Note 5.-To find the side of a square HEFG, whose area shall bear a certain proportion to that of a given square ABCD.—We may use either of the following methods (a or b) :

double that of ABCD, we may proceed thus

Since the square EFGH is double the square ABCD, and the area of ABCD is AB2 and that of EFGH is E F2, we have

So, also, the square roots of these quantities are equal; that is

EF=√2X AB2=AB√2.

Hence the side EF will be found by multiplying AB, the side of the given square, by √2.

Again: If the area of the square EFGH is treble that of the square ABCD, then the side EF will be found by multiplying AB by 3; and so on.

On the other hand, if the area of the required square is half that of the given square, we shall, in that case, find its side by dividing the side of the given square by 2; and if the area is one-third, then its side is found by dividing the side of the given square by 3; and so on.

(b) The following method will, however, be the easier, perhaps, for beginners :

:

If the area of the required square is double that of the given square, then multiply the area of the given square by 2, and the square root of this product is the side of the required square.

If the area is three times as much, then multiply the area of the given square by 3; and the square root of this product will be the side of the required square.

On the other hand, if the area is half as much, then divide the area of the given square by 2; and the square root of this quotient will be the side of the required square. The same remarks will apply to the case when the area is one-third, or one-fourth, &c.

[√2=1·414+; √3=1·732+;

√4=2, &c.]

Example 1.-If the side of a square is 3 ft. 5 in., find its

area.

Here 3 ft. 5 in.=41 inches; square 41, and we shall have 1681 sq. in., or 11 sq. ft. 97 sq. in., as the area of the square.

Example 2.-How many square tiles, each 41 inches long, will be required for paving a square court which is 13 ft. 6 in. long?

tile

Now 13 ft. 6 in. 162 inches.

Then, area of court=1622=26244 sq. inches.

And area of each tile=square of 41=1 sq. inches.

Therefore, number of tiles area of court÷area of each

=26244-8=26244x=1296.

Example 3.-If the carpeting of a square room with carpet, at 7s. 6d. per sq. yard, costs £16 17s. 6d., find the length of the room.

First, if we divide £16 17s. 6d. by 7s. 6d., we have 45 sq. yds., or 405 sq. ft., as area of the floor.

Then, taking the square root of 405, we have 20·12+ft.; the length of the room.

EXAMPLES.

Find the area of the square, when the length of each of its sides is

(1) 97 in.

(2) 2 ft. 7 in.

(3) 6.25 ft.

Determine the area of a square field, when the length of each of its sides is

(4) 160 yds. (5) 3 ch. 25 lks. (6) 1 fur. 20 poles 3 yds. Find the area of a square field, when its diagonal is(7) 162 yds. (8) 7 chains 30 lks. (9) 30 poles 3 yds. Find the length of each side of a square, when its area is(10) 77841 sq. yds. (11) 32 ac. 1 rood 24 sq. poles. (12) 86 sq. ft. 92.89 sq. in.

(13) What is the length of a square field which contains 15.625 acres?

(14) Find the side of a square field which contains 3 ac. 1 rood 13 poles 52 sq. yds.

(15*) Find in yards the side of a square field which contains 10 acres.

(16) A square field contains 10 acres ; find the length of each side, and of its diagonal.

(17) The diagonal of a square court is 36 yds.; find its

area.

(18) What is the diagonal of a square court, the length of whose side is 35 yds?

(19) A square plot of ground, which is 127 yds. long, has a path 1 yd. wide running round the inside of it; what will be the expense of gravelling it, at 6d. per square yard?

(20*) Determine the side of a square field which cost £57 158. 2 d. trenching, at 2ąd. per square yd.

(21*) What is the diagonal of a square whose area is 7 sq. inches ?

(22) What will be the expense of building a wall round a square plot of ground, which contains 3 acres 1 rood 4 poles 25 sq. yds., at 10s. 6d. per yd. ?

(23) A

square room is 15 ft. 6 in. long; what must be the length of a square piece of carpet that shall cover half the room?

(24) A walk 2 ft. 6 in. wide, round the outside of a square court, which is 30 yds. long, is to be tiled; find the expense, at 9d. per sq. ft.

(25) A path 2 yds. wide surrounds a square field containing 27 ac. 12 poles 1 sq. yd.; determine the area of the field, if the path were ploughed up and enclosed in the field.

(26) The tiling of a square court, at 1s. 9d. per sq. ft., costs £22 8s.; what is the length of the court?

(27) The perimeter of a square field is 684 yds.; find its area.

(28) 2,500 square tiles are required for paving a square courtyard which is 16 ft. 8 in. long; find the length of each tile.

(29) Find the cost of building a wall round a new burialground, in the shape of a square, which contains 2 ac. 10 poles 17 sq. yds., at 17s. 6d. per yd.

(30) A square field, containing 6 ac. 2 roods 31 poles 21 sq. yds., is to be planted with trees, to the depth of 10 yds., running all round the inside of the field, at 1s. 3d. per sq. yd. Find the cost.

(31) A square field 210 yds. long is to be planted all round the inside of it to an uniform depth; the plantation is to occupy just one-seventh of the whole field. Find its width.

(32) A square plot of ground is 35 yds. 1 ft. long; what is the side of another plot, which is of the same shape, but which contains four times the amount of land?

The perimeter is the sum of all the sides=2 length +2 width.

RULE. To find the area of a rectangle.

Multiply the length by the breadth.

Note 1.-To find either the length or the breadth of a rectangle, when the area and one dimension are given.The area divided by the breadth will give the length. The area divided by the length will give the breadth.

Note 2.-To find the area of a rectangle, when one dimension, either the length or the breadth, and the diagonal, are given. Since BD is the hypothenuse of a right-angled triangle; and one side also is given, the other side must be found by Rule 2, Prob. 1. We shall then have found the length and breadth of the rectangle; and these, being multiplied together, will give the area of the rectangle.

Note 3.-The superficial area of the walls of a room, or of the sides of a rectangular box or cistern, may be found by adding double the length to double the breadth, and then multiplying this sum by the height of the room, or by the depth of the box or cistern.

The floor of a room, or the bottom of a box or cistern, will be found by multiplying the length by the breadth.