Question

b) Consider the following cross section of a cylinder of radius BC = 2R and height 2R, a sphere of center K and radius R that is inside the cylinder, and a cone of the same base as the cylinder and of height 2R. Choose a point Don the symmetry axis AC. The perpendicular to the axis at D intersects the surface of the cone at G, the surface of the sphere at F, and the surface of the cylinder at E. (i) Prove that ED - DG =CA.DA. (I) Prove that FD2 = DADC. (iii) Is it true that FAP = FD2 + DAP? OK (iv) Consider the segment AH = 2R, on the line AC. Prove that AH. (FD2 +DG2) = DA. ED. E B

Answer 1

sol Ž BC=2R AC = 2R LBACELCBA-45 OK in A ADG CD=90 LA=45 <6=45 AD=GO CA.AD= ED.GD ED=BC=AC-2R (6) AC diameter of circular section LA FC 90° ( Diameter sublends go at circumference) Let' take LEAD=4 LAID = gora in AA FO is in AEDC LCFD = <AFC-LAFD = x AED and AFCD similar DC =) FD>=DA.DC. FD. FO DA = (C) <D=90 in right ang A Yes it's true right AAFD FA²= FBTDA ノ (d) AH=aR=BCEED =AC L.H.S AH(FO²+DG2) DG = DA AS <GAD =45° ao at D AH (FD² +DA²) Fo²= DA .DL or (6) AH (DA.DC+DA) DA. AH. (D(IDA) = DA AH. AC = DA. ED. ED = DA.ED? -proved from (ii) =R.H.S

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