PROPOSITION III. THEOREM.

319. The area of a rectangle is equal to the product of its base and altitude.

Let R be the rectangle, b the base, and a the altitude; and let U be a square whose side is the linear unit.

(two rectangles are to each other as the product of their bases and altitudes).

320. SCHOLIUM. When the base and altitude are exactly divisible by the linear unit, this proposition is rendered evident by dividing the figure into squares, each equal to the unit of

measure. Thus, if the base contain seven linear units, and the altitude four, the figure may be divided into twenty-eight squares, each equal to the unit of measure; and the area of the figure equals 7 × 4.

PROPOSITION IV. THEOREM.

321. The area of a parallelogram is equal to the product of its base and altitude.

Let A EFD be a parallelogram, A D its base, and C D its altitude.

We are to prove the area of the

AEFD=ADX CD.

From A draw A B to DC to meet FE produced.

Then the figure ABCD will be a rectangle, with the same

base and altitude as the ☐ A EF D.

In the rt. A ABE and CDF,

(two rt. ▲ are equal, when the hypotenuse and a side of the one are equal respectively to the hypotenuse and a side of the other).

Take

Take

away the ACDF and we have left the rect. A B C D. away the AABE and we have left the

AEFD.

Ax. 3

But the area of the rect. A B C D = ADX CD, § 319 (the area of a rectangle equals the product of its base and altitude). .. the area of the AEFD=ADX C D.

Ax. 1

Q. E. D.

322. COROLLARY 1. Parallelograms having equal bases and equal altitudes are equivalent.

323. Cor. 2. Parallelograms having equal bases are to each other as their altitudes; parallelograms having equal altitudes are to each other as their bases; and any two parallelograms are to each other as the products of their bases by their altitudes.

PROPOSITION V. THEOREM.

324. The area of a triangle is equal to one-half of the product of its base by its altitude.

Let ABC be a triangle, AB its base, and CD its

=

.. ΔΑΒΟ AAHC,

(the diagonal of a divides it into two equal ▲ ).

The area of the

base by its altitude.

§ 136

$133

ABCH is equal to the product of its

§ 321

.. the area of one-half the, or the ▲ ABC, is equal to one-half the product of its base by its altitude,

or,

ABX CD.

Q. E. D.

325. COROLLARY 1. Triangles having equal bases and equal altitudes are equivalent.

326. COR. 2. Triangles having equal bases are to each other as their altitudes; triangles having equal altitudes are to each other as their bases; any two triangles are to each other as the product of their bases by their altitudes.

PROPOSITION VI. THEOREM.

327. The area of a trapezoid is equal to one-half the sum of the parallel sides multiplied by the altitude.

H

E

C

Let ABCH be a trapezoid, and EF the altitude.

We are to prove

area of ABC H= } (HC + A B) E F. Draw the diagonal A C.

Then the area of the ▲ AHC = {HCX EF,

$ 324

(the area of a A is equal to one-half of the product of its base by its altitude),

328. COROLLARY. The area of a trapezoid is equal to the product of the line joining the middle points of the non-parallel sides multiplied by the altitude; for the line OP, joining the middle points of the non-parallel sides, is equal to (HC + A B).

.. by substituting OP for (HC+ A B), we have,
the area of ABCH:

329. SCHOLIUM. The area of an irregular polygon may be found by dividing the polygon into triangles, and by finding the area of each of these triangles separately. But the method generally employed in practice is to draw the longest

=

OP XEF.

$142

diagonal, and to let fall perpendiculars upon this diagonal from the other angular points of the polygon.

The polygon is thus divided into figures which are right triangles, rectangles, or trapezoids; and the areas of each of these figures may be readily found.

PROPOSITION VII. THEOREM.

330. The area of a circumscribed polygon is equal to onehalf the product of the perimeter by the radius of the inscribed circle.

Let ABSQ, etc., be a circumscribed polygon, and C

the centre of the inscribed circle.

Denote the perimeter of the polygon by P, and the radius of the inscribed circle by R.

We are to prove

the area of the circumscribed polygon = } P × R.

Draw CA, CB, CS, etc.;

also draw CO, CD, etc.,

to A B, BS, etc.

ABX CO,

The area of the ACAB = § 324 (the area of a ▲ is equal to one-half the product of its base and altitude).

The area of the ACBS

=

BS X CD, $ 324

..the area of the sum of all the ACA B, CBS, etc.,

= 1 (A B+ B S, etc.) CO,

(for CO, CD, etc., are equal, being radii of the same O).

Substitute for AB+BS+ SQ, etc., P, and for CO, R; then the area of the circumscribed polygon = P × R.

Q. E. D.