Proposition XIII

403. Problem. To inscribe a square in a circle.

Datum: Any circle.

Required to inscribe a square in the circle.

Solution. Draw any two diameters at right angles to each other, as AB and CD.

Draw AD, DB, BC, and CA.

Then, ADBC is the required square.

Proof. By the student.

Q.E.F.

Proposition XIV

404. Problem. To inscribe a regular hexagon in a circle.

Solution. From 4, any point in the circumference, as a center, and with a radius equal to the radius of the circle, describe an arc intersecting the circumference as at B.

From B as a center with the same radius, describe another arc intersecting the circumference as at c.

In like manner determine the points D, E, and F.
Draw chords connecting these points in succession.
Then, ABCDEF is the required hexagon.

Proof. By the student.

Ex. 670. To circumscribe an equilateral triangle about a circle.
Ex. 671. To circumscribe a square about a circle.

Q.E.F.

Proposition XV

405. Problem. To inscribe a regular decagon in a circle.

Solution. Draw any radius, as 04, and divide it in extreme and mean ratio as at P; that is, so that 40 : PO= PO : AP.

From 4 as a center and with PO as a radius, describe an arc intersecting the circumference as at B. Draw AB.

From B as a center with the same radius, or AB, describe an arc intersecting the circumference as at C.

In like manner determine the points D, E, F, G, H, J, and K.
Draw chords connecting these points in succession.

Then, ABCD-K is the required decagon.

Draw BP and BO.

Q.E.F.

Proof.

Const.,

and the chord AB, which subtends the arc AB, is a side of the

regular inscribed decagon ABCD-K.

Proposition XVI

406. Problem. To inscribe a regular pentadecagon in a

Solution.

Draw the chord AB equal to a side of the regular inscribed hexagon, and from a draw the chord AC equal to a side of the regular inscribed decagon. Draw CB.

From B as a center with CB as a radius, describe an arc intersecting the circumference as at D.

In like manner determine the points E, F, G, H, etc.
Draw chords connecting these points in succession.
Then, CBDEF etc., is the required pentadecagon.

Q.E.F.

and the chord CB, which subtends the arc CB, is a side of the regular inscribed pentadecagon CBDEF etc.

407. Cor. I. By joining the alternate vertices of any regular inscribed polygon of an even number of sides, a regular polygon of half the number of sides is inscribed.

408. Cor. II. By joining to the vertices of any regular inscribed polygon the middle points of the arcs subtended by its sides, a regular polygon of double the number of sides is inscribed.

Ex. 672. To inscribe a regular octagon in a circle.

Ex. 673. To inscribe a regular dodecagon in a circle.

Ex. 674. To circumscribe a regular hexagon about a circle.
Ex. 675. To circumscribe a regular octagon about a circle.

Ex. 676. To inscribe a regular hexagon in an equilateral triangle.

Ex. 677. To divide an angle of an equilateral triangle into five equal parts. Ex. 678. The segment of a circle is equal to of a similar segment. What is the ratio of their radii?

Ex. 679. How many degrees are there in an arc 18 in. long on a circumference whose radius is 5 ft.?

Ex. 680. The radii of two similar segments are as 35. What is the ratio of their areas?

Ex. 681. In a circle 3 ft. in diameter an equilateral triangle is inscribed. What is the area of a segment without the triangle ?

Ex. 682. Two chords drawn from the same point in a circumference to the extremities of a diameter of a circle are 6 in. and 8 in. respectively. What is the area of the circle ?

MAXIMA AND MINIMA

409. Of any number of magnitudes of the same kind the greatest is called the Maximum, and the least is called the Minimum.

Of all chords of any circle the diameter is the maximum; and of all lines from any point to a given line the perpendicular is the minimum.

410. Figures which have equal perimeters are called Isoperimetric.

Proposition XVII

411. Theorem. Of all triangles having two given sides, that in which these sides are perpendicular to each other is the maximum.

Data: Any two triangles, as ABC D and ABD, such that AC AD, AB is

=

common, and AC and AB are perpen-
dicular to each other.

To prove
A ABC the maximum.
Proof. Draw EDL AB.

C

Proposition XVIII

412. Theorem. Of all isoperimetric triangles which have the same base, the isosceles triangle is the maximum.

Data: Any two isoperimetric triangles

upon the same base AB, as ABC and ABD,

of which ABC is isosceles.

To prove ▲ ABC the maximum.

E

Proof. Produce AC to E, making CE equal to AC. Draw EB. Since c is equidistant from A, B, and E, ZABE may be inscribed in a semicircumference;

Draw DF equal to DB meeting EB produced in F; CG and DH parallel to AB; CJ and DK perpendicular to AB; and draw AF.

413. Cor. Of all isoperimetric triangles, the equilateral triangle is the maximum.

Ex. 683. Of all equivalent parallelograms having equal bases, the rectangle has the least perimeter.

A ABC is the maximum.

Q.E.D.