rohit_oberoi Posted June 20, 2007 Share Posted June 20, 2007 http://www.maxmytest.com/FileStore/31.jpg The diameter of the circle is 20 and the area of the shaded region is 80 pi. What is the value of a + b + c + d A) 144 B) 216 C) 240 D) 270 E) 288 http://www.maxmytest.com/FileStore/21.jpg How do you solve these fast? Quote Link to comment Share on other sites More sharing options...
zymeth02 Posted June 20, 2007 Share Posted June 20, 2007 wow #1 seems pretty hard, i dont know how to solve it either. for #2, represent the radius of circle Q with r. col A = area of circle O - area of white region col B = area of white region * 4 so it's best to start with getting the area of white region. area of white region = Pi * (2r)^2 - Pi * r^2 = 3r^2 * Pi col A = Pi * (4r)^2 - area of white region = 16r^2 Pi - 3r^2 Pi = 13r^2 Pi col B = 3r^2 Pi * 4 = 12r^2 Pi Therefore, col A > col B Quote Link to comment Share on other sites More sharing options...
rohit_oberoi Posted June 21, 2007 Author Share Posted June 21, 2007 Btw OA for first is 288 Quote Link to comment Share on other sites More sharing options...
kumaraguru Posted June 21, 2007 Share Posted June 21, 2007 Its better that we assume values for these type of problems. Let OA = 4; so OP = 2; PQ = 1 Now comparison is between pi(4)^2 + pi (1)^2 - pi ( 2^2 - 1^2) and 4[ pi(2^2 - 1^2] 14pi Vs 12pi col A is greater Quote Link to comment Share on other sites More sharing options...
Sandeep Bansal Posted June 27, 2007 Share Posted June 27, 2007 What is the solution for the first one?can some one explain? My GRe is on 3rd...please help Quote Link to comment Share on other sites More sharing options...
rohit_oberoi Posted June 27, 2007 Author Share Posted June 27, 2007 What is the solution for the first one?can some one explain? My GRe is on 3rd...please help Solution is 288. How is still a mystery :( Quote Link to comment Share on other sites More sharing options...
KBTA Posted June 27, 2007 Share Posted June 27, 2007 Hi, The question needs some clarifications. It should mention that two vertices of the traingles meet at the centre of the circle. Total area of the circle=100pi Area of the two traingle=100pi-80pi=20pi Angle at the vertex (of one traingle)= (360*20pi*1/2)/100pi = 36 degree so, a+b+c+d=(180-36)+(180-36)=2800 Regards, Jakir Quote Link to comment Share on other sites More sharing options...
Sandeep Bansal Posted June 27, 2007 Share Posted June 27, 2007 Answer is 288 There are two ways to get it easy by approximation as angle subtended at the center has nothing to do with the area of traingle , but with the sector.... 1) See it is told that area of shaded region is 80pi and area of circle is pi*(R^2)=pi*100 so the area of traingle is 100pi - 80 pi= 20pie now the two chords are diameters if they intersect, they at the centre and bisect equally and engendering two equal trianlges.... approx(area of one trinagle will be ) 10pi in circle total area is find out by sector ;;;; if we can fine the sector we can get the answer... ratio of the areas=20pi/100pi(100pi is the total area , full circle) that is ratio of angles should be for two triangles (or so called sector) --->360*(2/10)=72 for 1 sector angle should be 360*(1/10)=36 A+B+C+D+2x(say)=180+180 A+B+C+D= 360-2*36 A+B+C+D=288 Hence proved without interring further into the formula..... Other probable solution is Let E be the thrid angle in the traingle 1/2*radius*radiusSinA=area of traingle area of unshaded cirle comes out to be=100pie-80pie=20pie so, E=180-(c+d),180-(a+b) 1/2*100*SinE + 1/2*100*SinE =20pie E comes out to be 39 near about so 360 -(39+39) = 282 (approx) = 288 :):grad: Quote Link to comment Share on other sites More sharing options...
zymeth02 Posted June 27, 2007 Share Posted June 27, 2007 Hi, KBTA, i see that u used the angle and area relationship to solve question 1. I have a question though. Let's say angle E is the the third angle of the triangle with angles C and D. The equation should be E/360 = area of circular region covered by E / area of circle, right? The area of the circular region covered by E includes the small unshaded triangle, and a small shaded region. However, in your equation, you seem to have ignored the area of the shaded part, and only considered the area of the small triangle (20pi*1/2). Could you explain why? Quote Link to comment Share on other sites More sharing options...
Sandeep Bansal Posted June 27, 2007 Share Posted June 27, 2007 I have not ignored this section that is why wrote, by approximation , the angle of the sector(which we need to know) has nothing to do with the area of traingle...... so by approximation the answer should be.... as there is no way except ,you do lenghty calculations........ the 2nd solution to this problem....... please let me know other solution for this, I am not able to descry the figure, and found it a little ambigous.....i guess we have to reach the best possible answer to get it going.....:hmm: Hi, KBTA, i see that u used the angle and area relationship to solve question 1. I have a question though. Let's say angle E is the the third angle of the triangle with angles C and D. The equation should be E/360 = area of circular region covered by E / area of circle, right? The area of the circular region covered by E includes the small unshaded triangle, and a small shaded region. However, in your equation, you seem to have ignored the area of the shaded part, and only considered the area of the small triangle (20pi*1/2). Could you explain why? Quote Link to comment Share on other sites More sharing options...
KBTA Posted June 28, 2007 Share Posted June 28, 2007 Hi, KBTA, i see that u used the angle and area relationship to solve question 1. I have a question though. Let's say angle E is the the third angle of the triangle with angles C and D. The equation should be E/360 = area of circular region covered by E / area of circle, right? The area of the circular region covered by E includes the small unshaded triangle, and a small shaded region. However, in your equation, you seem to have ignored the area of the shaded part, and only considered the area of the small triangle (20pi*1/2). Could you explain why? Hi zymeth02, Yes, you are right. I have wrongfully neglected the shaded area behind the tringle. We should wait to get the better solution. Regards, Jakir Quote Link to comment Share on other sites More sharing options...
rohit_oberoi Posted June 28, 2007 Author Share Posted June 28, 2007 Answer is 288 There are two ways to get it easy by approximation as angle subtended at the center has nothing to do with the area of traingle , but with the sector.... 1) See it is told that area of shaded region is 80pi and area of circle is pi*(R^2)=pi*100 so the area of traingle is 100pi - 80 pi= 20pie now the two chords are diameters if they intersect, they at the centre and bisect equally and engendering two equal trianlges.... approx(area of one trinagle will be ) 10pi in circle total area is find out by sector ;;;; if we can fine the sector we can get the answer... ratio of the areas=20pi/100pi(100pi is the total area , full circle) that is ratio of angles should be for two triangles (or so called sector) --->360*(2/10)=72 for 1 sector angle should be 360*(1/10)=36 A+B+C+D+2x(say)=180+180 A+B+C+D= 360-2*36 A+B+C+D=288 Hence proved without interring further into the formula..... Other probable solution is Let E be the thrid angle in the traingle 1/2*radius*radiusSinA=area of traingle area of unshaded cirle comes out to be=100pie-80pie=20pie so, E=180-(c+d),180-(a+b) 1/2*100*SinE + 1/2*100*SinE =20pie E comes out to be 39 near about so 360 -(39+39) = 282 (approx) = 288 :):grad: now the two chords are diameters if they intersect,? pl explain how u concluded the chords are diameters Quote Link to comment Share on other sites More sharing options...
Sandeep Bansal Posted June 28, 2007 Share Posted June 28, 2007 It is totally based on assumption there is no way out , for this problem ...... otherwise , they should have provided with another option....... f)Information given is incomplete :) or if we can prove these two traingles equal........the only way... Regards Sandeep Quote Link to comment Share on other sites More sharing options...
KBTA Posted July 7, 2007 Share Posted July 7, 2007 Hi Everybody, The solution of no. 1 problem I have given before is totally correct. Thebasic principle: The angle subtended at the center is same for arc as well as the triangle formed with radius and chord connecting the arc end points. Kapil leo, a newcomer of our GMAT Testmagic forum has helped me to clarify the mystery that was us before. Cheers Kapil leo. For further clarification we can solve the following problem that I have borrowed from TM GMAT Problem Solving p> 4.The shape of a tunnel entrance. is of a circle which is 3/4th of the full circle and the base of the entrance is 12 feet across, what is the perimeter, in feet, of the curved portion of the entrance'? (A) 9π (B) 12π © (D) 18π (E) 9π/sqrt2 Rgds, Jakir Quote Link to comment Share on other sites More sharing options...
sakeba Posted December 12, 2017 Share Posted December 12, 2017 http://www.maxmytest.com/FileStore/31.jpg The diameter of the circle is 20 and the area of the shaded region is 80 pi. What is the value of a + b + c + d A) 144 B) 216 C) 240 D) 270 E) 288 http://www.maxmytest.com/FileStore/21.jpg How do you solve these fast? Could you please insert the figure of this problem Quote Link to comment Share on other sites More sharing options...
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