(6) A room, which is 20 ft. long and 15 ft. wide, has an opening or recess at one end, in the shape of a segment of a circle, the opening or chord of the recess being 15 ft. and its greatest width 4 ft.; find the area of the whole room.

(7) An arched gateway is 30 ft. wide, and measures 20 ft. from the ground to the spring of the arch; and the distance also from the ground to the crown of the arch is 30 ft. Find how many square feet of timber will be required for blocking it up.

(8) Find how many sq. ft. of brickwork are used in blocking up one of the arches in a railway viaduct :

The span of the arch is 60 ft., the height above the piers is 20 ft., and the distance from the ground to the spring of the arch is 20 ft.

XVII. THE ELLIPSE.

Definitions. An ellipse, in simple language, may be defined to be a plane figure bounded by a curved line, A called the circumference, and which has a longer and a

shorter diameter.

B

F

F

D

The longer diameter AB is called the transverse diameter, or major axis.

The shorter diameter CD is called the conjugate diameter, or minor axis.

F and F are the foci of the ellipse.

RULES.-(1) To find the circumference or perimeter of the ellipse.

Multiply half the sum of the two diameters by 22.

(2) To find the area of an ellipse.

Multiply the product of the two diameters by 14.

Or, multiply the product of half the diameters by 22.

Note 1.-To find either the longer or the shorter diameter, when the circumference and the other diameter are given.-Subtract the given diameter from of the eircumference.

Note 2.-To find either diameter, when the area and the other diameter are given.-Divide of the area by the given diameter.

Note 3.-An ellipse is equal to a circle whose diameter is a mean proportional between the two diameters; that is, the area of the ellipse ACBD is equal to the area of a circle

Or, in other words, the ellipse is a mean proportional between its inscribed and circumscribed circles; that is, between the circles about the transverse and conjugate diameters. For instance, if a circle were described having AB for its diameter, and another circle were described having CD for its diameter, then the area of the ellipse is equal to half the sum of the areas of these two circles.

Note 4.-By the rules given above, the circumference and the area of an ellipse will be found with sufficient accuracy for all practical purposes. If, however, greater accuracy is desirable, then, in Rule 1, substitute 3.1416 as multiplier, instead of 22; and in Rule 2, 7854 for 11.

Note 5.-The area of an elliptical ring is the difference between the areas of the outer and inner ellipse.

Example 1.-Find the perimeter or circumference of an

ellipse whose transverse diameter is 28 ft. 6 in., and conjugate diameter is 20 ft. 6 in.

By Rule 1: Circumference=sum of diameters 27 = (28 ft. 6 in. +20 ft. 6 in.) x 22×49 ft. x 22=77 ft.

Example 2.-Find the area of an ellipse, whose major axis is 156 ft., and minor axis is 120 ft.

By Rule 2: Area=156 × 120 × 11=18480 sq. ft.

Example 3.-The perimeter of an ellipse is 28 yds. 1 ft. 3 in., and its minor axis is 22 ft. 9 in.; find the major axis. Now, 28 yds. 1 ft. 3 in.=1023 in. ; and 22 ft. 9 in.=273 in. By Note 1:

The major axis=1 of the circumference-minor axis =( of 1023)-273-651-273=378 in. =31 ft. 6 in.

Example 4.-The area of an ellipse is 28 sq. ft. 60 sq. in., and its transverse diameter is 7 ft. 9 in. ; find the conjugate diameter.

Now, 28 sq. ft. 60 sq. in.=4092 sq. in.; and 7 ft. 9 in.= 93 in.

By Note 2:

The conjugate diameter=14 of the area÷given diameter of 4092-93-5208-93-56 in. =4 ft. 8 in.

Formulæ

I. Circumference (sum of the two diameters) × 22.

II. Area=transverse diameter x conjugate diameter x 14.

III. Required diameter of the circumference-given diameter.

IV. Required diameter=1 of the area given diameter.

EXAMPLES.

Find the circumference of the ellipse whose transverse and conjugate diameters are, respectively

(1) 48 in., and 36 in. (2) 13 ft., and 8 ft. (3) 87 yds. 1 ft., and 34 yds.

(4) 3 ch. 80 lks., and 3 ch. 20 lks.

Find the minor axis of an ellipse, when its circumference and its major axis are, respectively

(5) Circumference 286 ft., and major axis 121 ft.

(6) Circumference 385 ft., and major axis 145 ft. (7) Circumference 47 ft. 8 in., and major axis 16 ft. 8 in. (8) Circumference mile, and major axis 165 yds.

Find the area of the ellipse, when its major and minor axes are, respectively

(9) 110 ft., and 98 ft. (10) 40 ft. 10 in., and 29 ft. 2 in. (11) 42 yds., and 32 yds. 2 ft.

(12) 102 yds. 2 ft., and 93 yds. 1 ft.

(13) 2 ch. 38 lks., and 2 ch.

Find the conjugate diameter of the ellipse, when its area and its transverse diameter are, respectively—

(14) Area 550 sq. ft., and transverse diameter 35 ft. (15) Area 15 sq. ft. 40 sq. in., and transverse diameter 5 ft. 10 in.

(16) Area 7 ac. O rood 20 poles 11 yds., and transverse diameter 224 yds.

(17) Area 2 roods 8 poles, and transverse diameter 2 ch. 80 lks.

(18) In a rectangular plot of land, which is 100 yds. long and 70 yds. broad, is dug a fish-pond, in the shape of an ellipse, whose greater diameter is 98 yds. and less diameter is 60 yds.; the remaining part is to be gravelled, at 3d. per sq. yd. Find the expense.

(19) The rental of a field, in the shape of an ellipse, whose greater diameter is 140 yds., at £6 ls. per acre, is £15 88.; find the shorter diameter.

(20) The building of a wall round an elliptical plot of land, whose greater diameter is 106 yds., at 2s. 9d. per yard, costs £36 6s.; find the length of the shorter diameter.

(21) A lawn, in the shape of an ellipse, whose greater and less diameter are, respectively, 98 ft. and 58 ft., is surrounded by a walk 1 yd. wide; find the cost of gravelling it, at 6d. per sq. yd.

SOLIDS.

DEFINITION S.

1. The volume, solidity, or solid content of a solid is the space (cubic yds., ft., or in.) that it occupies.

2. The surface of a solid is its outside.

3. The different solids have been so fully explained in the following chapters, that it has not been thought necessary to give any definition of them here. The Student must try to understand the explanations given of each solid before attempting the Examples.

4. Tables:

1728 cubic in.=1 cubic ft.

27 cubic ft.=1 cubic yd.

1 cubic ft. of water weighs 1000 oz. avoir., very nearly. 1 cubic ft. contains 6 gallons, very nearly.

1 gallon contains 277 cubic in., or, more nearly, 277-274 cubic in., or 277 cubic in.