numbers. In the rectangle ABCD, divide AD into a, and AB into b, equal parts. Through the points of division draw lines parallel to the sides of the rectangle. These lines divide the rectangle into a number of squares, each of which equals the assumed unit of surface. In the bottom row, there are b such squares; and, since there are a rows, we have b squares repeated a times, which gives, in all, ab squares.

NOTE. The composition of ratios includes numerical multiplication as a particular case.

But ordinary multiplication is also an independent growth from addition.

In this latter point of view, the multiplier indicates the number of additions or repetitions, while the multiplicand indicates the thing added or repeated. This is not a mutual operation, and the product is always in terms of the unit of the multiplicand. The multiplicand may be any aggregate; the multiplier is an aggregate of repetitions. To repeat a thing does not change it in kind, so the result is an aggregate of the same sort exactly as the multiplicand.

But if the multiplicand itself is also an aggregate of repetitions, the two factors are the same in kind, and the multiplication is commutative.

This is the only sort of multiplication needed in mensuration; for all ratios are supposed to be expressed exactly or approximately in numbers, and in our rules it is only of these numbers that we speak. Thus, when the rule says, "Multiply the base by the altitude," it means, Multiply the number taken as the length of the base, by the number which is the measure of the altitude in terms of the same linear unit. The product is a number, which we prove to be the area of the rectangle ; that is, its numerical measure in terms of the superficial unit. This is the meaning to be assigned whenever in mensuration we speak of the product of one sect by another.

GENERAL PROOF. If L represent the unit for length, and S the unit of surface, and ab the rectangle, the length of

whose base is b and the length of whose altitude is a, then, by 542,

S

L L

But the first member of this equation is the area of the rectangle, which number we may represent by R; and the second member is equal to the product of the numbers a and b;

EXAMPLE I. Find the area of a ribbon rm. long and rcm. wide.

Answer. 1.2 =100cm.2

Since a square is a rectangle having its length and breadth equal, therefore

789. To find the area of a square.

RULE. Take the second power of the number denoting the length of its side.

NOTE. This is why the product of a number into itself is called the square of that number.

area.

790. Given, the area of a square, to find the length of a side. RULE. Extract the square root of the number denoting the

EXAMPLE 2. How many square centimeters in 10 millimeters square? Answer. (10mm.) 2 = 100mm.2 = 1cm.2.

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

We say 10 square kilometers (10km.2), meaning a surface which would contain 10 others, each a square kilometer; while the expression "5 kilometers square" (5km.)2 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. c2 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.