Page images
PDF
EPUB
[blocks in formation]

These three expressions for the area are comprised in the single statement

Again,

1

2

(product of two sides) × (sine of included angle).

[merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][subsumed][merged small]

=√s (s− a) (s—b) (s — c),

which gives the area in terms of the sides.

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small]

which gives the area in terms of one side and the functions of the adjacent angles.

NOTE. Many writers use the symbol S for the area of a triangle, but to avoid confusion between S and s in manuscript work the symbol A is preferable.

Example 1. The sides of a triangle are 17, 25, 28: find the lengths of the perpendiculars from the angles upon the opposite sides.

[blocks in formation]

it is evident that the three perpendiculars are found by dividing 24 by the three sides in turn.

Now A=√s (sa) (s — b) (s – c) = √35 × 18 × 10 × 7

=5×7×6=210.

[blocks in formation]

Example 2. Two angles of a triangular field are 22.5° and 45°, and the length of the side opposite to the latter is one furlong: find

the area.

Let A=2230, B=45°, then b=220 yds., and C=1121°.

[blocks in formation]

204. To find the radius of the circle circumscribing a triangle.

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

Example. Shew that 2R2 sin A sin B sin C=A.

[merged small][merged small][merged small][ocr errors][merged small][merged small][merged small]

205. From the result of the last article we deduce the following important theorem:

If a chord of length 1 subtend an angle 0 at the circumference of a circle whose radius is R, then 1=2R sin 8.

206. For shortness, the circle circumscribing a triangle may be called the Circum-circle, its centre the Circum-centre, and its radius the Circum-radius.

The circum-radius may be expressed in a form not involving the angles, for

[blocks in formation]

207. To find the radius of the circle inscribed in a triangle.

Let I be the centre of the circle inscribed in the triangle ABC, and D, E, F the points of contact; then ID, IE, IF are perpendicular to the sides.

E

B

Now A=sum of the areas of the triangles BIC, CIA, AIB

[merged small][merged small][merged small][merged small][merged small][subsumed][merged small][ocr errors][merged small][merged small][merged small]

208. To express the radius of the inscribed circle in terms of one side and the functions of the half-angles.

In the figure of the previous article, we know from Euc. IV. 4 that I is the point of intersection of the lines bisecting the angles, so that

B

LIBD: =3, LICD=0.

2,

[blocks in formation]

209. DEFINITION. A circle which touches one side of a triangle and the other two sides produced is said to be an escribed circle of the triangle.

Thus the triangle ABC has three escribed circles, one touching BC, and AB, AC produced; a second touching CA, and BC, BA produced; a third touching AB, and CA, CB produced.

We shall assume that the student is familiar with the construction of the escribed circles.

[See Hall and Stevens' Euclid, p. 255.]

For shortness, we shall call the circle inscribed in a triangle the In-circle, its centre the In-centre, and its radius the Inradius; and similarly the escribed circles may be called the Ex-circles, their centres the Ex-centres, and their radii the Ex-radii.

210. To find the radius of an escribed circle of a triangle.

Let I1 be the centre of the circle touching the side BC and the two sides AB and AC produced. Let D1, E1, F1 be the points of contact; then the lines joining I to these points are perpendicular to the sides.

Let Τι be the radius; then

A area ABC

= area ABIС- area BIC
=area BI1A+area CIA

[merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small]

F

..

A

s-a

Similarly, if r2, " be the radii of the escribed circles opposite to the angles B and C respectively,

[ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small]
« PreviousContinue »