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:he same number of sides. Now whatever be the number of sides of the polygons, their perimeters will be to each other as the radii of the circumscribed circles (Prop. VIII.). Conceive the arcs subtended by the sides of the polygons to be continually bisected, until the number of sides of the polygons becomes indefinitely great, the perimeters of the polygons will ultimately become equal to the circumferences of the circles. and we shall have

C:c::R: r.

Again, the areas of the polygons are to each other as the squares of the radii of the circumscribed circles (Prop. VIII.). But when the number of sides of the polygons is indefinitely increased, the areas of the polygons become equal to the areas of the circles, and we shall have

A : a : : R2 : r2.

Cor. 1. Similar arcs are to each other as their radii; and similar sectors are as the squares of their radii.

For since the arcs AB, ab are similar, the angle C is equal to the angle c (Def. 5, B. IV.). But the angle C is to four right angles, as the arc AB is to the whole circumference described with the radius AC (Prop. XIV., B. III.); and the

A

B

b

angle c is to four right angles, as the arc ab is to the circumference described with the radius ac. Therefore the arcs AB, ab are to each other as the circumferences of which they form a part. But these circumferences are to each other as AC, ac; therefore,

Arc AB: arc ab:: AC: ac.

For the same reason, the sectors ACB, acb are as the en tire circles to which they belong; and these are as the squares of their radii; therefore,

Sector ACB: sector acb :: AC2 : ac2.

Cor. 2. Let π represent the circumference of a circle whose diameter is unity; also, let D represent the diameter, R the radius, and C the circumference of any other circle; then, since the circumferences of circles are to each other as their diameters,

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that is, the circumference of a circle is equal to the product of its diameter by the constant number π.

Cor. 3. According to Prop. XII., the area of a circle is equal to the product of its circumference by half the radius If we put A to represent the area of a circle, then

A=C×R=2πR × ¦R=πR2;

that is, the area of a circle is equal to the product of the square of its radius by the constant number п.

Cor. 4. When R is equal to unity, we have A=π; that is, π is equal to the area of a circle whose radius is unity. According to Prop. XI., is therefore equal to 3.14159 nearly. This number is represented by π, because it is the first letter of the Greek word which signifies circumference.

SOLID GEOMETRY,

BOOK VII.

PLANES AND SOLID ANGLES

Definitions.

1. A STRAIGHT line is perpendicular to a plane, when it is perpendicular to every straight line which it meets in that plane. Conversely, the plane in this case is per pendicular to the line.

The foot of the perpendicular, is the point in which it meets the plane.

2. A line is parallel to a plane, when it can not meet the plane, though produced ever so far.

Conversely, the plane in this case is parallel to the line.

3. Two planes are parallel to each other, when they can not meet, though produced ever so far.

4. The angle contained by two planes which cut each other, is the angle contained by two lines drawn

from any point in the line of their common

section, at right angles to that line, one in each of the planes.

This angle may be acute, right, or obtuse. If it is a right angle, the two planes are perpendicular to each other.

5. A solid angle is the angular space contained by more than two planes which meet at the same point.

PROPOSITION I. THEOREM

One part of a straight line can not be in a plane, and another part without it.

For from the definition of a plane (Def. 6, B. I.), when a

straight line has two points common with a plane it lies wholly in that plane.

Scholium. To discover whether a surface is plane, we apply a straight line in different directions to this surface, and see if it touches throughout its whole extent.

PROPOSITION II. THEOREM.

Any two straight lines which cut each other, are in one plane, and determine its position.

Let the two straight lines AB, BC cut each other in B; then will AB, BC be in the same plane.

B

A

Conceive a plane to pass through the straight line BC, and let this plane be turned about BC, until it pass through the point A. Then, because the points A and B are situated in this plane the straight line AB lies in it (Def. 6, B. I.). Hence the position of the plane is determined by the condition of its containing the two lines AB, BC. Therefore, any two straight lines, &c.

Cor. 1. A triangle ABC, or three points A, B, C, not in the same straight line, determine the position of a plane. Cor. 2. Two parallel lines AB, CD determine the position of a plane. For A if the line EF be drawn, the plane of the two straight lines AE, EF will be C the same as that of the parallels AB,

-B

F

-D

CD; and it has already been proved that two straight lines which cut each other, determine the position of a plane.

PROPOSITION III. THEOREM.

If two planes cut each other, their common section is a straight line.

Let the two planes AB, CD cut each

other, and let E. F be two points in their common section. From E to F draw the straight line EF. Then, since the points E and Fare in the plane AB, the straight line EF which joins them, must lie wholly in that plane (Def. 6, B. I.). For the same reason, EF must lie wholly in the plane

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CD.

Therefore the straight line EF is common to the two planes AB, CD; that is, it is their common section. Hence, if two planes, &c.

PROPOSITION IV. THEOREM.

If a straight line be perpendicular to each of two straight lines at their point of intersection, it will be perpendicular to the plane in which these lines are.

Let the straight line AB be perpendicular to each of the straight lines CD, EF which intersect at B; AB will also be perpendicular to the plane MN M, which passes through these lines.

Through B draw any line BG, in the plane MN; let G be any point of this line, and through G draw DGF, so that DG shall be equal to GF (Prob. XXI., B. V.). Join AD, AG, and AF.

E

A

F

B

D

Then, since the base DF of the triangle DBF is bisected in G, we shall have (Prop. XIV., B. IV.),

BD2+BF2=2BG2+2GF2.

Also, in the triangle DAF,

AD2+AF2=2AG2+2GF2.

Subtracting the first equation from the second, we have

AD-BD+AF-BF-2AG2 — 2BG2.

But, because ABD is a right-angled triangle,

AD-BD-AB';

and, because ABF is a right-angled triangle,

AF-BFAB2.

Therefore, substituting these values in the former equation, AB2+AB2-2AG'-2BG';

whence

or

AB'-AG'-BG",
AG' AB'+BG'.

=

Wherefore ABG is a right angle (Prop. XIII., Sch., B. IV.) · that is, AB is perpendicular to the straight line BG. In like manner, it may be proved that AB is perpendicular to any other straigh line passing through B in the plane MN; hence it is perpendicular to the plane MN (Def. 1). Therefore, if a straight line, &c.

Scholium. Hence it appears not only that a straight line may be perpendicular to every straight line which passes through its foot in a plane, but that it always must be so whenever it is perpendicular to two lines in the plane, which shows that the first definition involves no impossibility.

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