14. The base of a triangle is 3 in. longer than the altitude, and the area is 44 sq. in. Find the lengths of the base and altitude.

15. The altitude of a triangle is 3 times the base, and the area is 37 sq. in. Find the lengths of the base and altitude. Ans. 5 in. and 15 in.

16. The difference between the base and the altitude of a triangle is 6 in., and their sum is 36 in. Find the area of the triangle. Ans. 157 sq. in.

If the

17. The altitude of a triangle is 7 in. longer than the base. altitude is decreased by 4 in. and the base increased by 6 in., the area is increased by 25 sq. in. Find the altitude and D

base.

100t

*18. The figure is a trapezoid with dimensions as given. Find the distance from the longer base DE to the line AB which is drawn parallel to DE and G divides the area into two equal parts.

90

A

1500'

B

E

CONSTRUCTIONS

1. Construct a triangle equivalent to a given triangle and having one of its sides equal to a given line.

2. Construct a triangle equivalent to a given triangle and having one of its angles equal to a given angle.

3. Divide a triangle into four equivalent triangles by lines through one of its vertices.

4. Construct a triangle equivalent to a given square.

5. Construct a triangle equivalent to a given triangle AHK and having its base double the length AH.

6. Construct a triangle equivalent to of a given triangle.

7. On the base of a given triangle, construct a right triangle equal in area to the given triangle.

8. On the base of a given triangle, construct an isosceles triangle equal in area to the given triangle.

9. On the base of a given triangle, construct a rectangle equal in area to the triangle.

PRACTICAL METHODS

372. Practical Methods for Finding the Area of a Polygon. The area of any polygon may be found by dividing it into triangles and finding the sum of the

areas of the triangles.

The area of a polygon may also be found by drawing the longest diagonal and drawing A perpendiculars to this diagonal from the vertices. This will divide the polygon into triangles and trapezoids or rectangles, of which

B

H

G

F

E

the areas can be found. The sum of these areas is the area of the polygon. 373. Trapezoidal Rule. An area between a curved line and a straight line is sometimes found approximately as follows:

(1) Divide the straight line AB, called the base line, into a number of equal parts of length d.

(2) Measure the lines, called offsets, drawn perpendicular to AB from the points of division to the curved line, their lengths being y1, Y2, y3, etc. (3) Consider each part, as ACDF, a trapezoid and find the sum of their areas, which is the area required, approximately.

Show that this gives the following rule which is known as the trapezoidal rule:

To half the sum of the first and last offsets add the sum of all the other offsets, and multiply this sum by the common distance between the offsets.

By drawing a straight line, as AB in the figure, through a surface bounded by an irregular line, the area of each

part can be computed approximately by the trapezoidal rule.

374. Areas on Square-ruled Paper.

A very useful practical way to estimate

B

the area of any plane surface is to draw it on square-ruled paper. The figure will usually be drawn to some scale that uses a side of one of the squares as a unit of length. The squares

enclosed in the figure can then be counted, and thus the area of the surface may be estimated.

A convenient way is to count as whole squares those which are entirely within or more than half within the figure, to disregard those squares which are more than half without the figure, and to count every other one of the squares that are half within and half without.

EXERCISES

1. Draw a rectangle 2 in. by 3 in. with a semicircle on one of the long sides. Measure 6 offsets from the other long side, and compute the area by the trapezoidal rule. Meas

ure 12 offsets and compute. If the area of the figure is 9.534 sq. in., find the per cent of error in each of your results.

2. The figure is an ellipse that is 48 mm. long and 38 mm. wide; find its area by the trapezoidal rule. Find the per cent of error in your result if the area of the ellipse is 1432.57 sq. mm.

3. Draw an irregular poly

gon and compute its area.

4. Estimate the area of this figure if it is drawn on a scale of 1 in. to a side of a square.

5. Take leaves of various trees and find their areas by drawing their outlines on square-ruled paper.

6. Estimate the area of this figure if it is drawn on a scale of 1 ft. to a side of a square. If the area of this figure is 150.8 sq. ft., what is the per cent of error in your estimate?

D

C

B

*7. By the trapezoidal rule determine the area of a lake, using a map of the lake. It is necessary to know the scale to which the map is drawn. Draw the outline of the lake on square-ruled paper and determine its area. *8. To determine the amount of flow of water in a river that is 120 ft. wide, soundings were made for each 10 ft.

of width. Thus 13 measurements were made as shown in the figure. Find the area of the cross section in square feet.

3.9'

120

5.1

5.5'

5.7

5.8'

5.7

5.4'

5.1'

4.71

375. Theorem. Two triangles that have an angle of one equal

to an angle of the other are to each other as the products of the sides including the equal angles.

Proof. Place ADEF upon ▲ABC with ZF upon ≤C. Then D will fall at D' and E at E', and ADEF=AD'E'C.

1. Given two triangles with an angle of one equal to an angle of the other, and the including sides of one 10 in. and 15 in., and of the other 5 in. and 9 inches. Find the ratio of their areas.

2. If an angle of a triangle is unchanged, but each of the two including sides is doubled, how is the area changed?

THEOREM OF PYTHAGORAS

376. Theorem. The square constructed on the hypotenuse of a right triangle is equivalent to the sum of the squares constructed on the other two sides.

Given rt.AABC, the hypotenuse c, and the other two sides, a and b, with the squares EB, CK, and FC constructed on the three sides.

H

G

a

x

CA

B บ

In A ABK and DBC, AB= DB and BK = BC.

Why?

In a similar manner, it may be proved that

377. Theorem. The square on either side of a right triangle is equivalent to the square on the hypotenuse minus the square on the other side. a2=c2-b2, and b2 = c2-a2.