A complete and practical solution book for the common school teacher . E, FC : FE : (7) AE=V(AF2— FE2)=3V2. (8) AG=AF+FG=4£. (9) From AGE and AOF, AG : AF(10) From FBC and CIK, CF : CK :: EF : CP= : FG=f AE : AO ¥V2. FB IK=¥ HK. (11) OH = HK-OK=|. (12) From geometry, OM=V(OHxIO)=-f. (13) AMDN = AOxMOX)r={f7rV2. (14) The cone ACD isequal toff7rV2x PC-h3=+ (15) Cone ABC—ACD = cu. in., the amount of wine that will run out, NOTE.—This solution was prepared for the School Visitor by H. Sunkle. PROBLEM 445. Required the largest cube that can be placed in a conical c
A complete and practical solution book for the common school teacher . E, FC : FE : (7) AE=V(AF2— FE2)=3V2. (8) AG=AF+FG=4£. (9) From AGE and AOF, AG : AF(10) From FBC and CIK, CF : CK :: EF : CP= : FG=f AE : AO ¥V2. FB IK=¥ HK. (11) OH = HK-OK=|. (12) From geometry, OM=V(OHxIO)=-f. (13) AMDN = AOxMOX)r={f7rV2. (14) The cone ACD isequal toff7rV2x PC-h3=+ (15) Cone ABC—ACD = cu. in., the amount of wine that will run out, NOTE.—This solution was prepared for the School Visitor by H. Sunkle. PROBLEM 445. Required the largest cube that can be placed in a conical cup 20 and 12 in. in diameter. Solution. (1) Let ACB represent a section of the conical cup andcube, where EF representsthe diameter of the cup onwhich the lower side of thecube rests and also thediagonal of the bottom ofthe cube. (2) DC = 20 in., DB=5 in. (3) Let the edge of the cube=#. (4) Then DC : DB :: TC : TF, or 20 : 6 :: 20—* : (120—6*)-r-20. (5) But the diagonal of the side of thecube is (120—6*)4-10—x^/2f whence we find x=+ FIG. 114. MENSURA TION. 225
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