A path of uniform width runs round inside the rectangular field of 38

1.	A path of uniform width runs round inside the rectangular field of 38 m long and 32 m wide. if the path occupies 600 m2, then the width of the path is 1. 30 m2. 5 m3. 10 m4. 15 m
Answer» Correct Answer - Option 2 : 5 m Given: A path of uniform width runs round inside the rectangular field of 38 m long and 32 m wide. if the path occupies 600 m². Concept used: Area of rectangle = Length × Breadth Calculation: 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 ⇒ 4x² - 140x + 1216 Area of path ⇒ 1216 - (4x² - 140x + 1216) ⇒ 140x - 4x² The path of the rectangular field occupies is 600 m² ⇒ 140x - 4x² = 600 ⇒ 4x² - 140x + 600 ⇒ x² - 35x + 150 ⇒ x² - 30x - 5x + 150 ⇒ x(x - 30) - 5(x - 30) ⇒ (x - 5) (x - 30) ∴ x = 5 because \(x \ne 30\) ∴ x = 5m is the width of the path.

A path of uniform width runs round inside the rectangular field of 38 m long and 32 m wide. if the path occupies 600 m2, then the width of the path is 1. 30 m2. 5 m3. 10 m4. 15 m

Answer» Correct Answer - Option 2 : 5 m

Given:

A path of uniform width runs round inside the rectangular field of 38 m long and 32 m wide. if the path occupies 600 m².

Concept used:

Area of rectangle = Length × Breadth

Calculation:

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

⇒ 4x² - 140x + 1216

Area of path

⇒ 1216 - (4x² - 140x + 1216)

⇒ 140x - 4x²

The path of the rectangular field occupies is 600 m²

⇒ 140x - 4x² = 600

⇒ 4x² - 140x + 600

⇒ x² - 35x + 150

⇒ x² - 30x - 5x + 150

⇒ x(x - 30) - 5(x - 30)

⇒ (x - 5) (x - 30)

∴ x = 5 because \(x \ne 30\)

∴ x = 5m is the width of the path.

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