(8) Find the cost of a plot of land, in the shape of a trapezium, whose diagonal is 108 ft., and the perpendiculars upon it 55 ft. 3 in. and 60 ft. 9 in. respectively, at 2s. 6d. per sq. yd.

(9) The area of a trapezium is 759 sq. yds., its diagonal is 108 ft. 6 in., and one of its perpendiculars is 65 ft. 3 in. ; find the other perpendicular.

(10) ABCD (see fig.) is a trapezium having AB=87 ft., BC= 119 ft., AD = 169 ft., CD = 41 ft., and the diagonal AC=200 ft.; find the area.

(11) Find the area of the four-sided figure ABCD (see fig.), whose dimensions are: AB = 145 ft., CD = 135 ft., AE=87 ft., FC 81 ft., and the diagonal Ac=368 ft.

(12) Find the area of a trapezium, inscribed in a circle, whose sides are, respectively, 48 ft., 52 ft., 56 ft., and 60 ft.

(13) Find the area of a trapezium whose opposite angles are together equal to two right angles, and whose sides are, respectively, 80 ft., 110 ft., 120 ft., and 150 ft.

(14*) The diagonal of a trapezium is 50-08 ft., and the perpendiculars upon it from the two opposite angles are 10.12 ft. and 8·4 ft.; find the area.

(15**) The length of the diagonal of a four-sided figure is 54 ft., and the lengths of the perpendiculars upon the diagonal from the opposite corners are 23 ft. 9 in. and 18 ft. 3 in.; how many square yards are there in the field?

(16**) ABCD is a quadrilateral field; AB = 48 chains, BC=20 chains, the diagonal AC=52 chains, and the perpendicular from D upon AC=30 chains. Find the area of the field.

(17**) One diagonal of a quadrilateral field is 10 chains 14 links, and the perpendiculars upon it from the angles are 6 chains 27 links, and 8 chains 6 links. How many acres does the field contain ?

VII. THE TRAPEZOID.

Definition. The trapezoid is a four-sided figure, having

Multiply the sum of the parallel sides by the perpendicular distance between them, and divide the product by 2.

Or, multiply half the sum of the parallel sides by the perpendicular distance between them.

Note 1.-To find the sum of the parallel sides, when the area and the perpendicular distance are given.—Divide double the area by the perpendicular distance.

Note 2.-To find the perpendicular distance, when the area and the sum of the parallel sides are given.-Divide double the area by the sum of the parallel sides.

Note 3.-Whenever we find the word distance mentioned in questions on the trapezoid, we must understand it to mean the perpendicular distance between the parallel sides.

Example 1.-Find the area of a field, in the shape of a trapezoid, whose parallel sides are 12 ch. 75 lks. and 8 ch. 25 lks., and the perpendicular distance between them is 6 ch. 20 lks.

Sum of the parallel sides 12 ch. 75 lks. +8 ch. 25 lks. =21 ch.=2100 lks.

Example 2.-The area of a field, in the shape of a trapezoid, whose parallel sides are 325 yds. and 215 yds., is 8 ac. 3 roods 28 poles 3 sq. yds.; find the perpendicular distance between them.

=

Now, the double of 8 ac. 3 roods 28 poles 3 sq. yds.: 17 ac. 3 roods 16 poles 6 sq. yds.=86400 sq. yds; and the sum of the parallel sides=325 yds. +215 yds.=540 yds.

Find the area of the trapezoid whose dimensions are, respectively

(1) Parallel sides 50 ft. and 34 ft., and the perpendicular distance 25 ft.

(2) Parallel sides 24 ft. 9 in. and 15 ft. 3 in., and the perpendicular distance 12 ft. 6 in.

Find the area of a field, in the shape of a trapezoid, when the dimensions are

(3) Parallel sides 256 yds. and 144 yds., and perpendicular distance 85 yds.

(4) Parallel sides 12 ch. 25 lks. and 7 ch. 75 lks., and nerpendicular distance 10 ch. 40 lks.

Find the distance between the parallel sides in a trapezoid whose area and parallel sides are, respectively

(5) Area 311 sq. yds. 1 sq. ft., and parallel sides 14 yds. 2 ft. and 12 yds.

(6) Area 3 ac. 0 rood 34 poles, and parallel sides 7 ch. 50 lks. and 5 ch.

(7) Find the rental of a field, in the shape of a trapezoid, when its parallel sides are 6 ch. 45 lks. and 3 ch. 55 lks., and the distance between them 4 ch. 20 lks., at £3 5s. per acre.

(8) The cost of a field, in the shape of a trapezoid, at £60 per acre, is £312 7s. 6d. ; the parallel sides are 17 ch. and 12 ch. 75 lks. Find the distance between them.

(9) The area of a field, in the shape of a trapezoid, is 2 ac. 2 roods 20 poles; the perpendicular distance between the parallel sides, of which one is 115 yds., is 70 yds. Find the other parallel side.

(10**) A field is bounded by four straight lines, of which two are parallel. If the sum of the parallel sides is 1235 lks., and the perpendicular distance between them is 240 lks., determine the area of the field.

(11) The rental of a field, which is in the shape of a trapezoid, is £9 8s. 6d.; the parallel sides are 200 yds. and 119 yds., and the perpendicular distance between them is 110 yds. Find the charge per acre.

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VIII. THE REGULAR POLYGON.

E

Definitions.-A polygon is a figure contained by three or more sides; and it is also called regular when all its sides and angles are equal.

o is the centre of the polygon.

OH is the radius of the inscribed circle.

OA is the radius of the circumscribed circle.

F

C

The perimeter of a regular polygon is the length of each side multiplied by the number of sides.

If lines be drawn from centre o to the angles at A, B, C, &c., the figure will then be divided into equal triangles OAB, OBC, &c.

RULES.-(1) To find the area of a regular polygon, when a side, and the perpendicular upon it from the centre, are given.

Multiply the length of each side by the number of sides, and this product again by the perpendicular drawn from the centre to the middle of one of the sides; and half the product is the area of the polygon.

(2) To find the area, when only the length of each side is given.

Multiply the square of the side given by the number standing opposite the name of the polygon in the subjoined table; and the product is the area of the polygon.