Calculate areas and perimeters of rectangles, squares and tiles
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Math - Class V Area (IGCSE) International Board
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Calculate areas and perimeters of rectangles, squares and tiles
- 1. A B
- 5. Length Length Breadth Breadth Area of Rectangle = Length X Breadth Perimeter of Rectangle = 2 x (Length + Breadth) If Length of rectangle = 4 cm Breadth of rectangle = 2 cm Area of rectangle = 4 cm X 2 cm = 8 Sq. cm Perimeter of rectangle = 2 x (4 cm + 2 cm) = 2 x 6 = 12 cm
- 6. Length Length Breadth Breadth Area of Square = Length X Breadth = (πΊππ π)π Perimeter of Square = 2 x (Length + Breadth) (Length = Breadth) Length of square = 4 Breadth of square = 4 Area of square = 4 X 4 = 16 Perimeter of square = 2 x (4+4) = 2 x 8 = 16
- 7. 40 m 20 m
- 8. Length Length Breadth Breadth Area of Rectangle = Length X Breadth Perimeter of Rectangle = 2 x (Length + Breadth) If Length of rectangle = 40 m Breadth of rectangle = 20 m Area of rectangle = 40 m X 20 m = 800 Sq. m Perimeter of rectangle = 2 x (40 m + 20 m) = 2 x 60 = 120 m
- 9. A B D C
- 10. Findings: 1. AB= 7 boxes 2. BC= 8 boxes 3. Perimeter = 8 + 7 = 15 A D C B
- 11. Total No. of Boxes= (No. of Box in Length) X (No. of Box in breadth) Findings: 1. AB= 7 boxes 2. BC= 8 boxes 3. Area of shape = 8 x 7 = 56 4. 56 boxes A D C B
- 15. 6 squares along the vertical Length = 6 Squares 10 squares along the horizontal Breadth = 10 Squares Total Squares = (6 X 10) Squares = 60 Squares
- 16. 1.5 cm 1.5 cm
- 17. Area of square = Side X Side 1. 5 cm 1. 5 cm Area of square = 1.5 X 1.5 Area of square = 2.25 sq. cm
- 19. Number of stamps will required to fill this post card = 7 X 4 = 28 stamps Number of stamps will required to fill this post card = 28 So, we have 28 squares, each of area 2.25 sq. cm So, area of postcard = 28 x 2.25 = 63 sq. cm
- 20. β’ Saniya plans to tile his courtyard floor with white tiles. Each side of one tile is 30 cm. the courtyard 240 cm in length and 150 cm in breadth. How many tiles will be required to cover the entire courtyard. 30 cm
- 21. 30 cm Area of the tile = (πΊππ π)π = 30 X 30 = 900 Sq. cm
- 22. Area of the courtyard The courtyard is the rectangular in shape length = 240 cm and breadth = 150 cm Area of rectangle = Length x Breadth Area of courtyard = 240 x 150 = 36,000 sq. cm
- 23. Let the number of tiles required to cover the courtyard , Then N x (area of tile) = (area of courtyard) Or number of tiles = Area of courtyard / area of tile = 36000 sq. cm / 900 cm = 40 Hence, the required number of tiles is 40
- 24. MEMORY MATCH W L L L 2 cm 2 cm 4cm 4 sq. cm 32 sq. cm 8 cm L X L L X W L X W L X L 32 sq. cm 4 sq. cm
- 25. QUIZ 1. Area of rectangle is _________ a) Length x Breadth b) Square units c) Square centimetres d) Area Ans. a) Length x Breadth 2. The unit used to express area is ______ a) Square units b) Length x Breadth c) Area d) Square kilometres Ans. a) Square units
- 26. QUIZ 3. The area of a book can be calculated in _________ a) Square units b) Square centimetres c) Area d) Square kilometres Ans. b) Square centimetres 4. The area of a city can be calculated in ______ a) Square units b) Area c) Square kilometres d) Square centimetres Ans. c) Square kilometres
- 27. QUIZ 5. ______ is the amount of surface enclosed within the boundary of a figure. a) Square units b) Area c) Length x Breadth d) Square kilometres Ans. b) Area 6. A rectangle has an area of 200m. The length of the rectangle is 20 m. Find the perimeter of the rectangle and choose the right answer. a) 60 m b) 59 m c) 58 m d) 57 m Ans. a) 60 m
- 29. Its your turn