Jump to content

Puzzles For Timepass


bheemavaram_bullodu

Recommended Posts

tapppaaphoto-thumb-8066.jpg?_r=1424905253

Straight Line equations 

101.png

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

Link to comment
Share on other sites

how ? gallery_12949_2_16397.gif

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 photo-thumb-8066.jpg?_r=1424905253

Link to comment
Share on other sites

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 photo-thumb-8066.jpg?_r=1424905253

R2:r2 = 100:9

correct ee.... but ratio ki R:r chalu emo = 10:3

Link to comment
Share on other sites

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 photo-thumb-8066.jpg?_r=1424905253

:(
Link to comment
Share on other sites

cpotg_img8.png

 

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?

Link to comment
Share on other sites

 

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

Link to comment
Share on other sites

 

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

Link to comment
Share on other sites

×
×
  • Create New...