This section is from the book "The American House Carpenter", by R. G. Hatfield. Also available from Amazon: The American House Carpenter.

and, putting r for the radius of the inscribed circle, we have -

(132.)

Or: The radius of the inscribed circle of a regular trigon equals the half of a side of the trigon divided by the square root of 3. To obtain the area of a trigon or equilateral triangle; we have (Art. 408) the area of a parallelogram by multiplying its base into its height; and (Arts. 341 and 342) the area of a triangle is equal to half that of a parallelogram of equal base and height, therefore, the area of the triangle B D C(Fig. 287) is obtained by multiplying B C, the base, into the half of ED, its height. Or, when N is put for the area -

N=BCxED/2,

N=bxR/4;

or-

substituting: for R its value (131.) -

This is the area of the triangle B D C.

The triangle A B C is compounded of three equal triangles, one of which is the triangle B D C; therefore the area of the triangle ABC equals three times the area of the triangle BDC; or, when A represents the area -

(133)

Or: The area of a regular trigon or equilateral triangle equals three fourths of the square of a side of the triangle divided by the square root of 3.

434. - Tetragon: Radius of Circumscribed and Inscribed Circles: Area. - Let A B CD (Fig. 288) be a given tetragon or square, with its circumscribed and inscribed circles, of which A E is the radius of the former and E F that of the latter. The point F bisects A B, the side of the square. A F equals E F and equals half A B, a. side of the square. Putting R for the radius of the circumscribed circle and b for A B, we have (Art. 416) -

Fig. 288.

AE2 = AF2 + EF2

R2 = (b/2)2+(b/2)2=2(b/2)2=b2/2,

(134.)

Or: The radius of the circumscribed circle of a regular tetragon equals a side of the square divided by the square root of 2.

By referring to the figure it will be seen that the radius of the inscribed circle equals half a side of the square -

r = b/2 (135.)

The area of the square equals the square of a side -

A = b2 (136.)

435. - Hexagon: Radius of Circumscribed and Inscribed Circles: Area. - Let A B C D EF(Fig. 289) be an equilateral hexagon with its circumscribed and inscribed circles, of which EG is the radius of the former, and GH that of the latter. The three lines, A D, BE, and C F, divide the hexagon into six equal triangles with their apexes converging at G. The six angles thus formed at G are equal, and since their sum about the point G amounts to four right angles (Art. 335), therefore each angle equals 4/6 or 2/3 of a right angle. The sides of the six triangles radiating from G are the radii of the circle, hence they are equal; therefore, each of the triangles is isosceles (Art. 338), having equal angles at the base. In the triangle EGD, the sum of the three angles being equal to two right angles (Art. 345), and the angle at G being, as above shown, equal to 2/3 of a right angle, therefore the sum of the two angles at E and D equals 2 - 2/8 = 4/3 of a right angle; and, since they equal each other, therefore each equals 2/8 of a right angle and equals the angle at G; therefore E G D is an equilateral triangle. Hence E D, a side of a hexagon, equals E G, the radius of the circumscribing circle -

Fig. 289.

R=b. (137.)

As to the radius of the inscribed circle, represented by G H, a perpendicular from the centre upon ED, the base; the point H bisects E D. Therefore, E H equals half of a side of the hexagon, equals half the radius of the circumscribing circle. Let R = this radius, and r the radius of the inscribed circle, while b = a side of the hexagon; then we have (Arts. 353 and 416) -

GH2 = EG2-EH2,

r2 = R2 - (R/2)2,

r2 = R2-1/4R2 = 3/4R2,

Now, R - b, therefore -

(138.)

Or: The radius of the inscribed circle of a regular hexagon equals the half of a side of the hexagon, multiplied by the square root of 3.

As to the area of the hexagon, it will be observed that the six triangles, A B G, B G C, etc., converging at G, the centre, are together equal to the area of the hexagon. The area of E G D, one of these triangles, is equal to the product of E D, the base, into the half of G H, the perpendicular; or, when N is put to equal the area -

N=EDxGH/2

N=bxr/2

and, since r, as above, equals

This is the area of one of the six equal triangles; therefore, when A is put to represent the area of the hexagon, we have -

(139.)

Or: The area of a regular hexagon equals three halves of the square of a side multiplied by the square root of 3.

Continue to: