MENSURATION. 
Ill 
proportion^ between OAx BO and SA x BS; that is, the 
fquare of the area =OAxBO x SAxBS. 
Cor. Hence the perimeter of the triangle ABDwiU 
always be equal to both the tangents BS, BT, whatever 
may be the pofition of the fide AD, provided it is drawn 
to touch the circle whole centre is G. 
Let BD=42, AD=3 o, and BA=22, as in the pre¬ 
ceding example: then 42+30+22+2=47, which is halt 
the perimeter. The three remainders are 5, 17, 25; and 
47X5X17X25=99875, the continued product; and the 
Iquare-root of 99875 is 316*03 nearly, the area of the 
triangle. ■ , . , 
To find the Area of a Trapezium.— Let the trapezium be 
divided into two triangles by a diagonal, then the aieas of 
the triangles added together will be the content ot the 
trape7.ium. . 
Ex. 1. What is the area of the trapezium ABCD, 
fig. 5. when the diagonal BD=49'7, and the perpendicu¬ 
lars on BD are 00=33*5, anc ^ AP=22 ? 
49 7 x 33 5 „ 832*475 area of triang. B D C. 
2 
49*7X11 = 546*7 area of triang. DBA. 
1379*175 area of ABCD. 
Ex. 2. Let the meafured Tides of the quadrangular field 
ABCD, fig. 6. be, AB=i5 chains 24 links, AD=n 
chains 14 links, CB=i8 chains 86 links, CD=9 chains 
qo links *. and fuppofe the angles at A and C, taken with 
a theodolite, are, DAB=io5° 28', and DCB_89 54. 
What is the content in acres?—The Tides AB and AD 
with the included angle give the area of the triangle 
BDA=8i*8i4 chains; and CB and CD with the included 
angle give 9.3*3.55 chains, the area of CDB; the Turn is 
375*169 chains, =17*5169 acres. 
To find the Area of an irregular Polygon of any number 
of fides— Divide the figure into triangles and trapeziums 
in the moft convenient manner; and find the area ot each 
feparately; then the fum of thefe areas will be the area of 
the polygon. . 
Ex. It is required to lay down the irregular figure 
ABCDEFGA, fig. 7. and to find its area from the fol- 
ronals. Feet. 
Perpendiculars. Feet. 
A»n=25 1 
Cm— 8 
AB = 35 * 5 _J 
Go— 7*5 \ 
A 0 = 12 1 
Cn = 8*6/ 
An = 24*8 V 
E p =10*4! 
AD = 36 J 
G> = 9*2 J 
T)p— 16 ] 
D r =23*5 V 
D F = 34 J 
Draw the line AB, which make =35*5; and lay off 
-25 from A to m , at which point ere< 5 t the perpendicular 
Cm »=8 ; join AC and BC, and you will have the tri¬ 
angle ABC. With C as a centre, and the radius C n, de- 
feribe an arc ; and with A as a centre, and the radius A n, 
deferibe another arc cutting the former in n. Through n 
draw the diagonal A D= 36, upon which lay off A 0=12. 
At 0 ereft the perpendicular G0=7*5 5 join CD, DG, 
and G A, and the trapezium A C D G will be completed. 
The trapezium DEFG may be conflru&ed in a iimilar 
manner. 
Here 35*5X 8=284, double the area of the triangle ABC. 
Again, 7*5+8 6X 36 =i6*iX 36 = 579 ‘ 6 » do uble the area of 
the trapezium ACDG. Alfo, io*4+9*2X 34=19*6 x 34 
=666*4, double the area of the trapezium DEFG.. Then, 
. — - -=-= 765 feet, the area of the lrre- 
2. 2 
jular polygon required. 
Given the fide of a Regultir Polygon, to find the radius 
of its inferibed or circumfcribing circle .—Multiply the given 
tide by the number Handing opjmfite to the name of the 
polygon, in the third or fourth column of the following 
Table, as the cafe requires; and the product will be the 
radius of the inlcribed or circumfcribing circle. 
Table of Polygons. 
No. of 
Sides. 
Names. 
Rad.of the 
inferibed 
Circle. 
Rad. of the 
circums. 
Circle. 
Multipliers 
or Areas. 
3 
Trigon, or 
0*2886751 
°' 57735°3 
0*43301 27 
4 
Tetragon, or O 
0*5000000 
0*7071068 
I *0000000 
5 
Pentagon 
0*6881910 
0*8506508 
17204774 
6 
Hexagon 
0*8660254 
I'OOOOOOO 
2*5980762 
7 
Heptagon 
1*0382617 
1*1523825 
3*6339124 
8 
Ottagon 
1*2071068 
1*3065630 
4*8284271 
9 
Nonagon 
I ' 37373 s 7 
1*4619022 
6*i 818242 
IO 
Decagon 
1*5388418 
1*6186340 
7*6942088 
I I 
Undecagon 
1*7028437 
1*7747329 
9*3656399 
I z 
Duodecagon 
1*8660254 
1*93*18516 
11 *1961 524 
Ex. The fide of a regular pentagon, fig. 8. is 21.75; 
what are the radii of the inferibed and circumfcribing 
circles ?—Here *688191x21*75 = 14*96815425 = D C, the 
radius of the inferibed circle; and *8506508X21*75 = 
i8*5oi 6549=AC, the radius of the circumfcribing circle. 
To find the Area of a Regular Poh/gon, when the fide 
only is given. —Multiply the fquare of the given fide by 
the number or area Handing oppofite to the name of the 
polygon in the laH column of the preceding Table, and 
the produft will be the area. 
Ex. If the fide of a pentagon be 8 feet 4 inches, what 
is its area ?—Here 8 feet 4 inches = 8*^ feet ——; and 
3 
25 25 625 , . r , r , , 625 
—X—=—, the fquare-of the fide; then 17204774X—- 
3 3 9 9 
1075*298375 . , . , 
=---- =119*4775972 feet, the area required, 
9 
The Diameter of a Circle being given, to find the Circum¬ 
ference ; or, the Circumference being given, to find the Dia¬ 
meter. 
Rule I. As 7 is to 22, fo is the diameter to the circum¬ 
ference ; or, as 22 is to 7, fo is the circumference to the 
diameter. 
Rule II. As 113 is to 355, fo is the diameter to the 
circumference; or, as 355 is to 113, fo is the circumfe¬ 
rence to the diameter. 
Rule III. Multiply the diameter by 3*1416, and the 
product will be the circumference; or, divide the cir¬ 
cumference by 3* 1416, and the quotient will be the dia¬ 
meter. 
Ex. If the diameter of a circle be 12, whatis the circum¬ 
ference ? 
By Rule I. As 7:22:: 
12 : 
264 
-=37714285, the cir- 
7 
cumference required. 
By Rrde II. As 113 : 355 :: 12 : 37*699115, the cir¬ 
cumference required. 
By Rule III. Here 3*1416X12=37*6992, the circumfe¬ 
rence required. 
To find the Length of any Arc of a Circle .—From 8 times 
the chord of half the arc fubtrafl the chord of the whole 
arc, and A of the remainder will be the length of the arc, 
nearly. 
Ex. The chord, A B, fig. 9, of the whole arc, is 24, and 
the verfed fine CD=9 ; what is the length of the arc ABC ? 
•—Here i2 2 +9 2 =i44+8i=225 ; and 3/225 = 15 = AC, 
the chord of half the arc ; then.-=---= 
3 3 
—=32, the length of the arc required. 
3 
To find the Area of a Circle. 
Rules, i. Multiply half the circumference by half the 
diameter, and the product will be the area. Or, divide the 
product of the whole circumference and diameter by 4, and 
the quotient will be the area. a. Multiply the iquare of 
the diameter by 7854, gnd the produft will be the area. 
J 3. Mul- 
