ANSWER KEY
Multiple Choice questions:
1. (c) 2 × (Length + Breadth)
2. (a) 4 × Length of a side
3. (b) 3 × Length of a side
4. (a) Length × Breadth
5. (a) side × side
6. (d) 5 × Length of a side.
7. (d) 6 × Length of a side.
8. (b) 60 m
Hint:
Distance covered = 2(20 + 10) = 60 m
9. (b) 14m
Hint:
Perimeter = 2(4 + 3) = 14m
10. (a) 8m
Hint:
Perimeter = 4 × 2 = 8m
11. (a) 90 cm
Hint:
Perimeter = 2(25 + 20) = 90 cm
12. (d) 20 cm.
Hint:
Perimeter = 4(1 + 2 + 2) = 20cm
13. (a) 20 cm
Hint:
Perimeter = 4 + 3 + 2 + 2 + 3 + 6 = 20 cm
14. (d) 6m.
Hint:
Length of the lace = 2(2 + 1) = 6 m
15. (b) 20 cm
Hint:
Perimeter = 2(9 + 1) = 20 cm
Match The Following:
Fill in the blanks:
1. Squareis a rectangle whose all sides are equal.
2. The amount of surface enclosed by a figure is called its area.
3. For fencing the plot, we need to calculate its Perimeter.
4. Perimeter is the sum of all sides.
True /False:
1. True
2. True
3. True
4. True
Very Short Answer:
1. Area of the rectangle = Length × Breadth = 3 × 4 cm = 12 sq cm
2. Area of the rectangle = Length × Breadth = 12 m × 21 m = 252 sq m
3. (i) Perimeter = 3 cm + 3 cm + 3 cm + 3 cm + 3 cm
= 15 cm
(ii) Perimeter = 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm
= 15 cm
(iii) Perimeter = 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm
+ 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm
= 52 cm
4. Perimeter
= AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA
= 1 cm + 3 cm + 3 cm + 1 cm + 3 cm + 3 cm + 1 cm + 3 cm + 3 cm + 1 cm + 3 cm + 3 cm
= 28 cm
5. Perimeter
= AB + BC + CD + DA
= 5 cm + 5 cm + 5 cm + 5 cm = 20
6. Full-filled squares = 2
half-filled squares = 4
Area covered by full squares = 2 × 1 sq unit = 2 sq units
Area covered by half squares = 4 × 12
12 sq unit = 2 sq units
∴ Total Area = 2 sq units + 2 sq units = 4 sq units
7. Let the breadth of rectangle = b
length of rectangle = 3b
Perimeter of rectangle =2 × (3b + b)
48 = 2 × (3b + b)
48/2=4b
24 = 4b
24/4=b
⇒ 6 = b ⇒ breadth = 6cm
length = 3b = 3 × b = 18cm
8. Perimeter of the square = 64 cm
9. Length of the rectangular table-top = 36 cm
and its breadth = 24 cm.
10. ∴ Perimeter of the table-top = 2 [length + breadth]
= 2 [36 cm + 24 cm]
= 2 × 60 cm = 120 cm.
Fig. (i) Perimeter of the square = 4 × side
= 4 × 4 cm = 16 cm
Fig. (ii) Perimeter of the rectangle
= 2 [length + breadth]
= 2[8 cm + 3 cm]
= 2 × 11 cm = 22 cm
Since 22 cm > 16 cm
∴ Rectangle has greater perimeter than the square.
Short Answer:
1. Distance travelled in going around Fig. (i)
= 12 cm + 3 cm + 12 cm + 3 cm = 30 cm
Distance travelled in going around Fig. (ii)
= 6 cm + 4 cm + 4 cm + 4 cm = 18 cm
2. Side of the square = 15 cm
∴ Perimeter of the square = 15 cm × 4 = 60 cm
3. Length of the park = 300 m
Breadth = 200 m
∴ Perimeter of the park = 2 [length + breadth]
= 2 [300 m + 200 m]
= 2 × 500 m = 1000 m.
Cost of fencing the rectangular park = 1000 × 4 = ₹ 4000
4. Side of the square field = 150 m
∴ Area of the square field = Side × Side
= 150 m × 150 m
= 22500 sq m.
5. Length of the rectangular paper = 22 cm
Breadth = 10 cm
∴ Area of the rectangular paper = length × breadth
= 22 cm × 10 cm
= 220 sq cm
6. Perimeter of the rectangle = 2 [length + breadth]
∴ 2 [length + breadth] = 880
length + breadth = 880 ÷ 2 = 440
∵ Breadth = 88 m
∴ Length = 440 m – 88 m = 352 m
Hence, the required length = 352 m.
7. Length of the rectangular plot = 120 m
Breadth = 90 m
∴ Perimeter of the rectangular plot
= 2 [length + breadth]
= 2 [120 m + 90 m]
= 2 × 210 m = 420 m
Now distance between two trees = 6 m
∴ Number of trees around the rectangular plot = 420 m ÷ 6 m = 70
Long Answer:
1. Length of the rectangular park = 30 m
Breadth = 20 m
∴ Perimeter of the rectangular park = 2(length + breadth)
= 2 [30 + 20] = 2 × 50 m = 100 m
∴ Cost of fencing all around the park = ₹15 × 100 = ₹1500
2. (A) Counting the squares, we have 8 squares
∴ Area = 8 sq units
(B) Counting the squares, we have 4 squares
∴ Area = 4 sq units
(C) Counting the squares, we have 5 squares
∴ Area = 5 sq units
(D) Counting the squares, we have 7 squares
∴ Area = 7 sq units
3. Perimeter of the square = 100 cm
Perimeter 100
25 cm.
∴ Breadth of the rectangle = 25 cm – 2 cm = 23 cm
Now perimeter of the rectangle = 100 cm
∴ 2 [length + breadth] = 100
length + breadth = 100 ÷ 2 = 50 cm
But breadth = 23 cm
∴ Length = 50 cm – 23 cm = 27 cm
Now, Area of the rectangle
= length × breadth = 27 cm × 23 cm
= 621 sq cm.
4. Cost of fencing the compound = ₹5452
nd the rate of fencing = ₹94 per metre
∴ Perimeter of the compound = 5452 ÷ 94 = 58 metres
Now breadth of the compound = 10 m.
2 [length + breadth] = 58 m
∴ length + breadth = 58 + 2 m = 29 m
∴ Length of the compound = 29 m – 10 m = 19 m.
Assertion and Reason Answers:
(1)(a) A is false but R is true
(2)(b) A is false but R is true