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18. If such a polygon be described about a circle, the sums of the alternate sides are each equal to half the perimeter of the polygon.

19. If a polygon be inscribed in a circle, the sum of the angles in the segments exterior to the polygon, together with two right angles, is equal to twice as many right angles as the polygon has sides.

20. Draw the common tangents to two given circles.

21. From a given point draw a straight line cutting a given circle, so that the intercepted segment of the line may have a given length.

22. The straight line which joins the extremities of equal arcs towards the same parts are parallel.

23. Any parallelogram described about a circle is equilateral, and any parallelogram inscribed in a circle is rectangular.

24. Two opposite sides of a quadrilateral circumscribing a circle touch the circle at extremities of a diameter. Show that the area of the quadrilateral is equal to one-half the rectangle contained by the diameter, and the sum of the other sides.

PROP. 31–37. 25. A tangent is drawn to a circle of 21 inches diameter from a point 17.5 inches from the centre. Find the length of the tangent.

26. Show that a man 6 feet high, standing at the sea level, has a view of 3 miles (approximately) in every direction, along a horizontal plane passing through his eye.

27. The angle between a tangent to a circle and the chord through the point of contact is equal to half the angle which the chord subtends at the centre.

28. From a given point P, within or without a circle, draw a straight line cutting the circle in A and B such that PA shall be three-fourths of PB.

Ex. Let the circle be of 1.5 inches radius, and point P 3.5 inches from its centre. Prove your construction by scale.

29. The greatest rectangle which can be inscribed in a circle is a square whose area is equal to half that of the square described upon the diameter as side.

30. If the base and vertical angle of a triangle remain constant in magnitude while the sides vary, show that the locus of the middle point of the base is a circle.

31. Given the vertical angle, the difference of the two sides con. taining it, and the difference of the segments of the base made by a perpendicular from the vertex, to construct the triangle.

32. Show that the locus of the middle point of a straight line, which moves with its extremities upon two straight lines at right angles to each other, is a circle.

33. Show how to produce a straight line, that the rectangle contained by the given line, and the whole line thus produced, may be equal to the square of the part produced.

Ex. If the length of the given line be 2 inches, show geometrically that the length of the part produced is (V5 + 1) inches.

34. Given the height and chord of a segment of a circle to find the radius of the circle.

Ex. If the chord be 24 inches, and the height of the segment be 4 inches, show that the radius of the circle is 20 inches.

35. Show that the locus of the middle points of chords which pass through a fixed point is the circle described as diameter upon the line joining the fixed point and the centre of the given circle.

36. Let ACDB be a semicircle whose diameter is AB, and AD, BC any two chords intersecting in P; prove that

ABP = DA AP + CB.BP.

MATHEMATICS.

SECOND STAGE.

SEOTION 'II.

ALGEBRA.

CHAPTER I.

QUADRATIC EQUATIONS.

1. Equations of this class, when reduced to a rational integral form, contain the square of the unknown quantity, but no higher powers.

When the equation contains the square only of the unknown quantity, and not the first power, it is called a pure quadratic.

Thus, ac – 25 = 0, 4 ca + 10 = 19, 5 c2 = 180, are pure quadratics. When the equation contains the square of the unknown quantity, as well as the first power, it is called an adfected quadratic.

Thus, ä - 5 x = 6, x2 – 20 – 30 = 0, 2 x2 + x + 3 = 6, are adfected quadratics.

!

Pure Quadratics. 2. To solve these, we treat them exactly as we do simple equations, until we obtain the value of the square of the unknown quantity; then, taking the square root of each side, we obtain the value of the unknown quantity. It will be

seen (Stage I., Alg. ; Art. 35) that the unknown quantity in a pure quadratic has always two values, equal in magnitude, but of opposite sign. Ex. 1. Given 3 x + 12 = 687, find x. We have 3 x2 = 687 - '12 = 675, or x2 = 225.

i. W = † 15. 2x + 7 2x – 7 56 Given 2x2 – 7 x - 2xco + 7 ac = 6 x? – 99' Bringing the fractions on the first side to a common denominator, we have

(2 x + 7)2 – (2 x – 7)2 56
x (4 x? -- 49) = 6 x– 99'
56 x

56
Or, « (4 x* -. 49) = 6 22 - 99'
or, 4x* – 49 = 6 x – 99, clearing of fractions ;
then, 6 x* – 99 = 4x2 – 49, from which

202 = 25
8C = £ 5.

Adfected Quadratics. 3. Solution by completing the square.

Suppose we have given the equation wca + 2 ax = 3 a-, to find x.

It will be remembered that (ac2 + 2 ax + a?) is a perfect square, viz., the square of (c + a). It is evident, then, that the first side of our equation will become a perfect square by the addition of a’ as a third term.

Adding, then, ato each side of the given equation, we have

aca + 2 ax + a2 = 4 a, or

(x + a)= 4 aʻ.
Taking now the square root of each side, we have

ac + a = 2 a.
1 X = + 2 a a.

= a or - 3 a.

We may remark that the quantity a?, added to the expression x + 2 ax in order to make it a perfect square, is the square of half the coefficient of X. The operation itself is called completing the square.

An adfected quadratic may therefore be solved as follows:

1. Reduce and arrange it until all the terms involving x are on the first side, and the coefficient of xo is unity.

2. COMPLETE THE SQUARE by adding to each side the SQUARE OF HALF the coefficient of x.

3. Take the square root of each side, put a double sign to the second side, and transpose the term of the first side not involving x.

Ex. 1. Solve the equation 3 202 + 18 x + 4 = 52.

We have 3 c2 + 18 x = 52 - 4 = 48; or, dividing each side by 3, 2c + 6 x = 16.

Here, the coefficient of x is 6, the half of which is 3. Adding then the square of 3 to each side, to complete the square, we have

20+ 6 x + 32 = 16 + 9 = 25. Taking the square root of each side, we have

2 + 3 = $ 5. 1. = 5 – 3 = 2 or - 8. 2. clicron 2C + 3 C + 1 42c + 9 12 x + 17

x + 4 + 2 = 2 x + 7 - 6x + 16' find X.

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15 or; simplifying; -

=
3 x + 2 - 2 + 4 = 6x + 16 - 2x + 7

- - i hi (c + 4) – (ac + 2) 15 (2 x + 7) – 5 (6 x + 16). Or? (x + 2) (a + 4) (6 x + 16) (2x + 7) 1 2

25 ôr, simplifying,

ac + 6 30 + 8 - 12 ** + 74 2 + 112'

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