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I * ^ f B ® 0R EM VII. Rule 5. —The sine, s, and versed sine,
half the are Kpirur n-jo o
the arc being given :* = 2s +
8 v a s 6 s ! + 5
, or 2 s-
Tbe
6 s ! + v 2
6 s 2 + v
1 W Be ° Re m VIII. Rule 6. —The diameter, d, the sine, s,Ve fsed sine, ®, of half the arc, being given :
Z =
22 d s
11 d — 8 v
, or 22 s
Jbeorem IX.
1 Slne , ®, being given :
11 s J -f 3
-^ DLE 7.—The radius, r, the chord, c, and
c v v s
Z = C + - (- —.
or c r
sis, t j® 0RR M X. Rule 1.—To rectify the curve of an ellip-e major axis, t, and the minor axis, c, being given :
22
3d 3.5 d 5.7 d' 44 B _ 6.6 ° a?
-, &c., where
—; and a, b, c, &c., are the preceding terms.
Theorem XI. Rule 2.
23 t + 21 c
z = -•
14
la
’Tjjp e following theorems, let A be the proposed area.j«tigth , REl1 —To find the area of a parallelogram ; the
Hi,, ’ an d the breadth, b, at right angles to the length,0 shven ;
Ijjeo a — lb.
ba Sfi P ^' M H- Rule 1.—To find the area of a triangle ;"he at) ’ "> and the perpendicular height, h, from the oppo-® e > being given :
Areas of Plane Surfaces.
A =
bh
tt>, l^ Elt Rule 2. —The three sides, a, b, c, being* he the half sum of the three sides, then
A -~
f " (* X s — a x S — b X s — c) 2 .Vlf BE ° R EST Tvr
* 8u m of H ULE 3.—Given, b, the base, and s, thethe two opposite sides :
A — ^ (s 2 b 2 X b 2 d 2 )\
• T 'Urnvf it ^—To find the area of a polygon ; the side, s,
• ’ tl le . er > n > of the sides of the polygon, and the radius,mscribed circle, being given :
A =
VT ^
^ V I- Rule 1.—To find the area of a circle; the
’ atl d circumference, c, being given :d c dc
%
0R Sjt ■
2 _X 2’ ° r 4
Rule 2. —The diameter, d, being given:- 7854rf -
* v HI. Rule 3.—The circumference, c, being
.0796 c*.
Theorem IX.—To find the area of the sector of a circlethe radius, r, and the arc, a, being given :
' a r a
A = fX 2- 0r T
Theorem X.—To find the area of a frustum sector, orpart of a circular ring, contained between two radii; thebreadth, b, the exterior arc, e, and interior arc, a, of the ring,being given:
, a + ea = 5 X -f-.
Theorem XI. Rule 1. —To find the area of the seg-ment of a circle ; the diameter, d, and the versed sine, v,being given ;
X 1 /2 v \
A = 2 ^ 2 " 2 X (s-rd-28d-72d^ &C -
Or thus, Rule 2.
D, &C.^
3®
5 v
A_ 7 Al*
7.3®
9.5 ®
9.6 d
11.8 d
A, B, C, &c. being the first, second, third, &c. terms.
Or thus, Rule 3.
v being the supplemental versed sine, and v = d — v,
A A , \
therefore d = v + v; therefore A = 2 ® v 2 ® 2 x ( — ■
'1.1.0
V ® S V * c.„
I 13 5 y I g 5 7 yl I"* ^ ijr q ^. 3 ) y
Or thus, Rule 4.— Where a, b, c, denote the precedingterms:
4 | | v 1® 3® 5®
A =: -®® 2 v 2 + — a — -—b + — c-—— d, &c.
3 5 v 7v 9 v 11 v
Theorem XII. Rule 5. — The chord, c, and versed sine,®, being given :
2c® ® s /2 c ®» \
+ T? orv h + 2-J? r
A =
2 cv v 3
A = _ T + 2~c'
20 c
still nearer.
Theorem XIII. Rule 6.—The chord, c, of the half are,and the versed sine, ®, of the same, being given :
/ ^ ^ 1 1 4
A = (c + —j X iqV,otc + -d 2 v 2 X — r, by Sir Isaac
Theorem XIV. Rule 7.—The radius, r, the sine, s, andthe versed sine, ®, of half the arc, being given ;
a = 11 rs x
s 1 4- ®*
11 s’ + 3 ® a
Theorem XV.—To find the area of an ellipsis; the majoraxis, M, and the minor, m, being given :
A = .7854 Mm.
Theorem XVT.—To find the area of an elliptic segment;the length, l, and the altitude, a, being given: also the breadth,b, of the segment of a circle of the same altitude and of thesame diameter as the axis of the ellipsis, perpendicularto b.
I (2ab a 3 \A = b X V3“ + 2c)‘