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PROPOSITION XXIX. PROBLEM.

362. To construct a polygon similar to a given polygon, and having two and a half times its area.

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It is required to construct a polygon similar to P, and equivalent to 2] P.

Let A B be a side of the given polygon P.
Then VĪ : V21 :: A B : x,
V2 : V5 :: A B : x,

$ 345 (the homologous sides of similar polygons are to each other as the square roots

of their areas). Take any convenient unit of length, as MC, and apply it six times to the indefinite line M N.

On MO = 3 M C) describe a semi-circumference ; and on MN (= 6 M C) describe a semi-circumference.

At C erect a I to MN, intersecting the semi-circumferences at D and H.

Then C D is the V2, and C H is the V5. § 360 Draw C Y, making any convenient { with C H.

On C Y take C E = A B.

From D draw D E,
and from H draw H Y |to D E.

Then C Y will equal X, and be a side of the polygon required, homologous to A B. For CD:CH :: CE : CY,

$ 275 (a line drawn through two sides of a A, Il to the third side, divides the two

sides proportionally).
Substitute their equivalents for C D, C H, and C E;
then V2 : V5 :: AB :CY.

On C Y, homologous to A B, construct a polygon similar to the given polygon P; and this is the polygon required.

Q. E. F.

Ex. 1. The perpendicular distance between two parallels is 30, and a line is drawn across them at an angle of 45° ; what is its length between the parallels ?

2. Given an equilateral triangle each of whose sides is 20; find the altitude of the triangle, and its area.

3. Given the angle A of a triangle equal to s of a right angle, the angle B equal to } of a right angle, and the side a, opposite the angle A, equal to 10; construct the triangle.

4. The two segments of a chord intersected by another chord are 6 and 5, and one segment of the other chord is 3; what is the other segment of the latter chord ?

5. If a circle be inscribed in a right triangle : show that the difference between the sum of the two sides containing the right angle and the hypotenuse is equal to the diameter of the circle.

6. Construct a parallelogram the area and perimeter of which shall be respectively equal to the area and perimeter of a given triangle.

7. Given the difference between the diagonal and side of a square; construct the square.

BOOK V.

REGULAR POLYGONS AND CIRCLES.

363. DEF. A Regular Polygon is a polygon which is equilateral and equiangular.

PROPOSITION I. THEOREM. 364. Every equilateral polygon inscribed in a circle is a regular polygon.

Let A B C, etc., be an equilateral polygon inscribed

in a circle.
We are to prove the polygon A B C, etc., regular.

The arcs A B, BC, C D, etc., are equal, § 182
(in the same O, equal chords subtend equal arcs).
... arcs A B C, BC D, etc., are equal, Ax. 6
.. the 6 A, B, C, etc., are equal,

(being inscribed in equal segments). .. the polygon A B C, etc., is a regular polygon, being equilateral and equiangular.

Q. E. D.

PROPOSITION II. THEOREM. 365. I. A circle may be circumscribed about a regular polygon.

II. A circle may be inscribed in a regular polygon.

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F Let A B C D, etc., be a regular polygon. We are to prove that a O may be circumscribed about this regular polygon, and also a O may be inscribed in this regular polygon. Case I. — Describe a circumference passing through A, B, and C.

From the centre 0, draw 0 A, O D,

and draw Os I to chord BC.
On Os as an axis revolve the quadrilateral 0 A B s,
until it comes into the plane of Os C D.

The line s B will fall upon s C,
(for 20 s B = 20 s C, both being rt. 4).
The point B will fall upon C,

§ 183
(since s B = 8C).
The line B A will fall upon CD, § 363
(since Z B= 2 C, being 6 of a regular polygon).
The point A will fall upon D,

§ 363 (since B A= C D, being sides of a regular polygon). .:. the line 0 A will coincide with line 0 D,

(their extremitics being the same points).

.. the circumference will pass through D. In like manner we may prove that the circumference, passing through vertices B, C, and D will also pass through the vertex E, and thus through all the vertices of the polygon in succession. CASE II. — The sides of the regular polygon, being equal chords of the circumscribed O, are equally distant from the centre, § 185

... a circle described with the centre () and a radius Os will touch all the sides, and be inscribed in the polygon. § 174

Q. E. D.

366. DEF. The Centre of a regular polygon is the common centre () of the circumscribed and inscribed circles.

367. DEF. The Radius of a regular polygon is the radius O A of the circumscribed circle.

368. DEF. The Apothem of a regular polygon is the radius Os of the inscribed circle.

369. DEF. The Angle at the centre is the angle included by the radii drawn to the extremities of any side.

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PROPOSITION III. THEOREM. 370. Each angle at the centre of a regular polygon is equal to four right angles divided by the number of sides of the polygon.

n

Let A BC, etc., be a regular polygon of n sides.

- 4 rt. Is We are to prove ZA OB== Circumscribe a O about the polygon.

The Is A OB, BO C, etc., are equal, § 180

(in the same o equal arcs subtend equal & at the centre). .. the ZA OB = 4 rt. Is divided by the number of /s about 0.

But the number of Ís about (=n, the number of sides of the polygon.

4 rt. A ..ZA OB=

Q. E. D. 371. COROLLARY. The radius drawn to any vertex of a regular polygon bisects the angle at that vertex.

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