A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. The area of the lawn is 2109 sq. m. what is the width of the road?
Options:
A .  5 m
B .  4 m
C .  2 m
D .  3 m
Answer: Option D
Please refer the diagram given above. Area of the park = 60 — 40 = 2400 m2 Given that area of the lawn = 2109 m2 ˆ´ Area of the cross roads = 2400 - 2109 = 291 m2 Assume that the width of the cross roads = x Then total area of the cross roads = Area of road 1 + area of road 2 - (Common Area of the cross roads) = 60x + 40x - x2 (Let's look in detail how we got the total area of the cross roads as 60x + 40x - x2 As shown in the diagram, area of the road 1 = 60x. This has the areas of the parts 1,2 and 3 given in the diagram Area of the road 2 = 40x. This has the parts 4, 5 and 6 You can see that there is an area which is intersecting (i.e. part 2 and part 5) and the intersection area = x2. Since 60x + 40x covers the intersecting area (x2) two times ( part 2 and part 5) ,we need to subtract the intersecting area of (x2) once time to get the total area. . Hence total area of the cross roads = 60x + 40x - x2) Now, we have Total areas of cross roads = 60x + 40x - x2 But area of the cross roads = 291 m2 Hence 60x + 40x - x2 = 291 => 100x - x2 = 291 => x2 - 100x + 291 = 0 => (x - 97)(x - 3) = 0 => x = 3 (x can not be 97 as the park is only 60 m long and 40 m wide)
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