Mensuration
- A street of width 10 metres surrounds from outside a rectangular garden whose measurement is 200 m × 180 m. The area of the path (in square metres) is
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Area of garden without street = 200 × 180 = 36000 sq.metre
Area of garden with street = 220 × 200 = 44000 sq.metre
∴ Area of the path = 44000 – 36000 = 8000 sq.metre
Aliter : Using Rule 3,
Here, L = 200 m, B = 180 m
w = 10 m,
Area of path = 2w [L + B + 2w] = 2 × 10 (200 + 180 + 2 × 10)
= 20 (400) = 8000 m²Correct Option: A
Area of garden without street = 200 × 180 = 36000 sq.metre
Area of garden with street = 220 × 200 = 44000 sq.metre
∴ Area of the path = 44000 – 36000 = 8000 sq.metre
Aliter : Using Rule 3,
Here, L = 200 m, B = 180 m
w = 10 m,
Area of path = 2w [L + B + 2w] = 2 × 10 (200 + 180 + 2 × 10)
= 20 (400) = 8000 m²
- The length of a room floor exceeds its breadth by 20 m. The area of the floor remains unaltered when the length is decreased by 10 m but the breadth is increased by 5 m. The area of the floor (in square metres) is :
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Let the breadth of floor be x metre.
∴ Length = (x + 20) metre
∴ Area of the floor = (x + 20) x sq. metre
According to question,
(x + 10) (x + 5) = x (x + 20)
⇒ x² + 15x + 50 = x² + 20x
20x = 15x + 50
⇒ 5x = 50
⇒ x = 10 metre
∴ Length = x + 20 = 10 + 20 = 30 metre
∴ Area of the floor = 30 × 10 = 300 sq.metreCorrect Option: C
Let the breadth of floor be x metre.
∴ Length = (x + 20) metre
∴ Area of the floor = (x + 20) x sq. metre
According to question,
(x + 10) (x + 5) = x (x + 20)
⇒ x² + 15x + 50 = x² + 20x
20x = 15x + 50
⇒ 5x = 50
⇒ x = 10 metre
∴ Length = x + 20 = 10 + 20 = 30 metre
∴ Area of the floor = 30 × 10 = 300 sq.metre
- The length and breadth of a rectangle are increased by 20% and 25% respectively. The increase in the area of the resulting rectangle will be :
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Using Rule 10,
Net Effect on area of rectangle = 
20 + 25 + 20 × 25 
% = 50% 100 
∵ Net % change = a + b + ab 
% 100 Correct Option: B
Using Rule 10,
Net Effect on area of rectangle = 20 + 25 + 20 × 25 % = 50% 100 ∵ Net % change = a + b + ab % 100
- A path of uniform width runs round the inside of a rectangular field 38 m long and 32 m wide. If the path occupies 600 m², then the width of the path is
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Let the width of path be x m.
Area of rectangular field = 38 × 32 = 1216 m²
Area of rectangular field without path = (38 – 2x) (32 – 2x)
= 1216 – 64x – 76x +4x²
= 1216 – 140x + 4x²
∴ Area of the path = 1216 – 1216 + 140x – 4x²
= 140x – 4x²
&there44; 140x – 4x² = 600
⇒ 4x² – 140x + 600 = 0
⇒ x² – 35x + 150 = 0
⇒ x² – 30x – 5x + 150 = 0
→ x (x– 30) –5 (x –30) = 0
→ (x– 5)(x –30) = 0
⇒ x = 5 as x ≠ 30
Aliter : Using Rule 3,
Here, L = 38 m,
B = 32 m w = ?,
Area of path = 600 m²
Area of path = 2w [L + B – 2w]
600 = 2w [38 + 32 – 2w]
300 = w (70 – 2w)
2w² – 70w + 300 = 0
w² – 35w + 150 = 0 (w – 30) (w – 5) = 0
⇒ Eitherw – 30 = 0, w = 30
But w ≠ 30
or, w – 5 = 0, w = 5
∴ w = 5 is the width of path.Correct Option: B
Let the width of path be x m.
Area of rectangular field = 38 × 32 = 1216 m²
Area of rectangular field without path = (38 – 2x) (32 – 2x)
= 1216 – 64x – 76x +4x²
= 1216 – 140x + 4x²
∴ Area of the path = 1216 – 1216 + 140x – 4x²
= 140x – 4x²
&there44; 140x – 4x² = 600
⇒ 4x² – 140x + 600 = 0
⇒ x² – 35x + 150 = 0
⇒ x² – 30x – 5x + 150 = 0
→ x (x– 30) –5 (x –30) = 0
→ (x– 5)(x –30) = 0
⇒ x = 5 as x ≠ 30
Aliter : Using Rule 3,
Here, L = 38 m,
B = 32 m w = ?,
Area of path = 600 m²
Area of path = 2w [L + B – 2w]
600 = 2w [38 + 32 – 2w]
300 = w (70 – 2w)
2w² – 70w + 300 = 0
w² – 35w + 150 = 0 (w – 30) (w – 5) = 0
⇒ Eitherw – 30 = 0, w = 30
But w ≠ 30
or, w – 5 = 0, w = 5
∴ w = 5 is the width of path.
- If the length and breadth of a rectangle are in the ratio 3 : 2 and its perimeter is 20 cm, then the area of the rectangle (in cm²) is :
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Using Rule 9,
Let the length and breadth of the rectangle be 3x and 2x cm respectively. Then, 2(3x + 2x) = 20
⇒ 10x = 20⇒ x = 20 = 2 10
∴ Length = 3x = 3 × 2 = 6 cm
Breadth = 2x = 2 × 2 = 4 cm
∴ Area = 6 × 4 = 24 cm²Correct Option: A
Using Rule 9,
Let the length and breadth of the rectangle be 3x and 2x cm respectively. Then, 2(3x + 2x) = 20
⇒ 10x = 20⇒ x = 20 = 2 10
∴ Length = 3x = 3 × 2 = 6 cm
Breadth = 2x = 2 × 2 = 4 cm
∴ Area = 6 × 4 = 24 cm²
