Latest Elitmus Aptitude Question SOLUTION: 3 spheres of radios 1 unit are kept inside a triangle. a big sphere of diameter 8 unit is kept on top of center of those 3 balls. find the height of the top of the ball from ground

10 unit
height of small sphere+height of bigger sphere

height of the smaller sphere+height of the largest sphere=2 unit+16 unit=18..i dont know the correct answer...

ans =5+2root6

ans=17+root3

f we consider the centre of each sphere to be the corner of a tetrahedron, then we have the following shape:

tetrahedron

We can easily see that AB=BC=CA=2AB=BC=CA=2, AH=BH=CH=9AH=BH=CH=9 - if you can't, then try placing three spheres next to each other and seeing what the distance between their centres is. We also have D,E,FD,E,F are the midpoints of AC,BC,ABAC,BC,AB respectively.

We want to find the distance GHGH - we know that HH is directly above GG as △ABC△ABC is an equilateral, and so the lines DB,CF,AEDB,CF,AE intersect at the centre of the triangle.

To find GHGH, we can consider the triangle AGHAGH, where AG^H=90∘AG^H=90∘ and AH=9AH=9. To solve this triangle, we need to find AGAG. We can do this by looking at the triangle ACGACG.

first triangle

We know the length of one side, all the angles, and we want to find another side length and so we can use the sine rule:

AGsin(AC^G)AGsin(30∘)AG=ACsin(AG^C)=2sin(120∘)=2sin(30∘)sin(120∘)=23–√
AGsin⁡(AC^G)=ACsin⁡(AG^C)AGsin⁡(30∘)=2sin⁡(120∘)AG=2sin⁡(30∘)sin⁡(120∘)=23
We now know 2 sides of the right angle triangle AGHAGH and so we can find the third side

second triangle

AG2+GH2(23–√)2+GH2GH2GH2GH=AH2=92=81−43=2393=2393−−−−√
AG2+GH2=AH2(23)2+GH2=92GH2=81−43GH2=2393GH=2393
Now we can see that the base of the tetrahedron is 11 unit off the ground, and the top point is 88 units below the top of the top ball. So, the total height of the top of the top ball off the ground is
2393−−−−√+9≈17.93
2393+9≈17.93

Anybody have cryptography question then plzz share...6 Dec

10+root(3)

height of small sphere + height of big sphere =9unit...

sm1 plz explain me with dig

Prashant y u consider only two 1 unit sphere +big sphere
1+1+1+8=11,give 3 spheres of 1 unit and a big sphere is put on the top of those 3 balls

sry 2+2+2+8=14

8.5 unit
radius of small sphere(as half height created by single sphere)+height of bigger sphere
8+(1/2)=8.5

I think the answer is 33underroot2/3.....d+(n-1)dunderroot2/3 is formula to calculate height for such question when triangle is not equilateral.