A complete and practical solution book for the common school teacher . FIG. 105. FIG. 106. (1) 304 sq. ft.—1 sq. ft.—303 sq. ft. (2) 6x5 = 300 squares as shown in Fig. 105. (3) 6+5=11 rectangles 1 ft. wide. (4) 30 squares+11 rectangles = 303 sq. ft. (5) 303+30=10^0 sq. ft., area of each square. (6) 11 rectangles+30=4^ sq. ft., addition to each square. (7) i-i+2=fL ft., width of the small rectangle in Fig. 106. (8) (U)2=-ii(h SQ- ft-» area of square O in Fig. 106. (9) W0-=WoV- (10) -^VoV+aVV^WA1 sq. ft. (11) Extracting the square root of this fraction, we have J^ ft., side of square enla
A complete and practical solution book for the common school teacher . FIG. 105. FIG. 106. (1) 304 sq. ft.—1 sq. ft.—303 sq. ft. (2) 6x5 = 300 squares as shown in Fig. 105. (3) 6+5=11 rectangles 1 ft. wide. (4) 30 squares+11 rectangles = 303 sq. ft. (5) 303+30=10^0 sq. ft., area of each square. (6) 11 rectangles+30=4^ sq. ft., addition to each square. (7) i-i+2=fL ft., width of the small rectangle in Fig. 106. (8) (U)2=-ii(h SQ- ft-» area of square O in Fig. 106. (9) W0-=WoV- (10) -^VoV+aVV^WA1 sq. ft. (11) Extracting the square root of this fraction, we have J^ ft., side of square enlarged, Fig. 106. (12) ^ ft. = the addition. (13) VV—H=W-=^ ftri side of squares in Fig. 105, room 5 by 6 squares. (14) 5X3=15 ft, one dimension. (15) 6x3 = 18 ft., the other dimension. Note.—Solved for the Teachers Review bv N. D. Moser. PROBLEM 433. What is the area of a trapezium the diagonal of which is 110 the perpendiculars to the diagonal 40 and 60 ft. respectively? MENSURA TION. !19. FIG. 107. Solution. (1) Let ABCD be the trapez- ium, AC the diagonal. (2) DE=40ft., and FB = 60 ft., the perpendiculars to AC (3) 110 ft. = AC=base of the triangle ACD. (4) ^(ACxDE)=2200 sq. ft., area of the triangle. (5) AC=the ba^-e of the tri- angle ACB. (6) FB=altitude of ACB. (7) .. \ of (ACX FB)=330O sq. ft. = area of ACB. (8) Then ACD + ACB=5500 sq. ft., area of trapezium. PROBLEM 434. Find the perimeter of a rhombus, area 21(j sq. ft., and one diagonal24 ft. Solution. (1) Let ABCD be the rhombus. (2) AC=24 ft., the given diag- onal, and DB is the otherdiagonal. (3) The triangle DBC=| of 216, or 108 sq. ft. (4) The area of the right triangle DFC=* of 108=54 sq. ft. (5) DF=(54x2) — FC, or 12= 9 ft. (6) Then DB = 9x2=18 ft. (7) DC= VDF* + FC* = 15 ft., the side of the rhombus. (8) .. The perimeter is 15x4, or 60 ft.
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Keywords: ., bookcentury1800, bookdecade1890, booksubject, booksubjectgeometry
