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PROBLEM XV.

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A

B

To find the Side of a Square nearly equal in Area to a

given Circle. Draw the two diameters A B and C D at right angles to each other, bisect the radius 0 C by a line from one end of the diameter at A, meeting the circumference in E, then will the line A E be the side of a square nearly equal in area to the given circle.

And if the line E F be drawn parallel to CD, it will be of the circumference nearly.

Or three times the diameter A B or CD, and once the versed sine Q H, of the angle A OD, will be the circumference nearly.

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PROBLEM XVI.

D

To find a right Line that shall be nearly equal to any

given Arc of a Circle. Divide the chord A B into four equal parts, set one part on the circumference from B to D, draw a line from C, the first division on the chord ; and twice the length of the line CD will be the length of the arc nearly.

A

PROBLEM XVII.

B В

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To describe an Ellipsis, the transverse and conjugate

diameters being given. From o as a centre, with the difference of the transverse and conjugate semi-diameters, set off o c and od; draw the diagonal cd, and continue the line oc to k, by the addition of half the diagonal c d, then will the distance o k, be the

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H

radius of the centres that will describe the ellipsis; draw the lines A B, C D, C E, and B H, cutting the semi-diameters of the ellipsis in the centres k l mn; then with the radius m s, describe the arcs D H and A E; also, with the radius n r, describe the arcs ED and A H, which will be the ellipsis required.

PROBLEM XVIII.

To describe a Parabola, any ordinate to the axe and its

abscissa being given.

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Let V R and R o be the given abscissa and ordinate, bisect Ro in m, join V m, and draw mn perpendicular to it, meeting the axe in n; make V C and V F each equal to R n, then will F be the focus of the curve.

Take any number of points, r, r, &c. in the axis, and draw the double ordinates of an indefinite length.

From F as a centre, with the radii C F, C r, &c., describe arcs cutting the corresponding ordinates in the points o 0 0 0, &c., and the curve o V o, drawn through all the points of intersection, will be the parabola required.

R

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MENSURATION OF SUPERFICIES.

The area or superficial content of any figure, is the space contained within length and breadth, without having any regard to the thickness.

PROBLEM I.

To find the Area of any Parallelogram, whether it be a

a square, a rectangle, a rhombus, or a rhomboid.

RULE.—Multiply the length by the breadth or height, and the product will be the area.

Square. Rectangle.

Rhomboid.

Rhombus.
A B

А

B

A

B

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D C

D C

D

a DC a EXAMPLE.-Required the area of a rhomboid, whose length, A B, = 20.5, and breadth @ A, = 11.75.

20.5 x 11.75 = 240.875 the area.

PROBLEM II.

To find the Area of a Trapezoid. RULE.—Add together the two parallel sides, multiply their sum by the breadth or height, and half the product is the area.

Examplr.-Required the area of a trapezoid whose sides, A B and CD, are 14.5 and 10.25, and breadth, a A, = 7.25.

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PROBLEM III.

To find the Area of a Triangle.

19

RULE.-Multiply one of its sides as a base by a perpendicular let fall from the opposite angle, and take half the product for the area,

Or, Subtract each side separately from half the sum of the sides, multiply the half sum of the sides by the three remainders, and the square root of the product will be the area.

Example 1.-Required the area of a triangle, ABC, whose base, A B, = 16.5, and perpendicular, D C, = 10.25.

16.5 x 10.25

= 84.5625 2

the area

А

D

B

EXAMPLE 2.—What is the area of that triangle whose three sides are 8, 12, and 16 respectively ? 8 + 12 + 16_

= 18, the half sum of the sides, 2 then, 18 18 18

8 16

12

6

10 the area.

2 and V 18 x 10 x 6 x 2 = 46.47

PROBLEM IV.

If any two sides of a right-angled Triangle be given, the third side may be found by the following rules :

1.–To the square of the base add the square of the perpendicular; and the square root of the sum will be the hypotenuse or longest side.

2.-Multiply the sum of the hypotenuse, and one side by their difference; and the square root of the product will be the other side.

EXAMPLE 1.–Given the base A B = 16, and per. pendicular BC = 12; required the length of the hypotenuse A C.

✓ 162 + 122 = 20 the length of the hypotenuse A C.

A

B

EXAMPLE 2.-Given the base À B = 16, and hypotenuse AC= 20: required the length of the perpendicular B C.

✓20 + 16 x 4 = 12, length of the perpendicular B C.

PROBLEM V.

To find the Area of any regular Polygon. RULE.—Multiply the sum of its sides by a per: pendicular drawn from its centre to one of its sides, and take half the product for the area.

Or, Multiply the square of the side of a polygon (from three to twelve sides) by the numbers in the fourth column of the table for polygons, opposite the number of sides required, and the product will be the area nearly.

EXAMPLE 1.-Required the area of the regular pentagon A B C D E, each side being 7.5, and perpendicular F G = 6.4.

D

7.5 X 5 X 6.4

= 120 the area. 2

E

A G B

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