It will be interesting for the teacher to calculate the areas of such triangles by the usual trigonometrical formulæ, that he may inform the class as to the closeness of their approximations reached by instrumental methods.

6. In the adjoining figure BED is the diagonal of each of the parallelograms ABCD, FBHE, KEGD, and therefore bisects each of them. Hence we have

B

F

A

K

H

G

Therefore the parallelogram AFEK is equal to the parallelogram CHEG. Note the position of these parallelograms with respect to the diagonal BD.

7. In constructing a rectangle equal to a given triangle (Ch. VIII., 4), one of the sides of the rectangle is half the base of the triangle. We may, however, construct a rectangle equal to any triangle, and give to one of the sides of the rectangle any length we choose.

Thus having constructed FECG equal to the triangle ABC, suppose we wish to make a rectangle equal to the triangle, with one of its sides of length CH. Complete the rect

B

angle EKHC, and let KC, FG meet in L. Draw the remain

IF

G

N

ing lines as indicated in the figure.

angle CM, whose side CH is of the required length, is

equal to the rectangle FC (Ch. VIII., 6), which is equal to the original triangle ABC.

Construct rectangles, each with a side of 50 millimetres, equal to the triangles in 4.

The sides of a triangle are 2, 3 and 4 inches. Construct a rectangle equal to it, having one side of 21 inches. Measuring the other side of the rectangle, calculate approximately the area of the rectangle, i.e., of the triangle.

The sides of a triangle are 3 and 4 inches, and the included angle is 50°; construct a rectangle equal to it, one of whose sides is 2 inches. Measuring the other side of the rectangle to the nearest sixteenth of an inch, calculate approximately the area of the rectangle, i.e., of the triangle.

8. If we wish to construct a rectangle equal in area to a polygon, and thence, if necessary, calculate the area of the polygon, it is well first to construct a triangle equal to the polygon by the following method: Let ABCD be a quadrilateral

whose area we wish to calculate. Place the edge of the parallel

rulers along AC, and slide one

bar out until the edge reaches B

E

D, and mark the point E in BC produced. AC is parallel to DE, and therefore the triangle ACE is equal to the triangle ACD (Ch. VIII., 3); and therefore the triangle ABE is equal to the quadrilateral ABCD.

We may then measure the perpendicular height of ABE and its base BE: their product divided by 2 gives the area of ABE (Ch. VIII., 5), and therefore of ABCD.

Suppose we wish to find the area of the pentagon ABCDE. Place the edge of the parallel rulers along CE, and slide one bar out until the edge reaches D, and mark the point F in BC produced. DF is parallel to CE, and therefore the triangle ECF is equal to the triangle ECD. Thus

B

the quadrilateral ABFE is equal to the pentagon ABCDE.

Again, place the edge of the parallel rulers along AF, and slide one bar out until the edge reaches E, and mark the point & in BC produced. EG is parallel to AF, and therefore the triangle GAF is equal to the triangle EAF; and therefore the triangle ABG is equal to the quadrilateral ABFE, and to the the pentagon

ABCDE.

We may then measure the perpendicular height of ABG and its base BG: their product divided by 2 gives the area of ABG and therefore of ABCDE.

In the preceding figures the dotted lines need not be drawn. The point E in the former figure, and the points F and G in the latter, are where the edge of the parallel rulers cuts BC.

Had we selected AB as our base, instead of BC, our resulting triangle would have had a different height and base, but would necessarily have been of the same area as ABE or ABG.

The sides AB, BC, CD of a quadrilateral are 70, 60 and 50 millimetres; the angles ABC, BCD are 70° and Construct a triangle equal to it in area, and

thence calculate its area. Here use BC and next AB as bases; and, by comparing the areas of the resulting triangles, obtain a test of the accuracy of your construction.

Construct several quadrilaterals and pentagons, and find triangles equal to them in area. In each case construct the triangle in two different ways (as in the preceding example) and, by comparing the areas of such triangles, obtain a test of the accuracy of your construction.

In all cases where numerical measurements are made, such measurements are necessarily approximate, and therefore in examples such as the preceding the areas will be found only approximately. Hence where a numerical area has been reached by two different ways, we are to expect only approximate agreement.

Chapters XIX., XX., and XXI., relating to similar triangles, may now be taken up if thought desirable.

Exercises.

1. If two triangles have the same base and equal areas, what relation exists between their altitudes?

If their vertices be joined, what position does it occupy with respect to the common base?

2. If D and E be the middle points of the sides AB, AC of the triangle ABC, what relation exists between the areas of the triangles DBC, EBC? What do you infer as to the position of DE with respect to BC?

3. Construct a quadrilateral, and bisect the sides. What positions do the lines joining the bisections of adjacent sides occupy with respect to the diagonals?

What is the figure formed by joining in succession the points of bisection?

4. Construct a quadrilateral, and bisect the sides. How do the lines joining the bisections of opposite sides divide each other?/ Give

reason.

5. Two sides of a quadrilateral are parallel and of lengths 23 and 3 inches. The distance of these sides apart is of an inch. What is the area of the quadrilateral? (Join two opposite corners, and find area of each triangle.)

6. The sides of a rectangle are 2 and 3 inches. Find by geometrical construction a rectangle equal to it in area, one of whose sides is 2 inches. Test by measurement and numerical calculation the accuracy of your construction.

7. The sides of a triangle are 3, 4 and 5 inches. Construct a rectangle equal to it in area with one side 2 inches. Construct also a rectangle equal to it in area, one of whose sides is 3 inches.

8. The base of a triangle is 70 millimetres, and the angles at the base 30° and 50°. Construct a rectangle equal to it in area, one of whose sides is 45 millimetres.

Construct a

9. The sides of a rectangle are 30 and 40 millimetres. parallelogram equal to it in area, one of whose sides is 30 millimetres, and one of whose angles is 60°.

10. On a base of 35 millimetres construct two parallelograms of equal area, one having a side of 55 millimetres and an angle of 75°, and the other an angle of 120°.

11. The sides of a triangle are 2 and 3 inches, and the included angle 45°. Construct a rectangle equal to it in area, one of whose

sides is 2 inches.

12. In the previous question, construct a parallelogram equal to the triangle, with one of its angles 45°.

13. In a quadrilateral ABCD, AB=35, BC=45, CD=55 millimetres; ABC=60°, BCD=75°. Construct a triangle and also a rectangle equal to it in area. Hence calculate its area, approximately, in square millimetres.

14. A quadrilateral ABCD has AB (2 in.) and CD (31⁄2 in.) parallel, and 1 in. apart. Construct a rectangle equal to it in area, one of whose sides is 11⁄2 in.

15. ABC is a triangle, and D, E the middle points of AB, AC. BE, CD intersect in O. Join AO, and show that the triangles OAB, OBC, OCA are equal in area.