AE EB, by th. 83. But AE: EB:: M: N, by construction; therefore AC: CBM N. Q. E. D.

Ex. 2. From a given circle to cut off an arc such, that the sum of m times the sine, and n times the versed sine, may be equal to a given line.

FM

But

Anal. Suppose it done, and that AEE'B is the given circle, BE'E the required arc, ED its sine, BD its versed sine; in DA (produced if necessary) take BP an nth part of the given sum; join PE, and produce it to meet BF D to AB or | to ED, in the point F. Then, since B m ED + n BD = n, BP = n. PD+n. BD; consequently m EDN. PD; hence PD: ED: m : n. PD ED (by sim. tri) PB: BF; therefore PB BF: m : n. Now PB is given, therefore BF is given in magnitude, and, being at right angles to PB, is also given in position; therefore the point is given and consequently PF given in position; and therefore the point E, its intersection with the circumference of the circle AEE B, or the arc BE is given. Hence the following.

Const. From B, the extremity of any diameter AB of the given circle, draw Bм at right angles to AB; in AB (produced if necessary) take BP an nth part of the given sum; and on BM take BF so that BF: BP: :n: m. Join PF, meeting the circumference of the circle in E and E', and BE or BE' is the arc required.

Demon. From the points E and E' draw ED and E' D'at right angles to AB. Then, since BF: BP::n: m, and (by sim. tri.) BF BP :: DE: DP; therefore DE: DP: n: m. Hence m DE = N DP; add to each n. BD, then will m. DE + n. BD=n. BD + a DP = n. PB, or the given

sum.

Ex. 3. In a given triangle ABH, to inscribe another triangle abc, similar to a given one, having one of its sides parallel to a line men given by position, and the angular points a, b, c, situate in the sides AB, BH, AH, of the triangle ABH respectively.

H

Analysis. Suppose the thing done, and that abc is inscribed as required. Through any point c in BH draw CD parallel to mвn or to ab, and cutting AB in D; draw CE parallel to be, and A DE to ac, intersecting each other in E The triangles DEC, acb, are similar, and DC : ab also BDC, Bab, are similar, and DC: ab:: BC: Bb.

E

a D

:: CE : bc ; Therefore

BC:

BC: CE :: Bb: be; and they are about equal angles, consequently B, E, C, are in a right line.

Construc. From any point c in вH, draw CD parallel to nm; on CD constitute a triangle CDE similar to the given one; and through its angles E draw BE, which produce till it cuts AH in through e draw ca parallel to ED and co parallel to Ec; join ab, then abc is the triangle required, having its side ab parallel to mn, and being similar to the given triangle.

Demon. For, because of the parallel lines ac, DE, and cb, Ec. the quadrilaterals BDEC and Bacb, are similar; and therefore the proportional lines Dc, ab, cutting off equal angles BDC, Bab; BCD, Bba; must make the angles ED, ECD. respectively equal to the angles cab, cba; while ab is parallel to DC, which is parallel to men, by construction.

Ex. 4. Given, in a plane triangle, the verticle angle, the perpendicular, and the rectangle of the segments of the base, made by that perpendicular; to construct the triangle.

F

OLI

C

ED

G

Anal. Suppose ABC the triangle required, BD the given perpendicular to the base AC, produce it to meet the periphery of the circumscribing circle ABCH, whose centre is o, in н; then, by th. 61 Geom. the rectangle BD DHAD DC, the given rectangle hence, since BD is given, DH and BH are given; therefore BI = HI is given as also ID = OE and the angle EOC is ABC the given one, be cause Eoc is measured by the arc KC, and ABC by half the arc AKC or by KC. Consequently EC and AC = 2Ec are given. Whence this

Construction.

K

Find DH such, that DB DH = the given

rectangle, or find DH =

AD. DC

=

BD

; then on any right line

=

GF take FE = the given perpendicular, and EG DH; bisect VG in o, and make EOC the given vertical angle; then will oc cut Ec, drawn perpendicular to OE, in c. With centre o and radius oc, describe a circle, cutting CE produced in a through r parallel to ac draw FB, to cut the circle in B; join AB, CB, and ABC is the triangle required.

Remark. In a similar manner we may proceed, when it is required to divide a given angle into two parts, the rect

angle

angle of whose tangents may be of a given magnitude. See prob. 40, Simpson's Select Exercisis.

Note

For other exercises, the student may construct all the problems, except the 24th, in the Application of Algebra to Geometry, at page 369, &c. of this volume. And that he may be the better able to trace the relative advantages of the ancient and the modern analysis, it will be adviseable that he solve those problems both geometrically and algebraically.

149

PRACTICAL EXERCISES IN MENSURATION.

QUEST. 1

WHAT difference is there between a floor 28 feet long by 20 broad, and two others, each of half the dimensions; and what do all three come to at 45s. per square, or 100 square feet?

Ans. diff. 280 sq. feet. Amount 18 guineas.

QUEST 2. An elm plank is 14 feet 3 inches long, and I would have just a square yard slit off it; at what distance from the edge must the line be struck? Ans. 71 inches.

QUEST 3. A ceiling contains 114 yards 6 feet of plastering, and the room 28 feet broad; what is the length of it?

Ans. 369 feet.

QUEST. 4. A common joist is 7 inches deep, and 21 thick; but wanting a scantling just as big again, that shall be 3 inches thick; what will the other dimensions be?

Ans. 113 inches.

QUEST. 5. A wooden cistern cost me 38. 2d painting within, at 6d. per yard; the length of it was 102 inches, and the depth 21 inches; what was the width ?

Ans. 271 inches.

QUEST. 6. If my court-yard be 47 feet 9 inches square, and I have laid a foot-path with Purbeck stone, of 4 feet wide, along one side of it, what will paving the rest with flints come to, at 6d. per square yard? Ans. 5 168. 0žd.

QUEST 7. A ladder, 263 feet long, may be so planted, that it shall reach a window 22 feet from the ground on one side of the street; and. by only turning it over, without moving the foot out of its place, it will do the same by a

window 14 feet high on the other side; what is the breadth of the street? Ans. 37 feet 9 inches.

QUEST. 8. The paving of a triangular court, at 18d. per foot, came to 100.; the longest of the three sides was 88 feet; required the sum of the other two equal sides? Ans. 106-85 feet.

QUEST. 9. There are two columns in the ruins of Persepolis left standing upright: the one is 64 feet above the plain, and the other 50 in a straight line between these stands an ancient small statue, the head of which is 97 feet from the summit of the higher, and 86 feet from the top of the lower column, the base of which measures just 76 feet to the centre of the figure's base. Required the distance between the tops of the two columns ? Ans. 157 feet nearly.

QUEST. 10. The perambulator, or surveying wheel, is so contrived, as to turn just twice in the length of 1 pole, or 16 feet; required the diameter ? Ans. 2.626 feet.

QUEST. 11. In turning a one-horse chaise within a ring of a certain diameter, it was observed that the outer wheel made two turns, while the inner made but one: the wheels were both 4 feet high; and supposing them fixed at the distance of 5 feet asunder on the axletree, what was the circumference of the track described by the outer wheel?

Ans. 62.83 feet.

QUEST. 12. What is the side of that equilateral triangle, whose area cost as much paving at 8d. a foot, as the pallisading the three sides did at a guinea a yard?

Ans. 72-746 feet.

QUEST. 13. In the trapezium ABCD, are given, ab = 13, BC = 31, CD= 24, and DA = 18, also в a right angle; required the area? Ans. 410 122.

QUEST. 14. A roof which is 24 feet 8 inches by 14 feet 6 inches, is to be covered with lead at 8lb. per square foot : what will it come to at 18s. per cwt. ? Ans. 221. 19s. 104d.

QUEST. 15. Having a rectangular marble slab, 58 inches by 27, I would have a square foot cut off parallel to the shorter edge; I would then have the like quantity divided from the remainder parallel to the longer side; and this alternately repeated, till there shall not be the quantity of a foot

left:

left: what will be the dimensions of the remaining piece? Ans. 20 7 inches by 6.086.

QUEST. 16. Given two sides of an obtuse-angled triangle, which are 20 and 40 poles; required the third side, that the triangle may contain just an acre of land?

Ans. 58.876 or 23.099.

QUEST. 17. The end wall of a house is 24 feet 6 inches in breadth, and 40 feet to the eaves; of which is 2 bricks thick, more is 1 brick thick, and the rest 1 brick thick. Now the triangular gable rises 38 courses of bricks, 4 of which usually make a foot in depth, and this is but 4 inches, or half a brick thick what will this piece of work come to at 57. 108. per statute rod? Ans. 201. 118. 7 ¿d.

QUEST. 18. How many bricks will it take to build a wall, 10 feet high, and 500 feet long, of a brick and half thick: reckoning the brick 10 inches long, and 4 courses to the foot in height? Ans. 72000.

QUEST. 19. How many bricks will build a square pyramid of 100 feet on each side at the base, and also 100 feet perpendicular height: the dimensions of a brick being supposed 10 inches long, 5 inches broad, and 3 inches thick? Ans. 3840000;

QUEST. 20. If, from a right-angled triangle, whose base is 12, and perpendicular 16 feet, a line be drawn parallel to the perpendicular, cutting off a triangle whose area is 24 square feet; required the sides of this triangle ?

Ans. 6, 8, and 10.

QUEST. 21. The ellipse in Grosvenor-square measures 840 links across the longest way, and 612 the shortest, within the rails now the walls being 14 inches thick, what ground do they enclose, and what do they stand upon?

Ans.

enclose 4 ac.0r 6 p. stand on 1760 sq. feet.

QUEST. 22. If a round pillar, 7 inches over, have 4 feet of stone in it of what diameter is the column, of equal length, that contains 10 times as much?

:

Ans. 22.136 inches.

QUEST. 23. A circular fish-pond is to be made in a gar den, that shall take up just half an acre; what must be the . length of the chord that strikes the circle? Ans. 273 yards. Eeee QUEST.

VOL. I.