Imágenes de páginas
PDF
EPUB

7. What is the area of a triangular field, of which the base is 1236 links, and the vertical height 731 links? Ans. 4 ac. 2 roods 2.8 + perches.

8. What is the surface of a piece of land of triangular shape, the base being 1589 links, and the vertical height 3568 links?

Ans. 28 ac. 1 rood 15 6 perches.

III. To find the area of a parallelogram, having given the base and the perpendicular height.

This is equal to the area of a rectangle described on the same base, and having the other side equal to the perpendicular height of the parallelogram,

:. Area = Base × Height, as before.

IV. To find the area of a trapezium or trapezoid.

A trapezium, or, as it is defined by some, a trapezoid, is a four-sided figure, of which two of the opposite sides are parallel.

RULE. Area one-half sum of parallel sides × perpendicular. Of the two parallel sides one is longer than the other, but the half-sum is the average of the two.

What is the area of a trapezium, having the parallel sides 24 +18, and the perpendicular distance between them 8 feet?

[ocr errors]

21

24+18 == 42. 1 × 42
Area: 21 x8 = 168 sq. feet.

1. Find the area of a trapezium having the parallel sides 16 and 24 feet, and the perpendicular distance between them 10 feet. Ans. 200 sq. feet.

2. What is the area of a figure of which the two opposite parallel sides are 25 and 20 feet, and the shortest distance between them 62 feet?

Ans. 1395 sq. feet. 3. Find the area of a trapezium, the perpendicular between the parallel sides being 24 feet, and these sides respectively 16 and 20 feet. Ans. 432 feet.

4. What is the perpendicular distance between the two parallel sides of a four-sided figure, if the area be 680 square yards, and the parallel sides 20 and 24 feet respectively? Ans. 278 feet.

V.

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

RULE. From half the sum of the sides subtract each side separately. Multiply the half-sum of the sides by the differences thus found, and the area will be the square root of the product, or

Areas (-a) (s-b) (s-c).

where s = half the sum of three sides, a, b, and c. What is the area of a field of which the sides are 24, 12, and 18 feet?

[blocks in formation]

Areas (s-a) (s—b) (s—c) = √√27 × 3 × 15×9

= 81 x 15 x 9

But square root of 81

[blocks in formation]

= 9, and of 9 = 3.

9x3/1527/15 104.5 nearly.

= =

1. What is the area of a triangle, of which the sides are 4, 6, and 8 feet? Ans. 11.6 nearly.

2. What is the area of a triangle, of which the sides are 3, 6, and 8 feet? Ans. 7.64 nearly.

3. What is the area of a triangle, of which the sides are 7, 8, and 9 feet? Ans. 26 8 nearly. triangle, of which the Ans. 30 6 nearly. triangle, of which the Ans. 31 42 nearly.

4. What is the area of a sides are 7, 9, and 10 yards?

5. What is the area of a sides are 7, 9, and 11 feet?

6. What is the area of a triangle, the sides being

4, 7, and 6 feet?

Ans. 11.97.

7. Required the same, if the sides be 6, 5, and 10 feet? Ans., 11.398. 8. What is the area of a triangular field, of which the sides are respectively 2139, 2345, and 2564 links? Ans., 25 ac. 2 roods ⚫584 perches.

In the case of right-angled triangles, sometimes the vertical height is not given; but this can be determined if the hypotenuse, or side subtending the right angle, be given instead. From Euclid 47, Book I., the square of the side subtending the right angle is equal to the sum of the squares of the sides containing the right angle. Let A B C be the rightangled triangle, having the right angle at B.

Then AC2=AB2+BC2

:: AB AC-BC

and AB=√AC-BC

Similarly BC2 = AC2-AB2 :: BC = √AC2—AB2

EXERCISES.

1. Find the perpendicular of a right-angled triangle of which the hypotenuse is 30, and the other side 24 feet? Ans., 18. 2. What is the breadth of a street if the houses be 51 feet high, and a ladder from the top of the house to the foot of the opposite house measures 160 feet? Ans., 151 4851478 feet.

3. The hypotenuse of a right-angled triangle is 5 feet, what is the length of the perpendicular if that of the base be 4 feet? Ans., 3 feet. 4. What is the length of the base of a right-angled triangle if the hypotenuse be 6 feet, and the other side 4 feet? Ans., 20. 5. What is the length of the hypotenuse of a right-angled triangle if the base be 5, and the perpendicular 5 feet? Ans., 61. A polygon is a figure containing more than four sides.

Polygons are regular or irregular. A regular polygon is one made up of symmetrical parts, or of triangles of the same size and shape, the bases of which are the sides of the polygon, and the vertices of which meet in a common point.

The area of such regular polygons will manifestly be the sum of the areas of the respective triangles, and as these are determined by the product of the perpendicular and the base divided by 2, the

Area:

Perimeter x Perpendicular,

2

where the perimeter is the sum of the sides of the regular polygon.

Regular polygons are

1. The Pentagon, with 5 sides.

2. The Hexagon, 3. The Heptagon, 4. The Octagon, 9. The Nonagon, 10. The Decagon,

[ocr errors]
[ocr errors]

7

[ocr errors]

8

[ocr errors]

9

[ocr errors]

10

&c.

[ocr errors]
[ocr errors]
[ocr errors]

EXERCISES.

1. What is the area of a regular pentagon, of which the side is 25 feet and the perpendicular 17.204774 feet? Ans., 1075+ feet.

2. What is the area of a hexagon of which the side is 10 feet and the perpendicular 8 feet? Ans. 255 ft. 3. What is the area of a regular hexagon of which one side is 14-6 feet, and the perpendicular from the centre 12.64 feet? Ans., 553 632 feet.

4. What is the area of a regular polygon of seven sides, if one of these measure 19.38 inches, and the perpendicular from the centre be 20 inches in length? Ans., 1356 6 square inches.

5. A regular octagon contains 477 168 square inches; the length of one of the equal sides is 9.941 inches, what is the length of the perpendicular? Ans., 12 inches. 6. The area of a regular heptagon is 1356 6 square inches; the side is 19 38 inches, what is the length of the perpendicular? Ans., 20 inches.

To find the area of a circle.

RULE.-Area

Diameter squared multiplied by

⚫7854, or A=D2x •7854.

As the diameter is twice the radius .. diameter= radius × 2; or d=rx2.

But

..D2=rx2xrx2=r2x4.

7854x4=3·1416.

..Area 2x3.1416.

=

3 1416 is the ratio between the diameter and the circumference, or

Circumference-diameter x 3.1416.

We can, therefore, find the area when the circumference alone is given for

circumference

circumference

diameter=

or radius=

3.1416

3.1416 x 2.

EXERCISES.

1. What is the area of a circle of which the

diameter is 20 feet?

Ans., 314 16 square feet.

« AnteriorContinuar »