578. The square described on the hypothenuse of a rightangled triangle, is equal to the sum of the squares described on the other two sides. (Thomson's Legendre, B. IV. 11, Euc. I. 47.)

The truth of this principle may be seen from the following geometrical illustration. Thus,

Let the base AB of the rightangled triangle ABC be 4 feet, 'he perpendicular AC, be 3 feet; hen will the squares described on the base AB, and the perpendicular AC, contain as many square feet as the square described on the hypothenuse BC. Now (4)2+(3)2=25 sq. ft.; and the square described on BC also contains 25 sq. ft. Hence, the square described on the hypothenuse of any right-angled triangle, is equal to the sum of the squares described on the other two sides.

OBS. Since the square of the hypothenuse BC, is 25, it follows that the √25, or 5, must be the hypothenuse itself. Hence,

579. When the base and perpendicular are given, to find the hypothenuse.

Add the square of the base to the square of the perpendicular, and the square root of the sum will be the hypothenuse.

Thus, in the right-angled triangle ABC, if the base is 4 and the perpendicular 3, then (4)+(3)=25, and √25=5, the hypothenuse.

580. When the hypothenuse and base are given, to find the perpendicular.

From the square of the hypothenuse subtract the square of the base, and the square root of the remainder will be the perpendicular.

QUEST.-576. What is a triangle? What is a right-angle? 577. What is a rightangled triangle? What is the side opposite the right-angle called? What are the other two sides called? 578. What is the square described on the hypothenuse equal to? 579. When the base and perpendicular are given, how is the hypothenuse found 580. When the hypothenuse and base are given, how is the perpendicular found?

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Thus, if the hypothenuse is 5 and the base 4, then (5) *—(4)% =9, and √93, the perpendicular.

581. When the hypothenuse and the perpendicular are given, to find the base.

From the square of the hypothenuse subtract the square of the perpendicular, and the square root of the remainder will be the base. Thus, if the hypothenuse is 5 and the perpendicular 3, then (5)2——(3)2=16, and 16-4, the base.

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OBS. 1. From the preceding principles it is manifest that the area of a square may be found by dividing the square of its hypothenuse by 2. (Arts. 285, 578.) 2. The areas of all similar figures are to each other as the squares of their homologous sides, or their like dimensions. (Leg. IV. 25, 27. V. 10.) Hence,

The st..n of the areas of equilateral or other similar triangles, also of similar polygons, circles and semicircles described on the base and perpendicular of a right-angled triangle, is equal to the area of a similar figure described on the hypothenuse.

3. The square of a simple ratio is called a duplicate ratio; the cube of a simple ratio, a triplicate ratio.

The ratio of the square roots of two numbers is called a sub-duplicate ratio; that of the cube roots, a sub-triplicate ratio.

Ex. 1. If a street is 28 feet wide, and the height of a tower is 96 feet, how long must a rope be to reach from the top of the tower to the opposite side of the street?

Solution.—(96)2+(28)2=10000, and √10000=100 ft. Ans.

2. A ladder 40 feet long being placed at the opposite side of a street 24 feet wide, just reached the top of a house: how high was the house ?

3. Two ships, one sailing 7 miles, the other 12 miles an hour, spoke each other at sea; one was going due east, the other due south how far apart were they at the expiration of 12 hours?

4. What is the length of the side of a square farm which contains 360 acres; and how far apart are its opposite corners?

582. A mean proportional between two numbers is equal to the square root of their product. (Arts. 494, 498. Obs. 2.)

QUEST.-590 When the hypothenuse and perpendicular are given, how is the base

found?

5. Find a mean proportional between 2 and 8. Solution.-8X2=16; and √16=4. Ans.

Find a mean proportional between the following numbers:

583. To find the side of a square equal in area to any given

surface.

Extract the square root of the given area, and it will be the side of the square sought.

OBS. When it is required to find the dimensions of a rectangular field, equal in area to a given surface, and whose length is double, triple, or quadruple, &c., of its breadth, the square root of 1,,, of the given surface, will be the width; and this being doubled, tripled, or quadrupled, as the case may be, will be the length.

18. What is the side of a square equal in area to a rectangular field 81 rods long, and 49 rods wide?

19. What is the side of a square equal in area to a triangular field which contains 160 acres?

20. What is the side of a square equal in area to a circular field which contains 640 acres?

21. What are the length and breadth of a rectangular field which contains 480 acres, and whose length is triple its breadth? 22. A general arranged 10952 soldiers, so that the number in rank was double the file: how many were there in each?

584. When the sum of two numbers and the difference of their squares are given, to find the numbers.

Divide the difference of their squares by the sum of the numbers, and the quotient will be their difference; then proceed as in Art. 155.

23. The sum of two numbers is 42, and the difference of their squares is 756: what are the numbers ? Ans. 12 and 30. 24. The sum of two numbers is 65, and the difference of their squares is 975: what are the numbers?

585. When the difference of two numbers and the difference of their squares are given, to find the numbers.

Divide the difference of the squares by the difference of the numbers, and the quotient will be their sum; then proceed as in Art. 155.

24. The difference of two numbers is 29, and the difference of their squares is 1885: what are the numbers?

EXTRACTION OF THE CUBE ROOT.

586. To extract the cube root, is to resolve a given number into three equal factors; or, to find a number which being multiplied into itself twice, will produce the given number. (Art. 564.)

Ex. 1. What is the cube root of 64?

Solution.-Resolving the given numbers into three equal factors, we have 64-4X4X4. Ans. 4.

2. What is the cube root of 12167?

two figures, (Art. 562. Obs. 3,) and thus enables us to find part of it at a time.

Beginning with the left hand period, we find the greatest cube of 12 is 8, the root of which is 2. Place the 2 on the right of the given number for the first part of the root, and also in Col. I. on the left of the number. Multiplying the 2 into itself, write the product 4 in Col. II.; and multiplying 4 by 2, subtract its product from the period, and to the right of the remainder bring down the next period for a dividend. Then adding 2, the first figure of the root, to the first term of Col. I., and multiplying the sum by 2, we add the product 8 to the 1st term of Col. II., and to this sum

QUEST.-586. What is it to extract the cube root?

annex two ciphers, for a divisor; also add 2, the first figure of the root, to the 2d term of Col. I. Finding the divisor is contained in the dividend 3 times, we place the 3 in the root, also on the right of the 3d term of Col. I. Then multiply the 3d term thus increased, by 3, the figure last placed in the root, and add the product to the divisor. Finally, we multiply this sum by 3, and subtract the product from the dividend. Ans. 23.

587. Hence, we derive the following general

RULE FOR EXTRACTING THE CUBE ROOT.

I. Separate the given number into periods of three figures each, placing a point over units, then over every third figure towards the left in whole numbers, and over every third figure towards the right in decimals.

II. Find the greatest cube in the first period on the left hand; place its root on the right of the number for the first figure of the root, and also in Col. I. on the left of the number, Then multi

plying this figure into itself, set the product for the first term in Col. II.; and multiplying this term by the same figure again, subtract this product from the period, and to the remainder bring down the next period for a dividend.

III. Adding the figure placed in the root to the first term in Col. I., multiply the sum by the same figure, add the product to the first term in Col. II., and to this sum annex two ciphers, for a divisor; also add the figure of the root to the second term of Col. I. IV. Find how many times the divisor is contained in the dividend, and place the result in the root, and also on the right of the third term of Col. I. Next multiply the third term thus increased by the figure last placed in the root, and add the product to the divisor; then multiply this sum by the same figure, and subtract the product from the dividend. To the remainder bring down the next period for a new dividend.

V. Find a new divisor in the same manner that the last divisor was found, then divide, &c., as before; thus continue the operation tril the root of all the periods is found.

QUEST.-587. What is the first step in extracting the cube root? The second Third 1 Fourth? Fifth? How is the cube root proved?