Circle

1. Area of a circle = \pi r ^{2}, where ‘r’ is the radius. If d stands for diameter then d =2r.           1. Circumference of a circle = 2 \pi r 3. Length of an arc = \frac{ ^{2\pi r}}{360} \theta where \theta is the central angle 4. Area of a sector \frac{1}{2}(arc * r) =  \frac{\pi r^{2}}{360}\theta 1. Circumference of a semi-circle =\pi r 2. Area of semi-circle = \frac{\pi r^{2}}{2}
1. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is: A. 15360 B. 153600 C. 30720 D. 307200 Answer: Option B Explanation: Perimeter = Distance covered in 8 min. = _{{[\frac{12000}{60}*8]m=1600m.}} Let length = 3x metres and breadth = 2x metres. Then, 2(3x + 2x) = 1600 or x = 160. Length = 480 m and Breadth = 320 m. Area = (480 x 320)  m^{2} = 153600  m^{2}.(Ans) 2. 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: A. 2% B. 2.02% C. 4% D. 4.04% Answer: Option D Explanation: 100 cm is read as 102 cm. A_{{1}}= (100 x 100)  cm^{2} and A_{{2}} (102 x 102)  cm^{2}.  (A_{{1}}-A_{{2}}) =  [(102) ^{2}-(102)^{2}] = (102 + 100) x (102 - 100) = 404  cm^{{2}}. Percentage error =[\frac{404}{{100*100}}*100]_{{%}}=4.04% 3. The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle? A. 16 cm B. 18 cm C. 24 cm D. Data inadequate E. None of these Answer: Option B Explanation: \frac{2(l + b)}{B}=\frac{5}{1} \Rightarrow2l + 2b = 5b \Rightarrow3b = 2l  b=_{3}^{2}\textrm{1} Then, Area = 216  cm^{2} \Rightarrow l\times b= 216 \Rightarrow l x \(\frac {2}{3}\)1= 216 \Rightarrow \rho = 234 \Rightarrow l = 18 cm.(Ans) 4. The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is: A. 40% B. 42% C. 44% D. 46% Answer: Option C Explanation: Let original length = x metres and original breadth = y metres. Original area =  (xy)^{2}. New length =  (_{100}^{120}\textrm{x}) _{{m}} = (_{5}^{6}\textrm{x}) _{{m}} New breadth =  (_{100}^{120}\textrm{y}) _{{m}} = (_{5}^{6}\textrm{y}) _{{m}} New Area =  (\frac{5}{6}\textrm{x}\times\frac{5}{6}) _{m^{2}} = (\frac{36}{25}{xy}) _{m^{2}} The difference between the original area = xy and new-area 36/25 xy is = (36/25)xy - xy = xy(36/25 - 1) = xy(11/25) or (11/25)xy Increase % = (\frac{11}{26}\textrm{xy}\times\frac{1}{xy}\textrm\times{100}) _{%} = 44%. 5. A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road? A. 2.91 m B. 3 m C. 5.82 m D. None of these Answer: Option B Explanation: Area of the park = (60 x 40)  m^{{2}} = 2400  m^{{2}}. Area of the lawn = 2109  m^{{2}}. Area of the crossroads = (2400 - 2109)  m^{{2}} = 291  m^{{2}}. Let the width of the road be x metres. Then, 60x + 40x -  x^{{2}} = 291 \Rightarrow x^{2}- 100x + 291 = 0 \Rightarrow(x - 97)(x - 3) = 0 \Rightarrowx = 3. 6. The diagonal of the floor of a rectangular closet is 7\frac{1}{2} feet. The shorter side of the closet is 4\frac{1}{2} feet. What is the area of the closet in square feet? A.   5_{4}^{1} B.   13_{2}^{1} C. 27 D. 37 Answer: Option C Explanation: Other side =   (\frac{15}{2})^{2} - (\frac{9}{2})^{2} ft =  \frac{225}{4} - \frac{81}{4}  ft   =  \frac{144}{4}   ft   =6 ft Area of closet = (6 x 4.5) sq. ft = 27 sq. ft. 1. What is the area of the hall? I. Material cost of flooring per square metre is Rs. 2.50 II. Labour cost of flooring the hall is Rs. 3500 III. Total cost of flooring the hall is Rs. 14,500. A. I and II only B. II and III only C. All I, II and III D. Any two of the three E. None of these Answer: Option C Explanation: I. Material cost = Rs. 2.50 per  m^{2} II. Labour cost = Rs. 3500. III. Total cost = Rs. 14,500. Let the area be A sq. metres. Material cost = Rs. (14500 - 3500) = Rs. 11,000. \frac{5A}{2} = 11000  \Leftrightarrow A = (\frac{11000\times2}{5})

= 4400  m^{2}. Thus, all I, II and III are needed to get the answer. Correct answer is (C). (Ans) 2. What is the area of a right-angled triangle? I. The perimeter of the triangle is 30 cm. II. The ratio between the base and the height of the triangle is 5 : 12. III. The area of the triangle is equal to the area of a rectangle of length 10 cm. A. I and II only B. II and III only C. I and III only D. III, and either I or II only E. None of these Answer: Option A Explanation: From II, base : height = 5 : 12. Let base = 5x and height = 12x. Then, hypotenuse = (5x)^{{2}} +  (12x)^{{2}} = 13x. From I, perimeter of the triangle = 30 cm. 5x + 12x + 13x = 30 \Leftrightarrow x = 1. So, base = 5x = 5 cm, height = 12x = 12 cm. Area =\ (_{2}^{1}\times5\times12) _{{cm^{2}}}  = 30 {{cm^{2}}} . Thus, I and II together give the answer. Clearly III is redundant, since the breadth of the rectangle is not given. Correct answer is (A). 3. What is the area of rectangular field? I. The perimeter of the field is 110 metres. II. The length is 5 metres more than the width. III. The ratio between length and width is 6 : 5 respectively. A. I and II only B. Any two of the three C. All I, II and III D. I, and either II or III only E. None of these Answer: Option B Explanation: I.   2(l + b) = 110 \Rightarrow l + b = 55. II.  l = (b + 5) \Rightarrow l - b = 5. III.  \frac{l}{b}=\frac{6}{5}\Rightarrow 5l - 6b = 0. These are three equations in l and b. We may solve them pairwise. Any two of the three will give the answer. Correct answer is (B). (Ans) 1. What is the area of the given rectangle? I. Perimeter of the rectangle is 60 cm. II. Breadth of the rectangle is 12 cm. III. Sum of two adjacent sides is 30 cm. A. I only B. II only C. I and II only D. II and III only E. II and either I or III Answer: Option E Explanation: From I and II, we can find the length and breadth of the rectangle and therefore the area can be obtained. So, III is redundant. Also, from II and III, we can find the length and breadth and therefore the area can be obtained. So, I is redundant. Correct answer is "II and either I or III". 2. What is the cost painting the two adjacent walls of a hall at Rs. 5 per m2 which has no windows or doors? I. The area of the hall is 24 sq. m. II. The breadth, length and height of the hall are in the ratio of 4 : 6 : 5 respectively. III. Area of one wall is 30 sq. m. A. I only B. II only C. III only D. Either I or III E. All I, II and III are required. Answer: Option C Explanation: From II, let l = 4x, b = 6x and h = 5x. Then, area of the hall = (24 x^{2})  m^{2}. From I. Area of the hall = 24  m^{2}. From II and I, we get 24 x^{2} = 24 \Leftrightarrow x = 1. l = 4 m, b = 6 and h = 5 m. Thus, area of two adjacent walls = [(l x h) + (b x h)]  m^{2} can be found out and so the cost of painting two adjacent walls may be found out. Thus, III is redundant. Correct answer is (C). (Ans)