Proposition 3. Theorem.

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

Hyp. Let R be the rectangle, the base, and a the altitude expressed in numbers of the same linear unit; and let S be the square whose side is the linear unit.

To prove area of Ra × b.

But since S is the unit of surface,

...R÷S= the area of R.

R

S

(358)

(355)

Q.E.D.

361. SCH. 1. The statement of this Proposition is an abbreviation of the following:

The number of units of area in a rectangular figure is equal to the product of the number of linear units in its base by the number of linear units in its altitude. When the base and altitude can be D expressed exactly in terms of some common unit, this proposition is rendered evident by dividing the figure into squares, each equal to the unit of measure. Thus, if AB contain 4 linear units and AD 3, and if through the A

points of division parallels are drawn, it is seen that the rectangle is divided into three rows, each having four squares, or into four columns, each having three squares. Hence, the whole rectangle contains 3 x 4, or 12 squares, each equal to the unit of measure.

Similarly, if AB and AD contain m and n units of length respectively, the rectangle will contain mn units of area. 362. SCH. 2. The area of a square is equal to the second power of its side, being the product of two equal sides.

In Geometry, the expression "rectangle of two lines" is frequently used for the "product of two lines," meaning the product of their numerical measures.

Proposition 4. Theorem.

363. The area of a parallelogram is equal to the prod

uct of its base and altitude.

Hyp. Let ABCD be a, AB its FD base, and BE its altitude.

To prove ABCD = AB × BE.

E

ing the same base and altitude as the ABCD.

Then AD = BC, and AF

BE,

being opp. sides of a □ (129).

Δ ADF = Δ BEC,

B

having the hypotenuse and a side respectively equal in each (110).

Take the ADF from the trapezoid ABCF, and there is left the

ABCD.

Take the ABEC from the trapezoid ABCF, and there is left the rect. ABEF.

But area of rect. ABEF AB × BE.

(360)

(Ax. 1)

area of ABCD = AB X BE.

Q.E. D.

364. COR. 1. Parallelograms having equal bases and equal altitudes are equivalent, because they are all equivalent to the same rectangle.

365. COR. 2. Any two parallelograms are to each other as the products of their bases by their altitudes; therefore parallelograms having equal bases are to each other as their altitudes, and parallelograms of equal altitudes are to each other as their bases.

Proposition 5. Theorem.

366. The area of a triangle is equal to half the product

of its base and altitude.

Hyp. Let ABC be a ▲, BC its

base, and AD its altitude.

BCAH is a

But

having the base BC and the altitude AD. (124)

.. A BAC = Δ ΒΑΗ,

A diagonal of a

divides it into two equal ▲s (130).

BCAH = BC × AD.

The area of a is equal to the product of its base and altitude (363).

367. COR. 1. Triangles having equal bases and equal altitudes are equivalent; and therefore triangles on the same base, and having their vertices in the same straight line parallel to the base, are equivalent.

368. COR. 2. If a triangle and a parallelogram have the same base and are between the same parallels, the area of the triangle is half that of the parallelogram.

369. COR. 3. Any two triangles are to each other as the products of their bases by their altitudes; therefore triangles having equal bases are to each other as their altitudes, and triangles of equal altitudes are to each other as their bases.

370. COR. 4. Of two parallelograms or two triangles on equal bases, that is the greater which has the greater altitude; and of two parallelograms or triangles of equal altitudes, that is the greater which has the greater base.

Proposition 6. Theorem.

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

Hyp. Let ABCD be a trapezoid, AB and CD the || sides, and DH the altitude.

To prove

area ABCD (AB + CD) DH. Proof. Draw the diagonal BD, dividing the trapezoid s ABD and DCB, having the common altitude

into two

The area of a A equals the product of its base and alt. (366).

since the two As make up the area of the trapezoid.

=

Q.E. D.

372. COR. Since the median EF (AB+ CD) (156), therefore, the area of a trapezoid is equal to the product of the median joining the middle points of the non-parallel sides by the altitude.

... area ABCD = FE × DH.

373. SCH. The area of any polygon may be found by dividing it into triangles, and finding the areas of the several triangles. But in practice the method usually employed is to draw the longest diagonal AF of the polygon, and upon AF let fall the perpendiculars BM, CN, DP, EO, GQ, thus decomposing the polygon into right triangles and

A

M

E

trapezoids. By measuring the lengths of the perpendiculars, and the distances between their feet upon AF, the areas of these figures are readily found, and their sum will be the area of the polygon.

COMPARISON OF AREAS.

Proposition 7. Theorem.

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

Hyp. Let ABC be a rt. A,

rt. angled at A, and BE, AK, AF squares on BC, AC, AB.

H

K

B

... CAG is a st. line. (52)

... sq. BG is double the ▲ FBC,

having the same base and between the same || s (368),

LE

each being the sum of a rt. L and the common ▲ ABC,

... ▲ FBC = ^ ABD,

(Cons.)

having two sides and the included equal, each to each (104).

.. sq. BG rect. BL.

(Ax. 6)

In like manner, by joining BK, AE, it may be proved that sq. HC rect. CL.

•*. whole sq. BDEC = sum of sqs. BG and HC. (Ax. 1) sq. on AB + sq. on AC. Q.E.D.

... sq. on BC =

NOTE.-This proposition is commonly called the Pythagorean Proposition, because it is said to have been discovered by Pythagoras (born about 600 B.C.). The above demonstration of it was given by Euclid, about 300 B.C. (Prop. 47, Book I. Euclid).