greater base 9 inches, and each side of the less 6 inches; how much must be cut off from the less end to contain a solid foot? Ans. 3.39214 ft.

89. What must be the inside dimensions of a cubical box to hold 200 balls, each 2 inches in diameter ?

90. Near my house I intend making a hexagonal or sixsided seat around a tree, for which I have procured a pine plank 16 feet long and 11 inches broad; what must be the inner and outer lengths of each side of the seat, that there may be the least loss in cutting up the plank ?

Ans. 26.64915 in. inner, and 39.35085 in. outer length. 91. Required the capacity of a tub in the form of a frustum of a cone, of which the greatest diameter is 48 inches, the inside length of the staves 30 inches, and the diagonal between the farthest extremities of the diameters 50 inches. Ans. 165.34 gals.

92. The front of a house is of such a height, that, if the foot of a ladder of a certain length be placed at the distance of 12 feet from it, the top of the ladder will just reach to the top of the house; but if the foot of the ladder be placed 20 feet from the front, its top will fall 4 feet below the top of the house. Required the height of the house, and the length of the ladder.

Ans. 34 feet, the height of the building; 36.0555 feet, the length of the ladder.

93. A sugar-loaf in form of a cone is 20 inches high; it is required to divide it equally among three persons, by sections parallel to the base; what is the height of each part?

Ans. Upper 13.8672, next 3.6044, lowest 2.5284 in. 94. Within a rectangular court, whose length is four chains, and breadth three chains, there is a piece of water in the form of a trapezium, whose opposite angles are in a direct line with those of the court, and the respective distances of the angles of the one from those of the other are 20, 25, 40, and 45 yards, in a successive order; required the area of the water. Ans. 960 sq. yd.

95. What will the diameter of a sphere be, when its solidity and the area of its surface are expressed by the same numbers ? Ans. 6.

96. There is a circular fortification, which occupies a quarter of an acre of ground, surrounded by a ditch coinciding with the circumference, 24 feet wide at bottom, 26 at top, and 12 deep; how much water will fill the ditch, if it slope equally on both sides? Ans. 135483.25 cu. ft.

97. A father, dying, left a square field containing 30 acres to be divided among his five sons, in such a manner that the oldest son may have 8 acres, the second 7, the third 6, the fourth 5, and the fifth 4 acres. Now, the division fences are to be so made that the oldest son's share shall be a narrow piece of equal breadth all around the field, leaving the remaining four shares in the form of a square; and in like manner for each of the other shares, leaving always the remainders in form of squares, one within another, till the share of the youngest be the innermost square of all, equal to 4 acres. Required a side of each of the enclosures.

Ans. 17.3205, 14.8324, 12.2474, 9.4868, and 6.3246 chains.

98. Required the dimensions of a cone, its solidity being 282 inches, and its slant height being to its base diameter as 5 to 4.

Ans. 9.796 in. the base diameter; 12.246 in. the slant height; and 11.223 in. the altitude.

99. A gentleman has a piece of ground in form of a square, the difference between whose side and diagonal is 10 rods. He would convert two thirds of the area into a garden of an octagonal form, but would have a fish-pond at the centre of the garden, in the form of an equilateral triangle, whose area must equal five square rods. Required the length of each side of the garden, and of each side of the pond.

Ans. 8.9707 rods, each side of the garden, and 3.398 rods, each side of the pond.

BOOK XIV.

APPLICATIONS OF ALGEBRA TO GEOMETRY.

692. WHEN it is proposed to solve a geometrical problem by aid of Algebra, draw a figure which shall represent the several parts or conditions of the problem, both known and required.

Represent the known parts by the first letters of the alphabet, and the required parts by the last letters.

Then, observing the geometrical relations that the parts of the figure have to each other, make as many independent equations as there are unknown quantities introduced, and the solution of these equations will determine the unknown quantities or required parts.

To form these equations, however, no definite rules can be given; but the best aids may be derived from experience, and a thorough knowledge of geometrical principles. It should be the aim of the learner to effect the simplest solution possible of each problem.

PROBLEM I.

693. In a right-angled triangle, having given the hypothenuse, and the sum of the other two sides, to determine these sides.

Let A B C be the triangle, right-an

gled at B. Put A C

=

a, the sum AB

+ BC=s, AB = x, and B C = y.

If AC 5, and the sum A B+BC=7, y = 4 or 3,

and x =

3 or 4.

PROBLEM II.

694. Having given the base and perpendicular of a triangle, to find the side of an inscribed square.

Let ABC be the triangle,

and H E F G the

the inscribed

square. Put A B = b, C D = a,

C

and GF or G H

DI = x;

that is, the side of the inscribed square is equal to the product of the base by the altitude, divided by their sum.

PROBLEM III.

695. Having given the lengths of two straight lines drawn from the acute angles of a right-angled triangle to the middle of the opposite sides, to determine those sides.

Let ABC be the given triangle, and AD, BE the given lines.

A

Put A D

= BE:
a,

=

b, CD or

by subtracting the first equation from four times the second,

which values of x and y are half the base and perpendiculars of the triangle.

PROBLEM IV.

696. In an equilateral triangle, having given the lengths of the three perpendiculars drawn from a point within to the three sides, to determine these sides.