Question Bank  Geometry  Hemant Malhotra

2set can be triangle formation

The insect can go from A to B either in exactly 3 steps or in exactly 5 steps.
No: of ways to go from A to B in exactly 3 steps = (3C1) * (2C1) = 6 ways.because at vertex A it has to choose 1 edge out of 3 which it will do in 3C1 ways. and at each vertex 1 or 3 or 5 it has to choose 1 edge out of 2 edges which it will do in 2C1 ways.
No: of ways to go from A to B in exactly 5 steps = No: of ways to go from A to B in exactly 5 steps when no edge is traversed more than once + No: of ways to go from A to B in exactly 5 steps when one of the edge is traversed more than once.
No: of ways to go from A to B in exactly 5 steps when no edge is traversed more than once = 6 ways.
now lets find out the number of ways when insect goes from A  1  A, that is it has completed 2 steps and is back to A from which it has to reach B in remaining 3 steps which it can do in 6 ways.
so when edge A1 is repeated = 6 ways. when edge A3 is repeated = 6 ways, when edge A5 is repeated = 6 ways.
hence (6 + 6 + 6) = 18 ways
now say the insect go from vertex A to any of the three vertex . lets say it goes from A to 1 here when edge 12 is repeated then insect has 2 ways and when 16 is repeated insect has 2 ways. so total 4 ways, hence 3C1*4 = 12 ways.one more case is there when insect travels from A1232B third side is repeated here also 6 ways will be there.
hence total number of ways = 6 + 6 + 18 + 12 + 6 = 48 ways

Let A be the point at which the goat is tied.
Now the radius of the circle is 21
So Total area of the circle will be pie r^2 = pie * 21 *21
But it covered 3/4 th of the circle + Some extra part
So it will be 3/4 (pie * 21^2) + Red part + Yellow part
= 3/4( pie * 21^2) + 1/4(pie * 14^2) + 1/4(pie * 7^2)
= 1232
OA = A[credits : @hemant_malhotra]

@hemant_malhotra 218............

Funda 1 : In any triangle, the orthocenter, circumcenter and centroid are collinear (Euler line)
Funda 2 : On the Euler line the centroid G is between the circumcenter O and the orthocenter H and is twice as far from the orthocenter as it is from the circumcenterC(x, y) ........ G(1, 1) ................. O (5, 3)
CG = 2GOSo point G divides CO in the ratio 1 : 2. Time for the next funda!
Funda 3 : Section Formula
If point P lies on segment AB and satisfies AP : PB = m : n then P = ( (mx2 + nx1)/(m + n), (my2 + ny1)/(m + n) )
Here,
1 = (5 + 2x)/3 => x = 1
1 = (3 + 2y)/3 => y = 3So C is (1, 3)
Now try this.
If the orthocentre of a triangle is (3,5) and the circumcentre is (6,2) then what is the centroid?

@hemant_malhotra wouldn't 57 be the maximum common median ?

@hemant_malhotra Ans 7.4?
r= sqrt(6.75^2+3.05^2)= 7.41 approx.

@hemant_malhotra Ans 26 ?

@hemant_malhotra Sir, 49? Two triangles can be formed, one with ABC 24 area and another with ACD 25 area.

@hemant_malhotra By similarity d.5/4

@hemant_malhotra b. 18rt2

@hemant_malhotra a+p = 2l+2b+lb = (l+2)(b+2)4 = 100/102/104/106/108 only 102+4 = 106= 53*2 not possible . so 102 ?

@hemant_malhotra Is 43 the maximum perimeter possible ?

@hemant_malhotra using mass point geometry i got answe as 1/2 (144)=72 ?

@hemant_malhotra 50 intersection points ??

@hemant_malhotra 7.4 is the answer !!