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Ex. 4. If a room be 163 feet wide and 18 feet long,

how many yards of carpeting will cover the floor, allowing for a chimney 3 feet wide and 4 feet long?

16×181305 3x41-131/1

3051÷9=33||
131÷9= 111⁄2

325, Ans.

Ex. 5. How many pieces of paper that are yard wide and 9 yards in a piece, will it require to paper a room 16 feet wide, 18 feet long, and 11 feet high, deducting for three doors, 7 feet 8 inches high and 4 feet 2 inches wide; also, two windows, 7 feet high and 4 feet 2 inches wide, and one fire-place, 6 feet long and 5 feet high? Ans. 138.

NOTE. First find the whole contents of the room as if there were no windows and doors; theu find the contents of the windows and doors, and deduct the amount from the whole: the remainder will be the true answer in square feet. Next, divide the number of square feet by the number of square feet in a piece, and the quotient will be the number of pieces required.

Ex. 6. What is the side of a square which is equal to a parallelogram 936 feet long and 104 feet broad?

Ans. 312.

Ex. 7. What is the side of a piece of land containing square rods?

169

PROBLEM II.

To find the area of a Triangle.

Ans. 13.

A RT.6. Rule.-Multiply the length of one of the sides by the perpendicular falling upon it, and half the product will be the area. Or multiply half the side by the perpen

dicular.

NOTE. In a right angled triangle the longest side is called the hypotenuse, the next longest the base, and the shortest side the perpendi

cular.

The truth of this rule is evident, because any triangle is half of a parallelogram of the same base and altitude.

Thus, the area of the right D

angled triangle ABC, contains

precisely half as much surface
as would be contained in a
square or parallelogram ABCD,
two of whose sides are formed
by the base and perpendicular
of the triangle. Therefore, the A

C

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area of a right angled triangle is found by multiplying together half the base AB and the perpendicular BC, or the side AB by half of BC.

And whatever may be the form of a triangle, if it have not a right angle, it must be cut into two right angled triangles before it can be measured.

[graphic]

P B

This is done by letting fall a A perpendicular from the opposite angle to the base, as CP in the subjoined figure, and then the area is found as before stated.

Ex. 1. What is the area of a triangle whose base is 20 feet, and altitude 10.25 feet? Ans. 102.5 sq. ft.

Once more: The area of the triangle ABC is precisely half of the parallelogram ABDF; for the surface contained in the right angled triangle APC is precisely equal to the surface contained in the right angled triangle CFA, and so BPC-CDB.

F

C

D

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Ex. 2. What is the area of a triangular board, whose base is 4 feet 3 inches, and height 2 feet 11 inches?

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Ex. 3. What is the area of a triangle whose base is 18 feet 4 inches, and height 11 feet 10 inches?

Ans. 108 ft. 52 in.

Ex. 4. How many square rods of land are there in a lot which is laid out in a right angled triangle, the base measuring 19 rods, and the perpendicular breadth 15 rods?

Ans. 142.5.

Ex. 5. Find the number of square yards in a triangle whose base is 40 and altitude 30 feet.

Ans. 662 sq. yds.

Ex. 6. What is the area of a triangle whose base is 72.7 yards, and altitude 36.5 yards?

Ans. 1326.775 sq. yds.

ART. 7. If the three sides of a triangle are given, the area may be directly obtained by the following method :

To find the area of a Triangle from the length of its sides. Rule.-I. Add together the lengths of the three sides, and take half their sum.

II. From this half sum subtract each side separately. III. Multiply together the half sum and each of the three remainders, and extract the square root of the product; the quotient will be the required area of the triangle.

Ex. 1. If the sides of a triangle are 134.108 and 80 rods,

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Then to obtain the products, we have 161x27x53x81 -18661671: from which we find area=18661671,-4319 square rods.

Ex. 2. What is the area of a triangle whose three sides are 52, 39 and 65 feet?

Ans. 1014 sq. ft.

An Isosceles Triangle is that which has only two sides equal; as ABC in the adjoining figure.

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Ex. 3. What is the area of an isosceles triangle whose base is 20, and each of the equal sides 15? Ans. 111.803.

Ex. 4. How many square yards of plastering are there in a triangle whose sides are 30, 40, and 50 feet?

Ans. 663 sq. yds.

Ex. 5. Required the area of a triangular field, whose

sides are 49, 501, and 25.69 chains?

Ans. 61 A, 1 R. 39.68 P.

ART. 8. To find the Hypotenuse of a right angled Triangle, when the base and perpendicular are known.

I. Square each of the sides separately.

II. Add together these squares.

III. Extract the square root of the sum, which will be the hypotenuse.

One of the properties of a right angled triangle is, that the square of either side is equivalent to the square of the hypotenuse diminished by the square of the other side; and the square of the hypotenuse is equal to the sum of the squares of the other two sides, usually called the legs of the triangle. This property is of great use, for by this means any two sides of a triangle being given, the other may be found by common Arithmetic.

Thus, in the right angled triangle ABC, the perpendicular BC and the base AB being given, the hypotenuse AC may be found by extracting the square root of the sum of the squares of the base and perpendicular.

Hipotkanuse

A

Leg

B

C

Ex. 1. Let AB be 18 feet 8 inches, and BC 12 feet 6

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From which we find the hypotenuse-512.20-22.63+

NOTE. We first square each side, and then take the sum, of which

we extract the square root, which gives AB=22.63.

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