Area Problems An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is: 2% 2.02% 4.04% 4% 2% 2.02% 4.04% 4% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP 100 cm is read as 102 cm. ∴ A1 = (100 x 100) cm2 and A2 (102 x 102) cm2. (A2 - A1) = [(102)2 - (100)2] = (102 + 100) x (102 - 100) = 404 cm2. ∴ Percentage error = ❨ 404 x 100 ❩% = 4.04%
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? 50 cm 20 cm 40 cm 30 cm 50 cm 20 cm 40 cm 30 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 the sides of a rectangle are increased by 10% find the percentage increased in its diagonals. 15% 18% 20% 10% 15% 18% 20% 10% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP According to the formula,? Percentage increase in sides = 10%? Percentage increase diagonals = 10%
Area Problems The area of a square is equal to the area of a rectangle. The length of the rectangle is 5 cm more than a side of the square and its breadth is 3 cm less than the side of the square. What is the perimeter of the rectangle ? 34 cm 26 cm 18 cm 15 cm 34 cm 26 cm 18 cm 15 cm ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Let the side of the square be 's' cm length of rectangle = (s+5) cm breadth of rectangle = (s-3)cm (s+5) (s-3) = s 2 - 5s - 3s - 15 = s 2 2s = 15 Perimeter of rectangle = 2(L+B) = 2(s+5 + s?3) = 2(2s + 2) = 2(15 + 2) = 34 cm
Area Problems The circumferences of two concentric are 176 m and 132 m respectively. what is the difference between their radii? 7 meter 5 meter 8 meter 44 meter 7 meter 5 meter 8 meter 44 meter ANSWER EXPLANATION DOWNLOAD EXAMIANS APP ? 2?R - 2?r = (176-132)? 2?(R-r) = 44? R-r = ( 44 x 7 )/ (2 x 22)= 7 m
Area Problems A wheel make 4000 revolution in moving a distance of 44 km. Find the radius of the wheel. 28 m 1.75 m 27 m 25 m 28 m 1.75 m 27 m 25 m ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Distance covered in 1 revolution = (44 x 1000) / 4000 = 11 mAccording to the question, 2?r = 11 ? (44/7) x r = 11? r = (11 x 7) / 44 = 1.75 m
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