EXAMPLE 2. How many square centimeters in 10 millimeters square? Answer. (10mm.) 2 = 100mm.2 = [cm.3 ̧

792. REMARK. Distinguish carefully between square meters and meters square.

We say 10 square kilometers (10km.), meaning a surface which would contain 10 others, each a square kilometer; while the expression "5 kilometers square" (5km.) means a square whose sides are each 5 kilometers long, so that the figure contains 25km.2.

EXAMPLE 3. A square is 1000m.2. Find its side.

Answer. V1000m. 31.623m.

793. Because the sum of the squares on the two sides of a right-angled triangle is the square of the hypothenuse, therefore, also,

Given, the hypothenuse and one side, to find the other side. RULE. Multiply their sum by their difference, and extract the square root.

FORMULA. - a2 = (c + a) (c — a) = b2.

From this it follows, that, in an acute-angled triangle, if we are given two sides and the projection of one on the other, or two sides and an altitude, we can find the third side.

EXERCISES.

112. What must be given in order to find the medials of a triangle?

113. If on the three sides of any triangle squares are described outward, the sects joining their outer corners are twice the medials of the triangle, and perpendicular to them.

CHAPTER II.

RATIO OF ANY CIRCLE TO ITS DIAMETER.

PROBLEM I.

794. Given, the perimeters of a regular inscribed and a similar circumscribed polygon, to compute the perimeters of the regular inscribed and circumscribed polygons of double the number of sides.

Take AB a side of the given inscribed polygon, and CD a side of the similar circumscribed polygon, tangent to the arc AB at its midpoint E.

Join AE, and at A and B draw the tangents AF and BG; then AE is a side of the regular inscribed polygon of double the number of sides, and FG is a side of the circumscribed polygon of double the number of sides.

Denote the perimeters of the given inscribed and circumscribed polygons by p and q respectively, and the required perimeters of the inscribed and circumscribed polygons of double the number of sides by and respectively.

Since OC is the radius of the circle circumscribed about the polygon whose perimeter is 9,

:. 9 :p :: OC : OE.

(548. The perimeters of two similar regular polygons are as the radii of their circumscribed circles.)

But, since OF bisects the

. OC

COE,

OE :: CF : FE,

(523. The bisector of an angle of a triangle divides the opposite side in the ratio of the other two sides of the triangle.)

whence, by composition,

.:: CF : FE,

+9 2p CF + FE: 2FE = CE: FG :: q: d',

since FG is a side of the polygon whose perimeter is q', and is contained as many times in as CE is contained in q.

If, now, the letters be taken to represent lengths in terms of the unit sect L, this proportion is

Again, right ▲ AEH ~ EFN, since acute & EAH = FEN,

:. AH : AE :: EN : EF,

:. Þ: p' :: p' : d,

since AH and AE are contained the same number of times in p and p respectively, and EN and EF are contained twice that number of times in and respectively. If, now, the letters be taken to represent length in terms of the same unit sect Z, this proportion is

which gives the number

pL: pL:: pl : d'L,

p' = √pd.

(2)

Therefore from the given lengths p and q we computed by equation (1), and then with p and q we compute p' by equation (2).

THEOREM I.

795. The length of a circle whose diameter is unity is 3.141592+.

The length of the perimeter of the circumscribed square is 4. A side of the inscribed square is √2, therefore its perimeter is 2√2. Now, putting = 2√2 and 9 = 4 in 794, we find, for the perimeters of the circumscribed and inscribed octagons,

9:

Then, taking these as given quantities, we put = 3.0614675, and 9 3.3137085, and find by the same formulæ, for the polygons of sixteen sides, d = 3.1825979, and = 3.1214452.

=

Continuing the process, the results will be found as in the following