For the triangle ABD is half the rectangle E BDA, and the triangle A D C is equal to half the rectangle ADCF. Therefore the whole triangle ABC is equal to half the rectangle EBCF. Hence it follows that the area of a triangle is equal to half the area of a rectangle having the same base and perpendicular height.

Ex. In the triangle A B C, the base B C 8 ft., and the perpendicular height A D=7 ft.; required the area.

Area of rectangle E B C F=8x7=56 sq. ft., but the triangle A B C contains half this surface, hence area of triangle A B C =56÷2-28 sq. ft.

Ex. XXIV.

Find the areas of the triangles having the following dimensions:

(1) Base, 24 ft.; height, 8 ft.

(2) Base, 13 yds. 2 ft.; height, 6 yds. 1 ft.

(3) Base, 3 yds. 2 ft. 6 in.; height, 2 yds. 1 ft. 9 in.

(4) Base, 7 yds.; height, 4 yds.

(5) Base, 15.625 ft.; height, 8.125 ft.

Find the area in acres, roods, and poles, of triangles having the following dimensions:

(6) Base, 750 lks.; height, 350 lks.

(7) Base, 10 ch. 50 lks.; height, 4 ch. 75 lks.

(8) Base, 18 ch. 20 lks.; height, 12 ch. 50 lks.

(9) Find the area of a triangle whose base is 10 and height 43.

(10) A triangular plot of ground whose base is 85 yds. 2 ft. and perpendicular height 72 yds. 1 ft. is to be paved at the rate of 1s. 6d. per sq. yd.; find the expense.

(11) A triangular field whose base is 850 links and perpendicular height 275 links sells for £1870; find the price per sq. pole.

(12) How much land is there in a triangular field whose base measures 4 ch. 25 lks. and height 2 ch. 50 lks., and what is the rental at 15s. per rood?

67. To find the base of a triangle when the area and perpendicular height are given.

RULE. Divide double the area by the height, and the quotient will be the base.

Ex. The area of a triangle is 648 sq. in. and the height is 27 in., find the base.

Base x height = twice the area.

..

base =

=twice the area÷height.

=1296÷27-48 in. [Compare Art. 51.]

68. To find the perpendicular height of a triangle when the area and base are given.

RULE. Divide double the area by the base and the quotient will be the perpendicular height.

Ex. The area of a triangle is 191.25 sq. ft. and the base is 25.5 ft., required the perpendicular height.

Base x height = twice the area.

height twice the area ÷ base.

= 382·5÷25·5-15 ft.

Ex. XXV.

(1) A triangular field contains 3 acres, and its base is 1750 links, find the height.

(2) An enclosure in the form of an equilateral triangle is paved at the rate of 9d. per sq. foot. Its perimeter is 63 yds., and the cost of paving the whole £35 18s. 9d. Find the height.

(3) The perimeter of an equilateral triangle is 24 yds. and its area is 24 sq. yds., what is the height?

(4) The base of a triangular field is 475 links; the field is let for £23 15s. a year, at the rate of £4 per acre. What is the height?

(5) The area of a triangle is 1056 sq. ft. and its height 11 yds., find the length of the base.

69. To find the area of a triangle when the three sides are given.

RULE. 1°. From half the sum of the three sides subtract each side separately.

2o. Multiply the half sum and the three remainders continually together.

3°. Take the square root of the last product.

Ex. What is the area of a triangle whose sides are 13, 14, 15?

Sum of three sides
Half sum

Rule 1°. First remainder

Second remainder

Third remainder

13+14+15=42.

=42221.
=21-13 8.

=21-14= 7.

=21-15= 6.

Find the areas of the triangles having the following sides :—

(1) 50, 40, 30.

(2) 13, 20, 21.

(3) 21, 10, 17.

(4) 35, 66, 53.

(5) 11, 13, 20.

(6) `10, 10, 12.

(13) Find the area of a triangle whose sides are 3.27 ft., 4.36 ft., and 5.45 ft.

(14) Find the area in acres, roods, and poles, of a triangular field, whose sides are 300 lks., 975 lks., and 1125 lks.

(15) Find the area of a triangle whose sides are 319 ft., 444 ft., and 455 ft.

(16) The sides of a triangular field are 5 ch., 16 ch. 25 lks., and 18 ch. 75 lks.; find the acreage.

(17) Required the expense of reaping the corn in a triangular field whose sides are 150, 200, and 250 lks. long, at 13s. 4d. per acre.

(18) A triangular field whose sides are 350, 440, and 750 yds. is let for £26 5s. a year. What is that per acre?

(19) What rent should be paid for a triangular field whose sides are 175, 220, and 375 yds. at £6 12s. per acre?

(20) A triangular piece of ground, whose sides are 800, 500, and 500 links, is let for £3; find the letting price per acre.

(21) Find the area of an isosceles triangle whose base is 10 ft. and each of whose equal sides is 13 ft.

(22) What is the rent of a triangular field whose sides are 550, 550, and 880 yds., at £2 15s. per acre?

(23) Find the side of a square field whose area is half as much again as that of a triangular field whose sides are 20, 65, and 75 yards.

70. A right-angled triangle

is one that has a right angle. AB is the hypotenuse, B O is the base, and A C the perpendicular.

AC and CB, which include the right angle, are also called. the sides of the right-angled triangle.

B

Hypotenuse.

Base.

Perpendicular.

71. To find the hypotenuse of a right-angled triangle when the sides are given.

RULE. Add the squares of the sides and extract the square root of the sum.

square, it will be found that 16 pieces are required for the square on AC, and 9 for the square on CB, or 25 pieces on both sides together. If the pieces be now removed, and taken to form a square on A B, it will be found that 25 pieces exactly are required to complete it. Thus showing that the square on A B, the hypotenuse, is equal to the sum of the squares on A C and CB, the two sides; and that the length of the hypotenuse is obtained by extracting the square root of the number representing the sum of the

squares.

Ex. The base of a right-angled triangle is 12, and its perpendicular 5; find its hypotenuse.

The square of 12 is 144; the square of 5 is 25.

The square on the hypotenuse=144+25=169.

The length of the hypotenuse=the square root of 169=13.

Ex. XXVII.

Find the hypotenuse from the given sides in the following

right-angled triangles:

(1) 7 ft., 24 ft.

(2) 8 ft., 15 ft.
(3) 30 yds., 40 yds.

(4) 13 yds., 84 yds.

(5) 3 ft. 8 in.; 40 ft. 3 in.
(6) 5 ft. 3 in.; 1 ft. 4 in.
(7) 3 yds. 1 ft.; 2 ft. 3 in.
(8) 3 yds. 2 ft. 11 in.; 2 ft.

(9) The sides of a right-angled triangle are 575, and 48; required the hypotenuse.

(10) Close by a tower 57 ft. high, flows a river 76 ft. broad; what length of cord will just reach from the top of the tower to the opposite bank of the river?

(11) Find the length of a cord attached to the top of a flag-staff 36 ft. high, and fastened in the ground at a distance of 15 ft. from the bottom of the staff.

Tower.

Cord.

River.

(12) At the distance of 10 ft. from a wall I placed a ladder which reached a window 24 ft. from the ground; find the length of the ladder.

(13) A house is 36 ft. high, and the street in front 27 ft. wide; find the length of a ladder which will just reach the top of the house from the opposite side of the street.

(14) The town C lies 21 miles east from B, and 72 miles distant from A, which lies north from C; required the distance from B to A. [See Fig. Art. 70.]

(15) Two persons start at the same time and from the same place; M due west, and N due north. M travels 3 miles an hour, and N 4 miles an hour. How far are they apart at the end of 5 hours?

(16) A starts on Monday and walks 10 miles a day due south; B starts from the same place on Tuesday and walks 20 miles a day due west. How far apart will the two travellers be on Wednesday night?

(17) The length A B of a rectangle D
ABCD is 40 feet, and the breadth
BC 30 feet; find the diagonal
A C.

[The diagonal of a rectangle or square is the straight line joining two of its opposite angles. It forms the hypotenuse of a right-angled triangle, of which the sides are the length and breadth of the rectangle or square.]

C