Questions and Answers #2 Circles

Questions and Answers Sheet 2 Circles Question #1 To construct: The angle equal to the average of measures of ∠𝐴 and ∠𝐵 . Answer: Construction: Draw two circles of equal radii with center at A and B name the intersection. Draw two circles of equal radii EF with center at F and H name the intersection points. Draw a ray ⃗⃗⃗⃗⃗ 𝐵𝑇 . Since 𝑚∠𝑇𝐵𝐻 = 𝑚∠𝐴 and hence 𝑚∠𝑇𝐵𝐻 = 𝑚∠𝐵 + 𝑚 ∠𝐴 Draw two circles of equal radii with center at G and T name the intersection points. Draw a line ⃗⃗⃗⃗ 𝐵𝐼 which bisects ∠𝑇𝐵𝐺 . The required angle ∠𝑇𝐵𝐺 has measure equal to the average of measures of ∠𝐴 and ∠𝐵 . Conclusion: Therefore, the angle equal to the average of measures of ∠𝐴 and ∠𝐵 is ∠𝑇𝐵𝐺 and shown in figure Question #2 Write the equation of a line passing through the centers of the two circles (𝑥 − 3)2 + (𝑦 + 2)2 = 16 and (𝑥 − 6)2 + (𝑦 + 1)2 = 25 Answer: Step 1 Given two circle with equations, 1) (𝑥 − 3)2 + (𝑦 + 2)2 = 16 2) (𝑥 − 6)2 + (𝑦 + 1)2 = 25 Compairing with standard equation of a circle is (𝑥 − ℎ)2 + (𝑦 + 𝑘)2 = 𝑠 2 whear (h, k): center of circle Fron equations (1), center: (ℎ1 , 𝑘1 ) = (3, −2) From equation (2), center: (ℎ2 , 𝑘2 ) = (6, −1) Step 2 Since equation of a line passing thsough two paints (𝑥1 , 𝑦1 ) and (𝑥2 , 𝑦2 ) is given by, 𝑦 − 𝑦1 = ( 𝑦2 −𝑦1 𝑥2 −𝑥1 ) (𝑥 − 𝑥1 ) \therefore equation of sequired line, 𝑦 − 𝑘1 = ( 𝑘2 −𝑘1 ℎ2 −ℎ1 ) (𝑥 − ℎ1 ) (𝑦 − (−2)) = ( 𝑦+2 =( −1+2 3 (−1)−(−2) 6−3 ) (𝑥 − 3) ) (𝑥 − 3) 1 𝑦 + 2 = (𝑥 − 3) 𝑦= 𝑦= 3 1 𝑥−1−2 3 1 𝑥−3 3 Question #3 What is the radius and center for (𝑥 − 2)2 + 𝑦 2 = 16? A. r=16; center is (2,0) B. r=4; center is (-2,0) C. r=16; center is (-2,0) D. r=4; center is (2,0) Answer: Step 1 Given equation is (𝑥 − 2)2 + 𝑦 2 = 16 find radius and center Step 2 given equation is a circle standard form of an equation of a circle is (𝑥 − ℎ)2 + (𝑦 − 𝑘)2 = 𝑟 2 center =(h,k) radius =r lets write the given equation in standard form (𝑥 − 2)2 + (𝑦 − 0)2 = 42 center =(2,0) radius =r=4 Thus D is a correct option. Question #4 Circle A has a diameter of approximately 20 inches and an area of 300 .. Circle B has a diameter of approximately 60 inches. Which of these could be the area of circle B. Explain your reasoning in question two. 100 ∈2 or 300 ∈2 or 900 or ∈22 , 700 ∈2 . Answer: Step 1 The area of a circle (A), measured in square inches, is directly proportional to the square of its diameter (d), measured in inches. Step 2 That is: 𝐴∞𝑑 2 Where is the constant of proportionality, dimensionless. In consequence, the following relationship between circles A and B is obtained: 𝐴𝐵 𝐴𝐴 = 2 𝑑𝐵 2 𝑑𝐴 The area of the circle B is now cleared: 𝑑 2 𝐴𝐵 = ( 𝐵) ⋅ 𝐴𝐴 𝑑𝐴 Given that 𝑑𝐴 = 20 ∈, 𝑑𝐵 = 60 ∈ and 𝐴𝐴 = 300 ∈2 , then: 60𝑖𝑛 2 𝐴𝐵 = ( ) ⋅ (300𝑖𝑛2 ) 20𝑖𝑛 𝐴𝐵 = 2700𝑖𝑛2 Therefore, the correct answer is 4. Question #5 Let C1 be the circle centre P = (-1, 1) with radius √17 17 and C2 be the circle centre Q = (1, -1) with radius √13. (a) Determine an equation for each of C1 and C2 in the form x 2+Ax+y 2+By+C = 0 . (b) Determine the two points of intersection R, S of the circles C1 and C2. Answer: Step 1 C1 be the circle center P=(-1,1) with radius √17 C2 be the circle center Q=(1, -1) with radius √13 a) To determine the equation of each circle. b) To find the points of intersection of C1 and C2. Step 2 2 a) 𝐶1: (𝑥 + 1)2 + (𝑦 − 1)2 = (√17) ⇒ 𝑥 2 + 2𝑥 + 1 + 𝑦 2 − 2𝑦 + 1 = 17 ∴ 𝑥 2 + 𝑦 2 + 2𝑥 − 2𝑦 − 15 = 0 2 𝐶2: (𝑥 − 1)2 + (𝑦 + 1)2 = (√13) ⇒ 𝑥 2 − 2𝑥 + 1 + 𝑦 2 + 2𝑦 + 1 = 13 ∴ 𝑥 2 + 𝑦 2 − 2𝑥 + 2𝑦 − 11 = 0 Step 3 b) For the points of intersection, we equate the equations of C1 and C2. ⇒ 𝑥 2 + 𝑦 2 + 2𝑥 − 2𝑦 − 15 = 𝑥 2 + 𝑦 2 − 2𝑥 + 2𝑦 − 11 ⇒ 4𝑥 − 4𝑦 = 15 − 11 ⇒ 4𝑥 − 4𝑦 = 4 ⇒𝑥−𝑦=1 ∴𝑥 =𝑦+1 Step 4 Put this value in the equation of C1, ⇒ (𝑦 + 1)2 + 𝑦 2 + 2(𝑦 + 1) − 2𝑦 − 15 = 0 ⇒ 𝑦 2 + 2𝑦 + 1 + 𝑦 2 + 2𝑦 + 2 − 2𝑦 − 15 = 0 ⇒ 2𝑦 2 + 2𝑦 − 12 = 0 ⇒ 𝑦2 + 𝑦 − 6 = 0 ⇒ 𝑦 2 + 3𝑦 − 2𝑦 − 6 = 0 ⇒ 𝑦(𝑦 + 3) − 2(𝑦 + 3) = 0 ⇒ (𝑦 + 3)(𝑦 − 2) = 0 ∴ 𝑦 = −3, 𝑦 = 2 𝑦 = −3 ⇒ 𝑥 = −3 + 1 = −2 𝑦 =2 ⇒𝑥 =2+1=3 Therefore points of intersection: 𝑅 = (−2, −3) 𝑆 = (3,2) Question #6 Let D be the region between the circles of radius 6 and radius 8 centered at the origin that lies in the third quadrant. Express D in polar coordinates. Select the correct answer below: 𝜋 1) 𝐷 = (𝑟, 𝜃)|6 ≤ 𝑟 ≤ 8,0 ≤ 𝜃 ≤ 2 2) 𝐷 = {(𝑟, 𝜃)) |6 ≤ 𝑟 ≤ 8, 𝜋 ≤ 𝜃 ≤ 3) 𝐷 = (𝑟, 𝜃)|0 ≤ 𝑟 ≤ 8, 𝜋 ≤ 𝜃 ≤ 3𝜋 2 } 3𝜋 2 4) 𝐷 = (𝑟, 𝜃)|6 ≤ 𝑟 ≤ 8, 𝜋 ≤ 𝜃 ≤ 2𝜋 Answer: Given: let D be the region between the circles of radius 6 and 8 centered at the origin that lies in the third quadrant express D in polar coordinates. Here the region D is between the circles with radius r=6,r=8 centered at the origin that is 6 ≤ 𝑟 ≤ 8 and that lies in quadrant 3 So, 𝜋 ≤ 𝜃 ≤ 3𝜋 2 Therefore, 𝐷 = (𝑟, ∴)|6 ≤ 𝑟 ≤ 8, 𝜋 ≤∴≤ 3𝜋2 Therefore the 2 nd option is the correct one Question #7 For the region R below, write ∫ ∫𝑍 𝑓 𝑑𝐴 as an integral in polar coordinates. Use t for 𝜃 in your expressions. Answer: From the given figure, it is seen that the region R is bounded by two circles, 𝑥 2 + 𝑦 2 = 9 and 𝑥 2 + 𝑦 2 = 16 Therefore, region R can be defined as 𝑅 = (𝑟, 𝜃) 3 ≤ 𝑟 ≤ 4,0 ≤ 𝜃 ≤ 𝜋 it is known that, 𝜃 2 ∫ ∫𝑅 𝑓 𝑑𝐴 = ∫𝜃 𝑓 (𝑟𝑐𝑜𝑠𝜃, 𝑟𝑠𝑖𝑛𝜃)𝑟𝑑𝑟𝑑 𝜃 1 Therefore, in polar co-ordinate, 𝜋 4 ∫ ∫ 𝑓 𝑑𝐴 = ∫ ∫ 𝑓 (𝑟𝑐𝑜𝑠𝜃, 𝑟𝑠𝑖𝑛𝜃)𝑟𝑑𝑟𝑑 𝜃(𝑖) 𝑅 0 3 𝜋 4 Thus, comparing (i) with ∫ ∫𝑅 𝑓 𝑑𝐴 = ∫0 ∫3 𝑓 (𝑟 𝑐𝑜𝑠 𝜃, 𝑟 𝑠𝑖𝑛 𝜃)𝑟𝑑𝑟𝑑 𝜃 it is inferred that a=0 𝑏 = 𝜋 c=3 d=4 With 𝑑𝐴 = 𝑟𝑑𝑟𝑑 𝜃 Question #8 Find the area of the region that lies inside both curves. 𝑟 2 = 50 sin(2𝜃) , 𝑟 = 5 Answer: Step 1 Consider the following curves: 𝑟 2 = 50 sin(2𝜃) , 𝑟 = 5 The objective is to find the area of the region that lies inside both the curves. Sketch the region as follows: Find the intersection points as follows: From the curves, 50 sin 2 𝜃 = 52 sin 2 𝜃 = sin 2 𝜃 = 25 50 1 2 𝜋 5𝜋 13𝜋 17𝜋 2𝜃 = , 𝜃= , , 6 6 6 6 𝜋 5𝜋 13𝜋 17𝜋 , 12 12 , 12 , 12 The formula for the area A of the polar region 𝑅 is, A = b 1  2[ f ( ] d . 2 Step 2 The desired area can be found by adding the area inside the cardioid between 𝜃 = 0 and area inside the circle from 0 to 𝜋 4 Since the region is symmetric about 𝜃 = 𝜋 4 𝜋 2 12 1 2 2 × ∫ 5 𝑑𝜃} 𝜋 4 1 𝜋 𝜋 4 from the 2 , you can write the area is, 𝐴 = 2{2 × ∫012 (√50 sin(2𝜃)) 𝑑𝜃 + 2 𝜋 12 𝜋 4 = 2{∫ 50 sin(2𝜃) 𝑑𝜃 + ∫ 25𝑑𝜃} 𝜋 12 0 = 2{[− 50 2 𝜋 12 = 2 [−25 (cos = 2 [−25 ( = 2 (− 𝜋 cos 2 𝜃] + 25[𝜃] 4𝜋 } √3 2 0 𝜋 6 12 𝜋 𝜋 4 12 − cos 0) + 25 ( − )] 2𝜋 − 1) + 25 ( )] 12 25√3 25𝜋 + 25 + ) 2 6 25𝜋 = −25√3 + 50 + 3 Thus, the area of the region that lies inside both the curves is −25√3 + 50 + 25𝜋 3 Question #9 Find the overlapping area of two equations also give its geometrical representation: 𝑢2 + 𝑣 2 = 𝑛 𝑎𝑛𝑑 𝑢2 + (𝑣 − 𝑛)2 = 1 put n = 516 Answer: Step 1 The equation of the form 𝑥 2 + 𝑦 2 = 𝑟 2 is the standard equation of a circle with radius r. To obtain the overlapping area the intersection of the circles should be obtained. Step 2 The function 𝑢2 + 𝑣 2 = 𝑛 for 𝑛 = 516 gives a circle with center at (0,0) and radius approximately 22.7156. The equation 𝑢2 + (𝑣 − 𝑛)2 = 1 for 𝑛 = 516 gives an circle with center at (0, 516) and radius 1. Therefore, these circles do not intersect and hence the overlapping area is 0. Step 3 These equations do not intersect and hence the overlapping area is 0. Question #10 The graph of a polar equation is given. Select the polar equation for graph from the following options. 𝑟 = 2 𝑠𝑖𝑛 𝜃, 𝑟 = 2 𝑐𝑜𝑠𝜃, 𝑟 = 1 + 𝑠𝑖𝑛 𝜃 𝑟 = 1 − 𝑠𝑖𝑛 𝜃, 𝑟 = 3 𝑠𝑖𝑛 2𝜃, 𝑟 = 3 𝑠𝑖𝑛 3𝜃 Answer: Step 1 The graph of a polar equation is given. Select the polar equation for graph from the following options. Step 2 From the given graph, it is in the form of a circle. Since the graphs of 𝑟 = 𝑎 𝑐𝑜𝑠 𝜃 and 𝑟 = 𝑎 𝑠𝑖𝑛 𝜃 are circles, it can be concluded that the possible options that matches with the general forms are 𝑟 = 2𝑐𝑜𝑠𝜃 and 𝑟 = 2𝑠𝑖𝑛𝜃. At 𝜃 = 0, the value r=0. 𝜋 At 𝜃 = 2, the value r=2 . 2 The only equation 𝑟 = 2𝑠𝑖𝑛𝜃 satisfies the above condition. Hence, the polar equation for graph is 𝑟 = 2𝑠𝑖𝑛𝜃. Question #11 Use a double integral to find the area of the region. The region inside the circle (𝑥 − 1)2 + 𝑦 2 = 1 and outside the circle 𝑥 2 + 𝑦 2 = 1 Answer: Step 1 A basic understanding of equations tells us that this equation is a circle shifted one unit to the right. I noticed that visualizing this in polar can be difficult so I laid out ever step for those that like to hand-jam. (𝑥 − 1)2 + 𝑦 2 = 1 (𝑟 cos 𝜃 − 1)2 + 𝑟 2 sin2 𝜃 = 1 (𝑟 2 cos 2 𝜃 − 2𝑟 cos 𝜃 + 1) + 1𝑟 2 sin2 𝜃 = 1 𝑟 2 cos 2 𝜃 + 𝑟 2 sin2 𝜃 = 2𝑟𝑐𝑜𝑠 𝜃 𝑥 2 + 𝑦 2 = 2𝑟 cos 𝜃 𝑟 2 = 2𝑟 cos 𝜃 𝑟 = 2 cos 𝜃 Step 2 Again, an understanding that this is the equation of a circle with radius 1 is useful, but if that cannot be seen you can convert to polar. 𝑥 2 + 𝑦2 = 1 𝑟2 = 1 r=1 Step 3 Graphing this is simple and recommended to better understand the limits of integration. To find your limits of integration for r make the appropriate substitution and solve for theta. 1 = 2cos cos 𝜃 = 1 2 𝜋 𝜋 𝜃=− , 3 3 Step 4 Our Region can now be defined. R = (r ,  ) |1  r  2 cos  , −  3    3 Step 5 Set up the integrals ⍁ 2 cos 𝜃 ∫−𝜋 𝜋3 ∫1 𝑟𝑑𝑟𝑑 𝜃 3 Step 6 Integrate with respect to r and run  1   2r 3 − 2 2cos 1 | d . 3 Step 7 At this point the power reduction formula is necessary for further integration.    2cos 3 − 3 2 1 2  − d . Step 8 Based on symmetry you can double the integral and run from 0 to 𝜋/3 but I will show it without.  1   2cos 2 + 2 d . 3 − 3 Step 9 Use trig to find the values of the inputs 1 2 1 𝜋 sin 2 𝜃 + 𝜃| 3 𝜋 2 − 3 Step 10 I went more in detail with the set up of the problem than I did with the integration because I believe that's where the most confusion comes from. Please let me know if there are any problems with this solution. 𝜋 3 + √3 2 Result: 𝜋 3 + √3 2 Question #12 A particle is moving along a circular path having a radius of 4 in. such that its position as a function of time is given by 𝜃 = cos 2 𝑡, where 𝜃 is in radians and t is in seconds. Determine the magnitude of the acceleration of the particle when 𝑢 = 30∘ . Answer: Given: r=4 in, 𝜃 = cos(2𝑡) , Acceleration in this case is equal to 𝑎 = 𝑎𝑟 ⃗⃗⃗⃗ 𝑢𝑟 + 𝑎𝜃 ⃗⃗⃗⃗ 𝑢𝜃 𝑎𝑟 = 𝑟̈ − 𝑟𝜃 2̇ = 𝑎 = 𝑟𝜃̈ + 2𝑟𝜃̇ ̇ 𝜃 Since the radius in circular motion is constant: 𝑟̇ = 𝑟̈ = 0, and time-derivatives of position are equal to: 𝜃̇ = −2 sin(2𝑡) , 𝜃̈ = −4 cos(2𝑡) 𝜋 If we want to find out in what time the particle will have 𝜃 = 30 = , we use following formula: 6 𝜃 = cos(2𝑡) 𝜋 = cos(2𝑡) 6 2𝑡 = 1.0197 𝑡 = 0.51𝑠 Acceleration is finally equal to: 𝑎 = 𝑎𝑟 ⃗⃗⃗⃗ 𝑢𝑟 + 𝑎𝜃 ⃗⃗⃗⃗ 𝑢𝜃 2 ̇ = −𝑟𝜃 ⃗⃗⃗⃗ 𝑢𝑟 + 𝑟𝜃𝑢 ⃗⃗⃗⃗̈ 𝜃 = −4 ⋅ (−2 sin(2 ⋅ 0.51))2 ⃗⃗⃗⃗ 𝑢𝑟 + 4 ⋅ −4 cos(2 ⋅ 0.51) ⃗⃗⃗⃗ 𝑢𝜃 = −11.62𝑢 ⃗⃗⃗⃗𝑟 − 8.37𝑢 ⃗⃗⃗⃗𝜃 Magnitude of the acceleration is: 𝑎 = √𝑎𝑟2 + 𝑎𝜃2 = √(11.62)2 + (8.37)2 𝑎 = 14.32 𝑖𝑛 𝑠2 Question #13 Determine how much air could fit inside a soccer ball with radius of 4.5 inches. Use 3.14 for 𝜋 Answer: That mean we have to find the volume of the Soccer ball. Since it is a Shere with radius 𝑟 = 4.5 inches \therefore volume of Soccer ball = 4 3 r 3 V= 4   (4.5)3 3 4 =   91.128 3 4 =  3.14  91.725 3 4  286.1325 = 3 1144.53 = 3 = 381.51in3 ∴ 𝑉 = 381.51𝑖𝑛3 Question #14 What is the perimeter of semicircle if 𝜋 = 3,142 and d=10 Answer: The perimeter of a circle: 𝑃 = 2𝜋𝑟 = 𝜋𝑑 , where r ia radius and d is diameter. Here, it is a semicircle, a half of a circle 1 ⇒ × 2𝜋𝑟 = 𝜋𝑟 is the perimeter 2 𝑑 = 10 ⇒ 𝑟 = 10 2 =5 𝑃 = (𝜋 × 5) + 10 = (3.142 × 5) 15.71 + 10 = 25.71𝑐𝑚 Question #15 Two circles have a scale factor of 3:2. What is the ratio of the areas of these circles? a) 3:2 b) 9:4 c) 12:26 d) 13:12 Answer: The area of the circle is = 𝜋𝑟 2 . Where r is the radius of the circle. Let the radius of circle A is 3 and the radius of the scale factor B is 2. Then, the ratio of areas of these circles 𝜋(3)2 𝜋(2)2 = 9 4 Hence the ratio of the areas of these circles is 9:4. Hence option (b) is true. Question #16 Find the two points of intersection of the circles: 𝑥 2 + 𝑦 2 + 5𝑥 + 𝑦 − 26 = 0 𝑥 2 + 𝑦 2 + 2𝑥 − 𝑦 − 15 = 0 Answer: Step 1 Multiply all terms in the first equation by -1 to obtain an equivalent equation and keeping the second equation unchanged. −𝑥 2 − 𝑦 2 − 5𝑥 − 𝑦 + 26 = 0 and 𝑥 2 + 𝑦 2 + 2𝑥 − 𝑦 − 15 = 0 Adding the same sides of the two equations to obtain a linear equation, −5𝑥 − 𝑦 + 26 + 2𝑥 − 𝑦 − 15 = 0 Simplifying, −3𝑥 − 2𝑦 + 11 = 0 3𝑥 + 2𝑦 − 11 = 0 Solving for x, 3𝑥 = 11 − 2𝑦 𝑥= 11 − 2𝑦 3 Step 2 Substituting the value of x in equation (1), 11 − 2𝑦 2 11 − 2𝑦 ( ) + 𝑦2 + 5 ( ) + 𝑦 − 26 = 0 3 3 Simplifying, (11−2𝑦)2 9 + 𝑦2 + 5(11−2𝑦) 3 + 𝑦 − 234 = 0 (11 − 2𝑦)2 + 9𝑦 2 + 15(11 − 2𝑦) + 9𝑦 − 234 = 0 121 − 44𝑦 + 4𝑦 2 + 9𝑦 2 + 165 − 30𝑦 + 9𝑦 − 234 = 0 13𝑦 2 − 65𝑦 + 286 − 234 = 0 13𝑦 2 − 65𝑦 − 52 = 0 𝑦 2 − 5𝑦 + 4 = 0 Solving quadratic equation, (𝑦 − 1)(𝑦 − 4) = 0 ⇒ 𝑦 = 1, 4 Question #17 The circumference of two circles are 12𝜋 and 36𝜋 . Find the ratio of the areas of the circles. a) 1:3 b) 1:9 c) 3:4 d) 2:5 Answer: Step 1 We know that circumference of circle = 2𝜋𝑟 For first circle 2𝜋𝑟 = 12𝜋 ⇒𝑟=6 For second circle, 2𝜋𝑟 = 36𝜋 ⇒ 𝑟 = 18 Step 2 Area of the circle = 𝜋𝑟 2 For first circle Area = 𝜋 × (6)2 [∵ 𝑟 = 6] = 36𝜋 For second circle Area = 𝜋 × (18)2 [𝑟 = 18] = 324𝜋 Ratio of the area of circle is 36𝜋: 324𝜋 =1:9 Hence the answer is 1:9 Question #18 Two circles touch internally. The sum of their areas is 116𝜋𝑐𝑚2 and distance between their centres is 6 cm. Find the radii of the circles Answer: Given: Let R and r be the radii of the circles having centres at O and O' respectively Finding the given Then, Sum of areas = 116𝜋𝑐𝑚2 𝜋𝑅2 + 𝜋𝑟 2 = 116 𝑅2 + 𝑟 2 = 116(1) Given that Distance between the centres =6cm OO'=6cm R-r=6cm (2) Now, (𝑅 + 𝑟)2 + (𝑅 − 𝑟)2 = 2(𝑅2 + 𝑟 2 ) (𝑅 + 𝑟)2 + 36 = 2 × 116 (𝑅 + 𝑟)2 = 2 × 116 − 36 (𝑅 + 𝑟)2 = 232 − 36 (𝑅 + 𝑟)2 = 196 𝑅 + 𝑟 = √196 𝑅 + 𝑟 = 14(3) Solving (2) and (3) R-r=6 𝑅 + 𝑟 = 14 2𝑅 = 20 𝑅= 20 2 R=10 Put R in (3) R+r=14 10+r=14 r=14-10 r=4 Therefore, radii of given circles are 10 cm and 4 cm Question #19 Circles A. B, and C cach have radius 1. Circles A and B share one point of tangency. Circle C has a point of tangency with the midpoint of 𝐴𝐵. What is the area inside circle C but outside circle A and circle B? a) 3 − b) 𝜋 2 𝜋 2 c) 2 d) 3𝜋 4 e) 1 + 𝜋 2 Answer: Step 1 Given, Circles A, B, and C each have radius 1. Circles A and B share one point of tangency. Circle C has a point of tangency with the midpoint of 𝐴𝐵. To find the are inside circle C but outside A and circle B, that is, the shaded region of the circle C. Now, Area of shaded region = Area of Circle C − Area shared between Circles A, B and C. . Let S be the mid point of 𝐴𝐵 and T be the point of intersection of Circles A and C. Join the points by joining the points ST Step 2 Now, the shared area of all three circles A, B, C = 4 X the region between line ST and arc ST. The area of the Circle 𝐴 = 𝜋𝑟 2 = 𝜋 The area of 1 △ 𝑆𝐴𝑇 = 𝑏ℎ 2 = 1 2 since b=h=r=1 1 So, the area between the line ST and the arc ST = × (𝐴𝑟𝑒𝑎 𝑜𝑓 𝐶𝑖𝑟𝑐𝑙𝑒 𝐴) − 𝐴𝑟𝑒𝑎 𝑜𝑓 △ 𝑆𝐴𝑇 4 1 1 𝜋−2 4 2 4 = ×𝜋− =( ) Step 3 Area of Shaded region in Circle C = (Area of Circle C ) − 4(𝐴𝑟𝑒𝑎 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 𝑆𝑇 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑎𝑟𝑐 𝑆𝑇)) = 𝜋(1)2 − 4 ( 𝜋−2 4 )= 𝜋−𝜋+2=2 Therefore, Area of Shaded region in Circle C is 2. Question #20 ⊙ 𝑄 and ⊙ 𝑅 are congruent circles that increasect at C and D. 𝐶𝐷 is called the common chors of the circles. a) What king of quadrilateral is QDRC? Why? b) 𝐶𝐷 must be the perpendicular bisector of 𝑄𝑅 . Why? c) If QC=17 and QR=30, find CD Answer: Since, ⊙ 𝑄 and ⊙ 𝑅 are congruent circles, Therefore, radius of each circle should be equal. a) Since, QD, DR, RC, CD are radius and are equal and also meet each other at right angle. Therefore, quadrilateral QDRC is square. b) 𝐶𝐷 must be the perpendicular bisector of 𝑄𝑅 because diagonal of square intersect each other orthogonally. c) 𝑄𝑅 will be equal 𝐶𝐷 since, diagonals of square are equal. Therefore, CD=30