FIG. 4.

A B C, Fig. 1, is a right-angled triangle, and also a scalene triangle.

ADC, Fig. 2, and DCB, Fig. 4, are obtuse-angled-triangles, and also scalene tri

angles.

A Quadrilateral is a surface bounded by four straight lines, Figs. 2, 3, and 4.

Quadrilaterals are subdivided into parallelograms, trapezoids, and trapeziums, that is, four-sided figures having their opposite sides parallel, ABEC, Fig. 3; having two sides parallel and two inclined to each other, Fig. 4; having no two sides parallel, Fig. 2, respectively.

Lines are said to be parallel when they will not meet, however far they may be produced, as CE and A B, Fig. 3; and DC and AB, Fig. 4.

a hexagon; by seven, a heptagon; by eight, an octagon; by ten, a decagon; by eleven, an undecagon; by twelve, a dodecagon.

536. The Base of a plane figure is the side on which it is supposed to rest, as BC, Fig. 1; A B, Figs. 2, 3, and 4.

537. The Altitude is the perpendicular distance between the base and the vertex of the opposite angle, or between the base and the opposite side; as A B, Fig. 1, and C D, Fig. 3, and D E, Fig. 4.

538. The Diagonal of a plane figure is the straight line joining the vertices of two angles not connected by one of the sides; as A C, Fig. 2, and D B, Fig. 4.

In Fig. 3, if we cut off the part A DCand place it to the right of the figure CDBE, so that A C will coincide with BE, we see that ABEC will be changed to a rectangle (207), which is equivalent to ABEC, and whose base is equal to A B, and altitude CD; and we find the area by multiplying together the two dimensions A B and CD (209). Therefore, to find the area of any parallelogram we have this

539. RULE.- Multiply the base by the altitude. And since a triangle is one-half of the parallelogram having the same base and altitude, (See A B C, Fig. 3,) we find its area by this 540. RULE. - A. Multiply the base by one-half the altitude.

And since a trapezoid is only two triangles, BCD and B A D, Fig. 4, and they have the same altitude, DE, and the base of one is A B, and DC may be regarded as the base of the other, we may take the sum of the areas of the two triangles as the area of the trapezoid. And as we have the area of one of the triangles equal to one-half its base multiplied by its altitude, and the area of the other triangle equal to one-half its base by the same altitude, we must have the area of the trapezoid equal to

541. One-half the sum of its two sides, multiplied by its altitude.

The area of a triangle is also found by this

542. RULE.-B. From the half sum of the three sides subtract each side separately. Then extract the square root of the continued product of these remainders and the half sum of the sides..

The area of any plane figure may be found

543. By dividing it into triangles, calculating the area of each triangle separately, and then finding the sum of these areas.

If a figure is regular, that is, has its sides and angles equal, each to each, its area may be found

544. By multiplying the distance around it (the perimeter), by one-half the perpendicular distance from its cente. to one of its sides.

If a regular figure has its sides exceedingly small and an exceedingly large number of them, the sides taken together form the circumference (223) of a circle, (222), and the perpendicular distance from the centre to one of its sides is the radius (223); and, therefore, we say that

545. The area of a circle equals the product of the circumference, by one-half the radius.

It is proven in Geometry, that the ratio of the circumference of a circle to its diameter is 3.14159, nearly. By means of this truth we deduce various rules for determining different parts of a circle when other parts are known.

Thus, since the circumference is 3.14159 times the diameter,
To find the circumference when the diameter is known,

546. RULE.- Multiply the diameter, or twice the radius, by 3.14159.

To find the diameter when the circumference is known,

547. RULE.-1. Divide the circumference by 3.14159.

Or, 2. Multiply the circumference by.31831

(=

3.14159).

additional rules for finding the area of a circle.

From the foregoing rules we may readily deduce the following

548. Multiply the square of the diameter by .7854.

Multiply the square of the radius by 3.14159.

Multiply the square of the circumference by.079577.

PROBLEMS.

Find the area of a triangle whose

1. Base is 25 ft., altitude 6.

2. Base is 100 rd., altitude 50.

3. Base is 16 yards, altitude 16.

4. Base is 374 ft., altitude 25. 5. Sides are 6, 10, and 12 rods.

6. Sides are 9, 15, and 20 yd.

Ans. 75 sq. ft.

Ans. 2500 sq. rd.

Ans. 128 sq. yd.

Ans. 468 sq. ft.

Ans. 29.933 sq. rd.

Ans. 63.277 sq. yd.

15. Are 75 ft. and 17 yd., altitude 10 ft. Ans. 70 sq. yd.

Find the area of a trapezuim whose diagonal

16. Is 21, perpendiculars 3 and 5 ft.

17. Is 36, perpendiculars 21 and 30 ft. 18. Is 31, perpendiculars 27 and 16 yd. 19. Is 15, perpendiculars 8 and 10 rd.

Ans. 84 sq. ft. Ans. 918 sq. ft. Ans. 666+ sq. yd.

Ans. 135 sq. rd.

D

A

Find the area of a regular figure of

20. Five sides, each 10 ft.; per. 6.88 ft.

21. Six sides, each 10 ft.; per. 8.66 ft.

Ans. 172 sq. ft. Ans. 259.8 sq. ft.

22. Eight sides, each 20 ft.; per. 24.14 ft.

E

Ans. 3848.46 sq. ft.

Ans. 14.5 sq. ft.

Ans. 12.434 sq. rods.

SOLIDS.

549. A Solid, or Body, is that which has length, breadth and thickness, or three dimensions.

A Prism is a solid, bounded by plane surfaces, two of which, the ends, or bases of the prism, are equal and similar plane figures ; and the sides, or faces, parallelograms; as Figs. 1, 2, and 3, and (215, Fig. 2.)