Area Problems A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required? 88 68 34 40 88 68 34 40 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP We have: l = 20 ft and lb = 680 sq. ft. So, b = 34 ft. ∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft
Area Problems If a regular hexagon is inscribed in a circle of radius, r then its perimeter? 6r 12r 9r 3r 6r 12r 9r 3r ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Length of each side of hexagoan = radius of circle ? its perimeter = 6r
Area Problems The length of a rectangular plot is twice of its width. If the length of a diagonal is 9?5 meters, the perimeter of the rectangular is? 54 m 27 m 81 m None of these 54 m 27 m 81 m None of these ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Let breadth = y meters,Then, length = 2y meters? Diagonal = ?y2 + (2y)2 = ?5y2 metersSo, ?5y2 = 9 ?5? y= 9Thus, breadth = 9 m and length = 18 m? Perimeter = 2 (18 + 9) m = 54m.
Area Problems The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and the breadth is increased by 5 cm, the area of the rectangle is increased by 75 cm 2 . Therefore , the length of the rectangle is? 40 cm 20 cm 30 cm 50 cm 40 cm 20 cm 30 cm 50 cm ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Let breadth = b, length = 2b? Area of rectangle = 2b x b = 2b2As per question. ? (2b - 5 ) (b + 5 ) = 2b2 + 75? 5b = 75 + 25? 5b = 100? b = 100 / 5 = 20Hence, length of the rectangle =2b = 2 x 20 = 40 cm.
Area Problems If 88 m wire is required to fence a circular plot of land, then the area of the plot is? 616 m2 None of these 526 m2 556 m2 616 m2 None of these 526 m2 556 m2 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Circumference of circular plot= 88 m? 2 x (22/7) x r = 88? r = 88 x (7/22) x (1/2) = 14 m Now area = ?r2=( 22/7) x 14 x 14 m2= 616 m2
Area Problems The Number of rounds that a wheel of diameter 7/11 m will make in going 4 km, is? 1700 1500 2000 1000 1700 1500 2000 1000 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Number of round = total distance / circumference of wheel.= (4 x 1000) / ?d= (4 x 1000) / ( 22/7 x 7/11) = 2000
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