Class 6 – Ganita Prakash – SOLUTIONS
Chapter 6: Perimeter and Area
6.1 Perimeter – Figure it Out (Page 132)
Question 1: Find the missing terms.
(a) Perimeter of a rectangle = 14 cm; breadth = 2 cm; length = ?
Answer: 5 cm
Explanation:
Perimeter of a rectangle = 2 × (Length + Breadth)
14 = 2 × (Length + 2)
14 ÷ 2 = Length + 2
7 = Length + 2
Length = 7 - 2 = 5 cm
(b) Perimeter of a square = 20 cm; length of a side = ?
Answer: 5 cm
Explanation:
Perimeter of a square = 4 × Side
20 = 4 × Side
Side = 20 ÷ 4 = 5 cm
(c) Perimeter of a rectangle = 12 m; length = 3 m; breadth = ?
Answer: 3 m
Explanation:
Perimeter of a rectangle = 2 × (Length + Breadth)
12 = 2 × (3 + Breadth)
12 ÷ 2 = 3 + Breadth
6 = 3 + Breadth
Breadth = 6 - 3 = 3 m
Question 2: A rectangle having side lengths 5 cm and 3 cm is made using a piece of wire. If the wire is straightened and then bent to form a square, what will be the length of a side of the square?
Answer: 4 cm
Explanation:
The length of the wire is the perimeter of the rectangle.
Perimeter of rectangle = 2 × (5 cm + 3 cm) = 2 × 8 cm = 16 cm.
This 16 cm is now the perimeter of the square.
Side of the square = Perimeter ÷ 4 = 16 cm ÷ 4 = 4 cm.
Question 3: Find the length of the third side of a triangle with a perimeter of 55 cm and two sides of length 20 cm and 14 cm, respectively.
Answer: 21 cm
Explanation:
Perimeter of a triangle = Sum of all three sides.
55 cm = 20 cm + 14 cm + Third Side
Third Side = 55 cm - 20 cm - 14 cm = 21 cm.
Question 4: What would be the cost of fencing a rectangular park whose length is 150 m and breadth is 120 m if the fence costs ₹ 40 per meter?
Answer: ₹ 21,600
Explanation:
Perimeter of the park = 2 × (Length + Breadth) = 2 × (150 m + 120 m) = 2 × 270 m = 540 m.
Cost of fencing = Perimeter × Cost per meter = 540 m × ₹ 40/m = ₹ 21,600.
Question 5: A piece of string is 36 cm long. What will be the length of each side, if it is used to form:
(a) A square
Answer: 9 cm
Explanation: Side = Perimeter ÷ 4 = 36 cm ÷ 4 = 9 cm.
(b) A triangle with all sides of equal length
Answer: 12 cm
Explanation: Side = Perimeter ÷ 3 = 36 cm ÷ 3 = 12 cm.
(c) A hexagon with sides of equal length
Answer: 6 cm
Explanation: Side = Perimeter ÷ 6 = 36 cm ÷ 6 = 6 cm.
Question 6: A farmer has a rectangular field having length 230 m and breadth 160 m. He wants to fence it with 3 rounds of rope. What is the total length of rope needed?
Answer: 2340 m
Explanation:
Perimeter of the field = 2 × (230 m + 160 m) = 2 × 390 m = 780 m.
Total rope for 3 rounds = 3 × Perimeter = 3 × 780 m = 2340 m.
6.1 Perimeter – Figure it Out (Pages 133-134)
Scenario: Akshi runs on an outer track (70 m by 40 m) and Toshi runs on an inner track (60 m by 30 m). Akshi completes 5 rounds. Toshi completes 7 rounds.
Question 1: Find out the total distance Akshi has covered in 5 rounds.
Answer: 1100 m
Explanation:
Perimeter of Akshi's track = 2 × (70 m + 40 m) = 220 m.
Total distance = 5 rounds × 220 m/round = 1100 m.
Question 2: Find out the total distance Toshi has covered in 7 rounds. Who ran a longer distance?
Answer: Toshi ran 1260 m. Toshi ran a longer distance.
Explanation:
Perimeter of Toshi's track = 2 × (60 m + 30 m) = 180 m.
Total distance = 7 rounds × 180 m/round = 1260 m.
Since 1260 m > 1100 m, Toshi ran a longer distance.
Question 3: Think and mark the positions...
(Conceptual answers based on calculations)
(a) Mark 'A' after Akshi runs 250 m: Akshi's track perimeter is 220 m. After 250 m, she has completed 1 round (220 m) and is 30 m into her second round.
(b) Mark 'B' after 500 m: 500 m is more than 2 rounds (440 m). She is 60 m into her third round (500 - 440 = 60).
(c) Mark 'C' after 1000 m: 1000 m ÷ 220 m/round ≈ 4.54 rounds. She has completed 4 full rounds (880 m) and is 120 m into her 5th round.
(d) Mark 'X' after Toshi runs 250 m: Toshi's track perimeter is 180 m. After 250 m, she has completed 1 round (180 m) and is 70 m into her second round.
(e) Mark 'Y' after 500 m: 500 m is more than 2 rounds (360 m). She is 140 m into her third round (500 - 360 = 140).
(f) Mark 'Z' after 1000 m: 1000 m ÷ 180 m/round ≈ 5.55 rounds. She has completed 5 full rounds (900 m) and is 100 m into her 6th round.
Split and Rejoin (Page 136)
A 6 cm × 4 cm paper is cut into two equal pieces (each 6 cm × 2 cm) and rearranged.
Question: Find the perimeter of the other arrangements.
Arrangement (b): 28 cm
Arrangement (c): 28 cm
Arrangement (d): 26 cm
Question: Arrange the two pieces to form a figure with a perimeter of 22 cm.
Answer: By arranging the pieces to form an irregular shape where the connecting sides create a less exposed boundary, a perimeter of 22 cm can be achieved. One way is to create a "stair-step" shape.
Explanation: The total perimeter changes based on how many sides are hidden when the pieces are joined. A more compact shape has a smaller perimeter than a more spread-out shape.
6.2 Area – Figure it Out (Page 138)
Question 1: The area of a rectangular garden 25 m long is 300 sq m. What is the width of the garden?
Answer: 12 m
Explanation:
Area of rectangle = Length × Width
300 sq m = 25 m × Width
Width = 300 sq m ÷ 25 m = 12 m.
Question 2: What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹ 8 per hundred sq m?
Answer: ₹ 8,000
Explanation:
Area of plot = 500 m × 200 m = 100,000 sq m.
Since cost is per hundred sq m, calculate the number of units: 100,000 ÷ 100 = 1,000 units.
Total cost = 1,000 units × ₹ 8/unit = ₹ 8,000.
Question 3: A rectangular coconut grove is 100 m long and 50 m wide. If each coconut tree requires 25 sq m, what is the maximum number of trees that can be planted in this grove?
Answer: 200 trees
Explanation:
Area of grove = 100 m × 50 m = 5000 sq m.
Number of trees = Total Area ÷ Area per tree = 5000 sq m ÷ 25 sq m/tree = 200 trees.
Question 4: By splitting the following figures into rectangles, find their areas.
(a)
Answer: 28 sq. cm
Explanation: The figure can be split into 4 rectangles.
I: 4 cm × 3 cm = 12 sq. cm
II: 3 cm × 2 cm = 6 sq. cm
III: 4 cm × 1 cm = 4 sq. cm
IV: 3 cm × 2 cm = 6 sq. cm
Total Area = 12 + 6 + 4 + 6 = 28 sq. cm.
(b)
Answer: 9 sq. cm
Explanation: The figure can be split into 3 squares/rectangles.
I: 3 cm × 1 cm = 3 sq. cm
II: 3 cm × 1 cm = 3 sq. cm
III: 3 cm × 1 cm = 3 sq. cm
Total Area = 3 + 3 + 3 = 9 sq. cm.
6.2 Area – Figure it Out (Page 139) – The Tangram Puzzle
Question 1: How many pieces have the same area?
Answer: Shapes A and B have the same area. Shapes C and E have the same area.
Question 2: How many times bigger is Shape D as compared to Shape C? What is the relationship between Shapes C, D and E?
Answer: Shape D is twice as big as Shape C. Shape D can be formed by combining Shape C and Shape E.
Question 3: Which shape has more area: Shape D or F?
Answer: They have the same area.
Explanation: Both are made from two small triangles (C or E).
Question 4: Which shape has more area: Shape F or G?
Answer: They have the same area.
Explanation: Both are made from two small triangles (C or E).
Question 5: What is the area of Shape A as compared to Shape G?
Answer: Shape A has twice the area of Shape G.
Explanation: Shape A is made of 4 small triangles, while Shape G is made of 2 small triangles.
Question 6: What is the area of the big square formed with all seven pieces in terms of the area of Shape C?
Answer: 16 times the area of Shape C.
Explanation: The entire tangram square is composed of 16 small triangles of the same size as Shape C.
Question 7: Arrange these 7 pieces to form a rectangle. What will be the area of this rectangle?
Answer: The area of the rectangle will be the same as the area of the big square, which is 16 times the area of Shape C.
Explanation: Rearranging the pieces into a different shape does not change the total area.
Question 8: Are the perimeters of the square and the rectangle formed from these 7 pieces different or the same?
Answer: They are different.
Explanation: For a given area, different shapes can have different perimeters. A square has the smallest possible perimeter for a given area. A rectangle that is long and thin will have a larger perimeter.
Let's Explore! (Page 141)
On squared paper, make rectangles of area 12 sq. units.
(a) Which rectangle has the greatest perimeter?
Answer: The 1 by 12 rectangle. Perimeter = 2(1+12) = 26 units.
(b) Which rectangle has the least perimeter?
Answer: The 3 by 4 rectangle (most square-like). Perimeter = 2(3+4) = 14 units.
(c) For a rectangle of area 32 sq cm, what will your answers be?
Greatest Perimeter: The 1 by 32 rectangle. Perimeter = 2(1+32) = 66 units.
Least Perimeter: The most square-like rectangle, which is 4 by 8 (since 4×8=32). Perimeter = 2(4+8) = 24 units.
General Rule: For a fixed area, the most elongated rectangle has the greatest perimeter, and the most square-like rectangle has the least perimeter.
6.3 Area of a Triangle – Figure it Out (Page 149)
Question 1: Give the dimensions of a rectangle whose area is the sum of the areas of these two rectangles having measurements: 5 m × 10 m and 2 m × 7 m.
Answer: Possible dimensions are 8 m × 8 m (or 16 m × 4 m, etc.).
Explanation:
Area of first rectangle = 5 × 10 = 50 sq m.
Area of second rectangle = 2 × 7 = 14 sq m.
Total Area = 50 + 14 = 64 sq m.
Any rectangle with length and breadth whose product is 64 is a valid answer (e.g., 8 m × 8 m, 16 m × 4 m, 32 m × 2 m).
Question 2: The area of a rectangular garden that is 50 m long is 1000 sq m. Find the width of the garden.
Answer: 20 m
Explanation:
Area = Length × Width
1000 sq m = 50 m × Width
Width = 1000 ÷ 50 = 20 m.
Question 3: The floor of a room is 5 m long and 4 m wide. A square carpet whose sides are 3 m in length is laid on the floor. Find the area that is not carpeted.
Answer: 11 sq m
Explanation:
Area of floor = 5 m × 4 m = 20 sq m.
Area of carpet = 3 m × 3 m = 9 sq m.
Area not carpeted = 20 - 9 = 11 sq m.
Question 4: Four flower beds having sides 2 m long and 1 m wide are dug at the four corners of a garden that is 15 m long and 12 m wide. How much area is now available for laying down a lawn?
Answer: 172 sq m
Explanation:
Area of garden = 15 m × 12 m = 180 sq m.
Area of one flower bed = 2 m × 1 m = 2 sq m.
Area of four flower beds = 4 × 2 sq m = 8 sq m.
Area for lawn = 180 - 8 = 172 sq m.
Question 5: Shape A has an area of 18 sq units and Shape B has an area of 20 sq units. Shape A has a longer perimeter than Shape B. Draw two such shapes.
Answer: Draw a very long and thin rectangle for Shape A (e.g., 1 unit by 18 units, Perimeter=38 units). Draw a more compact rectangle for Shape B (e.g., 4 units by 5 units, Perimeter=18 units). This shows that a smaller area can have a larger perimeter if its shape is more elongated.
Question 6: On a page in your book, draw a rectangular border that is 1 cm from the top and bottom and 1.5 cm from the left and right sides. What is the perimeter of the border?
Answer: The perimeter depends on the page size. For a standard A4 page (21 cm x 29.7 cm):
Border Length = 29.7 - 1 - 1 = 27.7 cm
Border Width = 21 - 1.5 - 1.5 = 18 cm
Perimeter of border = 2 × (27.7 + 18) = 2 × 45.7 = 91.4 cm.
Question 7: Draw a rectangle of size 12 units × 8 units. Draw another rectangle inside it, without touching the outer rectangle that occupies exactly half the area.
Answer: The outer rectangle's area is 12 × 8 = 96 sq units. Half of this is 48 sq units. Draw an inner rectangle with an area of 48 sq units (e.g., 8 units × 6 units) and place it centered inside the larger one so it doesn't touch the sides.
Question 8: A square piece of paper is folded in half and cut into two rectangles. Which statement is always true?
Which statement is true here?
(a) The area of each rectangle is larger than the area of the square.
(b) The perimeter of the square is greater than the perimeters of both the rectangles added together.
(c) The perimeters of both the rectangles added together are always 1 times the perimeter of the square.
(d) The area of the square is always three times as large as the areas of both rectangles added together.
Solution:
side of square = 1 unit
area of square = 1 × 1 = 1 sq. unit.
and perimeter of square = 1 + 1 + 1 + 1 = 4 units.
Perimeter of rectangle R1 = 1 + + 1 + = 3 units.
Area of rectangle R1 = × 1 = sq. unit.
Perimeter of rectangle R2 = 1 + + 1 + = 3 units.
Area of rectangle R2 = × 1 = sq. unit.
(a) area of rectangle R1 = area of rectangle R2 = < 1.
Hence, option (a) is not true
(b) perimeter of square = 4 units
and perimeters of both the rectangles = 3 + 3 = 6 units.
which is greater than 4 units.
Hence option (b) is not true.
(c)perimeters of both the rectangles = 6 units
and perimeter of square = 4 units × 1 = 6 The perimeters of both the rectangles added together are 1 times the perimeter of the square.
Hence, option (c) is true.
(d) Here, the area of the square = 4 units
and areas of both the rectangles = + = 1 unit.
The area of the square is four times the area of both rectangles.
Hence, option (d) is not true.
Answer: (c) The perimeters of both the rectangles added together are always 1½ times the perimeter of the square.
Explanation:
Let the square's side be 's'. Its perimeter is 4s.
Each rectangle has sides 's' and 's/2'. Perimeter of one rectangle = 2(s + s/2) = 3s.
Perimeter of both rectangles = 2 × 3s = 6s.
Ratio = 6s / 4s = 6/4 = 1.5 or 1½.
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