gallons; then 14.667 × 2.7 = 39.6009, the content in ale gallons,

EXAM. 4.

Here 108.4 x 70.6 = 7653.04, the area in square inches; and 7653.04 282 = 27.138, the area in ale gallons; then 27.138 × .8 = 21.7104, the content in ale gallons.

PROBLEM IV.

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

2

2

square

inches; and 5871.4282 = 20.82, the area in ale gallons; then 20.82 × 4.1 = 85.362, the content in ale gallons.

EXAM. 4.

Here

136.7 x 86.8
2

2

area in square inches; and 5932.78

2150.42 = 2.758, the area in malt bushels; then 2.758 × 42.3 = 116.6634, the content in malt bushels.

PROBLEM V.

To find the area of a triangle, when the three sides

are given.

EXAM. 3.

114112+ 98

Here

sides; then 162

= 162, half the sum of the 114 48, the first remainder;

162 112=50, the second remainder; and 162 9864, the third remainder; whence 162 × 48 x 50 x 64 =√√24883200 = 4988.306, the area in square inches; then 4988.306 ÷ 282 = 17.689, the are in ale gallons; 4988.306 ÷ 231 = 21.594, the area in wine gallons; and 4988.3062150.42 = 2.319, the area in malt bushels.

Here

EXAM. 4.

74+ 74+ 74
2

of the sides; then 111
mainder; 111

74 = 37, the first re74 37, the second remainder; and 111 – 74 = 37, the third remainder; whence √111 × 37 × 37 × 37 = √ 5622483 = 2371.177, the area in square inches; then 2371.177 ÷ 282 = 8.408, the area in ale gallons; and 8.408 x 28.6 = 240.4688, the content in ale gallons.

PROBLEM VI.

To find the area of a trapezium..

EXAM. 3.

Ale gallons.
14.967 area.
3.7 depth,

104769

44901

55.3779 Ans.

EXAM. 4.

Here 28.2+47 × 94.475.2 x 94.4 = 7098.88; and 7098.88 ÷ 2 = 3549.44, the area in square

inches; then 3549.44 231 = 15.365, the area in wine gallons.

EXAM. 5.

Here 58.4 + 46.8 × 126.6 = 105.2 × 126.6 = 13318.32; and 13318.32 ÷ 2 = 6659.16, the area in then 6659.162150.42 = 3.096, the square inches; area in malt bushels; whence 3.096 × 39.3 121.6728 bushels, the whole content; and 121.6728 121.6728

5

= 121.6728 24.3345 97.3383 bushels,

the answer required.

PROBLEM VII.

To find the area of a trapezoid.

EXAM. 3.

Here 214178 x 146 = 392 × 146 = 57232; and 57232 2 28616, the area in square inches; then 28616282 = 101.475, the area in ale gallons. EXAM. 4.

Here 101.475 x 1.7 172.5075 ale gallons, the content required.

PROBLEM VIII.

To find the area of an irregular polygon of any number of sides.

EXAM. 2.

Ale gallons.
24.414 area.
6.7 depth.

170898
146484

163.5738 Ans.

=

178 x 18 x 76 x 84 = √20434336 4522.647 inches, the area of the first triangle.

Again,

160 +130+ 64 354

2

= = 177, half the

2

sum of the sides; then 177 160 = 17, the first remainder; 177-13047, the second remainder; and 177 — 64 = 113, the third remainder; whence VITT x 17 x 47 x 11315980799 = 3997.599 inches, the area of the second triangle.

2

=

2

185, half the

sum of the sides; then 185 - 140=45, the first remainder; 185 — 130 = 55, the second remainder; and 185 — 100 = 85, the third remainder; whence √185 × 45 × 55 × 85 = √38919375 = 6238.539 inches, the area of the third triangle; hence we have 4522.647 + 3997.599 + 6238.539 = 14758.785 square inches, the area of the whole polygon.