378. Historical Note. The theorem of § 376 is known as the Pythagorean Proposition, and is so called because it is said to have been first proved by Pythagoras, a famous Greek mathematician who lived about 500 B. C. This proposition is the 47th of Euclid's Elements.

The proposition in a special form was known to the Egyptians as early as 2000 B. C. It was also known to the Chinese and Babylonians at a very early date.

It is not known how Pythagoras proved the theorem. The proof here given is attributed to Euclid, who lived about 300 B. C.

Because of its frequent use in mathematics and because of its numerous practical applications, the Pythagorean theorem is perhaps the most important one of plane geometry. A large number of different proofs have been devised for this famous proposition.

EXERCISES

THEOREMS

1. If a right triangle is isosceles, the area of the square on the hypotenuse is equal to four times the area of the triangle.

2. If one angle of a right triangle is 60°, the area of the square on the hypotenuse is equal to four times the area of the square on the shorter side.

3. If a perpendicular is drawn from the vertex of a triangle to its base, the difference of the squares on the segments of the base equals the difference of the squares on the other two sides.

4. The sum of the squares on the diagonals of a rhombus equals the sum of the squares on the four sides.

5. In an isosceles right triangle, prove that the sides have the ratios 1:1:√2

6. In a right triangle having acute angles of 30° and 60°, prove that the sides have the ratios 2 : 1 :√3.

7. Show that the following rules are true:

(1) The diagonal of a square equals a side multiplied by V2.

(2) The side of a square equals half of a diagonal multiplied by √2.

(3) The altitude of an equilateral triangle equals half of a side mul

tiplied by V3.

NOTE. Because of their frequent use in practical applications these three rules are well worth remembering.

*8. Prove the Pythagorean theorem by means of the adjoining figures. In Figs. 4, 5, and 6, use algebraic methods. In Fig. 6, note that ACDE is a trapezoid and equate areas.

NOTE. Fig. 3 is a proof by Bhaskara, a mathematician of India who lived about 1150 A.D. Fig. 6 is President Garfield's proof.

In the following Exercises, A, a, b, and c stand for the numerical measures of the area, altitude, base, and hypotenuse, respectively, of a right triangle. 1. Given c2=a2+b2, solve for each letter in terms of the others

2. Show that a2 = (c+b) (c—b), and b2 = (c+a) (c—a).

NOTE. The relations in Exercise 2 are often convenient in computation. 3. Given a =7 and b=9; find c and A.

4. Given c=17 and a = 13; find b and A.

5. Given A = 102.5 and a=13.4; find b and c.

6. Given A =60 and c = 13; find a and b.

7. Find the diagonal of a square whose sides are each 8 in.

8. Find the side of a square whose diagonal is 16 in.

9. Compute the altitude and area of an equilateral triangle of which a side is 3 in.

10. If the side of an equilateral triangle is s, find its area. Ans. 182√3. 11. If the area of an equilateral triangle is given by the formula s2 √3, find the side of an equilateral triangle which has an area of 100 sq. in.

12. In Fig. 1, find the length of the brace AB, and the distance from C to B.

13. The hypotenuse of a right triangle is 8 ft. and one side is 5 ft. Show that the equilateral triangle made on the hypotenuse is equal to the sum of the equilateral triangles made on the other two sides.

14. The area of a rectangular lawn is 5525 sq. m., and the length of one of its sides is 85 m. Find the length of its diagonal in meters to three decimal places. Ans. 107.005 m. *15. Find the diagonal of a cube 9 ft. on an edge. (Fig. 2.) Ans. 9√3 ft.

*16. In the gambrel roof shown in section in Fig. 3, find the lengths of rafters and parts whose dimensions are not given.

Ans. AB=15 ft. 7 in.; DB=15 ft. 2 in.; BC=14 ft. 3 in. (All to the nearest in.)

17. The height of the flagstaff shown in Fig. 4 is unknown; but it is noticed that the flag rope which is 4 ft. longer than the staff, when stretched out just reaches the ground at a point 25 ft. from the foot of the staff. If the ground is level, find the height of the staff.

Ans. 76 ft.

18. A tree, CD in Fig. 5, 75 ft. high, breaks at a point B, and the end D strikes the ground at A, a distance of 40 ft. from C. Find the length BD that was broken off. Ans. 48 ft.

19. The differences between the hypotenuse and the two sides of a right triangle are 3 and 6 feet respectively. Find the sides and the area of the triangle. Ans. 15 ft., 12 ft., 9 ft., and 54 sq. ft.

x ft.

---40 ft-·
FIG. 5

D

PROJECTION

379. If from a point P, Fig. 1, a perpendicular PQ is drawn to any straight line RS, the foot of the perpendicular, Q, is said to be the projection of P upon RS.

The projection of a line upon a given straight line is the portion of the given line lying between the projections of the ends of the line.

In Figs. 2 and 3, CD is the projection of AB upon OM.

380. Theorem. In any obtuse triangle, the square of the side opposite the obtuse angle is equivalent to the sum of the squares of the other two sides, plus twice the product of one of those sides and the projection of the other side upon it.

Given the obtuse triangle ABC, the angle ACB being obtuse, and d and AD being the projections of a and c, respectively, upon AC.

To prove c2=a2+b2+2bd

B

b

C

d

Proof.

Then

But

AD=b+d

AD2=b2+2bd+d2.

Adding h2 to both members of this equation,
AD2+h2= b2+2bd+d2+h2.

AD2+h2=c2, and d2+h2=a2.

.. c2=a2+b2+2bd.

Why?

$ 376

Why?

Similarly, if b' is the projection of b upon CB, it can be proved

that

c2=a2+b2+2ab'.

381. Theorem. In any triangle, the square of a side opposite an acute angle is equivalent to the sum of the squares of the other two sides, minus twice the product of one of these sides and the projection of the other side upon it.

Given the triangle ABC, having an acute angle C, and d the projection of a upon AC.

c2=a2+b2-2bd.

Proof. AD=b-d, Fig. 1, or AD=d—b, Fig. 2.

To prove

Similarly, if b' is the projection of b upon CB, it can be proved that

c2=a2+b2-2ab'.

NOTE. It should be noticed that the theorems of § 380 and § 381 become the same as the theorem of § 376, if the angle opposite side c changes into a right angle. Show that this is true.

382. Theorem. An angle of a triangle is acute, right, or obtuse according as the square of the opposite side is less than, equal to, or greater than the sum of the squares of the other two sides.