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 into two As ABD and DCB, having the common altitude

The area of 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

A

B

F

M

N

E

G

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 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. ▲ and the common ▲ ABC,

... A 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 BC = sq. on AB + sq. on AC. Q. E.D.

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).

Proposition 8. Theorem.

375. The areas of two triangles having an angle of the one equal to an angle of the other, are to each other as the products of the sides including the equal angles.

Hyp. Let ABC, ADE be the two As

having the common ▲ A.

Since the As ABC, ABE have their

bases AC, AE in the same straight line,

B

and the common vertex B, they have the same altitude.

As of the same altitude are to each other as their bases (369).

Also, the As ABE, ADE have the same altitude, since their bases AB, AD, opp. the common vertex E, are in the same straight line.

376. Cor. If the products of the sides including the equal angles are equal, the triangles are equivalent.

EXERCISE.

If two triangles have an angle of the one supplementary to an angle of the other, their areas are to each other as the products of the sides including these angles,

Proposition 9. Theorem.

377. The areas of similar triangles are to each other as the squares of their homologous sides.

homologous sides of similar ▲s are proportional (307).

378. Cor. The areas of two similar triangles are to each other as the squares of any two homologous lines.

EXERCISES.

1. ABC is a triangle. AE and BF, intersecting in G, are drawn to bisect the sides BC, AC in E and F. pare the areas of the triangles AGB, FGE.

Com

2. The base and altitude of a triangle are 18 and 8 respectively; what is the altitude of a similar triangle whose base is 12?

3. A has a triangular piece of ground, the base of the triangle being 40 rods; what is the base of a similarlyshaped lot containing 4 times as much? Ans. 80 rods.

Proposition 10. Theorem.

379. The areas of similar polygons are to each other as the squares of their homologous sides.

Hyp. Let S and S' denote the areas of the two similar polygons ABCDE and A'B'C'D'E', in which AB and A'B' are homologous sides.

similar As are to each other as the squares of their homologous sides (377).

and

=

380. COR. 1. The areas of similar polygons are to each other as the squares of any two homologous lines.

381. COR. 2. The homologous sides of similar polygons are to each other as the square roots of their areas.