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square
feet=9 square feet+16 square feet

=25 square feet,
.:: x2=25,

.: x=5. Therefore the length of the hypotenuse is 5 feet.

Example 2. Find the length of the perpendicular drawn from the vertex to the base of an isosceles triangle whose equal sides are 10 feet each, and whose base is 12 feet.

Let ABC be the isosceles triangle such that AB is 10 feet, AC is 10 feet and BC is 12 feet.

[blocks in formation]

Draw AD perpendicular to BC.

Then because the triangle ABC is isosceles AD will bisect the base BC in D ; therefore BD is 6 feet.

Let AD contain x feet.

Then by Euclid I. 47, the square on AB=the sum of the squares on BD and AD.

.: 102 sq. ft.=62 sq. ft. + x2 sq. ft.

.. 102=62 + x2,
... m2=100 – 36,
.. w=64,

.: X=8.
Therefore the required length of AD is 8 feet.

Example 3. Find the length of the diameter of the square one of whose sides contains a feet.

[blocks in formation]

Let ABCD be the square, so that AB is a feet, and AD is a feet.

Let the diameter BD be x feet.

Then the square on DB= the sum of the squares on DA and AB.

.: x2 sq. ft. =a2 sq. ft. + a2 sq. ft.

.:=a + a?,
.: x2=2a”,

.: x=2.a. Therefore the required length of the diameter is 12.a fcet.

EXAMPLES. III.

(1) Find the length of the hypotenuse of a right-angled triangle whose sides are 6 feet and 8 feet respectively.

(2) The hypotenuse of a right-angled triangle is 100 yards and one side is 60 yards : find the length of the other side.

(3) One end of a rope 52 feet long is tied to the top of a pole 48 feet high and the other end is fastened to a peg in the ground. If the pole be vertical and the rope tight, find how far the peg is from the foot of the pole.

(4) The houses in a certain street are 40 feet high and the street 30 feet wide : find the length of the ladder which will reach from the top of one of the houses to the opposite side of the street.

a ft.

.

(5) A wall 72 feet high is built at one edge of a moat 54 feet wide ; how long must scaling ladders be to reach from the other edge of the moat to the top of the wall ?

(6) A field is a quarter of a mile long and three-sixteenths of a mile wide : how many cubic yards of gravel would be required to make a path 2 feet wide to join two opposite corners, the depth of the gravel being 2 inches ?

(7) The sides of a rectangular field are 4a feet and 3a feet respectively. Find the length of its diameter.

(8) If the sides of an isosceles triangle be each 13a yards and the base 10a yards, what is the length of the perpendicular drawn from the vertex to the base ?

(9) Show that the perpendicular drawn from the right angle to the hypotenuse in an isosceles right-angled triangle, each of whose equal sides contains a feet, is

2 (10) If the hypotenuse of a right-angled isosceles triangle be a yards, what is the length of each side ?

(11) Show that the perpendicular drawn from an angular point to the opposite side of an equilateral triangle, each of whose

13 sides contains a feet, is

2 (12) If in an equilateral triangle the length of the perpendicular drawn from an angular point to the opposite side be a feet, what is the length of the side of the triangle ?

(13) Find the ratio of the side of a square inscribed in a circle to the diameter of the circle.

(14) Find the distance from the centre of a circle of radius 10 feet, of a chord whose length is 8 feet.

(15) Find the length of a chord of a circle of radius a yards, which is distant b feet from the centre.

(16) The three sides of a right-angled triangle, whose hypotenuse contains 5a feet, are in arithmetical progression : prove that the other two sides contain 4a feet and 3a feet respectively.

a ft.

* CHAPTER II.

ON INCOMMENSURABLE QUANTITIES.

9. Two numbers are said to be commensurable when their ratio can be expressed as an arithmetical fraction : that is, as a fraction whose numerator and denominator are both whole numbers. Example. 4.93 and 814 are two commensurable numbers.

444 x4 Their ratio is 484:814

90 x 327

10. Two numbers are said to be incommensurable when their ratio cannot be expressed as an arithmetical fraction.

Example. 12 and 1 are two incommensurable numbers.

N2 For cannot be expressed exactly as an arithmetical quantity.

1 So 13 and 12 are two incommensurable numbers.

11. One number alone is said to be an incommensurable number when it is incommensurable with unity. So that an incommensurable number cannot be expressed as an arithmetical fraction.

Example. 12 and 13, and all surd numbers, are incommensurable numbers.

12. Two quantities are said to be commensurable when their measures referred to a common unit are commensurable.

Example. A mile and a thousand yards are two commensurable quantities. Their measures with a yard for unit are 1760 and 1000; and these are commensurable numbers.

13. Two quantities are said to be incommensurable when their measures referred to a common unit are incommensurable.

Example. The side of a square and its diameter are two incommensurable quantities. For if the side of a square contain a feet, the diameter (see Example 3, p. 7) contains V2. a feet, and therefore the ratio of their measures is 1 : N2. So that their measures are incommensurable.

14. There is no practical difficulty in dealing with incommensurable quantities. We can always find for their measures arithmetical expressions sufficiently accurate for all practical purposes.

15. A little consideration will convince the student that no measurement can in practice be made with absolute accuracy.

For example : A skilful mechanic is probably satisfied if in measuring some material two or three feet in length the error in his measurement is less than the thirty-second part of an inch. (The thirty-second part of an inch is less than half the height of the smallest letter on this page.) That is to say, he is satisfied if he make no greater error than about a thousandth part of the whole length to be measured. He would record such a measurement thus,

2 ft. 37 inches, which is 891 thirty-second parts of an inch.

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