Let A and a be Perpendiculars to the two Bafes D and d. Then DA the Area of A B C D And d a the Area of A b c d 2 }By Lemma 3, Page 303. } &c. By Theorem 13. I 2 Ddd A. By Axiom 3. DD: dd: DA: da. And fo for other Sides. THEOREM XXIII. In every Obtufe-angled Triangle (as B CD) the Square of the Side fubtending the obtufe Angle (as D) is greater than the Squares of the other two Sides (B and C) by a double Rectangle made out of one of the Sides (as B) and the Segment or Part of that Side produced, (as a) until it meet with the Perpendicular (P) let fall upon it. (12. e. 2.). That is, DD=BB+CC+2 Ba. Hence 'tis evident, that, if the Sides of any Obtufe-angled Triangle are given, the Segment (a) of the Side produced (or the Perpendicular, P) may be eafily found. THEOREM XXIV. a Perpendicular (as P) be let fall in any Acute angled Triangle (as B CD), the Square of either of the Two Sides (as D) is lefs than the Squares of the other Side, and that Side upon which the Perpendicular falls (viz. C and B) by a double Rectangle made of the Side B, and that Segment or Part of it (viz. a) which lies next to the Side C. (13. e. 2.) That is, DD2Ba BB+ CC. Demon Hence it follows, that, if the Sides of any Acute-angled Triangle be known, the Perpendicular P, and the Segments of the Side whereon it falls (viz. a, e.) may be easily found. CHA P. IV. The Solution of several Easy Problems in Plain Geometry, whereby the Learner may (in Part) perceive the Application or Ufe of the foregoing Theorems. N OT E, when a Line, or the Side of any plain Triangle, is aný Way cut into two (or more Parts, either by a Perpendicular Line let fall upon it, or other wife, thofe Parts are usually call'd Segments; and fo much as one of thofe Parts is longer than the other, is call'd the Difference of the Segments. And when any Side of a Triangle, or any Segment of its Side is given, 'tis ufually mark'd with a small Line cross it, thus: -+and thofe Sides, or Parts of Sides, that are fought, are mark'd with four Points, thus: PROBLEM I. To cut or divide a given Right line (as S) into Extream and Mean Proportion. (11. e. 2.) That is, to divide a Line fo, that the Square of the greater Segment (or Part) a, may be equal to the Rectangle made of the whole Line S, and the leffer Segment e. S a aa 2 and 34 S-a. By Axiom 5. S 5aa SS-Sa 7a= √ SS + SS: SS. See Pages 195, 196. Note, The laft Problem cannot be truly anfwer'd by Numbers, but Geometrically it may be performed, thus: 1. Make a Square, whofe Side is S the given Line, and bifect one of its Sides in the Middle, as at C; upon the Point C defcribe fuch a Semicircle as will pafs through the remotest Points of the Square, and compleat its Diameter. a 2. Then will either Part of the Diameter, on each End of the Side S, bea, the greater Segment fought. But a+S:S:: S: a. By Theorem 13. Ergo, a a+Sa=SS. Which was to be done. PROBLEM II. The Bafe of any Right-angled Triangle, and the Difference between the Hypothenuse and Cathetus being given, to find the Cathetus, &c. 8 d d 6 2da = bb =2d7a= bb-dd dd =65 نک Or, 8b: d+2a:: db. By Theorem 13. dd2da. As before at the 5th Step. Here you fee that either Way raises the fame Æquation; neither is their any conftant Method or Road to be obferv'd in folving Geometrical Problems, but every one makes Ufe of such Ways and Theorems as happen to come first into their Mind, the Result being every Way the fame. PROBLEM III. The Difference between the Bafe and Hypothenufe of any Right-angled Triangle, and the Difference between the Cathetus and Hypothenuse being both given, to find the Triangle. 5 7xx+2xa+aa=ee 8 dd+2dx+2da+2xa+xx+aa=☐ Hypothenufe. ·6+79dd +2da+2xa+xx+2aa=yy+ee. :e The two laft Steps are equal, by Theorem 11. Confequently, if thofe Things that are equal in both be taken away, the Remainders will be equal. By Axiom 2. aa2dx 1600 12 13 d + a = 72 = y 1+2+11 14 | d + x + a = 97 PROBLEM IV. The Hypothenufe, and the Sum of the other two Sides, of any Rightangled Triangle, being given, thence to find the Sides. e 2ae-SS-HH S aa2ae +ee=2HH-SS By Fig. 3 aa+ee=HH 2 3456.7 35 a-e=√ HH-SS 27 82a=S+√2HH-SS=144 8 + 2 9 a=S+ √2HH-SS 2 72 The Bafe requir'd. 65 The Cathetus. PROBLEM V. The Hypothenufe, and the Difference of the other two Sides of any Right-angled Triangle being given, to find the Sides. Let Ilb=97 As before d=7 Quere a In any Right-angled Triangle, either the Bafe, or Cathetus, and the alternate Segment of the Hypothenufe made by a Perpendicular let fall from the Right-angle being given, to find the other Segment. 5 cc-aa-ee. By Theor. 11. 6 baccaa 7aa+ba = cc bb = cc + 4 bb 9a+b= √ cc + bb 1⁄2 610 a = √ cc +÷bb:-b=27 And fo on fore, &c. |