3×3 square is essentially the unique arrangement, apart from rotation/reflection.
A new isolated square has four sides. Every side that it shares with the old figure removes two boundary sides from the combined perimeter—one from the old figure and one from the new square.
where k is the number of sides shared by the new square.
- shares 1 side → perimeter increases by 2
- shares 2 sides → perimeter stays the same
- shares 3 sides → perimeter decreases by 2
The two rectangular plans are reconstructed proportionally and labelled with the solved dimensions.
The missing measurements from the supplied solution are:
| Region | Dimensions (ft) | Area (ft²) |
|---|---|---|
| Master bedroom | 15×15 | 225 |
| Toilet | 5×10 | 50 |
| Kitchen | 15×12 | 180 |
| Small bedroom | 15×12 | 180 |
| Utility | 15×3 | 45 |
| Hall | 20×12 | 240 |
| Parking | 15×3 | 45 |
| Garden | 20×3 | 60 |
The complete rectangular plan measures 35 ft × 30 ft.
Answer: 1050 ft².
The solved dimensions are:
| Region | Dimensions (ft) | Area (ft²) |
|---|---|---|
| Master bedroom | 12×15 | 180 |
| Toilet | 5×10 | 50 |
| Kitchen | 18×10 | 180 |
| Utility | 7×10 | 70 |
| Small bedroom | 12×10 | 120 |
| Hall | 23×15 | 345 |
| Entrance | 7×15 | 105 |
Sharan’s overall rectangle is 42 ft × 25 ft.
Charan’s overall rectangle is 35×30=1050 ft², so the areas are equal.
Sharan’s perimeter is 4 ft greater, even though both areas are 1050 ft².
Canvas redraws the four area puzzles and displays the missing values after the reasoning.
The top two rectangles have areas 13 and 26 but the same height, so the right rectangle is twice as wide as the left one.
The lower rectangles have the same respective widths. Therefore the lower-right area is twice the lower-left area:
Answer: 30 cm².
The lower 10 cm² rectangle is 2 cm high, so its width is 5 cm. The marked 3 cm portion leaves 2 cm for the width of the upright 10 cm² rectangle.
The pink rectangle begins 2 cm above the lower rectangle, so its height is 5−2=3 cm. Its width is also marked 3 cm.
Answer: 9 cm².
The middle rectangle has area 42 cm² and height 6 cm.
The bottom rectangle extends 5 cm farther, so its width is 7+5=12 cm.
Total height is 15 cm, so top height is:
The top rectangle is 3 cm narrower than the middle rectangle:
Answer: 16 cm².
The right rectangle has width 5 cm and area 18 cm².
The left rectangle is 4 cm taller, so its total height is 7.6 cm.
Answer: 5 cm.
First add the two areas.
Any rectangle with area 64 m² works. Source examples include:
- 16 m × 4 m
- 32 m × 2 m
- 8 m × 8 m
Width = area ÷ length.
Answer: 20 m.
Floor area =20 m² and carpet area =9 m².
Answer: 11 m².
Whole garden:
Four flower beds:
Answer: 172 m².
Canvas gives one valid construction for Q5 and Q7.
Choose rectangles:
- Shape A: 2×9 → area 18, perimeter 2(2+9)=22.
- Shape B: 4×5 → area 20, perimeter 2(4+5)=18.
This directly shows that a larger area does not necessarily mean a larger perimeter.
This answer depends on the actual dimensions of your page, so the source does not provide one fixed number.
If the page is W cm wide and H cm high, the inside border has:
Measure your page and substitute its width and height.
Outer area:
Half is:
One convenient inner rectangle is 8×6, since 8×6=48. Centre it: that leaves 2 units on the left and right and 1 unit on the top and bottom, so it does not touch the outer rectangle.
The square perimeter is:
Each half is s × s/2, so:
Compare 6s with 4s:
Therefore statement (c) is always true: the two rectangle perimeters added together are 1½ times the square’s perimeter.
SK Tuitions Challenge Questions
25 slightly difficult questions with full teacher-style solutions
Answer: 11 cm.
15 cm and 20 cm.
₹5,880.
12 cm.
54 cm.
6×8 has the smaller perimeter.
134 m².
108 cm².
12 cm.
30 square units each.
16 units.
The perimeter does not change.
It decreases by 2 units.
The pair with dimensions closest to each other is 8×9.
Least perimeter: 34 cm.
The square has 16 cm² more area.
280 m².
Let breadth=b, length=3b.
24 cm × 8 cm.
56 cm.
96 square units.
48 cm².
36 square units.
300 m.
The triangle is 60 square units smaller and has exactly half the area.
Possible fitting pairs include 8×12. Its perimeter is:
8×12 gives 40 units.
Same area, but A’s perimeter is 12 units larger.
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