PERIMETER AND AREA
6.1 Perimeter
The perimeter of a closed figure is the total length of its boundary. For a polygon, it’s simply the sum of the lengths of all its sides.
Perimeter of a Rectangle
For a rectangle with length l and breadth b:
Perimeter of a Square
For a square with side length s:
Perimeter of a Triangle
For a triangle with side lengths a, b, and c:
Figure it Out (Page 132)
1. Find the missing terms:
a. P = 14 cm, b = 2 cm. Formula: 14 = 2(l + 2) → 7 = l + 2 → length (l) = 5 cm.
b. P = 20 cm. Formula: 20 = 4s → side (s) = 5 cm.
c. P = 12 m, l = 3 m. Formula: 12 = 2(3 + b) → 6 = 3 + b → breadth (b) = 3 m. (This is a square!)
2. A wire is bent from a rectangle (5cm x 3cm) into a square. What is the side of the square?
The length of the wire is equal to the perimeter of the rectangle.
Perimeter of rectangle = 2(5 + 3) = 2(8) = 16 cm.
This 16 cm wire is now the perimeter of the square. Perimeter of square = 4s.
16 = 4s → Side of the square = 4 cm.
3. Find the third side of a triangle with P = 55 cm and two sides of 20 cm and 14 cm.
Sum of two sides = 20 + 14 = 34 cm.
Third side = Total Perimeter – Sum of two sides = 55 – 34 = 21 cm.
4. Cost of fencing a rectangular park (150m x 120m) at ₹40 per metre?
Perimeter of park = 2(150 + 120) = 2(270) = 540 m.
Total cost = Perimeter × Cost per metre = 540 × 40 = ₹21,600.
5. A 36 cm string is used to form a square, an equilateral triangle, and a regular hexagon. Find the side length of each.
The perimeter of each shape is 36 cm.
a. Square: 36 = 4s → Side = 9 cm.
b. Equilateral Triangle: 36 = 3s → Side = 12 cm.
c. Regular Hexagon (6 sides): 36 = 6s → Side = 6 cm.
6. A farmer fences a rectangular field (230m x 160m) with 3 rounds of rope. Find the total length of rope.
Perimeter of field (1 round of rope) = 2(230 + 160) = 2(390) = 780 m.
Total rope for 3 rounds = 3 × 780 = 2340 m.
Matha Pachchi! (Page 133-134)
Scenario: Akshi runs on an outer track (70m x 40m) for 5 rounds. Toshi runs on an inner track (60m x 30m) for 7 rounds.
Calculations:
Perimeter of Akshi’s track = 2(70 + 40) = 2(110) = 220 m per round.
Perimeter of Toshi’s track = 2(60 + 30) = 2(90) = 180 m per round.
1. Akshi’s total distance: 5 rounds × 220 m/round = 1100 m.
2. Toshi’s total distance: 7 rounds × 180 m/round = 1260 m.
Toshi ran a longer distance.
3. Marking Positions:
a. Akshi after 250m: One round is 220m. She runs 250 – 220 = 30m past the start. This is 30m along the first 70m side. (Position ‘A’)
b. Akshi after 500m: 500m is 2 full rounds (440m) plus 60m. This is 60m along the first 70m side. (Position ‘B’)
c. Akshi after 1000m: 1000m is 4 full rounds (880m) plus 120m. She finishes 4 full rounds. Her position (‘C’) is 70m (1st side) + 40m (2nd side) + 10m (along 3rd side) from the start.
d. Toshi after 250m: One round is 180m. He runs 250 – 180 = 70m past the start. This is 60m (1st side) + 10m (along 2nd side). (Position ‘X’)
e. Toshi after 500m: 500m is 2 full rounds (360m) plus 140m. This is 60m (1st side) + 30m (2nd side) + 50m (along 3rd side). (Position ‘Y’)
f. Toshi after 1000m: 1000m is 5 full rounds (900m) plus 100m. He finishes 5 full rounds. His position (‘Z’) is 60m (1st side) + 30m (2nd side) + 10m (along 3rd side).
Deep Dive: Staggered Start (Page 134)
Race: 350m. Inner track: 100m side. Outer track: 150m side. Common finish line at the center of one side.
Perimeter of inner track = 4 × 100 = 400 m.
Perimeter of outer track = 4 × 150 = 600 m.
The finish line is at the center of a side. Let’s say it’s 50m from a corner on the inner track and 75m from a corner on the outer track.
To run 350m and end at the finish line, we must find the starting point by measuring 350m backwards from the finish line.
- Inner runner (‘A’): Starts 350m before the finish line on the 400m track. This is 50m *after* the finish line on the previous lap.
- Outer runner (‘B’): Starts 350m before the finish line on the 600m track. This is 250m *after* the finish line.
This “stagger” ensures both runners cover the same distance. The runner on the outer track needs a significant head start.
6.2 Area
Area is the measure of the region enclosed by a closed figure. It is measured in square units (like sq cm or m²).
Figure it Out (Page 138)
1. Area = 300 sq m, Length = 25 m. Find width.
Area = l × w → 300 = 25 × w → w = 300 / 25 = 12 m.
2. Cost of tiling a plot (500m x 200m) at ₹8 per 100 sq m.
Area of plot = 500 × 200 = 100,000 sq m.
Number of “hundred sq m” units = 100,000 / 100 = 1000 units.
Total cost = 1000 units × ₹8/unit = ₹8,000.
3. Max trees in a coconut grove (100m x 50m) if each tree needs 25 sq m.
Area of grove = 100 × 50 = 5000 sq m.
Max number of trees = Total Area / Area per tree = 5000 / 25 = 200 trees.
4. Find areas by splitting figures.
Figure a: Split into two rectangles. Top (3×1) and Bottom (4×4). No, the dimensions are tricky. Let’s split it into a 3x(1+2) rectangle and a (4-1)x4 rectangle. Top: 3×3=9. Bottom: 3×4=12. Total = 21? Let’s re-examine the image.
Let’s split vertically: Left rectangle is 4×3=12. Right rectangle is 2×4=8. The numbers seem to be side lengths. Let’s assume the labels are lengths.
Figure a: Split into a 3×2 rectangle (Area=6) and a 3×4 rectangle (Area=12). Wait, the shape is complex. Let’s split it into a top 3×1 part, a middle 2x(3+1)=8 part, and a bottom 3×1 part. This is confusing.
Let’s assume the numbers are side lengths as labeled. The shape is composed of a 3x(2+4)=18 rectangle on the right, and a 2×4=8 rectangle on the left. No, that’s not right either.
Let’s assume the labels are areas. No, the question asks to find areas.
Final attempt by splitting into simpler rectangles:
Figure a: Split into a top rectangle 3×1, a middle rectangle (3+4)x2, and a bottom rectangle 3x(4-2-1). Let’s use the coordinates method. Assume bottom left is (0,0). The vertices are (0,4), (3,4), (3,3), (7,3), (7,1), (3,1), (3,0), (0,0). Let’s split differently.
Figure a: Vertical split. Left rectangle: 3×4=12. Right rectangle: 4×2=8. It seems to overlap.
Let’s split horizontally. Top rectangle: 3×1=3. Middle: 7×2=14. Bottom: 3×1=3. Total area = 3+14+3 = 20 sq m.
Figure b: Split into three rectangles. Left: 1×3=3. Middle: (5-1-1)x(3-2)=3×1=3. Bottom: 5×2=10. This is also confusing.
Let’s split differently. Left vertical bar: 1×3=3. Bottom horizontal bar: 5×1=5. Right vertical bar: 1×2=2. What about the middle? The shape is (5×3) – (3×1). Area = 15 – 3 = 12?
Let’s try another split for Figure b. Top rectangle: 5×1=5. Middle: 1×2=2. Left: 1×2=2. Right: 1×1=1. No.
Let’s assume the given lengths define the rectangles. Top left: 3x(3-2)=3. Top right: (5-3)x(3-2)=2. Middle: (5)x(2)=10. Total=15? Let’s use a simpler split:
Figure b: Left rectangle 3×1 = 3. Bottom rectangle (3+1)x1 = 4. Right rectangle 1x(3-1)=2. Total = 9? The provided diagram is ambiguous. Let’s assume standard L-shapes.
Figure b: Split into a vertical 3×1 bar (Area=3) and a horizontal 5×2 bar (Area=10). Total = 13? Let’s do a large rectangle minus a small one. Outer 5×3 rectangle minus a 3×1 missing piece. Area = 15 – 3 = 12 sq m. This is the most logical interpretation.
Tangram Puzzle (Page 139)
Let’s assume the area of the smallest triangle (Shape C or E) is 1 unit.
- Area C = 1, Area E = 1.
- Area D (medium triangle) = 2 units.
- Area G (square) = 2 units.
- Area F (parallelogram) = 2 units.
- Area A (large triangle) = 4 units.
- Area B (large triangle) = 4 units.
1. Pieces with the same area: {A, B}, {C, E}, and {D, F, G}.
2. Shape D vs Shape C: Area D (2 units) is twice as big as Area C (1 unit). Shapes C, E, and D are all triangles, and C and E together form D.
3. Shape D vs Shape F: They have the same area (2 units each).
4. Shape F vs Shape G: They have the same area (2 units each).
5. Shape A vs Shape G: Area A (4 units) is twice as big as Area G (2 units).
6. Area of the big square in terms of C: The total area is 4+4+2+2+2+1+1 = 16 units. This is 16 times the area of Shape C.
7. Area of the rectangle formed: The area remains the same. It is still 16 times the area of Shape C. The area of the pieces doesn’t change when rearranged.
8. Perimeters of the square and rectangle: They are different. For a fixed area, a square is the most compact shape and generally has the smallest perimeter. An elongated rectangle made from the same pieces will have a larger perimeter.
6.3 Area of a Triangle
The area of a triangle can be found by thinking of it as half of a rectangle.
For any triangle inside a rectangle that shares the same base and has the same height, its area will be exactly half the area of the rectangle.
Area Maze Puzzles (Page 148)
a. Top left area is 13, top right is 26. This means the right rectangle is twice as wide as the left one. If left width is w₁, right is w₂. So w₂ = 2w₁.
Bottom left area is 15. The height of the bottom row must be 15/w₁. The height of the top row is 13/w₁.
This means the area of the bottom right is related by the same ratio. Area = (15/w₁) * w₂ = (15/w₁) * (2w₁) = 15 * 2 = 30 sq cm.
b. Right column: A 2cm wide rectangle has area 10 sq cm, so its height is 10/2 = 5 cm. This 5cm is the combined height of the top two rectangles.
Top right: A 3cm wide rectangle has a missing area. Its height is unknown.
Top left: A 3cm wide rectangle has area 10 sq cm. Wait, the labels are side lengths.
Let’s re-read. “find the missing value of either the length of a side or the area”.
b. Top right square: 3cm side, so Area = 3×3 = 9. But there’s a smaller rectangle inside with 10 sq cm area. This puzzle is designed to be tricky. Let’s assume the outer lengths are for the combined shapes.
Let’s analyze the L-shape on the right. Total height is 3+2=5cm. Let the width of the top part be w₁ and bottom w₂. Area top = w₁*3. Area bottom = w₂*2.
Let’s assume the 10 sq cm and the `? sq cm` are two rectangles. And the 3cm and 2cm are their heights. The total width is 3+2=5cm. This is not logical.
Let’s try a different interpretation for (b).
There’s a rectangle of area 10 sq cm and height 2 cm. Its width must be 5 cm.
There’s a rectangle above it of height 3 cm and width 5 cm, so its area is 15 sq cm.
Next to it is a square of 3cm x 3cm. Area = 9 sq cm.
Below that is a rectangle of 2cm height and 3cm width. Area = 6 sq cm.
This is also inconsistent.
Let’s assume the simplest logic for (b): Top rectangle is 3cm wide, area is `? sq cm`. Right rectangle is 2cm wide, area is 10 sq cm. Height of right rect is 10/2 = 5cm. This height is shared. So the left rectangle height is also 5cm. Area of left rectangle = 3cm * 5cm = 15 sq cm.
c. Bottom rectangle: Area = 60, Width = 5. Height = 60/5 = 12 cm.
Middle rectangle: Area = 42, Height = 6. Width = 42/6 = 7 cm.
The total height on the left is 15 cm. Height of top rectangle = 15 – 12 = 3 cm.
The width of the top rectangle is the same as the middle one, 7 cm.
So, area of the top rectangle = 7 × 3 = 21 sq cm.
d. Right column: Bottom rectangle area = 18, width = 5. Height = 18/5 = 3.6 cm.
Top rectangle area = 38. Its height is 4 cm. Its width is shared with the bottom one, so 5 cm. This is not possible as 38 != 4*5.
Let’s assume the widths are different.
Let’s look at the horizontal split. Total width of the top part is `? cm`. Total width of the bottom part is `5 cm`.
Let’s assume the height of the bottom rectangle is H. Then 5H = 18. H = 3.6.
The total height of the right side is 4 + 3.6 = 7.6 cm.
The area of the top left shape is 38. It’s a rectangle. Let its width be W and height be 4. Then 4W = 38. W = 9.5.
The missing length `? cm` is W = 9.5 cm.
SUMMARY
- The perimeter of a polygon is the sum of the lengths of all its sides. For a rectangle, P = 2(l+b). For a square, P = 4s.
- The area of a closed figure is the measure of the region it encloses, measured in square units.
- For a rectangle, Area = l × b. For a square, Area = s².
- The area of a triangle is half the area of its enclosing rectangle, given by the formula Area = ½ × base × height.
- Two figures can have the same area but different perimeters, or the same perimeter but different areas.
- Generally, for a fixed area, a more compact shape (like a square) will have a smaller perimeter than an elongated shape (like a long, thin rectangle).