Puzzles For Timepass

[compose]

telidu nakem ardam kaledu, maths marchipoyanu :(

nenu math tappa anni marchipoyanu :(

[Kickuu]

apply pythagoras theorem....

 

then (p+q)^2 + (r+s)^2 : t^2

:3D_Smiles: :3D_Smiles:  Avi Circles

[chalkpiece]

:3D_Smiles: :3D_Smiles:  Avi Circles

[micxas]

nenu math tappa anni marchipoyanu :(

maree db ela gurthundi :(, maths cheppu Neeku rest of them cheptha.

[compose]

 

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

[compose]

maree db ela gurthundi :(, maths cheppu Neeku rest of them cheptha.

classes eppati nunchi start chedam :(

[k2s]

5/6 - sqrt(3)/2*pi

[micxas]

5/6 - sqrt(3)/2*pi

explain chey :(

[micxas]

classes eppati nunchi start chedam :(

Nee istam :)

[compose]

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  

[ipaddress0]

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:

[micxas]

 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  

 

[k2s]

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

[HarryPorter]

endi vaaya puzzles ani cheppi iit questions vestunnaru.

 

[k2s]

endi vaaya puzzles ani cheppi iit questions vestunnaru.

sorry sona dorling 

 

puzzles e estha aagu