Example. A square table has a side of 6 feet 6 inches: required its area.

feet.

Here 6.5

6.5

325

390

42.25

12

3.00

Ans. 42 feet, 3 inches.

Prob. II. To find the side of a square when the area is given.

Rule. Extract the square root of the area. The result is the required side.

Example. A court yard contains 1296 feet: required the length of its side.

Here 1296 feet 36 feet. Ans.

Prob. III. To find the area of a parallelogram or rectangle. Rule. Multiply the length by the breadth. The result is the required area.

Example. A dining-table is 15 feet 6 inches by 4 feet 3 inches required its area.

feet.

Here 15 5

4.25

775

310

620

65.875

12

10.500

12

6.000

Ans. 65 feet, 10 inches, 6 pa.

c 3

Prob. IV. To find the other side when the area and one side is given.

Rule. Divide the area by the given side. The result is the required side.

Example. A marble slab 8 feet long, was found to contain a surface of 5760 inches: required its width.

Here 5760 by 96 inches 60 inches, or 5 feet.

Ans. 5 feet. Prob. V. To find the solid contents of a cube. Rule. Multiply the height by the width, and that product by the height.

Example. A block of marble on measurement was found to be 10 feet long, 8 feet high, and 7 feet wide: required its solid contents.

Here 10 × 8 × 7560 feet.

Prob. VI. To find the solid contents of a parallelopipedon. Rule. Multiply the length by the width, and the product by the height.

Example. A rectangular stone pillar 25 feet 6 inches high, has a diameter of 15 inches: required the solid contents.

feet.

Here 25 5

1.25

1275

510

255

31.875

1.25

159375

63750

31875

39.84375

1728

675000

168750

590625

84375

1458-00000

Ans. 39 feet, 1458 inc

(D.)

1. Three lines at least are required to enclose a space. 2. If two angles of a triangle are equal, the sides opposed to them are also equal. (See EUCLID, Book I. Prop. vi.)

3. If two triangles have all their sides respectively equal, then the angles contained by those sides are also equal. (See EUCLID, Book I. Prop. viii.)

4. Any two sides of a triangle taken together are greater than the third side. (See EUCLID, Book I. Prop. xx.)

5. Any two angles of a triangle are together less than two right angles. (See EUCLID, Book I. Prop. xxii.)

6. Triangles on the same base and between the same parallels are equal. (See EUCLID, Book I. Prop. xxxvii.) 7. If a parallelogram and a triangle be on the same base, the parallelogram is double of the triangle. (See EUCLID, Book I. Prop. xli.)

8. In a right-angled triangle, the square of the hypothenuse is equal to the sum of the squares of the sides. (See EUCLID, Book I. Prop. xlvii.)

MENSURATION.

Prob. VII. To find the hypothenuse of a right-angled triangle when the other two sides are given.

Rule. Add the square of one side to the square of the other, the square root of which sum is the required hypothenuse.

Example. A rectangular room 20 feet by 15 feet was divided by a diagonal partition: required the length of the partition.

Here 202 + 152 = 400 + 225 = 625.

And 625 25 feet, the length of the partition.

Prob. VIII. To find the area of a triangle when the base and perpendicular are given.

Rule. Multiply one dimension by half the other. The result is the required area.

Example. The height of one of the Egyptian pyramids is about 600 feet, and its base 690 feet: required the area of its 4 sides.

Here 600

2

300 × 690207000 × 4828000 feet. Ans. 828000 feet.

Prob. IX. To find the area of a triangle when the three sides only are given.

Rule 1. Add the three sides, and take half the sum.

2. From the half sum subtract each side successively. 3. Multiply the half sum with the three remainders. 4. Take the square root.

Example. A churchyard of a triangular form was measured; the sides were found to be respectively 250 feet, 200 feet, and 150 feet: required the area.

Then 300 x 50 x 100 x 150225,000,000 feet

√225,000,000 = 15,000 feet, the required area. Prob. X. To find the solid contents of a triangular pyramid.

Rule. First find the area of the base, multiply the sum by one-third the perpendicular height: the product will be the solid contents.

Example. A triangular pyramid 21 feet high has sides of 5, 6, and 7 feet respectively.

By Prob. IX. 5+ 6+ 7 ÷ 2 = 18÷ 2 = 9

1. The angles in a regular rhombus are respectively 60 and 120 degrees.

MENSURATION.

Prob. XI. To find the area of a rhombus or rhomboid. Rule. Multiply the base by the perpendicular height. The product is the required area.

Example. The base of a square leaning tower was found to be 10 feet 5 inches, and its perpendicular height just 80 feet required the area of one side.

Prob. XII. To find the area of a trapezium.

Rule. Divide the trapezium into two triangles, find the area of these by Prob. VII., the sum of which is the required area.

Prob. XIII. To find the area of a trapezoid.

Rule. Multiply half the sum of the parallel sides by the distance between them.

Example. A yard in the form of a trapezoid was measured; its parallel sides were found to be 12 feet and 10 feet, the distance 15 feet: required the area.

Here (12 ft.+ 10 ft.) ÷ 2 = 22 ft. ÷ 2 = 11 ft.
And 11 ft. x 15 ft. 165 ft., the required area.

(F.)

1. Constructing or assisting lines will be found exceedingly useful in the delineation of polygons; these will be supplied by the previous investigation of the teacher.

Thus a pentagon of 10 inches in height will have the following proportionate dimensions.

Side in all cases, three-fifths of the total height.

MENSURATION.

Prob. XIV. To find the area of a polygon.

Rule. Multiply the square of the side by the integer and