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EXERCISES.

1. The altitude of an isosceles triangle is 3 feet, each of the equal sides is 5 feet; find the area.

2. The parallel sides of a trapezoid are 8 and 10 feet, and the altitude is 6 feet; what is the area?

3. The sides of a triangle are 60, 80, and 100 feet, the diameter of the inscribed circle is 40 feet; find the

area.

4. Construct a square equal to the sum of the squares whose sides are respectively 16, 12, 8, 4, and 2 units in length.

Show that the sum of the three perpendiculars drawn from any point within an equilateral triangle to the three sides is equal to the altitude of the triangle.

6. Show that the sum of the squares of two lines, drawn from any point in the circumference of a circle to twoman points on the diameter of the circle equidistant from the centre, will be always the same.

7. The distance of a chord, 8 feet long, from the centre of a circle is 3 feet; what is the diameter of the circle?

8. Construct a triangle, having given the vertical angle, the line bisecting the base, and the angle which the bisecting line makes with the base.

9. Show that if a line bisecting the exterior vertical angle of a triangle is not par

allel to the base, the distances of the point in which it meets the base produced, from the extremities of the base, are proportional to the other two sides of the triangle.

10. The segments made by a perpendicular, drawn from a point on the circumference of a circle to a diameter, are 16 feet and 4 feet; find the length of the perpendicular.

11. Two similar triangles, ABC and DEF, have the homologous sides AC and DF equal respectively to 4 feet and 6 feet, and the area of DEF is 9 square feet; find the area of ABC.

12. Two chords of a circle intersect; the segments of one are respectively 6 feet and 8 feet, and one segment of the other is 12 feet; find the remaining segment.

13. Two circles, whose radii are 6 feet and 10 feet, intersect, and the line joining their points of intersection is 8 feet; find the distance between their centres.

14. Find the area of a triangle whose sides are respectively 31, 28, and 20 feet.

A

15. Show that the area of an equilateral triangle is equal to one fourth the square of one side, multiplied by √3 ; or to the square of one side multiplied by .433. 16. From a point, O, in an equilateral triangle, ABC, the distances to the vertices were measured and found to be: OB =20, OA 28, OC = 31; 31; find the area of the triangle and the length of each side.

[AD is made equal to OA, CD to OB, CF to OC, BF to OA, BE to OB, AE to OC.]

B

BOOK V.

REGULAR POLYGONS.-AREA

OF THE CIRCLE.

DEFINITION.

1. A REGULAR POLYGON is a polygon which is both equilateral and equiangular.

PROPOSITION I. THEOREM.

Regular polygons of the same number of sides are similar.

Let ABCDEF and abcdef be regular polygons of the same number of sides: then they are similar.

For, the corresponding

angles in each are equal, because any angle in either polygon is equal to twice as many right angles as the polygon has sides, less four right angles, divided

by the number of angles (B. I., P. XXVI., C. 4); and further, the corresponding sides are proportional, because all the sides of either polygon are equal (D. 1): hence, the polygons are similar (B. IV., D. 1); which was to be proved.

PROPOSITION II. THEOREM.

The circumference of a circle may be circumscribed about any regular polygon; a circle may also be inscribed in it.

1°. Let ABCF be a regular polygon: then can the circumference of a circle be circumscribed about it.

For, through three consecutive vertices A, B, C, describe the circumference of a circle (B. III., Problem XIII., S.). Its centre O lies on PO, drawn perpendicular to BC, at its middle point P; draw OA and OD.

OPCD be

Let the quadrilateral OPCD turned about the line OP, until PC

falls on PB; then, because the angle

H

B

D

C is equal to B, the side CD will take the direction BA; and because CD is equal to BA, the vertex D, will fall upon the vertex A; and consequently, the line OD will coincide with OA, and is, therefore, equal to it: hence, the circumference which passes through A, B, and C, passes through D. In like manner, it may be shown that it passes through each of the other vertices: hence, it is circumscribed about the polygon; which was to be proved.

2o. A circle may be inscribed in the polygon.

For, the sides AB, BC, &c., being equal chords of the circumscribed circle, are equidistant from the centre O; hence, a circle described from O as a centre, with OP as a radius, is tangent to each of the sides of the polygon, and consequently, is inscribed in it; which was to be proved.

Scholium. If the circumference of a circle is divided into equal arcs, the chords of these arcs are sides of a regular inscribed polygon.

For, the sides are equal, because they are chords of equal arcs, and the angles are equal, because they are measured by halves of equal arcs.

If the vertices A, B, C, &c., of a regular inscribed polygon be joined with the centre O, the triangles thus formed will be equal, because their sides are equal, each to each: hence, all of the angles about the point O are equal to each other.

H

DEFINITIONS.

1. The CENTRE OF A REGULAR POLYGON is the common centre of the circumscribed and inscribed circles.

2. The ANGLE AT THE CENTRE is the angle formed by drawing lines from the centre to the extremities of any side.

The angle at the centre is equal to four right angles divided by the number of sides of the polygon.

3. The APOTHEM is the shortest distance from the centre to any side.

The apothem is equal to the radius of the inscribed circle.

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