87. To cut off a regular octagon from a given square. Obtain the inscribed square by joining the midpoints A, B, C, D of the sides of the given square.

Fig. 32.

Bisect the angles which the sides of the inscribed square make with the sides of the other. Let the bisecting lines meet in E. F,G, and H.

AEBFCGDH is a regular octagon.

The triangles AEB, BEC, CGD, and DHA are congruent isosceles triangles. The octagon is therefore equilateral.

The angles at the vertices, E, E, G, H of the same four triangles are each one right angle and a half, since the angles at the base are each one-fourth of a right angle.

Therefore the angles of the octagon at A, B, C, and D are each one right angle and a half.

Thus the octagon is equiangular.

The greatest breadth of the octagon is the side of the given square, a.

88. If R be the radius of the circumscribed circle, and a be the side of the original square,

89. The angle subtended at the center by each of the sides is half a right angle.

90. Draw the radius OE and let it cut AB in K (Fig. 33).

Then AK= OK= OA / √2 = a / 2√2.

KE = OA - OK= a / 2 - a / 2√2 = a/4. (2 - √2).

Now from triangle AEK, AE2 = AK2 + KE2

= a2 / 8 + a2 / 8 . (3 - 2√2)

= a2/8 (4-2√2) = a2/4 (4-2√2)

... ae = a/2.

91. The altitude of the octagon is CE (Fig. 33). But CE2 = AC2 - AE2

= a2 - a2 / 4 • (2 - √2) = a2 / 4 .(2 + √2).

Fig. 33.

... CE = a/2.

92. The area of the octagon is eight times the triangle AOE and

= 4OE . AK= 4. a/2. a/2√2 = a2 / √2.

93. A regular octagon may also be obtained by dividing the angles of the given square into four equal parts.

Fig. 34.

It is easily seen that EZ = WZ=a, the side of the square.

XZ=a√2;

XE = WH= WK; KX=a - a(√2 - 1) = a(2 - √2).

Now KZ2 = a2 + a2(√2 - 1)2 = a2 (4 - 2√2)

... KZ=a Also GE = XZ - 2XE

= a√2 - 2a(√2 - 1) = a(2 - √2).

... H0 = a/2 (2 - √2).

Again OZ=a/2√2, and HZ2 = HO2+ OZ2

= a2 / 4 (6 - 4√2 + 2)

= a2(2 - √2).

... HZ=

HK = KZ - HZ

•(√2 - l)

and HA = a/2

94. The area of the octagon is eight times the area of the triangle HOA,

= 8.½ HO. HO/√2

= HO2.2√2

= 2√2

= a2/4.2√2.(6 - 4√2)

= a2.(3√2 - 4)

= a2.√2.(√2 - l)2.

95. This octagon : the octagon in § 92

= (2 - √2)2 : 1 or 2 : (√2+ l)2; and their bases are to one another as

√2 : √2 + 1.