a rhomboides, I found the perpendicular to rise at 678, and its length 1264 links; the remainder of the base measured 2435 links; what is the area of the field? Ans. 39a. 1r. 153p.

5. A window contains 36 panes, each in the form of a rhombus, whose base measures 112 inches, and perpendicular 84 inches; what will it cost glazing at 1s. 11 d. per square foot? Ans. 2. 7s. 11d.

6. A grass-plot, in a gentleman's pleasure-ground, cost £3. 14s. 1d. making, at 4d. per square yard; what is the length of the base, the perpendicular being 40 feet, and the figure a rhombus ? Ans. 50 feet.

PROBLEM IV.

To find the area of a triangle, when the base and perpendicular are given.

RULE.

Multiply the base by the perpendicular, and half the product will be the area.

Or, multiply the base by half the perpendicular; or the perpendicular by half the base, and the product will be the area.

Note 1.-If double the area of a triangle be divided by one of its dimensions, the quotient will be the other dimension.

2. When the base, the perpendicular, and the place of the perpendicular upon the base are given, the triangle may be truly constructed; and its other two sides measured by the scale used in the construction.

3. If the area of a triangle be divided by half the perpendicular height, the quotient will be the base; and if the area be divided by half the base, the quotient will be the perpendicular height.

4. In order to find the true place of the perpendicular, in taking the dimensions of a triangle, or any other figure, it will be necessary to use an instrument called a "cross, or cross-staff," which may be made in the following manner: Procure a piece of board, either of sycamore, box, or mahogany, out of which form the quadrant of a circle, of about four inches radius. In it, with a fine saw, make two grooves at right angles to each other, parallel to the radii, and about half an inch deep. Through the centre of the quadrant make a circular hole, large enough to receive a staff of about three quarters of an inch in diameter. The cross or quadrant must rest and turn upon a shoulder made at a convenient distance from the top of the staff; and it may be prevented from coming off, by means of a screw passing through a hollow cylinder of wood or brass, made to fit the staff immediately above the cross. The staff must be piked with iron at the bottom, in order to enter the ground readily. It may be about eight feet in length; and if it be divided into feet and inches, or into links, it will be found very useful in taking dimensions. Crosses may be constructed in different ways; but the cross described above I find to be most convenient.

EXAMPLES.

1. Required the area of the triangle ABC, whose base AB measures, 21.68 feet, the perpendicular DC 9.46 feet, and AD 8.26 feet.

B

Feet.

21.68

9.46

13008

8672

19512

2)205.0928

102.5464 Answer.

2. What is the area of a triangular field, the base of which measures 3568 links, the perpendicular 1589 links, and the distance between one end of the base and the place of the perpendicular 1495 links?

Ans. 28a. 1r. 151p.

3. Required the area of the gable end of a house, the base or distance between the eaves being 22 feet 5 inches, and the perpendicular from the ridge to the middle of the base 9 feet 4 inches. Ans. 104 ft. 7 in. 4 parts. 4. After measuring along the base of a triangle 895 links, I found the place of the perpendicular, and the perpendicular itself: 994 links; the whole base measured 1958 links; what is the area of the triangle ?

Ans. 9a. 2r. 37p.

5. How many square yards of slating are there in the hipped roof of a square building; the base or length of the eaves from hip to hip being 23 feet 9 inches, and the distance between the middle of the base and the vertex of the roof 9 feet 6 inches? Ans. 50 yds. 1 ft. 3 in.

6. The area of a triangle is 6 acres 2 roods and 8 perches, and its perpendicular measures 826 links; what will be the expense of making a ditch, the whole length of its base, at 2s. 6d. per rood? Ans. £6. 4s. 74d.

PROBLEM V.

To find the area of a triangle, the three sides only of which

are given.

RULE.

From half the sum of the three sides subtract each side severally; multiply the half sum and the three remainders continually together, and the square root of the last product will be the area of the triangle.

Note 1.-If a triangle be accurately laid down, from a pretty large scale of equal parts, the perpendicular may be measured, and the area found by the last Problem.

2. If the rectangle of any two sides of a triangle be multiplied by the natural sine of their included angle, the product will be double the area of the triangle; consequently, if double the area of a triangle be divided by the rectangle of any two of its sides, the quotient will be the natural sine of their included angle.

3. If the area of a triangle be divided by half the sum of the sides, the quotient will be the radius of the inscribed circle.

4. Many calculations in Arithmetic, Mensuration, &c. may be greatly facilitated by the assistance of logarithms; and it is matter of surprise that their use is not more generally taught in public schools.

The method of working by logarithms is extremely simple, and may be applied to almost every kind of calculation; yet the advantages of this great discovery have hitherto been chiefly enjoyed by those only who have leisure and inclination to cultivate the higher branches of the Mathematics.

The following are the most useful Rules on this subject; they, however, apply only to numbers greater than unity; for if the numbers be less than unify, the indices of their logarithms will be negative, which requires a little more ingenuity in the management.

Multiplication by Logarithms.

RULE.

Add the logarithms of all the factors together, and their sum will be the logarithm of the product.

Division by Logarithms.
RULE.

From the logarithm of the dividend subtract the logarithm of the divisor; and the remainder will be the logarithm of the quotient.

Rule of Three by Logarithms.

RULE.

Add the logarithms of the second and third terms together, and from their sum subtract the logarithm of the first term; and the remainder will be the logarithm of the fourth term.

Involution, or Raising of Powers.

RULE.

Multiply the logarithm of the given number by the index of the proposed power; and the product will be the logarithm of the power sought.

Evolution, or Extraction of Roots.

RULE.

Divide the logarithm of the given number, by the index of the proposed root; and the quotient will be the logarithm of the root required.

Further directions on this head appear to be unnecessary, as it is presumed that few Teachers are without a treatise on logarithms; it may, however, be observed, that Dr. Hutton's Mathematical Tables are considered the most useful.

EXAMPLES.

1. What is the area of the triangle ABC, the side AB measuring 25, AC 20, and BC 15 chains?

C

Here (25 +20 + 15) ÷ 2 = 60 ÷ 2 = 30 = half the sum of the sides; then 30 - 25

5, the first remainder ; 30-20= 10, the second remainder; and 30 – 15 = 15,

=

the third remainder; whence ✓ (30 × 5 × 10 × 15): /22500: = 150 square chains = 15 acres, the area required.

Divide by the index of the root

The quotient is the log. of 150, the area

1.4771213

0.6989700

1.0000000

1.1760913

2)4.3521826

2.1760913

2. Required the area of a grass-plot in the form of an equilateral triangle, whose side is 36 feet.

Ans. 561.18446 feet. 3. What is the area of the gable end of a house forming an isosceles triangle, whose base is 25, and each of its other equal sides 20 feet? Ans. 195.15618 feet. 4. What is the area of a triangular field, whose three sides measure 2564, 2345, and 2139 links?

Ans. 23a. 2r. Op. 5. The three sides of a triangular fish-pond measure 293, 239, and 185 yards; what did the ground which it occupies cost, at £185 per acre?

PROBLEM VI.

Ans. £843. 7s. 8d.

Any two sides of a right-angled triangle being given, to find the third side.

RULE.

1. When the two legs are given, to find the hypothenuse. Add the square of one of the legs to the square of the other, and the square root of the sum will be the hypothenuse.

2. When the hypothenuse and one of the legs are given, to find the other leg.

From the square of the hypothenuse subtract the square of the given leg, and the square root of the remainder will be the required leg. (See Theorem 7.)

Note. When the area of a right-angled triangle and the hypothenuse are given, the legs may be found by the following General Rule: To the square of the hy pothenuse add four times the area of the triangle, and the square root of this num

ber will be the sum of the legs. From the square of the hypothenuse take four times the area of the triangle, and the square root of the remainder will be the difference of the legs. Add half the difference of the legs to half their sum, and you will obtain the greater leg; but if half the difference of the legs be taken from half their sum, the remainder will be the less leg.

Here 362 +272 = 1296 + 729 = 2025; and 2025 = 45, the hypothenuse AC.

2. If the hypothenuse AC be 60, and the perpendicular BC 36; what is the base AB?

=

=

Here 602 362 3600 - 1296 = 2304; and √2304 48, the base AB.

3. Required the length of a scaling ladder to reach the top of a wall whose height is 33 feet; the breadth of the moat before it being 44 feet. Ans. 55 feet. 4. What must be the length of a shore, which, strutting 10 feet 9 inches from the upright of a building, will support a jamb 18 feet 6 inches from the ground?

Ans. 21 ft. 4 in. 9 parts. 5. The distance from the ridge to the eaves of a building is 15 feet, and the perpendicular height of the gable end 9 feet; what is the breadth of the building?

Ans. 24 feet.

6. A ladder 46 feet in length, being placed in a street, reached a window 26 feet from the ground, on one side; and by turning it over, without removing the bottom, it reached another window 35 feet high, on the other side what is the breadth of the street? Ans. 67.79695 feet. 7. A castle wall there was, whose height was found To be 100 feet from th' top to th' ground; Against the wall a ladder stood upright, Of the same length the castle was in height: A waggish youngster did the ladder slide (The bottom of it) 10 feet from the side: Now I would know how far the top did fall, By pulling out the ladder from the wall?

Ans. 6 inches, nearly