Area Problems
A circular road runs rounds a circular ground. If the difference between the circumference of the outer circle and the inner circle is 66 meters, the width of the road is?
Let the parallel sides be 3a and 5a.So Area of trapezium = 1/2 x sum of parallel side x perpendicular distance between them.? 1/2 (3a +5a) x 12 = 384? 8a = 64? a =8? Smaller side = 3x = 3 x 8 = 24 cm.
Area of the room=(544 * 374) size of largest square tile= H.C.F of 544 & 374 = 34 cm Area of 1 tile = (34 x 34) c m 2 Number of tiles required== [(544 x 374) / (34 x 34)] = 176
Area of the park = (60 x 40) m2 = 2400 m2. Area of the lawn = 2109 m2. ∴ Area of the crossroads = (2400 - 2109) m2 = 291 m2. Let the width of the road be x metres. Then, 60x + 40x - x2 = 291 ⟹ x2 - 100x + 291 = 0 ⟹ (x - 97)(x - 3) = 0 ⟹ x = 3
EXAMIANS