163 EX AM PI 



1J. A rl.'d triangular lamina A HC- is sns].rn(Vd 



h in its hypothenuse u4A'; prove that in the 

 position of equilibrium Alt will be horizontal it' 

 J/>: ////:: .! -f AC 9 : AB* + 



1 .".. A irivrn isosceles triangle is inscribed in a circle; find 

 of -ra\ ity of the remaining area of the circle. 



11. If throe uniform rods be rigidly united so as to form 

 half of a regular hexagon, prove that if suspended from one 

 of the angles, one of the rods will be horizontal. 



15. If ABO be an isosceles triangle having a right angle 

 at C, and D, E be the middle points of AC, AB respecti 

 prove that a perpendicular from J^upon BD will pass through 

 the centre of gravity of the triangle BDC. 



1G. ABCD is any plane quadrilateral figure, and a, b, c, d 

 are respectively the centres of gravity of the triangles BCD, 

 CDA, DAB, ABC] shew that the quadrilateral abed is 

 similar to ABCD. 



1 7. A, B, C, D, E, .Fare six equal particles at th'- an-lrs 

 of nut/ plane hexagon, and a, I, c, d, e,f are the centres of 

 gravity respectively of ABC, BCD, CDE, DEF, EFA, and 

 FAB. Shi-w that the opposite sides and angles of the 

 hexagon abcdef are eaual, and that the straight lines joining 

 opposite angles pass through one point, whicn is the centre of 

 gravity of the particles A, B, C, D, E, F. 



18. A straight line ED cuts off - th part of the right- 



1 triangle ABC of which A is the right angle. A I> = ft, 

 b. Shew that the centre of gravity of CEDB describes 

 the curve whose equation is 



- = {3 (n - l)y-nl] (3 (n- 1) x - na). 

 n 



The distance of the centre of gravity of any number 

 of sides ^47?, BC, CD ...... KL of a regular polygon from the 



centre of the inscribed circle 



__ AL x radius _ 

 = AB + BC+CD + ...... + KL ' 



