71. To cut off a regular pentagon from the square ABCD.

Divide BA in X in median section and take M the mid-point of AX.

Fig. 27.

Then AB.AX=XB2, and AM=MX. Take BN=AM or MX. Then MN=XB.

Lay off NP and MR equal to MN, so that P and R may lie on BC and AD respectively.

Lay off RQ and PQ = MR and NP.

MNPQR is the pentagon required.

In Fig. 19, p. 22, AN, which is equal to AB, has the point N on the perpendicular MO. If A be moved on AB over the distance MB, then it is evident that N will be moved on to BC, and X to M.

Therefore, in Fig. 27, NR = AB. Similarly MP = AB. RP is also equal to AB and parallel to it.

RMA = 4/5 of a rt.

... NMR = 6/5 of a rt.

Similarly PNM= 6/5 of a rt..

From triangles MNR and QRP, NMR=RQP = 6/5 of a rt.

The three angles at M, N, and Q of the pentagon being each equal to 6/5 of a right angle, the remaining two angles are together equal to 12/5 of a right angle, and they are equal. Therefore each of them is 6/5 of a right angle.

Therefore all the angles of the pentagon are equal. ,

The pentagon is also equilateral by construction.

72. The base MN of the pentagon is equal to XB, i. e., to AB/2 (√5 - 1) =AB X 0.6180.... § 58.

The greatest breadth of the pentagon is AB.

73. If p be the altitude,

AB2 = P2 +

=p2 + AB2 • 3 - √5 / 8.

... p2 = AB2

=AB2.5 + √5 / 8.

... p = AB •

= AB X 0.9510....= AB cos 18°.

Fig. 28.

74. If R be the radius of the circumscribed circle,

= AB X 0.5257....

75. If r be the radius of the inscribed circle, then from Fig. 28 it is evident that r=p - R = AB. - AB.

= AB.

= AB.

= AB.

= AB X 0.4253....

76. The area of the pentagon is 5r X ½ the base of the pentagon, i. e.,

5AB. .AB/4.(√5-1)

= AB2.5/4. = AB2 X 0.6571....

77. In Fig. 27 let PR be divided by MQ and NQ in E and F.

Then ... MN = AB/2 • (√5 - 1) ... § 72 and cos 36°= ½ AB / ½ AB.(√5 - 1),

... RE = FP = MN/2 . 1/cos 36° = AB.√5 - 1 / √5+1

=AB.3 - √5 /2....(1)

EF=AB - 2RE = AB - AB(3 - √5)

= AB(√5 - 2)... (2) RF=MN.

RE: RE = RE : EF(by § 51)....................(3)

√5 - 1 : 3 - √5 = 3 - √5:2(√5 - 2)...........(4)

By § 76 the area of the pentagon

=AB2.5/4.

=MN2. .5/4.

=MN2. 1/4. since AB = MN • √5 + 1 / 2.

... the area of the inner pentagon

= EF2.1/4

= AB2 . (√5 - 2)2 • 1/4.

The larger pentagon divided by the smaller

= MN2 : EF2 = 2 : (7 - 3√5)

= 1 : 0.145898.. ..

78. If in Fig. 27, angles QEK and LFQ are made equal to ERQ or FQP, K, L being points on the sides QR and QF respectively, then EFLQK will be a regular pentagon congruent to the inner pentagon. Pentagons can be similarly described on the remaining sides of the inner pentagon. The resulting figure consisting of six pentagons is very interesting.