MENSURATION OF SUPERFICIES.

THE area of any figure is the measure of its surface, or the space contained within the bounds of that surface, without any regard to thickness.

A square whose side is one inch, one foot, or one yard, &c. is called the measuring unit, and the area or content of any figure is computed by the number of those squares contained in that figure.

N. B. In all questions involving decimals, they are carried out to the fourth place inclusive, and then taken to the nearest figure; that is, if it be found that, by extending the operation, the next figure would be 5, or upwards, the fourth decimal figure is increased by 1. The student by observing this rule will generally find his results to agree with those given in the book.

PROBLEM I.

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

RULE.*

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

D

* Take any rectangle ABCD, and divide each of its sides respectively, into as many equal parts as are 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 A

Note.-The perpendicular height of the parallelogram is equal to the area divided by the base.

EXAMPLES.

1. Required the area of the square ABCD whose side is 5 feet 9 inches.

Here 5 ft. 9 in.-5.75: and 5.75 -5.75× 5.75=33.0625 feet 33 fe. Oin. 9pa. area required.

=

2. Required the area of the rectangle ABCD, whose length AB is 13.75 chains, and breadth BC 9.5 chains.

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,) the rule is true for any parallelogram in general. Q. E. D.

RULE II. If any two sides of a parallelogram be multiplied together, and the product again by the natural sine of the included angle, the last product will give the area of the parallelogram. That is ABXBC nat. sine of the angle Barea.

Note.-Because the angles of a square and rectangle are each 90°, whose natural sine is unity, or 1, the rule in this case is the same as that given in the text.

Here 13,75 x 9.5=130.625; and

130.625
10

= 13.0625

ac.=13 ac. 0 ro. 10 po.=area required.

3. Required the area of the rhombus ABCD, whose length AB is 12 feet 6 inches, and its height DE 9 feet 3 inches.

Here 12 fe. 6 in.=12.5, and 9 fe. 3 in.=9.25.

Whence 12.5 x 9.25=115.625 fe.=115 fe. 7 in. 6 pa. area required.

4. What is the area of the rhomboides ABCD whose length AB is 10.52 chains, and height DE 7.63 chains?

acres=8 ac. 0 ro. 4 po.=area required.

5. What is the area of a square whose side is 35.25 chains?

ac. ro. po. Ans. 124 1 1

6. What is the area of a square whose side is 8 feet 4 inches?

fe. in. pa.

Ans. 69 5 4

7. What is the area of a rectangle whose length is 14 feet 6 inches, and breadth 4 feet 9 inches?

fe. in. pa. Ans. 68 10 6

8. Required the area of a rhombus, the length of whose side is 12.24 feet, and height 9.16 feet.

9. Required the area of a rhomboides 10.51 chains, and breadth 4.28 chains.

fe. in. pa.

Ans. 112 1 5 whose length is ac. ro. pa. Ans. 4 1 39.7

10. What is the area of a rhomboides whose length is 7 feet 9 inches, and height 3 feet 6 inches?

fe. in. pa.

Ans. 27 1 6

11. To find the area of a rectangular board, whose length is 12 feet, and breadth 9 inches.

PROBLEM II.

To find the area of a triangle.

RULE.*

Ans. 9 feet.

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

Note. The perpendicular height of the triangle is equal to twice the area divided by the base.

EXAMPLES.

1. Required the area of the triangle ABC, whose base AB is 10 feet 9 inches, and height DC 7 feet 3 inches.

* A triangle is half a parallelogram of the same base and altitude (Euc. I. 41,) and therefore the truth of this rule is evident.

68.875

10.509

6.000

Here 10 fe. 9 in.-10.75, and 7 fe. 3 in.-7.25;

Whence 10.75 × 7.25=77.9375,and

77.9375

= 38.96875

2

feet 38 fe. 11 in. 7 pa. area required.

71⁄2

2. What is the area of a triangle whose 4 inches, and height 11 feet 10 inches?

base is 18 feet fe. in. pa.

Ans. 108 5 8

3. What is the area of a triangle whose base is 16.75 feet, and height 6.24 feet?

fe. in. pa.

Ans. 52 3 1

4. Required the area of a triangle whose base is 12.25 chains, and height 8.5 chains.

ac. ro. po. Ans. 5 0 33

5. What is the area of a triangle whose base is 20 feet, and height 10.25? Ans. 102.5 fe.

PROBLEM III.

To find the area of a triangle whose three sides only are

given. RULE.*

1. From half the sum of the three sides subtract each side severally.

* Demon. Let Ac=a, AB=b, BC=c, and AD=x; (See preceding fig.) Then, since BD-b-x, we shall have c2-(b-x)2=CD2=a2x2, or c2—b2+2bx—x2—a2—x2,

a2+b2-c2

from which a is found =

by trans. and reduc

tion.

26

area ABX CD & √ (a + b2 — c2) × (c2 — a —_b2)