Here are some geometry related questions I need answers of by 3 hours. Read on in 'Details'.?

Question

The diameter of a wheel of a bus is 90cm which makes 315 revolutions per minute. Find its speed in km/h.

If AD, BE and CF are the medians of triangle ABC Show that 4(AD+BE+CF) > 3(AB+BC+CA)

In a right angled triangle, prove that the circumcentre is the mid-point of hypotenuse.


Please give detailed, step-by-step answers. This is urgent. Help with all questions wil be greatly appreciated. I"M NOT LOOKING FOR HINTS! I need ANSWERS.

Answer 1

1) All you need is the circumference, which is pi * diameter, to get the linear distance traveled per revolution. Then just divide by 100,000 to get it in terms of km, and multiply it by 60 to get it in terms of "per hour".

2) Are you sure you copied this one correctly? The way I figure it, AD = ½ of AB, BE is ½ of BC and CF is ½ of CA. This means that the left side can be written "4(½AB + ½BC + ½CA). You then factor out the ½, and you're left with 2(AB + BC + CA), which will always be less than 3(AB + BC + CA)...

3) Start by graphing a right triangle on an X-Y graph with one of the non-right angles at the origin (0,0). Then figure the circle that is formed by the three points of the triangle. When you are finished, the hypotenuse should be the same as a diameter of the circle.

I hope this helps.

Answer 2

one million) I deleted my grossly incorrect attempt to remedy this question and could attempt it lower back later. that is genuine that I mistook a and b as constants which replaced right into a gross blunders. could be that Falzoon is right in concluding it to be an ellipse which could be a circle as a particular case of an equilateral triangle, yet i could artwork out the splendid answer. Edit: My partial answer of one million) is as under. Lengths of factors of triangle ABC are consistent. enable A = (m, 0), B = (0, n) and C = (x, y) => x^2 + (y - n)^2 = a^2 ... ( one million ) (x - m)^2 + y^2 = b^2 ... ( 2 ) and m^2 + n^2 = c^2 ... ( 3 ) removing the variables m and n from the three equations could provide the equation of the locus. i could no longer try this final step. Edit: I genuinely have attempted to eliminate m and n as under: From eqn. ( one million ) and ( 2 ), m = x ± ?(b^2 - y^2) and n = y ± ?(a^2 - x^2) Plugging in eqn. ( 3 ), [x ± ?(b^2 - y^2)]^2 + [y ± ?(a^2 - x^2)]^2 = c^2 => x^2 + b^2 - y^2 ± 2x?(b^2 - y^2) + y^2 + a^2 - x^2 ± 2y?(a^2 - x^2) = c^2 => 2x?(b^2 - y^2) + 2y?(a^2 - x^2) = ± (c^2 - a^2 - b^2) => 4x^2 (b^2 - y^2) + 4y^2 (a^2 - x^2) + 8xy ?[(b^2 - y^2)(a^2 - x^2)] = (c^2 - a^2 - b^2)^2 => b^2x^2 + a^2y^2 - 2x^2y^2 + 2xy?[(b^2 - y^2)(a^2 - x^2)] = ok (a persevering with). Edit: Falzoon's prediction of the locus being ellipse would nicely be inferred as under. we can think of of the vertex C of the triangle ABC to be very close to to the line AB so as that C is fantastically much on the line AB. hence, the locus of C is an ellipse would nicely be proved easily. From this, that is inferred that the locus is elliptical or almost so. Edit: i attempted your suggestion for triangle, sq. and hexagon and curiously got here across the integer solutions that have been the comparable because of the fact the kind of factors, 3 for triangle, 4 for sq. and six for hexagon. thrilling!! I shall attempt this issue sometime as we communicate once I get sufficient time. Edit: I checked utilising Wolfram Alpha for n = 3 to 11 that the respond equals the kind of factors of the polygon, yet ought to no longer remedy a familiar evidence for any fee of n.