A complete and practical solution book for the common school teacher . F = 10 in. (3) The triangle BHC and BEV are similar. (4) BH : BE :: HC : EV, or 2\ : 4^ :: 10 : EV. (5) From this EV=20, AV= VXE2 + EV2 = (6) °=, solid contents of the vessel. (7) (4|)2X .7854 X ^ = cu. in., solid contents of the T\\Tr FIG. 112. cone DVC. (8) QZ=radius of the ball that will just go into the cone. (9) |(9x20)=90 sq. in, area of the triangle AVB. (10) 9+()=50 in, the perimeter of AVB. (11) Since the radius QZ of the inscribed circle is found by dividing the area by h
A complete and practical solution book for the common school teacher . F = 10 in. (3) The triangle BHC and BEV are similar. (4) BH : BE :: HC : EV, or 2\ : 4^ :: 10 : EV. (5) From this EV=20, AV= VXE2 + EV2 = (6) °=, solid contents of the vessel. (7) (4|)2X .7854 X ^ = cu. in., solid contents of the T\\Tr FIG. 112. cone DVC. (8) QZ=radius of the ball that will just go into the cone. (9) |(9x20)=90 sq. in, area of the triangle AVB. (10) 9+()=50 in, the perimeter of AVB. (11) Since the radius QZ of the inscribed circle is found by dividing the area by half the perimeter, we have QZ=90-^25 = in. (12) This is the radius of the inscribed sphere of the cone AVB, or = in., diameter of the ball that willjust go into the mouth of the vessel. (13) () = cu. in., solid contents of the ball Q. (14) — cu. in, solid contents of ADCB. (15) — cu. in, solid con- tents of whole cone—solid contents of a ball that justgoes into MENSURATION. 223 (16) \ of = cu. in. = amount of water given in the vessel. (17) + = cu. in. = amount of water given in the vessel-f-the solid contents of the smallcone DCV. (18) MN represents the surface of the water after the ball O is dropped in. (19) Now by similar solids, we have : :: ()3 : (?) or GK, from which GK = + in., thediameter of the ball O. NOTE.—If the ball O was 5 in. in diameter it would rest on the bot-tom of the vessel. PROBLEM 443. Find the diameter of the largest sphere that can be put in a hollowcone whose internal base is 2 ft. and altitude 3 ft. Solution. (1) Let MNV represent a middle section of the cone, and the inscribed sphere whose center is O. (2) Then MN = 2 ft., KV=3 ft., MV= V(32-fl2) = ft. (3) Dividing twice the area by the perimeter of MNV gives radius OG; orOG =(3x2)-=-2+ () =.720705ft. (4) He
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