Whence 262.3+255.2+49.6=567.1=area of the whole figure required.

2. *In a pentangular field, beginning with the south side and measuring round towards the east, the first, or south side, is 2735 links, the second 3115, the third 2370, the fourth 2925, and the fifth 2220; also the diagonal from the first angle to the third is 3800 links, and that from the third to the fifth 4010; required the area of the field.

Ans. 117 ac. 2 ro. 39 po.

Promiscuous Questions concerning Lines and Areas.

1. Given AC=32, AD=3, EC-8, the perpendicular DG=4, and the perpendicular EF=6, to find the area of the triangle ABC and the sides AB and BC.

To find the area of mixed or compound figures, or such as are composed of rectilineal and curvilineal figures together; the rule is to find the area of the several figures of which the whole figure is composed, then add all the areas together, and the sum will be the area of the whole compound figure. And in the same manner may any irregular field or piece of land be measured, by dividing it into trapeziums and triangles, and finding the area of each separately.

*Note. As this figure consists of three triangles, all of whose sides are given, by calculating their areas according to Problem III. the sum will be the area of the whole figure accurately, without drawing perpendiculars from the angles to the diagonals.

The same thing may also be done in most other cases of this kind.

Draw GH parallel to BC, then since the sides about the equal angles in equiangular or similar triangles are proportional, we have in the similar triangles CEF and HDG, FE: EC: GD: DH=1, which added to AD will give AH=5. Also in the sim. A's AGH, ABC, AH : AC :: GD: BI=15.36, and this multiplied by half the base will give the area of ABC=245.76.

Again in each of the right angled triangles ADG and CEF we have the two legs to find the hypothenuses AG= 5 and CF=10. Now by sim. A's ADG, AIB, GD : GA :: BI: BA=19.2; and FE : FC :: BI : BC.

2. If from the right angled triangle ABC, whose base is 12, and perpendicular 16 feet, be cut off, by a line DE parallel to the perpendicular, a triangle whose area is 24 square feet; what are the sides of this triangle?

C

E

A D B

The area of the triangle ABC=AB× BC=96, also having AB and BC, AC may be found=20.

Now it is evident that the triangles ABC and ADE are similar, and since the areas of sim. A's are as the squares of their like sides, we have,

Area ABC area ADE :: AC2: AE
Area ABC: area ADE :: BC2 : DE2

Area ABC: area ADE :: AB2 : AD2
From which AE=10, AB=6, and DE=8. Ans.

3. A gentleman in his yard has a circular grass-plot, the diameter of which is 25 yards. Query the length of the string that would describe a circle to contain nine times a much. Ans. 37.5 yards.

4. Suppose a ladder 100 feet long, placed against a perpendicular wall 100 feet high, how far would the top of the ladder move down the wall by pulling out the bottom thereof 10 feet? Ans. .5012563.

5. *There is a circular pond whose area is 50284 square feet, in the middle of which stood a pole 100 feet high: now the pole having been broken, it was observed that the top just struck the brink of the pond; what is the height of the pole? Ans. 41.9968.

6. In a level garden there are two lofty firs, having their tops ornamented with gilt balls; one is 100 feet high, the other 80, and they are 120 feet distant at the bottom; now the owner wants to place a fountain in a right line between the trees, to be equally distant from the top of each; what will be its distance from the bottom of each tree, and also from each of the balls?

Ans.

From the bottom of the lower tree 75 feet.
From the bottom of the higher tree 45 feet.
From each ball 109.6585 feet.

* This problem may be constructed by forming a right angled triangle, having the radius of the circle for the base, and the length of the pole for the perpendicular; and erecting a perpendicular on the middle of the hypothenuse to cut the perpendicular of the triangle; this will determine the place where the pole was broken.

+ The figure to this question is thus constructed. Draw AC=120, the distance of the trees at the bottom, and erect the perpendiculars

7. A person wishes to inclose 6ac. 1ro. 12po. in a triangle similar to a small triangle whose sides are 9, 8, and 6 perches respectively; required the sides of the triangle.

Ans. 59.029, 52.47, and 39.353 perches. 8. Required the sides of an isosceles triangle, containing 6ac. Oro. 12per. and whose base is 72 perches.

Ans. 45 perches each.

AE=height of the lower tree, and CD=the higher. Join ED, and from the middle of it draw the perpendicular GF, and F will represent the place of the fountain. Join EF and DF, and draw EI parallel to AC, and GB parallel to DC; then the triangles EID and GBF being similar, the calculation is evident.

OF THE

CONIC SECTIONS.

DEFINITIONS.

1. The conic sections are such plain figures as are formed by the cutting of a cone.

2. *A cone is a solid described by the revolution of a right-angled triangle about one of its legs, which remains fixed.

3. The axis of the cone is the right line about which the triangle revolves.

*This is Euclid's definition of a cone, and is that which is generally best understood by a learner; but the following one is more general.

Conceive the right line CB to move upon the fixed point C as a centre, and so as continually to touch the circumference of the circle AB, placed in any position, except in that of a plane which passes through the said point; and then that part of the line which is intercepted between the fixed point and the periphery of the circle will generate the convex superficies of a cone.