Area Problems
The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18, find its base and height.
Area of the field = Total cost/rate = (333.18/25.6) = 13.5 hectares 13 . 5 * 10000 m 2 = 135000 m 2 Let altitude = x metres and base = 3x metres.Then, 1 2 * 3 x * x = 135000 ? x 2 = 90000 ? x = 300 Base = 900 m and Altitude = 300 m.
Area of the field =1215/135 = 9 hec= 90000 m2 [1 hec =10000 m2]? Side of the field = ?90000 = 300 mPerimeter of the field = 4 x 300 = 1200 mNow, cost of putting a fence around field = (1200 x 75)/100 = ? 900
Let circumference = 100 cm . Then, ? 2?r = 100? r = 100/2? =50/?? New circumference = 105 cm Then, 2?R = 105? R = 105 / (2?)&rArr Original area = [ ? x (50/?) x (50/?) ] = 2500/? cm2? New Area = [? x (105/2?) x (105/2?)]= 11025 / (4?) cm2? Increase in area = [11025/(4?)] - 2500/? cm2= 1025 / 4? cm2Required increase percent [1025/(4?)] x 2500/? x 100 = 41/4%= 10.25%
In a triangleSum of two sides is always greater than 3rd side i.e., x < 25 + 15 = 40 .....(i)Difference of two sides is always less than 3rd side i.e., 25 - 15 = 10 < x ...(ii) From Eqs. (i) and (ii) , we get 10 < x < 40
Distance covered in one revolution = total distance travelled / total number of revolution. = ( 88 x 1000) / 1000 m = 88 mWe know that the distance covered in one revolution = circumference of the wheel.? ?d = 88? 22d / 7 = 88? d = 28 m
EXAMIANS