3. How many square yards in a right-angled triangle, whose base is 40 feet, and perpendicular 30 feet? Ans. 66 square yards.

4. How many square yards in the triangle, whose base is 49 feet, and height 254 feet?

Ans. 68, or 68.7361.

5. To find the area of the triangle, whose base is 18 feet 4 inches, and height 11 feet 10 inches.

To find one side of a right-angled triangle, having the other two sides given.

The square of the hypotenuse is equal to the sum of the squares of the two legs. Therefore,

1. To find the hypotenuse; add the squares of the two legs together, and extract the square root of the sum.

2. To find one leg; subtract the square of the other leg from the square of the hypotenuse, and extract the root of the difference.*

EXAMPLES.

1. Required the hypotenuse of a right-angled triangle, whose base is 40, and perpendicular 30.

40

*By Euc. 475 I. AB'+BC'=AC', or AC2-AB'= BC, and therefore AB+BC=AC, or √AÇ'—AB'— BC, and is the same as the rule.

2. What is the perpendicular of a right-angled triangle, whose base AB is 56, and hypotenuse AC 65 ?

3. Required the length of a scaling ladder, to reach the top of a wall, whose height is 28 feet, the breadth of the ditch before it being 45 fect. Ans. 53 feet.

4. To find the length of a shoar, which, strutting 12 feet from the upright of a building, may support a jamb 20 feet from the ground. Ans. 23 32380 feet.

5. A line of 320 feet will reach from the top of a precipice, standing close by the side of a brook, to the opposite bank; required the breadth of the brook, the height of the precipice being 103 feet.

Ans. 3029703 feet.

6. A ladder of 50 feet long, being placed in a street, reached a window 28 feet from the ground on one side; and by turning the ladder over, without removing the foot, it touched a moulding 36 feet high on the other side: required the breadth of the street.

Ans. 761233335 feet.

Multiply the sum of the two parallel sides by the perpendicular distance between them, and half the product will be the area.

EXAMPLES.

1. In a trapezoid, the parallel sides are AB 75, and DC 12:25, and the perpendicular distance DE or Bn is 154 chains required the area.

2. How many square feet are in the plank, whose length is 12 feet 6 inches, the breadth at the greater end 1 foot 3 inches, and at the less end 11 inches?

Ans. 131ft.

3. Required the area of a trapezoid, the parallel sides being 21 feet 3 inches and 18 feet 6 inches, and the distance between them 8 feet 5 inches.

Ans. 167 feet 3 inches 4′′ 6"".

PROBLEM

To find the area of a parallelogram; whether it be a square, a rectangle, a rhombus, or a rhomboid.

RULE.*

Multiply the length by the breadth, or perpendicular height, and the product will be the area.

EXAMPLES

Take any rectangle ABCD, and divide each of its sides. into as many equal parts as is expressed by the number of times they contain the linear measuring unit, and let all the opposite points of division be connected by right lines. Then, it is evident, that these lines divide the rectangle into a number of squares, each equal to the superficial measuring unit, and that the number of these squares, or the area of the figure, is equal to the number of linear measuring units in the length as often repeated, as there are linear measuring units in the breadth or height, that is, equal to the length multiplied by the height, which is the rule.

And since a rectangle is equal to an oblique parallelogram standing upon the same base, and between the same parallels ; (Euc. I. 35) therefore the rule is true for any parallelogram in, general. Q. E. D.

RULE