that is, the area of a triangle, which has one angle of 30°, is equal to one-fourth of the product of the two sides containing that angle.

Ex. Let AC-24 yds., and AB=176 yds.; then

area of ABC=

24×17.6
4

=105'6 sq. yds.

PROB. 4. To find the length of a side of the square 'inscribed' in a given circle.

Let ABCD be the given circle, and O its centre: AC and BD two diameters at right angles to each other. Join AB, BC, CD, DA; then we have a square inscribed (155, B Part 11.). Now,

AB2 = A02+BO2=2A02 = 2 × (rad.)2;

.. AB=rad.× √√2;

C

or the side of the inscribed square is equal to the radius multiplied by √2.

If rad.=1, the side of the square=√2.

PROB. 5. To find the length of a side of the equilateral triangle 'inscribed' in a given circle.

Let O be the centre of the given circle, and ABC an equilateral triangle inscribed in it. From O draw OD perpendicular to AB, and produce it to meet the circumference in E. Join AE,

E

B

D

:. AB=OA√3,

=rad.x√3.

If rad. 1, the side of the inscribed equilateral triangle=√3.

PROB. 6. To shew that the side of a square, together with the side of an equilateral triangle, both inscribed in the same circle, is equal to half the circumference of the circle, nearly.

By Prob. 4, the side of square

=rad. × √2=rad. ×1·414..............

By Prob. 5, the side of triangle

=rad.x√3=rad. x1·732.............. ;

.. sum of the two = rad. × (1·414+1·732),

— rad. × 3·146, nearly.

=

But half the circumference of the circle

=rad. x 3.14159;

therefore the side of the square added to the side of the triangle does not differ from the semi-circumference of the circle by a quantity so great as '005, or

is, the 200th part of a unit.

5

1000'

that

PROB. 7. To find the numerical value of the angle at the centre of a circle subtended by an arc equal to the radius.

By (240) we know, that the length of an arc subtending an angle of Ao at the centre

A

=2π × rad.x

360

therefore, in this case,

A

rad.=2 x rad. x

to find A.

PART III.

PROB. 8. To construct a rectangle, or triangle, which shall be equal to a given circle.

Assuming that the area of the given circle is

divide the given radius into 7 equal parts (168, Part 11.); then construct a rectangle, of which the base is 22 of such parts, and the height 7, that is, the radius. The area of this rectangle

For a triangle of equal area, make the base the same, and the height equal to double the radius, that is, to the diameter.

PROB. 9. To measure the area of a circular ring.

Let it be required to find the area of the ring enclosed by the two concentric circles ABC, DEF. It is plain, that this area will be found by subtracting the area of the smaller circle from that of the greater. So that, if R represents the radius of the greater, and r the radius of the smaller, circle, we have, by (238),

с

F

E

B

.. area of ring = π × R2 — π × r2 = π × (R2—r2);

that is, the square of the smaller radius must be subtracted from the square of the greater, and the difference multiplied by the number п, as given in (237).

Ex. Let the inner and outer radii of a ring be 30 feet, and 35 feet, respectively; then

R=1225, and r2=900;

and the difference =325; therefore area of ring

=3·1416 × 325 sq. feet,

=1021.02 sq. feet.

NOTE. By the well-known rule, which can be proved both geometrically and algebraically, viz. that the difference of the squares of any two numbers is equal to the product of their sum and difference, the trouble of squaring large numbers may, in this instance and in some others, be avoided. Thus R-r2 is equal to R+r multiplied by R-r, whatever numbers R and r stand for. And, taking the above Ex., R+r=65, R-r=5, therefore R2-r2=5×65=325, as before.

But the advantage of this device will be better seen in such an example as the following:

Ex. The outer and inner radii of a circular ring are 365 yards, and 355 yards, respectively; find the area. Here R+r=720, and R-r=10;

.. R2-r2=720×10=7200,

and area of ring =π×7200 sq. yards.

It is further to be observed, that, although in the above problem the circles were said to be concentric, this is not a necessary condition. The same results precisely will be obtained, provided one circle be wholly within the other.

PROB. 10. To measure the area of a portion of a circular ring, as ABCD in the diagram below.

It is plain, that the area of ABCD is equal to the difference of the sectors OAB, and OCD, O being the common centre of the two arcs AB and CD.

A

D

B

If, then, the radii are represented by R, r; we have, by (241),

and

sector OAB-AB× R,

..... OCD=1CD×r;

.. area ABCD=AB-R-CDxr.

If, however, the radii are not given and cannot easily be found, this result must be modified as follows:-(The process will be readily understood by those who have a little knowledge of Algebra:)

By (240), AB CD: R: r;

.. ABxr=CD× R, (74, Part I.),

or ABxr=CD×R;

and area ABCD=1AB× R−1CD×r,

=1⁄2AB×R−}AB×r+1AB×r−1CD×r,

=1AB×R-AB×r+1CD×R−1CD×r,

=&AB×(R−r)+1CD×(R−r),

=}(AB+CD)×(R−r),

= {(AB+CD) × AD;

that is, half the sum of the two arcs multiplied by the distance between them, as in an ordinary trapezium. And, since this is true for any portion of a ring, bounded as above, it follows, that it is true also for the complete ring, viz. the area of the ring is equal to half the sum of the circumferences multiplied by the distance between them.