fcrib'ds refpectively, and the Number of the Circumfcribings being (as manifeftly it is) = Number of equal Parts into which A B is, by Suppofition, divided = ВА (= -?--)and by one more than the infcrib'd's Number; the Sum therefore of the Area's of the Circumfcribings is, by the Area of the greatest of them, more than the Sum of the Area's of the infcrib'd s. Solution. The As BAC, BDD, BET, BFF, &c. are fimilar; wherefore (by 4. 6. Eucl. El.) ba Wherefore ×a× 1+ 2+ 3+ 4+ &c. continu'd to P the Circumfcribings; and therefore ; but bp + ba 2 - ba = bp — ba is the Sum of the Area's of the infcrib'd 2 s, and therefore → ; wherefore (by our Lemma 1.)oo Scholia. 1. Hence the Area of any A BCD may be had thus, viz. on enged silu jod ng a and of the Ld A BAD is ¡¿– DAX BA (by the foregoing 2 Propofition) wherefore the Area of the ABCD is → CA + Fig. 1 DA × BA 2. 'Tis plain, that ifs of equal Bafes be infcrib'd in, or circumfcrib'd about any Plane Figure terminating at one end in a Point, and thence towards the other end continually increafing, that the Number of the circumfcribings is to the Number of equal Parts into which the proper Line is fuppos'd to be divided, and by one more than the Number of the infcrib'd s. 3. But ifs of equal Bases be fuppos'd to be infcrib'd in, and circumfcrib'd about any Figure, neither of whofe Ends terminates in a Point (as CAFF.) [See Fig. to Prop. 1.] Or if A or Right Lines be infcrib'd in, and circumfcrib'd about any Figure whatfoever, the Number of the circumfcribing Os, As, or Right Lines is to the Number of equal parts into which the proper Line is fuppos'd to be divided, as alfo equal the Number of the infcrib'd s, As, or Is.. 4. [Obferve here.] But, if inftead of circumfcribing and infcribing the given Figure with indefinitely little Figures, In the foregoing manner you only fuppofe the proper Line; fuppofe AB to be divided into an indefinite Number of equal Parts, each Part being a, and through any two of thofe Divifions next to one another at the Distances 24; and IX a from (what you make to be) the beginning of the Line A B you draw two Lines, in fuch manner, as the Nature of the Problem requires, and then compleat one of the circumfcribing along with its refpective infcrib'd indefinitelylittle Figure: Then having drawn other Lines, fuch as are (if any be) needful, you'll, by the Property of the Given Figure, discover the Values (i. e. the Lengths, Area's, or Solidities) of the faid circumfcribing and inscrib'd indefinitely - little Figures AB feverally; then, by placing 1, 2, 3, 4, &c. to Terms conti a nu'd, instead of, and Univerfally 1", 2", 3", 4" &c. instead of ", in each of the laid Values, you will have the Values of the circumfcribing and infcrib'd indefinitely - little Figures feverally, in two or more Series, whofe Sums will be had exactly, or proxime, by our Lemma 2. (or by Part 13 or 14.) whence (by our Lemma 1.) the Value of the propos'd Figure will be found. By the help of this Schol. 4. Prop. 1. may be folv'd thus, 1 4 P.............. Suppofe the L BA to be divided into an indefinite Number of equal Parts, each part being a ( PQ;) then, at the indetermin'd Distances BQ= 4, and BP = 1×4 from B draw the Lines QqD, and Pp Is BA, and draw the Lines Pa, ps BA: Then the As BAC, BQ, and BP are similar; wherefore (AB) × a is = Area of PaQ; wherefore (by Scholium 4) *1+2+3+4+ &c. continued to ba ba P Terms= bp + ba Sum of fby our Lem. 2. or by Part 13 or 14.) is bp-ba 2 ·×0+1+2+3+ &c. continu'd to 2 *Book 1. (by our Lemma 2. or by Part 13 or 14.) = the Sum of the Area's of the infcrib'd, and therefore consequently (by our Lemma x.) bo = a. Problem II. ; Given the Abfciffe CA, Ordinate A B, and property of the Semi-Parabola ABC: Required to find the Area of the faid Semi-parabola ABC. Suppofe the Ordinate A B to be divided into an indefinite Number of equal Parts, each Part being a (= PQ) and through any two next to one another of thofe Divifions, as Q and P at the Distances a= AQ, and IX 4 = AP from A, draw the Lines Qq and Pp AC;) and draw IAB (or the Lines ap and Now fuppofe AB, AC = x, and AQ"=AC-DQ. Then L112 Solution, Solution. ́ ́ By the Property of the Parabola x — za" (= AC — AQI”) X m =DQ; and x − 7 −1×a" (AC-AP) = - 2 a " + x = 3 a 1" +x − 4 al′′ m + 1 &c. to Places continu'd = q × × × a m m+i my ફે×4Á (fince y is equal x) is the Sum of the Area's of the infcribeds, which is therefore Area of the Semi-parabola ABC: And a xx a m 0+x al” +x−2 a”+x-34"+ &c. to Places continu'd, is manifeftly the Sum of the infcrib'da myx m+s + xa (fincey is equal to x) m is the Sum of the Area's of the circumfcribings, which is therefore Area of the Semi-parabola ABC; confequently If the Line CA had been fuppos'd to be divided into an indefinite Number of equal Parts, each Part being 4, and thro' any two Points D and E, of thofe Divifions, at the Diftances 7 — 1 × 4, and a from C, the Lines DDd, ECe were |
