Then, AEXIO : BC × DH :: AE : BC.

But (387),

AE XIO BC × DH :: area AEI: area BCD.

Therefore (21),

2

2

area AEI area BCD :: AE: BC.

In a similar manner, prove that the areas have the same ratio as the squares of the altitudes IO and DH, or as the squares of any homologous lines.

AREA OF TRAPEZOIDS.

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

The trapezoid may be divided by a diagonal into two triangles, having for their bases the parallel sides.

The altitude of each of these triangles is equal to that of the trapezoid (264). The area of each triangle being half the product of the common altitude by its base, the area of their sum, or of the whole trapezoid, is half the product of the altitude by the sum of the bases.

EXERCISES.

393.-1. Measure the length and breadth, and find the area of the blackboard; of the floor.

2. To divide a given triangle into any number of equivalent triangles.

3. To divide a given parallelogram into any number of equivalent parallelograms.

4. To divide a given trapezoid into any number of equivalent trapezoids.

5. The area of a triangle is equal to half the product of the perimeter by the radius of the inscribed circle.

6. What is the radius of the circle inscribed in the triangle whose sides are 8, 10, and 12?

EQUIVALENT SURFACES.

394. ISOPERIMETRICAL figures are those whose perimeters have the same extent.

395. Theorem.—Of all equivalent triangles of a given base, the one having the least perimeter is isosceles.

D

B

The equivalent triangles having the same base, AE, have also the same altitude (388). Hence, their vertices are in the same line parallel to the base, that is, in DB.

Now, the shortest line that

E

can be made from A to E through some point of DB, will constitute the other two sides of the triangle of least perimeter. This shortest line is the one making equal angles with DB, as ACE, that is, making ACD and ECB equal (115). The angle ACD is equal to its alternate A, and the angle ECB to its alternate E. Therefore, the angles at the base are equal, and the triangle is isosceles.

396. Corollary. Of all isoperimetrical triangles of a given base, the one having the greatest area is isosceles.

397. To draw a square equivalent to a given figure, is called the squaring, or quadrature of the figure. How this can be done for any rectilinear figure, is shown in the following.

PROBLEMS IN DRAWING.

398. Problem.-To draw a rectangle with a given base, equivalent to a given parallelogram.

With the given base as a first term, and the base and altitude of the given figure as the second and third terms, find a fourth proportional (319). This is the required altitude (385).

399. Problem. - To draw a square equivalent to a given parallelogram.

Find a mean proportional between the base and altitude of the given figure (330). This is the side of the square (385).

400. Problem.To draw a triangle equivalent to a given polygon.

Let ABCDE be the given polygon. Join DA. Produce BA, and through E draw EF

parallel to DA. Join DF.

Now, the triangles DAF and DAE are equivalent, for they have the same base DA, and equal altitudes, since their vertices are in the line EF parallel to the base (264). To each of these equals, add

E

B

the figure ABCD, and we have the quadrilateral FBCD equivalent to the polygon ABCDE. In this manner, the number of sides may be diminished till a triangle is formed equivalent to the given polygon. In this diagram it is the triangle FDG.

401. Problem.-To draw a square equivalent to a given triangle.

Find a mean proportional between the altitude and half the base of the triangle. This will be the side of the required square.

EQUIVALENT SQUARES.

402. Having shown (379) how an area is expressed by the product of two lengths, it follows that an equa

tion will represent equivalent surfaces, if each of its terms is composed of two factors which represent lengths.

For example, let a and b represent the lengths of two straight lines. Now we know, from algebra, that whatever be the value of a and b,

(a+b)2=a2+2ab+b2.

This formula, therefore, includes the following geometrical

403. Theorem.-The square described upon the sum of two lines is equivalent to the sum of the squares described on the two lines, increased by twice the rectangle contained by these two lines.

Since the truths of algebra are universal in their application, this theorem is demon

strated by the truth of the above equation.

Such a proof is called algebraic. It is also called analytical, but with doubtful propriety.

ab

21

Let the student demonstrate the theorem geometrically, by the aid of this diagram.

ab

404. Theorem.—The square described on the difference of two straight lines is equivalent to the sum of the squares described on the two lines, diminished by twice the rectangle contained by those lines.

This is a consequence of the truth of the equation,

(a—b)2 — a2 — 2ab+b2.

405. Theorem.-The rectangle contained by the sum and the difference of two straight lines is equivalent to the difference of the squares of those lines.

Geom.-12

This, again, is proved by the principle expressed in the equation,

(a+b) (a—b)= a2 —b2.

406. These two theorems may also be demonstrated by purely geometrical reasoning.

The algebraic method is sometimes called the modern, while the other is called the ancient geometry. The algebraic method was invented by Descartes, in the seventeenth century, while the other is twenty centuries older.

THE PYTHAGOREAN THEOREM.

407. Since numerical equations represent geometrical truths, the following theorem might be inferred from Article 327.

This is called the Pythagorean Theorem, because it was discovered by Pythagoras. It is also known as the Forty-seventh Proposition, that being its number in the First Book of Euclid's Elements.

It has been demonstrated in a great variety of ways. One is by dividing the three squares into parts, so that the several parts of the large square are respectively equal to the several parts of the two others.

The fame of this theorem makes it proper to give here the demonstration from Euclid.

408. Theorem.—The square described on the hypotenuse of a right angled triangle is equivalent to the sum of the squares described on the two legs.

Let ABC be a right angled triangle, having the right angle BAC. The square described on the side BC is equivalent to the sum of the two squares described on BA and AC. Through A make AL parallel to BD, and join AD and FC.