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Question

(i) A footpath of uniform width runs all around the inside of a rectangular field 54 m long and 35 m wide. If the area of the path is 420 m2., find the width of the path.
(ii) A carpet is laid on the floor of a room 8 m by 5 m. There is a border of constant width all around the carpet. If the area of the border is 12 m2, find its width.

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Solution

(i) Area of the rectangular field = 54×35=1890 m2

Let the width of the path be x m. The path is shown in the following diagram:


Length of the park excluding the path = (54 - 2x) m
Breadth of the park excluding the path = (35 - 2x ) m

Thus, we have:
Area of the path = 420 m2
420=54×35- (54-2x)(35-2x) 420=1890-1890-70x-108x+4x2420=-4x2+178x4x2-178x+420=02x2-89x+210=02x2-84x-5x+210=02x(x-42)-5(x-42)=0(x-42)(2x-5) = 0x-42=0 or 2x-5=0 x=42 or x=2.5

The width of the path cannot be more than the breadth of the rectangular field.
∴ x = 2.5 m

​Thus, the path is 2.5 m wide.

(ii) Let the width of the border be x m.
The length and breadth of the carpet are 8 m and 5 m, respectively.
Area of the carpet = 8×5=40 m2
Length of the carpet without border = (8-2x)
Breadth of carpet without border = (5-2x)
Area of the border = 12 m2
Area of the carpet without border = (8-2x)(5-2x)
Thus, we have:12=40-(8-2x)(5-2x)12=40-(40-26x+4x2)12=26x-4x2

26x-4x2=12 4x2-26x+12=0 2x2-13x+6=0

(2x-1)(x-6)=02x-1=0 and x-6=0x=12 and x=6

Because the border cannot be wider than the entire carpet, the width of the carpet is 12 m, i.e., 50 cm.

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