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PROBLEM III. (87.) To find the area of a trapezoid.

RULE.

Multiply half the sum of the parallel sides into their per pendicular distance.

For demonstration, see Geometry, Prop. 7, B. IV.

Ex. 1. What is the area of a trapezoid whose parallel sides are 156 and 124, and the perpendicular distance between them 57 feet?

Ans., 7980 feet. Ex. 2. How many square yards in a trapezoid whose parallel sides are 678 and 987 feet, and altitude 524 feet?

Ans.

PROBLEM IV.

(88.) To find the area of an irregular polygon.

RULE.

Draw diagonals dividing the polygon into triangles, and find the sum of the areas of these triangles.

Ex. 1. What is the area of a quadrilateral, one of whose diagonals is 126 feet, and the two perpendiculars let fall upon it from the opposite angles are 74 and 28 feet?

Ans., 6426 feet. Ex. 2. In the polygon ABCDE, there are given EC=205, EB=242, AF=65, CG=114, and DH=110, to find the area.

Ans. 3.98

H (89.) If the diagonals of a quadrilateral are given, the area may be found by the following

D

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E

G

А.

B

RULE.

Multiply half the product of the diagonals by the sine of the angle at their intersection.

Demonstration.

The sines of the four angles at E are all equal to each other

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А

since the adjacent angles AED, DEC are the supplements of each other (Art. 27). But, according to

D the Rule, Art. 84, the area of

C the triangle ABE=AEXBE Xsine E;

AED= AE XDE X sine E;
BEC=BEX EC X sine E;

B
DEC=DEX ECX sine E.
Therefore,
the area of ABCD={(AE+EC)X(BE+ED) X sine E

== ACXBDX sine E. Ex. 1. If the diagonals of a quadrilateral are 34 and 56 rods, and if they intersect at an angle of 67°, what is the area?

Ans., 876.32. Ex. 2. If the diagonals of a quadrilateral are 75 and 49, and the angle of intersection is 42°, what is the area ?

Ans.

PROBLEM V.

(90.) To find the area of a regular polygon.

RULE I.

Multiply half the perimeter by the perpendicular let fall from the center on one of the sides.

For demonstration, see Geometry, Prop. 7, B. VI.

Ex. 1. What is the area of a regular pentagon whose side is 25, and the perpendicular from the center 17.205 feet?

Ans., 1075.31 feet. Ex. 2. What is the area of a regular octagon whose side is 53, and the perpendicular 63.977 ?

Ans. (91.) When the perpendicular is not given, it may be computed from the perimeter and number of sides. If we divide 360 degrees by the number of sides of the polygon, the quotient will be the angle ACB at the center, subtended by one of the sides. The perpendicular CD bisects the side AB, and the angle ACB. Then, in the triangle ACD, we have (Art. 42),

A
R: AD: cot. ACD: CD; that is,

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D

B

Radius is to half of one of the sides of the polygon, as the votangent of the opposite angle is to the perpendicular fron the center.

Ex. 3. Find the area of a regular hexagon whose side is 32 inches. The angle ACD is i'a of 360°=30°. Then R: 16 :: cot. 30° : 27.7128=CD, the perpendicular;

and the area=27.7128 x 16 x 6=2660.4288. Ex. 4. Find the area of a regular decagon whose side is 46 feet.

Ans., 16280.946. (92.) In this manner was computed the following table of the areas of regular polygons, in which the side of each polygro is supposed to be a unit.

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Sides.

3

Triangle,

0.4330127. Square,

4

1.0000000. Pentagon,

5

1.7204774. Hexagon,

6

2.5980762. Heptagon,

7

3.6339124. Octagon,

4.8281271. Nonagon,

9

6.1818242. Decagon,

10

7.6942088. Undecagon,

11

9.3656399. Dodecagon,

12

11.1961524. By the aid of this table may be computed the area of any other regular polygon having not more than twelve sides. For, since the areas of similar polygons are as the squares of their homologous sides, we derive

RULE II.

Multiply the square of one of the sides of the polygon by the area of a similar polygon whose side is unity.

Ex. 5. What is the area of a regular nonagon whose side is 63 ?

Ans., 24535.66. Ex. 6. What is the area of a regular dodecagon whose side is 54 feet?

Ans., 32647.98 feet. E

PROBLEM VI. (93.) To find the circumference of a circle from its diameter,

RULE.

Multiply the diameter by 3.14159.

For the demonstration of this rule, see Geometry, Prop. 13, Cor. 2, B. VI.

When the diameter of the circle is small, and no great accuracy is required, it may be sufficient to employ the value of 7 to only 4 or 5 decimal places. But if the diameter is large, and accuracy is required, it will be necessary to employ a corresponding number of decimal places of nr. The value of ar to ten decimal places is 3.14159,26536,

and its logarithm is 0.497150. Ex. 1. What is the circumference of a circle whose diameter is 125 feet?

Ans., 392.7 feet. Ex. 2. If the diameter of the earth is 7912 miles, what is its circumference ?

Ans., 24856.28 miles. Ex. 3. If the diameter of the earth's orbit is 189,761,000 miles, what is its circumference?

Ans., 596,151,764 miles. To obtain this answer, the value of a must be taken to at least eight decimal places.

.

PROBLEM VII. (94.) To find the diameter of a circle from its circumference.

RULE I.

Divide the circumference by 3.14159.

This rule is an obvious consequence from the preceding. To divide by a number is the same as to multiply by its reciprocal; and, since multiplication is more easily performed than division, it is generally most convenient to multiply by the reciprocal of 7, which is 0.3183099. Hence we have

RULE II.

Multiply the circumference by 0.31831.

Ex. 1. What is the diameter of a circle whose circumference is 875 feet?

Ans., 278.52 feet. Ex. 2. If the circumference of the moon is 6786 miles, what ís its diameter ?

Ans., 2160 miles, Ex. 3. If the circumference of the moon's orbit is 1,492,987 miles, what is its diameter ?

Ans., 475,233 miles.

PROBLEM VIII. (95.) To find the length of an arc of a circle.

RULE I.

As 360 is to the number of degrees in the arc, so is the circumference of the circle to the length of the arc.

This rule follows from Prop. 14, B. III., in Geometry, where it is proved that angles at the center of a circle have the same ratio with the intercepted arcs.

Ex. 1. What is the length of an arc of 22°, in a circle whose diameter is 125 feet?

The circumference of the circle is found to be 392.7 feet.

Then 360 : 22 :: 392.7 : 23.998 feet.

Ex. 2. If the circumference of the earth is 24,856.28 miles, what is the length of one degree?

Ans., 69.045 miles.

RULE II.

(96.) Multiply the diameter of the circle by the number of degrees in the arc, and this product by 0.0087266.

Since the circumference of a circle whose diameter is unity is 3.14159, if we divide this number by 360, we shall obtain the length of an arc of one degree, viz., 0.0087266. If we multiply this decimal by the number of degrees in any arc, we shall obtain the length of that are in a circle whose diameter is unity; and this product, multiplied by the diameter of any uther circle, will give the length of an arc of the given number of degrees in that circle.

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