Key to Milne's plane and solid geometry

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1899 Excerpt: ...const., EH = FG = EF = HG; hence, EFGH is a rhombus, and, as in Ex. 601, EFGH--ABCD. Sen. As in Ex. 601, there may be an indefinite number of solutions. Ex. 604. Construct a rhombus having a given altitude and equivalent to a given parallelogram. Solution. From data find the side of the required rhombus as the altitude was found in Ex. 603, and construct the rhombus as in that exercise. Ex. 605. Transform a triangle into an equivalent parallelogram, whose base shall be the base of the triangle and one of whose base angles shall be equal to a base angle of the triangle. Solution. Through E, the middle point of side AC of the given A ABC, draw ED II AB meeting BC in D; produce ED to F making DF = ED; and draw BF. Then, ABFE is the CD required. Proof. EFWAB, and since, § 158, ED = AB, EF = AB, and, § 150. ABCD is a CD. Cons'., DF= ED, § 158, DB = CD, and, § 59, Z.BDF-Z. EDC; heiiCL', ADFB = A EDC; consequently, ABDE + ADFB-ABDE + A EDC; that is,, ABFE A ABC. Ex. 606. Construct a triangle having a given angle, and equivalent to a given parallelogram. Solution. Draw the altitude h of the given CD ABCD. Draw EF = AB; at E erect the _L EH = 2 h; and draw HJW EF. Construct Z FEG = the given Z., the side EG meeting HJ, as at G. Draw FG. Then, A EFG is the A required. Proof. § 332, area ABCD = AB x h, and, § 335, urea. AEFG = EFx EH=AB x2h = ABxh; hence, AEFG-ABCD. See Sen., Ex. 518. Ex. 607. Construct a triangle equivalent to a given trapezium. Solution. Draw the diagonal AC of the given trapezium ABCD and the perpendiculars h and h from D and B respectively upon AC. Draw EF= AC, and on EF as base construct A EFG, having its altitude = h + h'. Then, A EFG is the A required. Proof. Area ABCZ) = area A.4CZ) +area A IGB = iACxh + iACx h' =...