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Areas of Triangles.

62. THEOREM V. The area of a triangle is equal to half the product of any two sides by the sine of their included angle.

Proof. It is shown in geometry

that the area is half the base into the

altitude. Now in the figure,

Altitude hb sin y.

Therefore, a being the base,

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Area = tahab sin y. Q.E.D.

Cor. 1. If two triangles have two sides of the one respectively equal to two sides of the other, and the angles which these sides form supplementary, the triangles will be equal in area.

For the sines of the supplementary angles are equal.

Cor. 2. Since we may take any one side as a base, if we call h, h', h" the altitudes above the respective sides a, b, and c, we shall have ah = bh' = ch",

and

ab sin y = bc sin a = ca sin ß.

For these expressions are each double the area of the triangle.

EXERCISES.
B

1. Given a = 75, b = 29, ẞ = 16° 15'.6; find the remaining parts and the areas of the two triangles which may be formed.

2. Express the area of a parallelogram of which two adjoining sides are of given length, a and b, and make with each other a given angle &.

3. Express the area of a triangle in terms of a base, c, and the two adjacent angles, a and B. sin a sin ẞ sin (a+B)*

Ans. c2

4. In a parallelogram is given a diagonal of length d, and the angles and which the diagonal makes with the two sides adjoining it. Express the area of the parallelogram.

5. Express the area of the parallelogram in terms of the lengths, d and d', of its diagonals, and the angle & at which they intersect.

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6. In a quadrilateral are given the four sides, a 25.63,

b = 24.09, c = 9.92, d = 29.97, and the angle, 78° 25', which the sides a and b form with each other. Compute the angles and the area of the quadrilateral.

7. A triangle ABC is to be divided into two parts of equal area by a line parallel to AB. What will be the ratios of the segments into which the other two sides are divided? Ans. 1:1 1:1-3•

8. A city lot fronts 60 feet on a street, and the parallel sides run back, the one 100 and the other 135 feet. It is to be divided into two equal parts by a line parallel to its sides. What will be the frontage of each part, and the length of the dividing line?

Remark. The figure of the lot is a trapezoid, and the problem is that of dividing a trapezoid into two equal portions by a line parallel to the base. Let us put

a, b, the parallel sides;

m: 1m, the ratio in which the non parallel sides, and therefore the altitude, is divided by the dividing line;

k, the length of the dividing line.

If

The unknown quantities of the problem are then m and k. we put h for the altitude of the entire trapezoid, the altitudes of the two parts will be mh and (1 — m) h respectively.

the areas will be

a+k
2

mh and k+b (1 — m) h,

2

Therefore

the equality of which gives the first condition. For the second condition we have the geometrical theorem that the difference between the dividing line and either of the parallel sides is proportional to its distance from such side. This gives the proportion α ·k: k-b:: m : 1

whence

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m,

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The quotient of this equation by the preceding one gives an equation from which m is eliminated, and from which we find the value of k.

k = √ a2 + b2

We then find for m the equation

2

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Applying this method to the problem under consideration, we find

k118.796 feet;

m = 0.462 97;

frontages of lots, 27.778 and 32.222 feet.

CHAPTER VII.

THE THEORY OF POLYGONS.

63. A polygon is completely determined when the positions of its vertices, taken in regular order, are given. The polygon may then be formed by joining each pair of consecutive vertices. by a straight line.

The positions of the vertices may be defined by their co-ordinates, on a system now to be explained.

64. Co-ordinates of a point. In geometry the position of a point is fixed by assigning to it certain lines or numbers indicating its situation relative to a fixed line, and a point on that line.

Def. Any numbers or lines which determine the position of a point are called the co-ordinates of that point.

Rectangular co-ordinates. Let OX be the fixed line of refer

ence, and O a point of reference on that line

from which we measure.

Let P be a point whose position is to be

expressed. From P drop a perpendicular

PA upon OX. Then:

P

-X

The line PA is called the ordinate of the point P. The line OA is called the abscissa of the point P. The ordinate and abscissa are called rectangular co-ordinates of the point.

The indefinite line OX along which the abscissas are measured is called the axis of abscissas or the axis of X.

The zero point, O from which the co-ordinates are measured is called the origin.

When the rectangular co-ordinates are given the position of the point is completely determined.

To find the point when the ordinate and abscissa are given, we measure off from O, on the line OX, a distance equal to the given abscissa.

At the end of this distance we erect a perpendicular equal to the given ordinate.

The end of this perpendicular will be the required position of the point.

The abscissa of a point is represented by the symbol x, the ordinate by the symbol y.

If x is positive, its length is laid off from O toward the right; if negative, toward the left

If the ordinate y is positive, it is measured upward from the axis of X; if negative, downward.

EXERCISES.

Draw a line OX 4 or 5 inches long as a line of reference, and lay off points having the following co-ordinates from a zero point near the middle of the line:

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Polar co-ordinates. Draw a line from 0 to P, and call r its length and the angle XOP. Then

OD = x = cos P,

= y = r sin p,

P

DP

and Problem I. of Chapter V. reduces to:

Given the co-ordinates x and y of a o

D

point, to find the distance and direction of the point from the origin.

Since the quantities r and o completely determine the position of P, they are also co-ordinates. To distinguish them they are called polar co-ordinates.

EXERCISE.

Eight points have the following several co-ordinates. Find the values of p and r (the values of r being all equal), and note what relation exists among the values of p.

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65. Definition of direction of lines and angle between them. Two finite lines which do not meet are considered to form a certain angle with each other; namely, the angle which would be formed if they were continued until they met, or if a line parallel to the one were drawn through any point of the other.

Since at the point of crossing four angles are formed, we may, in the absence of any convention, regard either of these angles as that between the lines. But as opposite angles are equal, these angles only have two different values.

Also, in the absence of a convention, we may regard any angle as either positive or negative. Hence to a given inclination of the lines may be assigned any one of four different values, which values are divisible into two supplementary pairs.

Example. If two lines intersect at an angle of 85°, we may consider their inclination, or the angle which they form, to be

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