Schol. If AB and AD are incommensurable, that is, if no unit can be found into which they can both be divided without leaving a remainder in one of them, the theorem will still hold true; for if the unit be taken smaller and smaller, the remainder can be made less than any assignable quantity.

When the linear unit is one inch, the unit of area is a square inch; when the linear unit is one foot, the unit of area is a square foot, etc.

Cor. Since the base and altitude of a square are equal (Def. 2, Sec. VI), its area may be found by multiplying one side into itself.

THEOREM XVII.

The area of any parallelogram is equal to the area of a rectangle having the same base and altitude.

same base AB, and of the same altitude, namely the perpendicular distance between the parallels AB, DF. Then will ABFE be equivalent to ABCD.

Since AB and DC are opposite sides of a parallelogram they are equal (Theo. XIII); and for the same reason AB and EF are equal; therefore DC is equal to EF (Ax. 3). Taking away each of these in turn from the whole line DF, we have the remainder DE equal to the remainder CF. But DA is equal to CB (Theo. XIII), and the included angle ADE is equal to

the included angle BCF (Theo. III); therefore, the triangles ADE, BCF, are equal (Theo. VII); and hence if each of them be taken away in turn from the whole figure ABFD, the remainder ABFE will be equivalent to the remainder ABCD.

Therefore, the area of any parallelogram, etc.

Cor. 1. The area of any parallelogram is equal to the product of its base by its altitude.

Cor. 2. Since any triangle, as D ABC, is half of a parallelogram ABCD (Cor. 1, Theo. XIII) on the same base AB and having the same altitude CE (Def. 4, Sec.

BE

V), it follows that the area of a triangle is equal to half the product of its base by its altitude.

THEOREM XVIII.

The area of a trapezoid is equal to half the product of the sum of its parallel sides by its altitude.

Let ABCD be a trapezoid of D which AB and DC are the parallel sides. Then will its area "be equal to half the product of the sum of AB and DC into its altitude, namely, the perpendicular distance between. AB and DC.

B

Draw the diagonal AC. Now, the area of the triangle ABC (Cor. 2, Theo. XVII) is equal to half the product of its base AB into its altitude, which is the same as the altitude of the trapezoid; again, the area of the triangle ADC is equal to half the product of its base DC into its altitude, which is also the same

as the altitude of the trapezoid. Therefore, adding together the area of the two triangles, we have the area of the whole figure equal to half the product of the sum of AB and DC into the altitude.

Hence, the area of a trapezoid, etc.

THEOREM XIX.

The square described on the hypotenuse of a rightangled triangle is equivalent to the sum of the described on the other two sides.

Let ABC be a triangle right-angled at B. It is to be proved that the square AEDC is equivalent to the sum of the squares ABGF and BHIC.

E

B

squares

H

Join FC, BE, and draw BK parallel to AE. Also observe that since ABC and ABG are right angles, BC and BG form one straight line.

Now, in the triangles EAB, CAF, the side EA is equal to the side CA, since they are sides of the same square, and for the same reason the side AB is equal to the side AF. But the included angles EAB, CAF, are also equal; for each of them is composed of a light angle and the angle CAB. Therefore, the two triangles are equal (Theo. VII).

But since the triangle CAF and the square BAFG have the same base AF, and the same altitude, namely the perpendicular distance between the parallels AF,

CG, the square must be double of the triangle (Cors. 1 and 2, Theo. XVII). For like reason the parallelogram AEKL is double of the triangle EAB. But doubles of equals are equals; therefore, the parallelogram AEKL is equivalent to the square BAFG. In the same manner (by joining AI and BD), it may be shown that the parallelogram CDKL is equivalent to the square BCIH. Hence, the whole square ACDE is equivalent to the sum of the squares BAFG, BCIH.

Therefore, the square described, etc.

Cor. 1. The hypotenuse is equal to the square root of the sum of the squares of the other two sides.

Cor. 2. The square of either of the sides containing the right angle is equivalent to the square of the hypotenuse diminished by the square of the other side. By taking the square root of the remainder the side itself will be found.

Cor. 3. If two right-angled triangles have the hypotenuse and one side of the one respectively equal to the hypotenuse and one side of the other, the third sides will also be equal.

EXERCISES.

1. If the side of a square be 36 inches, what is its area in square inches? What in square feet?

2. If the base of a parallelogram be 3 feet and its altitude 4 feet and 6 inches, what is its area?

3. If the base of a triangle be 50 yards and its altitude 20 yards, what is its area?

4. If the parallel sides of a trapezoid be 12 rods

and 16 rods, and its altitude 8 rods, what is its area? 93

5. If the sides containing the right angle of a rightangled triangle be 3 and 4, what is the length of the hypotenuse?

6. If the hypotenuse be 10 and one of the sides 8, what is the length of the other side?

7. Prove that if a parallelogram has one right angle it is a rectangle.

8. Prove that the diagonals of a rectangle are equal

to each other.

9. Prove that a perpendicular is the shortest line that can be drawn to a straight line from a point without it; also, that of two oblique lines that which is furthest from the perpendicular is the longest (Theo. XIX).

SEC. VII.-OF POLYGONS IN GENERAL.

DEFINITIONS.

1. A polygon of five sides is called a PENTAGON ; one of six sides, a HEXAGON; of seven sides, a HEPTAGON; of eight sides, an OCTAGON; of ten sides, a DECAGON; of twelve sides, a DODECAGON.

2. A regular polygon is one which is both equilateral and equiangular. Squares and equilateral triangles are species of regular polygons.

3. The PERIMETER of a polygon is the sum of all its sides.