Fifthly. Turn the triangle I1 round until it is

placed like I 2 in the figure, and turn H1 until it is placed like H 2, and you will then have a rectangle equal to the triangle.

Now these last two figures have taught us two very useful rules:

1. That a parallelogram may be drawn equal in

area to a triangle by making its base half the length of the base of the triangle, and its height equal to the height of the triangle; for you will see that in Fig. 30 A B is half of A F, and A C is the same as B D.

2. That a parallelogram equal to a triangle may be drawn by taking the whole base and half the height of the triangle. This is shown in Fig. 31, where the base of the parallelogram is the same line which forms the base of the triangle, whilst the height D E is only half of D C.

Now, before we leave triangles, I must teach you how to draw one which shall be equal in size to any number of smaller triangles which you may wish to add together.

But, to do this, I must first help you to climb up just one little step, so that you may be able to reach the lesson.

Will you then draw three short lines, A B, C D,

same length, and must be so placed that a dotted line would pass through and unite them into one straight line G H.

Next draw a line, I J, parallel to G H.

Now, if I use the three little lines as the bases of triangles, so long as I place their apices (this word is the plural of apex, and therefore means their top points) in the line I J, they will be all equal in area. No matter where I place K, L, and M, so long as they are on I J, the triangles

M

4AN

B

E F

will be equal, and this brings me to tell you the

rule that

Triangles upon the same base and between the same parallels are equal.

And now, having reached this step, you will soon be able to perform the little task with which we set out; that is, to add several small triangles together, so as to form one equal to them all.

H

33

B

Let us suppose that you require to add together five triangles equal to A B C.

Produce A B on both sides, and mark off on it the lengths A D, D E, B F, and F G, all of them the same as A B, so that the line E G will be equal to the bases of the five triangles joined together. Now draw lines from each of the points in the line to meet C, the apex of the triangle A B C, and you will thus complete the triangle E C G, which you will see is made up of five triangles.

Now although these triangles are not all of the same shape, still they are equal in area. Although

the one in the middle is an isosceles triangle, and the other four are scalene, they are all equal, for they have all the same bases, and they are all between the same parallels, E G and HI; and therefore, as all the five equal triangles are united in the large triangle E C G, you will at once see that it is equal to all of them added together.

First, draw the square and the diagonals which will divide it into four isosceles triangles, meeting in the centre; divide each side of the square into five equal parts, and draw lines to the centre, and these will divide each of the four isosceles triangles into five triangles of equal area, and thus the whole square will be divided into twenty triangles. You will see that the two scalene triangles meeting, form a trapezium at each corner of the square.

OF ARCS, OR PARTS OF CIRCLES.

WE will now for awhile leave the figures which are composed of straight lines only, and study ARCS, that is, parts of circles, as you will remember from having already met with the term in our useful friend the kite, at page 37.

We will not for the present attempt a whole circle, but will content ourselves with a SEMICIR-CLE, that is, a half-circle.

The longest straight line which you could draw across a circle, passing through the centre, is called the DI-A-ME-TER, which means the "through