compose Posted May 19, 2015 Report Share Posted May 19, 2015 telidu nakem ardam kaledu, maths marchipoyanu :( nenu math tappa anni marchipoyanu :( Link to comment Share on other sites More sharing options...
Kickuu Posted May 19, 2015 Report Share Posted May 19, 2015 apply pythagoras theorem.... then (p+q)^2 + (r+s)^2 : t^2 :3D_Smiles: :3D_Smiles: Avi Circles Link to comment Share on other sites More sharing options...
chalkpiece Posted May 19, 2015 Report Share Posted May 19, 2015 :3D_Smiles: :3D_Smiles: Avi Circles Link to comment Share on other sites More sharing options...
micxas Posted May 19, 2015 Report Share Posted May 19, 2015 nenu math tappa anni marchipoyanu :(maree db ela gurthundi :(, maths cheppu Neeku rest of them cheptha. Link to comment Share on other sites More sharing options...
compose Posted May 19, 2015 Report Share 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? 5/6 - sqrt(3)/2*pi Link to comment Share on other sites More sharing options...
compose Posted May 19, 2015 Report Share Posted May 19, 2015 maree db ela gurthundi :(, maths cheppu Neeku rest of them cheptha. classes eppati nunchi start chedam :( Link to comment Share on other sites More sharing options...
k2s Posted May 19, 2015 Report Share Posted May 19, 2015 5/6 - sqrt(3)/2*pi Link to comment Share on other sites More sharing options...
micxas Posted May 19, 2015 Report Share Posted May 19, 2015 5/6 - sqrt(3)/2*piexplain chey :( Link to comment Share on other sites More sharing options...
micxas Posted May 19, 2015 Report Share Posted May 19, 2015 classes eppati nunchi start chedam :( Nee istam :) Link to comment Share on other sites More sharing options...
compose Posted May 19, 2015 Report Share Posted May 19, 2015 explain chey :( circle radius = 1 (say) => area of circle = pi JL is the diameter kabatti, area of the bottom part of the circle = area of the top part of the circle = pi/2 top part of circle lo the points where the lines JK and KL intersect the circle are their respective midpoints. so the area of triangle within the circle is divided into 2 smaller equilateral triangles and a sector. so aa chinna equilateral triangle angle at the center M is 60 degrees and hence the area of the sector it encompasses = 1/6 * area of circle = pi / 6 deenilo nundi smaller equilateral triangle area subtract cheyali. area of smaller equilateral triangle = 1/2 * b * h = 1/2 * 1 * sqrt(3)/2 = sqrt(3)/4 hence, area of green unoccupied sector on the left side = pi/6 - sqrt(3)/4 so area of two green unoccupied sectors combined (one on left and one on right side) = pi/3 - sqrt(3)/2 hence, the area of green area = pi/2 + pi/3 - sqrt(3)/2 = 5*pi/6 - sqrt(3)/2 hence required ratio = (5*pi/6 - sqrt(3)/2) / pi = 5/6 - sqrt(3)/2*pi Link to comment Share on other sites More sharing options...
ipaddress0 Posted May 19, 2015 Report Share Posted May 19, 2015 circle radius = 1 (say) => area of circle = pi JL is the diameter kabatti, area of the bottom part of the circle = area of the top part of the circle = pi/2 top part of circle lo the points where the lines JK and KL intersect the circle are their respective midpoints. so the area of triangle within the circle is divided into 2 smaller equilateral triangles and a sector. so aa chinna equilateral triangle angle at the center M is 60 degrees and hence the area of the sector it encompasses = 1/6 * area of circle = pi / 6 deenilo nundi smaller equilateral triangle area subtract cheyali. area of smaller equilateral triangle = 1/2 * b * h = 1/2 * 1 * sqrt(3)/2 = sqrt(3)/4 hence, area of green unoccupied sector on the left side = pi/6 - sqrt(3)/4 so area of two green unoccupied sectors combined (one on left and one on right side) = pi/3 - sqrt(3)/2 hence, the area of green area = pi/2 + pi/3 - sqrt(3)/2 = 5*pi/6 - sqrt(3)/2 hence required ratio = (5*pi/6 - sqrt(3)/2) / pi = 5/6 - sqrt(3)/2*pi vayyooo nuvvemmana state topper vaa endi fafa :3D_Smiles: Link to comment Share on other sites More sharing options...
micxas Posted May 20, 2015 Report Share Posted May 20, 2015 Ardham avvadanike 10 min paina pattindhi :police:.. cool, red-adhi midpoint how? circle radius = 1 (say) => area of circle = pi JL is the diameter kabatti, area of the bottom part of the circle = area of the top part of the circle = pi/2 top part of circle lo the points where the lines JK and KL intersect the circle are their respective midpoints. so the area of triangle within the circle is divided into 2 smaller equilateral triangles and a sector. so aa chinna equilateral triangle angle at the center M is 60 degrees and hence the area of the sector it encompasses = 1/6 * area of circle = pi / 6 deenilo nundi smaller equilateral triangle area subtract cheyali. area of smaller equilateral triangle = 1/2 * b * h = 1/2 * 1 * sqrt(3)/2 = sqrt(3)/4 hence, area of green unoccupied sector on the left side = pi/6 - sqrt(3)/4 so area of two green unoccupied sectors combined (one on left and one on right side) = pi/3 - sqrt(3)/2 hence, the area of green area = pi/2 + pi/3 - sqrt(3)/2 = 5*pi/6 - sqrt(3)/2 hence required ratio = (5*pi/6 - sqrt(3)/2) / pi = 5/6 - sqrt(3)/2*pi Link to comment Share on other sites More sharing options...
k2s Posted May 20, 2015 Report Share Posted May 20, 2015 circle radius = 1 (say) => area of circle = pi JL is the diameter kabatti, area of the bottom part of the circle = area of the top part of the circle = pi/2 top part of circle lo the points where the lines JK and KL intersect the circle are their respective midpoints. so the area of triangle within the circle is divided into 2 smaller equilateral triangles and a sector. so aa chinna equilateral triangle angle at the center M is 60 degrees and hence the area of the sector it encompasses = 1/6 * area of circle = pi / 6 deenilo nundi smaller equilateral triangle area subtract cheyali. area of smaller equilateral triangle = 1/2 * b * h = 1/2 * 1 * sqrt(3)/2 = sqrt(3)/4 hence, area of green unoccupied sector on the left side = pi/6 - sqrt(3)/4 so area of two green unoccupied sectors combined (one on left and one on right side) = pi/3 - sqrt(3)/2 hence, the area of green area = pi/2 + pi/3 - sqrt(3)/2 = 5*pi/6 - sqrt(3)/2 hence required ratio = (5*pi/6 - sqrt(3)/2) / pi = 5/6 - sqrt(3)/2*pi :buttkick: cofy faster googleress Link to comment Share on other sites More sharing options...
HarryPorter Posted May 20, 2015 Report Share Posted May 20, 2015 endi vaaya puzzles ani cheppi iit questions vestunnaru. Link to comment Share on other sites More sharing options...
k2s Posted May 20, 2015 Report Share Posted May 20, 2015 endi vaaya puzzles ani cheppi iit questions vestunnaru. sorry sona dorling puzzles e estha aagu Link to comment Share on other sites More sharing options...
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