(25-x)^2 + y^2 = 20^2 solve
1st:
what is the relation between x and y
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(25-x)^2 = 625 - 50x + x^2
(25-x)^2 + y^2 = 20^2
625 - 50x + x^2 + y^2 = 400
y^2 + x^2 - 50x = 400 - 625
y^2 + x^2 - 50x = -225
y^2 + x^2 -50x + 225 = 0
we can solve upto this because there is not given any relation about x & y
(25-x)^2 + y^2 = 20^2
625 - 50x + x^2 + y^2 = 400
y^2 + x^2 - 50x = 400 - 625
y^2 + x^2 - 50x = -225
y^2 + x^2 -50x + 225 = 0
we can solve upto this because there is not given any relation about x & y
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hypotenuse. Find the Volume and Surface Area of the double cone so formed.this is the actual question. im sorry but i didnt get how to find the height and base of the bottom cone.
hypotenuse. Find the Volume and Surface Area of the double cone so formed.this is the actual question. im sorry but i didnt get how to find the height and base of the bottom cone.
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