OF THE RIGHT ANGLED TRIANGLE.

A right angled triangle is a plane superficies, having one right angle: the side opposite the right angle is called the hypothenuse, the side on which the triangle stands is called the base, and the other side the perpendicular. The base and perpendicular are sometimes called the legs.

The legs given, to find the area and hypothenuse; or the area and one leg given, to find the other leg and hypothenuse.

RULE.

1. Multiply the base and perpendicular together, and half the product will be the area.

2. Divide the double area by the length of one leg, and the quotient will be the length of the other leg.

The hypothenuse may be found by the note at page 187. Examples.

1. The base AB of a right angled triangle is 60, and perpendicular BC 32-Required the area, and hypothenuse AC.

2. The area of a triangular field is 600 perches, and the base 50-I wish to know the length of the perpendicular and hypothenuse.

600×2=1200-double area,

and 1200÷50=24=perpendicular.
502+242=2500-576-3076,

and 3076 55.461 Answer.

8. What is the area and hypothenuse of a triangle whose base is 64 and perpendicular 47 perches ?

Ans. Area, 9 A. 1 R. 24 P. Hypothenuse, 54.845. 4. Required the area and base of a triangle whose perpendicular is 71 perches, and hypothenuse 2521 perches.

Ans. Area 559 A. 20 P. Base 2520 perches.

Note. When the hypothenuse and one leg are given, if you multiply their sum by their difference, the product will be the square of the other leg.

5. What is the base of a triangle whose hypothenuse is 45 and perpendicular 27 ?

45+27-72 sum

45-27-18 difference

1296 product; and 129636 Answer.

6. The hypothenuse of a triangle is 261, and the base 189; required the perpendicular. Ans. 180.

Note. When the area and hypothenuse are given, raise the hypothenuse to the 4th power; deduct 16 times the square of the area from this power, and extract the square root of the remainder, one half this root deducted from half the square of the hypothenuse, will give the square of the shorter leg, and if added thereto will give the square of the longer leg.

7. The area of a triangle is 84, and the hypothenuse 25— Required the other sides.

277729, and√277729=527, and 527+2=2631

252-625, and 625÷2-312/

312-263-49, and

49 7 shorter leg? Ans. 312+2631-576, and 576-24 longer leg S

8. The area of a triangle is 5280, and the hypothenuse 146; required the other sides.

Ans. 110 and 96.

To find the area of any triangle, having the three sides

given. RULE.

Add together the three sides and take half the sum, deduct each of the several sides from that half sum, then multiply that half sum and the three differences continually together, and extract the square root of the product, which will be the area of the required triangle.

Examples.

1. The sides of a triangle are as follow, viz. AB 90, BC 80 and AC 70-Required the area.

2.

B

The sides of a triangle are 189, 170 and 89-What is

the area?

Ans. 7560.

3. What is the area of a triangle whose sides are 48, 60 and 72 perches ? 4. The sides of a triangular perches-Required the area.

Ans. 8 A. 3 R. 28 P.+ garden are 8, 10 and 12 Ans. 39.68 perches +

OF THE TRAPEZIUM.

A trapezium is a plane superficies, bounded by four straight lines, no two of them being parallel to each other. A line connecting the opposite angles is called a diagonal.

The four sides and diagonal given, to find the area.

RULE.

Divide the trapezium into two triangles, by drawing a diagonal across it, then find the area of each triangle separately, as before taught, and the sum of these will be the area of the trapezium.

Examples.

1. Required the area of a four sided field, whose south side is 100 perches, east side 90 perches, north side 80 perches, and west side 70 perches; the diagonal from north-east to south-west being 120 perches.

155x35×65×55 = 19394375

and 19394375-4403.904 area of ABC.

2690.619 area of ADC.

Note. Irregular figures are such as have more than four sides, which, as well as their angles, are unequal. The superficial content of all such figures may be found by dividing them into trapeziums and triangles, by lines drawn from one angle to another, calculating these separately, as before taught, and adding all the areas together.

OF REGULAR POLYGONS.

A regular polygon is a plane figure, having all its sides and angles equal.

To find the superficial contents, having a side and perpendicular let fall from the centre to the middle of one of its sides given. RULE 1.

Multiply the sum of the sides by half the perpendicular; or multiply the whole perpendicular by half the sum of the sides, the product will be the area or superficial content.

Examples.

1. Required the area of an octagon, whose side is 4.9705, and perpendicular 6.

4.9705×8=39.764, and 39.764÷2=19.882 = half sum of the sides.

19.882×6=119.292 Answer.

2. Required the superficial content of a hexagon, whose side is 20 and perpendicular 17.320508. Ans. 103.923048.

A TABLE

For more readily finding the area and perpendicular of a regular Polygon.

Multiply the tabular area by the square of the side, and the product will be the area of the polygon; or multiply the tabular perpendicular by the side of the polygon, the product will be the perpendicular of the polygon: then proceed by Rule 1.