Whence, if the two fides AC, CB, of the triangle be equal to each other; the two lunes will, alfo, be equal, and each lune L1 equal to the ▲ ACD; and therefore the fegment si femicircle AQCA-AACD double the feg. AWDA, or double the feg. DVCD.

*SECT. II.

A PROMISCUOUS COLLECTION OF QUESTIONS CONCERNING AREAS.

QUESTION I.

a

'N the trapezium ABCD are given AB = 61⁄2, BC = 153, CD = 12, and DA 9, alfo в a right angle; to find the area of the trapezium.

✔ 37°9 × 4*1 × 13′9 × 19′9 = 4√42982°4279 = 51.8305098 the area of the triangle ADC.

And

*Some of the questions in this fection are taken from other books; but the methods of folution are, generally, different from those used in the books from which they were taken, they being there mostly folved by an analytical procefs. And I have conftructed thofe of which the conftructions do not appear to be felf-evident.

the triangle ABC.

The fum is 102.5305098, the area of the trapezium required.

QUEST. 2. To find the area of a trapezium, the length of its fides being as in the margin, and the fum of the two oppofite angles B and D equal to 180 degrees.

Chains AB = 15.6 BC= 13°2

CD 10°O

DA = 26.0

By rule 5 of prob. 3 of fect. 1. (32.4 being half the perimeter)

√(32·4—15⋅6) × (32·4—13·2) × (32′4—10) × (32′4—26)= ✓ 16.8 × 19.2 X 22·4 X 6·4 = 215.04 fquare chains 21504 acres 21 a. 2r. o 64 perches, the area required.

Note. A conftruction of this problem may be feen in Simpson's Select Exercifes, page 135.

QUEST. 3. In the pentangular field ABCDE are given AB 14, BC=7, CD = 10, DE 12, EA=5, and the diagonal AC 17 chains, alfo e a right angle; required the area.

E

A

D

B

First ADV DE2 + EA2 = √122 + 52 = 13. Then AEX ED = 5 × 6 = 30 ▲ADE. And 20X3X7X10=10/42=64.807407=A ADC. Alfo✓19×2×5×12=2/570=47°749345=▲ ABC. The fum of all three is 142 556752 fquare chains 14a. 10227 r. the area required.

QUEST.

QUEST. 4. Given the bafe Ac=32, AD=5, EC9, the perpendicular EF = 4, and the perpendicular DG3; required the area of the triangle

ABC.

F

A D HI

E

C

Then, by the fimilar
FE 4: EC 9: GD 3 :

Draw GH parallel to BC.

27

triangles GDH, FEC, we have DH ; and hence AH = 47 Again, from the fimilar triangles AGH, ABC, we have AH 7: AC 32:: DG 3 the perpendicular IB 334.

47.

47
4

Hence ACXIB 16X38413037=130*723447 = is the area of the triangle required.

QUEST. 5. What is the fide of that equilateral triangle, whofe area coft as much paving at 8d a foot, as the pallifading the three fides did at a guinea a yard?

The fides are 7s. a foot, and the areas. a foot. And that the produces may be equal, the quantities must be inversely as the prices; but, by rule 2, page 113, BC2/3 is the area; therefore:7:3BC: BC2 √3

B

A

P

QUEST. 6. Surveying a quadrangular field, I found the four fides to be 10, 9, 7, and 6 chains, in a fucceffive order: I likewife, at the two extremes of the longest fide, took the bearings of the oppofite angles, which were N. E. by E. and N. w. Hence the content of the field is required.

From

From the given bearings of the angles c and D, from A and B, it appears that the diagonals AC, BD, make the angle AID at their interfection of 7 points or 78 degrees; therefore, by rule 3, problem 3, fection 1, we have (102+7-92-62) × tang. 44 7810 = 32 × 1 × 50273395 = 8 × 5·0273395 = 40218716 fquare chains 4a. 3'499456 perches, the content required.*

QUEST.

the

With two of the given lines ab 10, and be 6, make a right angle b; and with the other two complete the trapezium Abc D. In the perpendicular Ec produced, take EF = 8.937492 double of the area 40.218716 divided by AD 9; and with the center F and radius equal to 6 a fourth proportional to DA, Ab, bc, defcribe an arc meeting, in c, another arc defcribed with the center D and radius Dc; then join D, C, and with the other two given lines, cb and ba, complete the trapezium ABCD, and it is done.

and

=

=

For, having drawn AC, AC, and CF, and let fall the perpendiculars, CP upon AD, CQ upon AB, and FG upon PCG; fince AD2 + DC2 + 2AD X DPAC AB2 + BC2+2AB X BQ, AD2+ DC2 + 2AD X DE = Ac2 = Ab2 + bc2, by taking thefe latter quantities from the former, it follows that 2AD XEP 2AD X DP2AD X DE = 2ABX BQ, and confequently BQ FG (EP) :: DA: AB :: BC: CF by the conftruction; whence the triangles CQB, CGF are fimilar, and therefore CQ: CG :: C: CF :: (by the conftruction) AD AB; and hence cox ABAD X CG; and, by adding CP X AD to each we have CP X AD + CQX AB = double the area ABCD CP X AD +CG X AD = EFX AD = (by the construction) double the given

area.

QUEST. 7. Given two fides of an obtuse-angled triangle, which are 20 and 40 poles; required the third fide, that the triangle may contain juft an acre

Now the area 160 20 (AB) = 8 the perpen÷ = dicular CD; but CB-CD2 = √202 - 82 = DB; hence AD AB ± BD = 40 ±√202 −82 and AC = ✔AD2+DC2=√82+402+202-8280/(202-82)

2

=4√125±20/21=4X√105 25=58.87634686 23*09925922, either of which may be the fide

and required.

QUEST. 8. Given the four fides AB = 9, BC = 10, CD11, and DA = 12, and the angle BBC = 30°, formed by the diagonal BD and the fide Dc; required the area of the trapezium.

I

First, by trigonometry, BC: CD :: S. 4 BDC

II

20

=55 the fine of the angle DBC = 33° 22′; whence 180°-30° — 33° 22′ = 180°— 63° 22′ = 116° 38′ = c; and s. 4BDC : s. LC :: BC BD = 10 X 8938936×2 = 17.877872.

Then BC XCD Xs. LC 10 X 11,X 4469468 = 49.164148 the arca of the triangle вCD.

And