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And can you find the area of any triangle, no matter what its shape is?

Again, could you find the area of a playground whose boundaries are not straight? You could replace the curved boundaries by one or more straight ones, placing the straight boundaries so that, as far as the eye can judge, as much playground is now left outside as other ground is brought in. Or, if you may not go beyond the boundaries of the playground, you could make a straight-sided figure by joining up a number of points in the boundary, and allow by eye or by further measurements for parts left over.

Ex. 2. Find in acres or hectares the area of a field represented in Fig. 3 on a scale of 1 to 10,000.

Ex. 3. Draw any irregular figure, and find its area. result by using squared paper.

Check your

Ex. 4. Find the area of a circle of radius 8 cm. by both methods.

Area of a triangle.

11. Let us return to a triangle of any form. Can you extend the methods used for the right-angled triangle to cover the general case?

We might, as in treating the triangle of Fig. 1, divide our triangle into two right-angled triangles. Suppose one side of the triangle b cm. long and the perpendicular from the opposite corner h cm. long, and that the perpendicular divides the base into two parts p and q cm. long. What are the areas of the two right-angled triangles ?—They are hxp 2 sq. cm. and And together ?

hx b


hx q

sq. cm.

sq. cm., since the side b is made up of the two parts p and q.

How must this be modified if the triangle is obtuse-angled * and the perpendicular h falls not on the side b, but on this side produced?-In this case the triangle dealt with is not the sum but the difference of two right-angled triangles.

* An angle less than 90° is called an acute angle, an angle greater than 90° and less than 180° is called an obtuse angle. A triangle whose angles are all acute is called an acute-angled triangle, and a triangle with one angle obtuse is called an obtuse angled triangle.

12. Again, take a rectangular sheet of paper (ABCD in Fig. 4) and make two cuts, ED and EC, from a point in




FIG. 4.


one side to the ends of the opposite side. Can you fit the corners cut off on to the remaining triangle CED? What then is the area of the triangle?—It is half the rectangle, that is, half the length of the rectangle multiplied by the breadth; the length and breadth and the area being expressed in corresponding units, e. g. inches and square inches or centimetres and square centimetres. And in terms of the parts of the triangle ?—The area is half the base CD multiplied by the height EF, as we found before. You assume that EF is equal to AD or BC. How do you know? By the experimental result that parallel lines are at the same distance apart all along.

How would you cut the rectangle to give an obtuseangled triangle? Cuts along AF and AC would do. And how far does the discussion differ for

this case?

13. Again, take any paper triangle CDE; Fig. 4 will serve. Fold the side DC on itself so that the crease passes through E, as EF in Fig. 4. Straighten out, and then fold the corners CDE down on to F. What figure results?—It is a rectangle exactly covered by the foldedin corners. What are the length and breadth of the rectangle?—They are half the base and height of the triangle. So that as before the area of the triangle is half the product of the base and the height.

Repeat this for the case in which the angle D is obtuse, noticing that till the folding is done an extra piece of paper to carry the point F must be left.

14. Once more draw any triangle, acute- or obtuseangled, and draw about it a rectangle of which it is half, and so proceed to the expression for the area of a triangle.

In this discussion we have chosen any side of the triangle and called it the base, while we called its distance from the opposite corner the height of the triangle. And as a triangle has three sides, the expression we have found gives three ways of finding the area of a triangle. Draw a triangle, and use the formula in the three ways, and compare your three results. Further, check your results by drawing the same triangle on squared paper and counting



15. The daily newspapers give curves showing the height of the barometer for the preceding day or the preceding week. In these, distances measured from an initial point along a base line running parallel to the lines of print (running horizontally, we may say by an obvious metaphor) represent time elapsed; and distances perpendicular to the base line, that is, parallel to the columns of print, or vertically, represent the height of the barometer. The length of the perpendicular from any point of the base line to the curve gives the height of the barometer at the moment denoted by that point of the base line. A second curve gives the height of the thermometer. To save space it is usual to use the same diagram for the two curves, and to show only the upper part of the diagram, thus omitting the base lines.

Such curves can be more quickly read than a table of numbers giving the same information. They are called graphs. The scales used are of course shown on the figure; without them the curves would lose much of their meaning.

16. Take a sheet of paper 20 cm. in breadth, and place a ruler across it at one end. Slide the ruler along, keeping it all the time parallel to the end of the sheet, till it has moved 20 cm. along the paper. Find and show in a table the area of the paper that the ruler uncovers by moving 2 cm., 4 cm., 6 cm., . . 18 cm., 20 cm.

Distance ruler has moved, in cm.

Area uncovered, in sq. cm.


4 6 8



120 160

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Now mark a line OX 20 cm. long on squared paper (if you have no paper squared in centimetres, use any convenient scale), and mark along it from 0 2 cm., 4 cm., . . At the point marked 2 cm. raise a perpendicular to OX to represent on a scale of 1 cm. to 40 sq. cm. the area uncovered by the ruler in moving 2 cm., that is, raise a perpendicular of length 1 cm. At the point marked 4 cm. raise a perpendicular to represent on the same scale the area uncovered by the ruler in moving 4 cm., and so on.

If you wanted to show on your diagram the area uncovered when the ruler had moved 1 cm. or 3 cm., how would you proceed?-Erect a perpendicular as before. And for any other distance, for instance,

3.4 cm.

? Through the ends of all these perpendiculars draw a smooth line. This line shows graphically how the uncovered area varies as the ruler moves.

17. Again, suppose the ruler to be returned to its position at the end of the sheet of paper, while a second ruler is placed along the side of the sheet. Let the two rulers now move away, each remaining parallel to its original position, in such a way that the area uncovered is always a square, that is, has its angles right angles and its sides all equal. As before, show in a table and in a graph the variation of the uncovered area as the rulers move. From your graph read off the area of a square whose side is 12.3 cm., and the length of the side of a square of area 178 sq. cm.

18. Two rods 13 cm. and 10 cm. long are hinged together at one pair of ends, and the other two ends are joined by an elastic string. Find by drawing, and show in a graph, how the area of the triangle formed by the two rods and the string varies as the angle between the rods increases, or, in mathematical language, show the area as a function of the angle.

Algebraic expressions.

19. In finding the area of a rectangular piece of ground we treated each dimension as made up of two parts, an integral part and a fractional part. Let us now suppose the

length made up of two parts a cm. long and b cm. long, where

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cm., and so find an expression for the whole area.

is in sq. cm.

axc + axd + bx c + b× d,

or, written in the shorter form,

ac + ad + bc + bd.


By considering the undivided rectangle you can give another expression for the area. It is the product of a+b and c+d, which we agreed to write in the form

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What then do we know of these two expressions?


they are equal, since each is the number of sq. cm. in a

certain area.

20. Is it true that whatever numbers a b c d are

(a + b) (c + d) = ac + ad + bc + bd?

Is it true only when these numbers are the lengths of lines, or is it true also when you buy a+b loaves at c+d pence each, or, as a particular case, 5 white and 3 brown loaves at 2 pence each ?-We are at liberty to buy the white loaves first and then the brown, and we are at liberty to pay down 2 pence on every loaf and then the remaining halfpenny; which gives us as the price in pence

5×2 + 5× + 3 × 2 + 3 × 1

in addition to the other form, 8x 2 pence. Or we could represent the whole price by a figure, such as Fig. 5, representing a loaf by a centimetre across the page and a

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