k2s Posted May 19, 2015 Report Posted May 19, 2015 tapppaa Straight Line equations 1) ax + by = c , then slope = -(a/b) x-intercept is when y = 0 => c/a y-intercept is when x = 0 => c/b 2) y = mx + c slope = m x-intercept = -c/m y-intercept = c
mtkr Posted May 19, 2015 Report Posted May 19, 2015 lekalu thappu maa oorila dosa lu baagaane thinnane....
k2s Posted May 19, 2015 Report Posted May 19, 2015 maa oorila dosa lu baagaane thinnane.... lekkallo thappulu chestey
compose Posted May 19, 2015 Report Posted May 19, 2015 how ? lets assume P,Q,R,S,T are located on the straight line in the same order based on the given info. ST=5x, PQ = 2x RT = 13y , PR = 7y since the bigger circle has center S and pass thru P and T , PS = ST = 5x ( say) since the smaller circle has center R and passes thru Q and S, QR = RS = z (say) PS = PQ + QR +RS => 5x= 2x+z+z => z=1.5x area of bigger circle = pi * 5x * 5x = 25x *x * pi area of smaller circle = pi * z* z = pi* 1.5x * 1.5x = 2.25x * x * pi hence area of bigger circle : area of smaller circle = 25:2.25 = 100 : 9
k2s Posted May 19, 2015 Report Posted May 19, 2015 lets assume P,Q,R,S,T are located on the straight line in the same order based on the given info. ST=5x, PQ = 2x RT = 13y , PR = 7y since the bigger circle has center S and pass thru P and T , PS = ST = 5x ( say) since the smaller circle has center R and passes thru Q and S, QR = RS = z (say) PS = PQ + QR +RS => 5x= 2x+z+z => z=1.5x area of bigger circle = pi * 5x * 5x = 25x *x * pi area of smaller circle = pi * z* z = pi* 1.5x * 1.5x = 2.25x * x * pi hence area of bigger circle : area of smaller circle = 25:2.25 = 100 : 9 R2:r2 = 100:9 correct ee.... but ratio ki R:r chalu emo = 10:3
compose Posted May 19, 2015 Report Posted May 19, 2015 R2:r2 = 100:9 correct ee.... but ratio ki R:r chalu emo = 10:3 u asked ratio of area... not radius
micxas Posted May 19, 2015 Report Posted May 19, 2015 lets assume P,Q,R,S,T are located on the straight line in the same order based on the given info. ST=5x, PQ = 2x RT = 13y , PR = 7y since the bigger circle has center S and pass thru P and T , PS = ST = 5x ( say) since the smaller circle has center R and passes thru Q and S, QR = RS = z (say) PS = PQ + QR +RS => 5x= 2x+z+z => z=1.5x area of bigger circle = pi * 5x * 5x = 25x *x * pi area of smaller circle = pi * z* z = pi* 1.5x * 1.5x = 2.25x * x * pi hence area of bigger circle : area of smaller circle = 25:2.25 = 100 : 9 :(
k2s Posted May 19, 2015 Report Posted May 19, 2015 In the diagram above, JKL is an equilateral triangle. Point M is the midpoint of segment JL, and M is the center of a circle that passes through points J and L. The shaded regions in the diagram indicate all the regions inside the circle that are outside the triangle. What fraction of the total area of the circle is outside the triangle?
mtkr Posted May 19, 2015 Report Posted May 19, 2015 Points P, Q, R, S, and T all lie on the same line. The larger circle has center S and passes through P and T. The smaller circle has center R and passes through Q and S. What is the ratio of the area of the larger circle to the area of the smaller circle? Statement #1: ST:PQ = 5/2 Statement #2: RT:PR = 13/7 apply pythagoras theorem.... then (p+q)^2 + (r+s)^2 : t^2
chalkpiece Posted May 19, 2015 Report Posted May 19, 2015 Points P, Q, R, S, and T all lie on the same line. The larger circle has center S and passes through P and T. The smaller circle has center R and passes through Q and S. What is the ratio of the area of the larger circle to the area of the smaller circle? Statement #1: ST:PQ = 5/2 Statement #2: RT:PR = 13/7 100/9
micxas Posted May 19, 2015 Report Posted May 19, 2015 taapaaa :( telidu nakem ardam kaledu, maths marchipoyanu :(
chalkpiece Posted May 19, 2015 Report Posted May 19, 2015 apply pythagoras theorem.... then (p+q)^2 + (r+s)^2 : t^2 Pythagoras circle ki kuda apply cheyocha
k2s Posted May 19, 2015 Report Posted May 19, 2015 apply pythagoras theorem.... then (p+q)^2 + (r+s)^2 : t^2 malli thappu seppav.......
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