Explanatory mensuration for the use of schools

Find the height of a triangle, when its area and base are, respectively

(6) Area 102 sq. ft. 72 sq. in., and base 10 ft. 3 in.

(7) Area 108 sq. ft. 68 sq. in., and base 9 ft. 2 in. (8) Area 5 ac. 0 rood 33 poles, and base 12 ch. 50 lks. (9) Area 1 ac. 2 roods 17 poles, and base 6 ch. 25 lks.

Find the area of a triangle whose sides are, respectively—
(10) 50, 40, and 30. (11) 61 ft., 91 ft., and 100 ft.
(12) 104 ft., 111 ft., and 175 ft.

(13) 260 yds., 287 yds., and 519 yds.
(14) 111 yds., 17.5 yds., and 17.6 yds.
(15) 119 yds., 150 yds., and 241 yds.
(16) 18-6 yds., 22.1 yds., and 27.5 yds.

(17) A plot of land, in the shape of a triangle, whose base is 120 yds. 2 ft., and perpendicular height 55 yds. 1 ft., is to be paved, at the rate of 1s. 6d. per sq. yd.; find the

expense.

(18) How many square feet of brickwork are there in the gable-top of a house, the breadth being 56 ft., and the lengths from the eaves to the ridge 39 ft. and 25 ft. respectively?

(19) The hypothenuse of a right-angled triangle is 185 yds. 2 ft., and the perpendicular height is 55 yds.; find the area.

(20) A plot of land, in the shape of a triangle, whose sides are, respectively, 25 yds., 101 yds., and 114 yds., sells for £1710; find the price per sq. yd.

(21) The perimeter of an equilateral triangle is 60 ft., and it is the same as that of a square; compare the areas of the two figures.

(22) The hypothenuse of a right-angled triangle is 76yds., and its other sides are equal; find the area of the triangle. (23) How much land is there in a triangular field whose sides measure, respectively, 2 ch. 73 lks., 4 ch. 25 lks., and 6 ch. 28 lks. ?

(24) Find the value of a triangular field, whose sides are, respectively, 113 yds., 225 yds., and 238 yds., at £60

per acre.

(25) The base of a right-angled triangle is 40 ft., and its perpendicular is 30 ft.; find its area. And if a perpendicular be drawn from the right angle upon the hypothenuse, find what is the length of the parts into which the hypothenuse is divided, and the area of each part of the triangle.

(26) Given AC (fig. 2)=24 ft. 7 in., BD=25 ft. 3 in., and CD=14 ft. 9 in.; find the area of the triangle.

(27) The three sides of a triangle (fig. 2) are AC=21 ft., CB=89 ft., and AB=100 ft.; required the length of the perpendicular from c on AB.

(28**) ABC is a triangle, and AD the perpendicular from A upon BC. If AD=13 ft., and the lengths of the perpendiculars from D on AB and AC be 5 ft. and 10% ft. respectively, find the lengths of the sides, and the area of the triangle.

(29) The perimeter of an isosceles triangle, when the base is half the length of each of its sides, is 120 ft.; find the side of a square equal in area to the triangle.

(30*) A field is in the form of a right-angled triangle, the two sides containing the right angle being 100 and 200 yards; how many acres does it contain? And if the triangle be divided into two parts by a line drawn from the right angle perpendicular to the opposite side, what is the area of each part?

(31*) Find the area of an isosceles triangle whose base is 3 ft., and each of whose equal sides is 5 ft.

(32) The breadth of the gable-end of a house is 48 ft., and the distances of its eaves from the ridge are 35 ft. and 29 ft. respectively; find the area of the gable-top; and also the height of the roof.

(33) Find the cost of a triangular plot of land, whose sides measure, respectively, 87 yds., 100 yds., and 143 yds., at 5s. 6d. per sq. yd.

(34) How many square feet of wood will be required for a triangular floor whose sides measure, respectively, 26 ft., 35 ft., and 51 ft. ?

(35) The sides of a triangular field are 65 yds., 119 yds., and 138 yds. respectively; find the side of a square field containing the same amount of land.

(36) What is the area of a right-angled triangle (fig. 1), when the angle at c is 45°, and the base BC is 40 ft. ?

(37) Find the area of a right-angled triangle (fig. 1), when the angle at c is 45°, and the hypothenuse is 98 ft. (38) The angle at c in a right-angled triangle (fig. 1) is 30°, and the perpendicular height is 40 ft; find the area. (39) The angle at c in a right-angled triangle (fig. 1) is 30°, and the hypothenuse is 60 ft.; find the area.

VI. THE TRAPEZIUM.

Definition. The trapezium is a four-sided figure having none of its sides parallel to each other.

AC is the diagonal, and BE, FD the perpendiculars upon it.

RULE.-To find the area, when the diagonal and the perpendiculars upon it are given.

Multiply the sum of the perpendiculars by the diagonal, and divide the product by 2.

given.-Divide twice the area by the sum of the perpendi

culars.

Note 3. The trapezium ABCD is made up of the triangle ABC and triangle ADC: hence the area of the trapezium may be found by adding together the areas of these two triangles.

Note 4.-If the sides of a trapezium and its diagonal are given, then its area is the sum of the areas of the two triangles ABC and ADC; and these areas can be found by Rule 2, Prob. V.

Note 5.-If the sides of a trapezium and its diagonal, and also the distances AE and FC, at which the perpendiculars rise, are given, then BE and FD must first be found by the Rule 2 in Prob. I. Having found these perpendiculars, we shall then be able to apply the rule given above for finding the area of the trapezium.

Note 6.—If the trapezium is inscribed in a circle, that is, if its opposite angles are equal to two right angles, then its area will be found thus :-Add together the four sides, and take half their sum. From this half sum subtract each side separately. Multiply these four remainders together, and the square root of the product is the area of the trapezium.

Example 1.-Find the area of a field, in the shape of a trapezium, whose diagonal is 10 chains 20 links, and the perpendiculars let fall upon it from the opposite angles are 6 ch. 30 lks. and 4 ch. 10 lks.

Now, 10 ch. 20 lks.=1020 lks.; 6 ch. 30 lks.=630 lks. ; 4 ch. 10 lks.=410 lks.