Page images
PDF
EPUB

PRACTICAL APPLICATIONS.

MENSURATION.

DEFINITIONS.

381. Mensuration is the process of determining the measurement of lines, surfaces and volumes.

382. The Measuring Unit assumed for a given line some linear unit, as 1 inch, 1 foot; for a given surface, some superficial unit, as 1 square inch, 1 square foot; and for a given volume, some cubic unit, as 1 cubic inch, 1 cubic foot.

NOTE. The following Problems, at the discretion of the teacher, may be studied, after the completion of Book III. of the Geometry, in connection with the text to which they refer, as the learner progresses in the remainder of the work.

MENSURATION OF LINES AND SURFACES.

PROBLEM I. Any two sides of a RIGHT-ANGLED TRIANGLE being given, to find the third side.

To the square of the base add the square of the perpendicular; and the square root of the sum will give the hypothenuse (Theo. IX. Bk. IV.).

From the square of the hypothenuse subtract the square of the given side, and the square root of the difference will be the side required (Theo. IX. Cor. 1, Bk. IV.).

1. The base, A B, of the triangle A B C is 48 feet, and the perpendicular, BC, 36 feet; what is the hypothenuse?

482+362 = 3600; √ 3600 = 60 feet, [the hypothenuse required.

A

C

B

2. The hypothenuse of a triangle is 53 feet, and the perpendicular 28 feet; what is the base ?

3. Two ships sail from the same port, one due west 50 miles, and the other due south 120 miles; how far are they apart? Ans. 130 miles.

4. A rectangular common is 25 rods long and 20 rods wide; what is the distance across it diagonally?

5. If a house is 40 feet long and 25 feet wide, with a pyramidal-shaped roof 10 feet in hight, how long is a rafter which reaches from the vertex of the roof to a corner of the building? PROBLEM II. To find the area of a PARALLELOGRAM.

Multiply the base by the altitude, and the product will be the area (Theo. V. Bk. IV.).

1. What is the area of a square, ABCD, whose side is 25 feet?

25 x 25 = 625 feet, Ans.

2. What is the area of a square field whose

side is 35.25 chains? Ans. 124 A. 1 R. 1 P.

D

C

A

B

3. How many square feet of boards are required to lay a

floor 21 ft. 6 in. square?

4. Required the area of a square farm, whose side is 3,525 links.

5. What is the area of the rectangle ABCD, whose length, A B, is 56 feet, and whose width, A D, is 37 feet?

56 × 37 = 2,072 sq. feet, Ans.

D

A

C

B

6. How many square feet in a plank, of a rectangular form, which is 18 feet long and 1 foot 6 inches wide?

7. How many acres in a rectangular garden, whose sides are 326 and 153 feet? Ans. 1 A. 23 P. 6+ sq. yd.

8. A rectangular court 68 ft. 3 in. long, by 56 ft. 8 in. broad, is to be paved with stones of a rectangular form, each 2 ft. 3 in. by 10 in.; how many stones will be required?

Ans. 2,062 stones.

9. Required the area of the rhomboid ABCD, of which the side A B is 354 feet, and the perpendicular distance, EF, between A B and the opposite side CD, is 192 feet.

354 × 192 = 67,968 feet, Ans.

[blocks in formation]

10. How many square feet in a flower-plat, in the form of a rhombus, whose side is 12 feet, and the perpendicular distance between two opposite sides of which is 8 feet?

PROBLEM III. The area of a RECTANGLE and either of its sides being given, to find the other side.

Divide the area by the given side, and the quotient will be the other side.

If the rectangle be a square, the square root of the area will be the side.

1. The area of a rectangle is 2,072 sq. feet, and the length of one of the sides is 56 feet; what is the length of the other

side?

207256 = 37 feet, the side required.

2. How long must a rectangular board be, which is 15 inches in width, to contain 11 square feet?

3. A rectangular piece of land containing 6 acres is 120 rods long; what is its width? Ans. 8 rods.

4. What is the side of a square containing 625 square feet?

√625 = 25 feet, the side required.

5. The area of a square farm is 124 A. 1 R. 1 P.; how many links in length is its side?

6. A certain corn-field in the form of a square contains 15 A. 2 R. 20 P. If the the corn is planted on the margin, 4 hills to a rod in length, how many hills are there on the margin of the field? Ans. 800 hills.

PROBLEM IV. The area of a RHOMBOID or RHOMBUS and the length of the base being given, to find the altitude; or the area and the altitude being given, to find the base.

Divide the area by the length of the base, and the quotient will be the altitude; or

Divide the area by the altitude, and the quotient will be the base.

1. The area of a rhomboid is 67,968 square feet, and the length of the side taken as its base 354 feet; what is the altitude?

67,968354 = 192 feet, the altitude required.

2. The area of a piece of land in the form of a rhombus is 69,452 square feet, and the perpendicular distance between two of its opposite sides is 194 feet; required the length of one of the equal sides. Ans. 358 ft.

3. On a base 12 feet in length it is required to find the altitude of a rhomboid containing 968 square feet.

4. The area of a rhomboidal-shaped park is 1 A. 3 R. 34 P. 51 yd.; and the perpendicular distance between the two shorter sides is 96 yards; required the length of each of these sides. Ans. 18 rods.

PROBLEM V. To find the area of a TRIANGLE.

If the base and altitude are given, multiply the base by half the altitude, or half the base by the altitude.

If the three sides are given, from half the sum of the three sides subtract each side; multiply the half sum and the three remainders together, and the square root of the product will be the area required.

1. Required the area of the triangle ABC, whose base, BC, is 210, and altitude, A D, is 190 feet.

210 × 10 = 19,950 square feet, the area

A

[required.

2. A piece of land is in the form of a

[blocks in formation]

right-angled triangle, having the sides about

the right angle, the one 254 and the other 136 yards; required the area in acres.

Ans. 3 A. 2 R. 10 P. 29 yd.

3. Required the number of square feet in a triangular board whose base is 27 inches, and altitude 27 feet.

4. What is the area of a triangle whose base is 15.75 chains, and the altitude 10.22 chains?

5. What is the area of a triangle, ABC, whose sides, AB, BC, CA, are 40, 30, and 50 feet?

30+40+50÷2=60, half the sum of

C

[blocks in formation]

60-30 = 30, first remainder.
60-40 20, second remainder.
60-50 = 10, third remainder.

60 × 30 × 20 × 10 = 360,000; √360,000 = 600, square feet, the area required.

6. How many square feet in a triangular floor, whose sides are 15, 16, and 21 feet?

7. Required the area of a triangular field whose sides are 834, 658, and 423 links. Ans. 1 A. 1 R. 20 P. 4 yd. 1.6 ft. 8. Required the area of an equilateral triangle, of which each side is 15 yards.

9. What is the area of a garden in the form of a parallelogram, whose sides are 432 and 263 feet, and a diagonal 342 Ans. 2 A. 10 P. 11.46 yd.

feet?

PROBLEM VI. The area and the base of a TRIANGLE being given, to find the altitude; or the area and altitude being given, to find the base.

Divide double the area by the base, and the quotient will be the altitude; or divide double the area by the altitude, and the quotient will be the base.

1. The area of a triangle is 1300 square feet, and the base 65 feet; what is the altitude?

1300 × 2 = 2600;2600÷65 = 40 ft., altitude required. 2. The area of a right-angled triangle is 17,272 yards, of which one of the sides about the right angle is 136 yards; required the other perpendicular side.

« PreviousContinue »