Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

н

616. A square is a parallelogram whose sides are equal, and whose angles are right angles ; as EFGH.

[blocks in formation]

617. A rhombus is a parallelogram whose sides are equal, and whose angles are not right angles; as IJKL.

I

618. A rhomboid is a parallelogram whose angles are not right angles; as MNOP.

N

M NOTE. — The altitude of a parallelogram is the perpendicular distance between any two of its parallel sides taken as bases, as the line P Q, drawn between two sides of the rhomboid MNOP, and perpendicular to the sides MN and OP.

[blocks in formation]

619. A trapezoid is a quadrilateral which has only two of its sides parallel; as RSTU.

[blocks in formation]

W

х

620. A trapezium is a quadrilateral which has no two sides parallel ; as WXYZ.

NoTE. — A diagonal of a quadrilateral, or of any polygon of more than four sides, is a straight line which joins the vertices of two opposite angles, or of two angles not adjacent; as the line X Z joining vertices of opposite angles of the trapezium W XYZ.

621. To find the area of a parallelogram. Multiply the base by the altitude, and the product will be the area. Ex. 1. What are the contents of a board 15 feet long and 2 feet wide ?

Ans. 30 feet. 2. A rectangular state is 128 miles long and 48 miles wide. How many square miles does it contain ?

Ans. 6144 miles. 3. The base of a rhomboid being 12 feet, and its height 8 feet, required the area.

Ans. 96 feet. 4. Required the area of a rhombus of which one of the equal sides is 358 feet, and the perpendicular distance between it and the opposite side is 194 feet.

Ans. 69452 sq. ft. 5. The largest of the Egyptian pyramids is square at its base, and measures 693 feet on a side. How much ground does it cover?

Ans. 11 acres 4 poles. 6. What is the difference between the area of a floor 40 feet square, and that of two others, each 20 feet square ? Ans. 800 feet.

7. There is a square whose area is 3600 yards; what is the side of a square, and the breadth of a walk along each side and each end of the square, which shall take up just one half of the whole ?

{

8.78+ yards, breadth of the walk. 622. To find the area of a trapezoid.

Multiply half of the sum of the parallel sides by the altitude, and the product is the area.

Ex. 1. If the parallel sides of a trapezoid are 75 and 33 feet, and the perpendicular breadth 20 feet, what is the area ?

Ans. 1080 sq. ft. 2. Required the area of a meadow in the form of a trapezoid, whose parallel sides are 786 and 473 links, and whose altitude is 986 links.

Ans. 6 acres 33 rods 3 yards. 623. To find the area of a trapezium.

Divide the trapezium into two triangles by a diagonal, and then find the areas of these triangles ; their sum will be the area of the trapezium.

Ex. 1. Required the area of a garden in the form of a trapezium, of which the four sides are 328, 456, 572, and 298 feet, and the diagonal, drawn from the angle between the first and second sides, 598 feet.

Ans. 3 acres 1 rood 31 rods 29 yards 3.85 feet. 2. Given one of the diagonals of a field, in the form of a trapezium, equal 17 chains 56 links, to compute the area, the perpendiculars to that diagonal from the opposite angles being 8 chains 82 links, and 7 chains 73 links.

Ans. 14 acres 2 roods 5 rods.

PENTAGONS, HEXAGONS, &c. 624. A pentagon is a polygon of five sides; a hexagon, one of six sides; a heptagon, one of seven sides; an octagon, one of eight sides; a nonagon, one of nine sides; and so on for a decagon, undecagon, dodecagon, &c.

D

E

625. A regular polygon is one whose sides and angles are equal; as the pentagon A B C D E.

A

B

626. To find the area of a regular polygon.

Multiply the perimeter by half the perpendicular let fall from the centre upon one of the sides. Or,

Multiply the square of one of the sides by the number against the polygon in the following

TABLE.
Pentagon,
1.720477

Nonagon, 6.181824
Hexagon,
2.598076

Decagon, 7.694209
Heptagon, 3.633913

Undecagon, 9.365641
Octagon,
4.828427

Dodecagon, 11.196152

Ex. 1. What is the area of a regular pentagon, of which the side is 250 feet, and the perpendicular from the centre to one side 172.05 feet?

Ans. 107531.25 sq.

ft. 2. What is the area of a regular hexagon whose side is 356 yards, and whose perpendicular is 308.305 yards? Ans. 329269.74yd.

3. The side of a regular octagonal enclosure is 60 yards; how many acres are included ? Ans. 3 acres 2 roods 14 rods 19 yards.

4. The side of a field, whose shape is that of a regular decagon, is 243 feet; what is its area?

Ans. 10 acres 1 rood 28 rods 24 yards 6.347 feet.

[blocks in formation]

627. A circle is a plane figure bounded by a line, every part of which is equally distant from a point within called the center; as A EFGBD.

[blocks in formation]

The circumference or periphery of a circle is the line that bounds it.

A radius is a line drawn from the center to the circumference; as CA, of C D

A diameter is a line which passes through the center, and is terminated by the circumference; as A B.

An arc is any portion of the circumference; as AD, A E, or EGF.

The chord of an arc is the straight line joining its extremities; as EF, which is the chord of the arc E GF.

628. The segment of a circle is the portion included by an arc and its chord; as the space included by the arc EGF and the chord E F.

629. The sector of a circle is the portion included by two radii and the intercepted arc; as the space A CD A.

630. A zone is the space between two parallel chords of a circle ; as the space A EFB A.

[ocr errors]

631. A lune, or crescent, is the space included between the intersecting arcs of two eccentric circles; as ACBĚ A.

E

D

632. A circular ring is the space included between the circumference of two concentric A circles; as the space between the rings A B and CD.

B

633. To find the circumference of a circle, the diameter being given.

Multiply the diameter by 3.141592.

Ex. 1. If the diameter of a circle is 144 feet, what is the cireumference ?

Ans. 452.389248 feet. 2. If the diameter of the earth is 7964 miles, what is its circumference?

Ans. 25019.6386887- miles. 3. Required the circumference of a circle, whose radius is 512 feet.

Ans. 4 furlongs 34 rods 5 yards 1 foot. 634. To find the diameter of a circle, the circumference being given.

Multiply the circumference by .318309.

Ex. 1. Required the diameter of a circle, whose circumference is 1043 feet.

Ans. 331.997+ feet. 2. It' the circumference of a circle is 25000 miles, what is its diameter ?

Ans. 7957.74+ miles. 3. If the circumference of a round stick of timber is 50 inches, what is its diameter ?

Ans. 15.91519+ inches. 635. To find the area of a circle, the diameter, or the circum. ference, or both, being given.

Multiply the square of the diameter by .785398. Or,
Multiply the square of the circumference by .079577. Or,
Multiply half ihe diameter by half the circumference.
Ex. 1. If the diameter of a circle is 761 feet, what is the area ?

Ans. 454840.475158 feet. 2. There is a circular island, three miles in diameter; how many acres does it contain ?

Ans. 4523.89+ acres. 3. Required the area of a circle, of which the circumference is 1284 yards.

Ans. 27 acres 17 rods 0.8+ yards. 4. Required the area of a circle, of which the diameter is 169, and the circumference 532 inches. Ans. 17 yards 3 feet 13 inches.

636. To find the area of a sector of a circle. Multiply the length of the arc by half the radius of the circle. Or,

As 360° are to the degrees in the arc of the sector, so is the area of the circle to the area of the sector.

Ex. 1. Required the area of a sector, of which the arc is 79 and the radius of the circle 47 inches.

Ans. 1856.5 inches. 2. Required the area of a sector, of which the arc is 26°, and the radius of the circle 25 feet.

Ans. 141.8 square feet. 637. To find the area of the segment of a circle.

Find the area of the sector which has the same arc with the segment ; and also the area of the triangle formed by the chord and the radii drawn to its extremities. The difference of these areas, when the segment is less, and their sum, when the segment is greater, than the semicircle, will be the area of the segment. Or,

To two thirds of the product of the height of the segment by the chord add the cube of the height, divided by twice the chord.

B

[blocks in formation]

Ex. 1. Required the area of the segment A B C A, of which the arc ABC is 49.250, the chord AC 10 feet, and the radii E A, E B, and EC, each 12 feet.

Ans. 7.35

sq.

ft.

E

639.

2. Required the area of a segment whose height is 15 rods and whose chord is 24 rods. Ans. 1 acre 3 roods 30 rods 9.4 yards.

638. To find the area of a zone of a circle.

From the area of the whole circle subtract the areas of the segments on the sides of the zone.

Ex. 1. Required the area of a zone whose parallel sides are 23.25 and 20.8 feet, in a circle whose radius is 12 feet. Ans. 206+ sq. ft.

2. Required the area of a zone included between two chords of 16 feet each, the diameter of the circle being 20 feet. Ans. 224.7 sq. ft.

To find the area of a lune or crescent. Find the difference of the areas of the two segments formed by the arcs of the lune and its chord.

Ex. 1. If the chord of two intersecting arcs is 72 feet, and the height of one of the segments is 30, and of the other 20 feet, what is the area of the crescent ?

Ans. 612 sq.

ft. 640. To find the area of a circular ring. Multiply the sum of the diameters of the two circles by the difference of the diameters, and that product by :7854.

Ex. 1. What is the area of the ring formed by two circles whose diameters are 10 and 20 yards ?

Ans. 235.62 sq. yd. 2. In the centre of a circular pond there is an island 128 yards in diameter; what is the area of the pond, provided the exact distance from any part of the outer side of the pond to the center of the island is 784 yards ? Ans. 1 acre 1 rood 14 rods 17 yards 7.4 feet.

641. To find the side of a square that shall equal the area of a circle of a given diameter or circumference.

Multiply the diameter of the circle by .886227. Or,
Multiply the circumference of the circle by .282094.

Ex. 1. I have a round field, 50 rods in diameter; what is the side of a square field that shall contain the same area ?

Ans. 44.31135+ rods. 2. I have a circular field 360 rods in circumference; what must be the side of a square field that shall contain the same area ?

Ans. 101.55+ rods. 3. John Smith had a farm which was 10,000 rods in circumference, which he sold at $ 71.75 per acre. He purchased another farm con

« ΠροηγούμενηΣυνέχεια »