AREA AND PERIMETER (CDS)

Total Questions: 116

31. There is a path of width 5 m around a circular plot of land whose area is 144π m². The total area of the circular plot including the path surrounding it is [2018 (II) Evening Shift]

Correct Answer: (b) 289π m²
Solution:Area of circular, Plot of land = 144π m²

32. An equilateral triangle, a square and a circle have equal perimeter. If T S, and C denote the area of the triangle, area of the square and area of the circle respectively, then which one of the following is correct? [2018 (II) Evening Shift]

Correct Answer: (a) T < S < C
Solution:An equilateral triangle, a square and a circle of perimeter are equal


33. The areas of two similar triangles are (7 - 4√3) cm² and (7 + 4√3) cm² respectively. The ratio of their corresponding sides is [2018 (II) Evening Shift]

Correct Answer: (a) 7 - 4√3
Solution:Ratio of area of two similar triangle = Ratio of square of their corresponding sides

34. The area of the region bounded internally by a square of side 2a cm and externally by the circle touching the four sides of the square is

Correct Answer: (a)
Solution:Side of square ABCD is 2a cm

35. In the figure given below, ABC is a right-angled triangle where ∠A = 90º, AB = p cm and AC = q cm. On the three sides as diameters semi-circles are drawn as shown in the figure. The area of the shaded portion, in square cm, is [2018 (II) Evening Shift]

Correct Answer: (d)
Solution:

36. In the figure given below, ABCD is the diameter of a circle of radius 9 cm. The lengths AB, BC and CD are equal. Semi-circles are drawn on AB and BD as diameters as shown in the figure. What is the area of the shaded region? [2018 (II) Evening Shift]

Correct Answer: (b) 27 π
Solution:Area of shaded regions = Area of semi-circle on AD + area of semicircle on AB – area of semi-circle on BD

37. In the figure given below, the diameter of bigger semi-circle is 108 cm. What is the area of the shaded region? [2018 (II) Evening Shift]

Correct Answer: (c) 405 π cm²
Solution:Given, AO = OB = OD = 54cm


38. In the figure given below, ABCD is a square of side 4 cm. Quadrants of a circle of diameter 2 cm are removed form the four corners and a circle of diameter 2 cm is also removed. What is the area of the shaded region? [2018 (II) Evening Shift]

Correct Answer: (c) 9⁵⁄₇
Solution:Area of shaded region = Area of square – 4 (area of quadrant)– area of circle

 

39. Walls (excluding roofs and floors) of 5 identical rooms having length, breadth and height 6 m, 4 m and 2.5 m respectively are to be painted. Out of five rooms, two rooms have one square window each having a side of 2.5 m. Paints are availbable only in cans of 1 L; and 1 L of paint can be used for painting 20 m² . The number of cans required for painting is [2018 (II) Evening Shift]

Correct Answer: (b) 12
Solution:Length (l), breadth (b) and height (h) of walls are 6 m, 4m and 2.5 m respectively,

40. Two equal circular regions of greatest possible area are cut off from a given circular sheet of area A. what is the remaining area of the sheet? [2018 (I) Morning Shift]

Correct Answer: (a) A/2
Solution:Let the radius of circular sheet be ‘R’