Perimeter and Area -- Exercise Solutions
Use this page alongside the official Samacheer Kalvi textbook. Solve each exercise first, then check the methods below.
Exercise 3.1 -- Rectangle Table (Q1)
Find the unknown values for each rectangle:
Q1(i). l = 5 cm, b = 8 cm
Step-by-step:
- Perimeter = 2(l + b) = 2(5 + 8) = 2 x 13 = 26 cm
- Area = l x b = 5 x 8 = 40 cm squared
Q1(ii). l = 13 cm, Perimeter = 54 cm
Step-by-step:
- 2(13 + b) = 54
- 13 + b = 27
- b = 27 - 13 = 14 cm
- Area = 13 x 14 = 182 cm squared
Q1(iii). b = 15 cm, Perimeter = 60 cm
Step-by-step:
- 2(l + 15) = 60
- l + 15 = 30
- l = 30 - 15 = 15 cm
- Area = 15 x 15 = 225 cm squared
Q1(iv). l = 10 m, Area = 120 sq.m
Step-by-step:
- 10 x b = 120
- b = 120 / 10 = 12 m
- Perimeter = 2(10 + 12) = 2 x 22 = 44 m
Q1(v). b = 4 feet, Area = 20 sq.feet
Step-by-step:
- l x 4 = 20
- l = 20 / 4 = 5 feet
- Perimeter = 2(5 + 4) = 2 x 9 = 18 feet
Exercise 3.1 -- Square Table (Q2)
Find the unknown values for each square:
Q2(i). Side = 6 cm
Step-by-step:
- Perimeter = 4 x side = 4 x 6 = 24 cm
- Area = side squared = 6 x 6 = 36 cm squared
Q2(ii). Perimeter = 100 m
Step-by-step:
- Side = Perimeter / 4 = 100 / 4 = 25 m
- Area = 25 x 25 = 625 m squared
Q2(iii). Area = 49 sq.feet
Step-by-step:
- Side = square root of 49 = 7 feet
- Perimeter = 4 x 7 = 28 feet
Exercise 3.1 -- Right-Angled Triangle Table (Q3)
Find the unknown values for each right-angled triangle:
Q3(i). Base = 20 cm, Height = 40 cm
Step-by-step:
- Area = (1/2) x base x height = (1/2) x 20 x 40 = 10 x 40 = 400 cm squared
Q3(ii). Base = 5 ft, Area = 20 sq.ft
Step-by-step:
- (1/2) x 5 x h = 20
- 5h/2 = 20
- 5h = 40
- h = 8 feet
Q3(iii). Height = 12 m, Area = 24 sq.m
Step-by-step:
- (1/2) x b x 12 = 24
- 6b = 24
- b = 4 m
Exercise 3.1 -- Triangle Perimeter Table (Q4)
Q4(i). Sides: 6 cm, 5 cm, 2 cm
Step-by-step:
- Perimeter = 6 + 5 + 2 = 13 cm
Q4(ii). Side2 = 8 m, Side3 = 3 m, Perimeter = 17 m
Step-by-step:
- Side1 = 17 - 8 - 3 = 6 m
Q4(iii). Side1 = 11 ft, Side3 = 9 ft, Perimeter = 28 ft
Step-by-step:
- Side2 = 28 - 11 - 9 = 8 feet
Exercise 3.1 -- Unit Conversions (Q5)
Fill in the blanks:
1. 5 cm squared = 5 x 100 = 500 mm squared 2. 26 m squared = 26 x 10,000 = 2,60,000 cm squared 3. 8 km squared = 8 x 10,00,000 = 80,00,000 m squared
Remember: 1 cm squared = 100 mm squared; 1 m squared = 10,000 cm squared; 1 km squared = 10,00,000 m squared.
Exercise 3.1 -- Direct Calculation Questions (Q7-Q10)
Q7. Rectangle: l = 6 m, b = 4 m
Step-by-step:
- Perimeter = 2(6 + 4) = 2 x 10 = 20 m
- Area = 6 x 4 = 24 m squared
Q8. Square: side = 8 cm
Step-by-step:
- Perimeter = 4 x 8 = 32 cm
- Area = 8 x 8 = 64 cm squared
Q9. Right triangle with sides 6, 8, 10 feet
Step-by-step:
- Perimeter = 6 + 8 + 10 = 24 feet
- Area = (1/2) x 6 x 8 = (1/2) x 48 = 24 sq.feet
- Note: 6 and 8 are the two perpendicular sides (legs) of the right triangle.
Q10. Triangle perimeters:
(i) Scalene triangle (7, 8, 10 m):
- Perimeter = 7 + 8 + 10 = 25 m
(ii) Isosceles triangle (10, 10, 7 cm):
- Perimeter = 10 + 10 + 7 = 27 cm
(iii) Equilateral triangle (6 cm):
- Perimeter = 3 x 6 = 18 cm
Exercise 3.1 -- Word Problems (Q11-Q15)
Q11. A rectangular photo has an area of 820 cm squared and a width of 20 cm. Find its length and perimeter.
Step-by-step:
- Area = length x breadth
- 820 = l x 20
- l = 820 / 20 = 41 cm
- Perimeter = 2(41 + 20) = 2 x 61 = 122 cm
Q12. A square park has a perimeter of 40 m. Find its area.
Step-by-step:
- Side = 40 / 4 = 10 m
- Area = 10 x 10 = 100 m squared
Q13. A scalene triangle has a perimeter of 40 cm. Two of its sides are 13 cm and 15 cm. Find the third side.
Step-by-step:
- Third side = 40 - 13 - 15 = 12 cm
Q14. The base and height of a right triangular field are 25 m and 20 m. Find the cost of levelling at Rs. 45 per m squared.
Step-by-step:
- Area = (1/2) x 25 x 20 = 250 m squared
- Cost = 250 x 45 = Rs. 11,250
Q15. A square of side 2 cm is joined with a rectangle of 15 cm x 10 cm. Find the perimeter of the combined figure.
Step-by-step:
- Perimeter of rectangle = 2(15 + 10) = 50 cm
- Perimeter of square = 4 x 2 = 8 cm
- Common side = 2 cm (appears twice, once from each figure)
- Combined perimeter = 50 + 8 - 2(2) = 50 + 8 - 4 = 54 cm
Exercise 3.1 -- Objective Type Questions (Q16-Q20)
Q16. Which figure has the least perimeter among all shapes with equal area?
Q17. Two identical rectangles (perimeter 30 cm each) are joined together. The perimeter of the combined figure is:
Q18. If every side of a rectangle is doubled, the area becomes:
Step-by-step:
- Original area = l x b
- New length = 2l, new breadth = 2b
- New area = 2l x 2b = 4lb = 4 times the original area
Q19. A square has side 10 cm. If the side is tripled, the perimeter increases by:
Step-by-step:
- Original perimeter = 4 x 10 = 40 cm
- New side = 3 x 10 = 30 cm
- New perimeter = 4 x 30 = 120 cm
- Increase factor = 120 / 40 = 3 times
Q20. A rectangle (15 cm x 12 cm) has a piece cut from one corner. What happens to perimeter and area?
Exercise 3.2 -- Word Problems
Q1. A wire 36 cm long is bent to form:
(i) A square:
- Side = 36 / 4 = 9 cm
(ii) An equilateral triangle:
- Side = 36 / 3 = 12 cm
Q2. An equilateral triangle with side 40 cm has a small equilateral triangle (side 6 cm) removed from one vertex. Find the remaining perimeter.
Step-by-step:
- When the small triangle is removed, two new sides of 6 cm each appear, replacing 6 cm of the original side on each of the two adjacent sides.
- Remaining perimeter = (40 - 6) + (40 - 6) + 40 + 6 = 34 + 34 + 40 + 6 = 114 cm
Alternative reasoning: Original perimeter = 3 x 40 = 120. Remove 6 from two sides but add 6 + 6 from the cut. Perimeter = 120 - 6 - 6 + 6 + 6 = 120. Wait -- actually the small triangle's 3 sides of 6 cm replace 6 cm on each of two sides. Perimeter = (40 - 6) + (40 - 6) + 40 + 6 = 114 cm.
Q3. Rahim runs around a square field (side 50 m) and Peter runs around a rectangular field (40 m x 30 m), both taking 2 rounds. Who runs more and by how much?
Step-by-step:
- Rahim: Perimeter of square = 4 x 50 = 200 m. For 2 rounds = 2 x 200 = 400 m.
- Peter: Perimeter of rectangle = 2(40 + 30) = 140 m. For 2 rounds = 2 x 140 = 280 m.
- Difference = 400 - 280 = 120 m
Q4. The length of a rectangle exceeds its breadth by 14 m. The perimeter is 200 m. Find the length, breadth, and area.
Step-by-step:
- Let breadth = b, then length = b + 14
- 2(l + b) = 200, so l + b = 100
- (b + 14) + b = 100
- 2b + 14 = 100
- 2b = 86
- b = 43 m
- l = 43 + 14 = 57 m
- Area = 57 x 43 = 2,451 m squared
Q5. A square garden of side 5 m has fencing with 2 rows of wire at Rs. 10 per metre. Find the total cost.
Step-by-step:
- Perimeter = 4 x 5 = 20 m
- Total wire needed for 2 rows = 2 x 20 = 40 m
- Cost = 40 x 10 = Rs. 400
Q6. A closed shape has 20 equal sides, each 3 cm. Find the perimeter.
Step-by-step:
- Perimeter = 20 x 3 = 60 cm
Q7. A rectangle (40 cm x 20 cm) is divided into squares of side 10 cm. How many squares are obtained?
Step-by-step:
- Along the length: 40 / 10 = 4 squares
- Along the breadth: 20 / 10 = 2 squares
- Total squares = 4 x 2 = 8 squares
Q8. The length of a rectangle is 3 times its breadth. The perimeter is 64 cm. Find the dimensions.
Step-by-step:
- Let breadth = b, then length = 3b
- 2(3b + b) = 64
- 8b = 64
- b = 8 cm
- l = 3 x 8 = 24 cm
Q9. How many different rectangles (with integer sides, l >= b) can be made from a 48 cm string?
Step-by-step:
- 2(l + b) = 48, so l + b = 24
- Pairs (l, b) with l >= b: (23,1), (22,2), (21,3), (20,4), (19,5), (18,6), (17,7), (16,8), (15,9), (14,10), (13,11), (12,12)
- 11 different rectangles (12 if the square is counted separately)
Q10. Square B has a side that is twice the side of square A. Compare their perimeters.
Step-by-step:
- Let side of A = a, side of B = 2a
- Perimeter of A = 4a
- Perimeter of B = 4 x 2a = 8a = twice the perimeter of A
Q11. If the side of a square is made one-fourth of the original, what happens to its area?
Step-by-step:
- Original area = a squared
- New side = a/4
- New area = (a/4) squared = a squared / 16
- New area = 1/16 of the original area
Q12. A square (side 10 m) and a rectangle (breadth 8 m) have the same perimeter. Which has the greater area and by how much?
Step-by-step:
- Square perimeter = 4 x 10 = 40 m
- For rectangle: 2(l + 8) = 40, so l + 8 = 20, l = 12 m
- Square area = 10 x 10 = 100 m squared
- Rectangle area = 12 x 8 = 96 m squared
- Difference = 100 - 96 = 4 m squared
Common Mistakes to Avoid
1. Confusing perimeter and area formulas: Perimeter is the total boundary length (measured in cm, m, etc.). Area is the space inside (measured in cm squared, m squared, etc.). Never mix up P = 2(l + b) with A = l x b.
2. Forgetting units: Perimeter is in linear units (cm, m, km), while area is in square units (cm squared, m squared, km squared). Always write the correct unit.
3. Wrong area formula for triangle: Area of a triangle = (1/2) x base x height, NOT base x height. Forgetting the 1/2 is a very common error.
4. Using wrong sides for right triangle area: In a right triangle with sides 6, 8, 10, the base and height are the two perpendicular sides (6 and 8), NOT the hypotenuse (10).
5. Perimeter of combined figures: When two shapes are joined, the common side is NOT part of the external perimeter. Always subtract the overlapping sides.
6. Unit conversion errors: 1 m squared = 10,000 cm squared (NOT 100 cm squared). Remember to multiply by 100 twice when converting between m squared and cm squared.
7. Area scaling with side changes: If the side of a square doubles, the area becomes 4 times (not 2 times). Area scales with the square of the side ratio.
Exam Checklist
- [ ] Can you calculate the perimeter and area of a rectangle (P = 2(l+b), A = l x b)?
- [ ] Can you calculate the perimeter and area of a square (P = 4 x side, A = side squared)?
- [ ] Can you find the area of a right-angled triangle (A = 1/2 x base x height)?
- [ ] Can you find the perimeter of any triangle (sum of all three sides)?
- [ ] Can you convert between units of area (cm squared to mm squared, m squared to cm squared, km squared to m squared)?
- [ ] Can you solve word problems involving perimeter and area of combined figures?
- [ ] Do you know that a square has the least perimeter among shapes of equal area?
- [ ] Can you handle "find missing dimension" problems (given area, find length or breadth)?
- [ ] Can you determine the effect on area/perimeter when dimensions are scaled?
- [ ] Can you calculate the number of smaller shapes fitting inside a larger shape?