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equal to the square on the radius of the second circle, prove that D, 0, E are collinear. Apply (335).

27. If two circles intersect, prove that tangents drawn to them from any point in their common chord produced are equal to each other.

28. AEB, PEDQ are respectively a diameter and a chord of a circle at right angles to each other; ADC, BC are other chords. Prove that the product of AB, AE is equal to the product of AC, AD. (341).

29. If AB be a diameter of a circle, and APQ a straight line cutting the circle again at P and a fixed straight line perpendicular to AB at Q; the product of AP, AQ is constant. (341).

30. A, B, D are three points in order on a circle; C is the middle point of the arc AB; if AC, AD, BD, CD be joined, prove that the square on CD is equal to the square on AC with the product of AD, BD.

31. A circle is described round an equilateral triangle, and from any point in the circumference straight lines are drawn to the angular points of the triangle: show that one of these straight lines is equal to the sum of the other two.

32. From the extremities B, C of the base of an isosceles triangle ABC, straight lines are drawn at right angles to AB, AC respectively, and intersecting at D: show that BC X AD= 2 AD X DB.

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PROBLEMS.

33. AD bisects the angle BAC. From any point in AB draw a straight line to AC which shall be divided by AD in a given ratio, i.e., into parts which have the same ratio which two given straight lines have to each other.

34. Through a given point between two given straiglft lines draw a straight line which shall be terminated by the given lines and divided by the point in a given ratio.

35. Upon a given portion AC of the diameter AB of a semicircle another semicircle is described. Draw a line

through A so that the part intercepted between the semicircles may be of given length.

36. ABC is a segment of a circle; the chord AC is divided into two parts at D. Divide the arc ABC into two parts so that the chords of the parts shall have the same ratio which AD, DC have to each other. Complete the O ABCE; bisect arc AC opp. B in E; produce ED to B, etc.

37. Divide the straight line AB into two parts at C so that the square on AC may be equal to twice the square on BC. Make ZBAD = 4 rt. 2, draw BD I to AB, make ZADC = _DAB, etc. 38. From a given point as centre describe a circle cutting

a a given straight line in two points, so that the product of their distances from a fixed point in the straight line may be equal to a given square.

39. ABC is a given obtuse-angled triangle: find a point D in the side BC opposite the obtuse angle, so that BD X DC = AD.

Let O be the cent. of the o about a ABC. Join AO, on it describe : O ADO, cutting BC in D; join AD and produce it to E, etc.

40. Describe a circle which shall pass through two given points and touch a given straight line. Two solutions. (339). Let A, B be the pts., XY the line ; produce AB to meet XY in C, etc.

41. Describe a circle to pass through a given point and touch two given straight lines which are not parallel to each other. Two solutions. Let given pt. P be between the given lines AB, AC; bisect BAC, etc.

42. From a given point on the circumference of a given circle draw two chords so as to be in a given ratio and to contain a given angle. Let P be the given pt.; draw diameter PA ; make ZAPB = given 2, etc.

43. From any point A on one of two intersecting straight lines OA, OB draw two straight lines AB, AC cutting OB in B and C, such that the triangles AOB, OAC may be similar, and the sides AB, AC have a given ratio to each other.

44. To describe a circle which shall pass through two given points and touch a given circle.

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Book IV.

AREAS OF POLYGONS.

MEASUREMENT OF AREAS.

355. DEF. The area of a surface is its numerical measure (229), referred to the unit of surface.

To form a unit of surface, we take any unit of length, as an inch, a foot, etc., and construct a square on it. This square is the corresponding unit of surface, or the superficial unit.

Two surfaces are equivalent when they have equal areas and cannot coincide; they are equal when they can coincide.

Proposition 1. Theorem. 356. If two rectangles have equal altitudes, their areas are to each other as their bases.

Hyp. Let ABCD, AEHD be two rectangles, having equal altitudes. ABCD AB

A F
AEHD AE

D

H CASE I. When A B and AE are commensurable. Proof. Take AF, any common meas

AF

E ure of AB and AE, and suppose it to be contained 5 times in AB and 4 times in AE.

AB 5 Then

AE 4 Draw Is to AB and AE at the several points of division. The rectangle ABCD will be divided into 5 rectangles, and AEHD into 4 rectangles.

To prove

=

=

These rectangles are all equal.

(136)
ABCD 5
AEHD 4
ABCD AB
AEHD

(Az. 1)
AE
CASE II. When AB and AE are incommensurable.
Proof. In this case we know (232)

D that we may always find a line AG

KO as nearlg equal as we please to AB, and such that AG and AE are commensurable.

GB Draw GK || to BC.

H
Then, since AG and AE are com-
mensurable,
AGKD AG

(Case I)
A

E AEHD AE Now, these two ratios being always equal while the common measure is indefinitely diminished, they will be equal when G moves up to and as nearly as we please coincides with B.

(233) ABCD AB

Q. E. D. AEHD AE

=

357. COR. Since the altitudes of rectangles may be considered as their bases, and their bases as their altitudes, therefore :

The areas of rectangles having equal bases are to each other as their altitudes.

NOTE.- In these propositions the words “rectangle,” “triangle," etc., are often used to denote the “ area of the rectangle,” the "area of the triangle,"

etc.

EXERCISES.

The area of a rectangle is 1728 square feet, and its base is 2 yards : what is the area of a second rectangle whose base is 5 yards and whose altitude is the same as that of the first ?

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To prove

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Proposition 2. Theorem. 358. The areas of any two rectangles are to each other as the products of their bases by their altitudes.

Hyp. Let R and R' be two rectangles, b and V' their bases, a and a' their altitudes,

R

R
R

b'

b R'

a' x 0 Proof. Construct the rect

S angle S having the same base b as that of R, and the same altitude a' as that of R'.

R Then

S rectangles of equal bases are to each other as their altitudes (357);

S and

R'

Ő
rectangles of equal altitudes are to each other as their bases (356).

Multiplying these ratios, and canceling the common factor, we have R

Q. E.D. R' a' xb'

a

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=

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a X 5

359. Sch. By the product of two lines is meant the product of the numbers which represent those lines when they are measured by a common linear unit. (277)

EXERCISE.

Find the ratio of two rectangles, the base of the first being 3 yards and its altitude 13 feet, the base of the second being 8 feet and its altitude 39 inches. Reduce to the same unit, and compare.

Ans. 41.

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