EXAMPLE.1. The radius of a circle is 7 feet; required the side of the greatest regular octagon that may be inscribed therein.

7 X.7653: = 5.3571 feet, or 5 ft. 44 in., nearly. Ans.

2. The sides of a heptagon are to be, each, 5 inches; required the radius of circumscribing circle.

1.152 X 55.760 = 5 in. Ans.

3. Each side of a hexagon is 12 inches; required the distance (d c) from the centre of a side to the centre of the figure.

4. The side of an equilateral triangle is 12 inches; required its altitude, the perpendicular.

12 X 3 X .28868= 10.392. Ans.

5. A PERPENDICULAR, from the centre to either side of a hexagon, is required to be 12 inches; what must be the radius of circumscribing circle?

12 X 1.156 = 13.872 inches. Ans.

To find the area of a regular polygon, when the side only is given. RULE. - Multiply the square of the side by the number or factor in the TABLE — (column, Area) — opposite the name of the respective polygon, and the product will be the area.

EXAMPLE.

Each side of a nonagon is 6 rods; required its area. 6.52 × 6.181824 = 261.18 + square rods. Ans.

To find the area of an irregular polygon.

RULE. — Divide the figure into trapeziums and triangles, by drawing diagonals, and find the area of each, separately; the sum of the several areas will equal the area of the figure.

EXAMPLE. The outline of the above figure defines an irregular polygon; the enclosed lines divide it into three triangles; and the areas of the several triangles, taken collectively, are equal to, or constitute, the area sought. To find the areas of the triangles, see TRIMensuration of.

ANGLES -

Having the figure of a rhombus, rhomboid, or trapezoid, the perpendicular, whereby to find the area, may be found by the following method.

Suppose the diagram A B C D.

A D and B C sides parallel to each other.

A D base.

A B side inclining to the base and whose length is known. Then, at any convenient distance from the angle A, on A D,A erect a perpendicular to A D that

D

will cut the side A B, as i h; then will A i be to i h as A B to B E, perpendicular required; or A i will be to A h as A B to A E, distance from the angle A, on A D, at which a perpendicular dropped from B will fall; and

√ (A B2 — A E2) = BE, perpendicular required.

EXAMPLE.

The figure A B C D is that of a field whose side. A B is 68 rods: Ai is 4.2 feet, and i h is 3.7 feet; required the perpendicular distance from the side A D to the side B C.

4.2 3.7 68: 59.9 rods. Ans.

EXAMPLE. — A bin in the form of a trapezoid, A B C D (Fig.), has a side A B inclining to A D that is 12 feet in length, and a perpendicular to A D, erected on AD, that cuts the side A B, gives the segment A i 2 feet, and the segment A h 1 feet; required the perpendicular distance from the side A D to the side B C.

21.5

=

12 9, and (122 92) 7.94 feet. Ans. Ai:AB:: Ah: A E, diagram.

66

66

Ah: Eh: Ai: Bi,

ih: Ai::BE: AB,

Ahih:AE: BE, 66

To find a diagonal of the above figure A B C D.
√(BE2 + ED') = BD, diagonal.

Having the area of a rhombus, rhomboid, or trapezoid, and the sum of any two sides of the figure that are parallel to each other, the perpendicular to those sides will be found, if we divide twice the area by the sum referred to.

Suppose A D and B C the given sides;

BE the perpendicular required; then

If through any four-sided figure a diagonal be drawn or be supposed to be drawn, the figure will be converted into two triangles, each of which will have a side, the diagonal, that will be common to them both; and the length of that side or diagonal, which is an indispensable element in calculating the area of a trapezium, may be found, when more simple means may not be resorted to, by one or the other of the following methods:

-

Suppose the figure represented by the diagram A B C D.

AFCA, diagonal sought.

C to A, diagonal required.

Construct a partial diagonal, Ct, direction C A, and thereon erect a perpendicular to Ct that will cut an adjacent side, as rt cutting the side CD; then Cr: Ct::CĎ: C F, and Cr:rt::CD:DF; and √ (A D2 — D F3) = A F, and C F

EXAMPLE.A structure in the form of a trapezium, A B CD, (Fig.) has Cr 8 inches, Ct 6 inches, and rt 5 inches; the side CD is 16 feet, and the side A D is 20 feet; required the length of a diagonal C A.

861612 feet, C F, or distance from C, on a line C A, at which a perpendicular dropped from D will fall, and

=

300

8:5:: 16: 10 feet, D F, or length of that perpendicular, then 202 102 17.32+12=29.32 feet, length of a diagonal C A, or of the side A C of the triangle A C B, or of the side CA of the triangle C A D.

Again, suppose the same diagram, the side A D accessible, and D to B the diagonal required.

DnDE::Ds: D B, diagonal sought.

Of the Right-angled Triangle: -A D C, diagram.

The longest side of this figure, A C, is usually called the hypotenuse, and the other two sides, A D and D C, are called the sides or legs. The legs are perpendicular one to the other, and form the right-angle, D, or angle of 90°.

If one of the legs be made

D

base, the other will be perpendicular, and will be the altitude of the figure; for the altitude of any triangle is a perpendicular, dropped from the vertical angle to the opposite side or base, and either side may be made base; thus, if A D be made base, DC will be perpendicular, and will be the altitude of the figure, and if D C be made base, AD will be perpendicular, and will be the altitude of the figure.

If the hypotenuse be made base, the legs, A D and D C, will still be perpendicular one to the other, but the altitude of the figure, D E or ED, a perpendicular to the hypotenuse, will not be shown.

Whether the legs have equal lengths or unequal, so far as regards the principles of the figure, is immaterial.

√ (A D2 + D C2) = A C; √ (A C2 — D C3) = A D ;
√ (A C2 — A D3) = D C.

A D÷AC=AE,

Converting the right-angled triangle A D C into two right-angled tri√ (A D2 - A E2) = ED. angles, A E D and CED, ED a leg common to both, and perpendicular to A C, and the altitude of the triangle A D C, therefore, A C being base.

DC÷AC=CE. ACXAE= A D2.

2

√ (D C2 — C E2) = ED. ACXCE=D C2.
ADA EAC. A EXCEED2.

Twice the area of any right-angled triangle, divided by the hypotenuse of that triangle, equals the perpendicular to that hypotenuse, and the respective triangle is thereby divided into two rightangled triangles, having a leg common to both.

Half the product of the two legs of a right-angled triangle equals the area of that triangle.

Every triangle not a right-angled triangle is either an acuteangled triangle or an obtuse-angled triangle, and these two (the acute-angled and obtuse-angled) are classed under the general name oblique-angled. The following principles are alike applicable to either.

Let A B C be the triangle.

9.

h

D

AU-BC

2AB+AB=AD, distance along the base, from the angle formed by the base and longest vertical side, at which a perpendicular dropped from the vertical angle will fall; SB and

(AC-A D) =DC, the perpendicular alluded to; thus dividing the obtuse-angled triangle A B C into two right-angled triangles, A D C and B D C, D C a leg common to both.

Or,

(AC+ ADXAC

AD)

=

DC; for the sum of any

two quantities multiplied by their difference is equal to the difference of their squares.

A B2 - A C2

2 BC

+ B C = B g, and √ (A B2 — B g2) = A g,

perpendicular to B C produced.

AB-BO

2 AC

A C

+2 = A h, and √(A Ba — A h2) = B h, per

2