(47) A gardener has a circular garden plot, containing 3 roods 7 poles 84 sq. yds., which he desires to plant with three different kinds of trees, and all occupying the same extent of ground; one kind being planted so as to form a circle in the middle; round this a belt of a different kind; and outside of this another belt containing a third kind. What is the diameter of the middle part, and what is the width of the two belts ?

(48) A circular-fish pond, whose diameter is 98 yds., is to be planted all round it to an uniform width, so that the plantation may contain the same area as the fish-pond; find the width of the plantation.

(49*) A gravel walk, 4 ft. wide, is made round a circular court 90 yds. in diameter; the making of the path costs 8d. per sq. yd., and a border along its inner edge costs 3d. a yard. Find the total cost.

(50**) Calculate the expense of making a moat round a circular island, at 2s. 6d. per square yd.; the diameter of the island being 525 ft., and the breadth of the moat being 21 ft. 6 in.

XIII. THE CHORDS OF A CIRCLE.

Definition. Any part of the circumference of a circle, as ACB, is called the arc ACB.

A B is the chord of the arc ACB.
AD or BD is half the chord of the

arc ACB.

AC is the chord of half the arc AC B. CD is the height or versed sine of the arc ACB.

CF is the diameter of the circle, and E the centre of the circle.

с

B

D

E

F

N.B.-The following rules are derived from these two

important formula, which, if thoroughly impressed upon the mind, will enable the student to work the questions in this chapter without referring to the rules given below:AD2=FDX CD;

AC2=CDX CF.

RULES.-(1) To find (CF) the diameter of a circle, when (AB) the chord of the arc, and (CD) the height of the arc are given.

Divide the square of (AD) half the chord by (CD) the height of the arc; and the quotient will be (DF) the part of the diameter wanting; to which add (CD) the height of the arc, and the sum will be the diameter (CF).

(2) To find (CF) the diameter of a circle, when (AC) the chord of half the arc and (CD) the height are given.

Divide the square of (AC) the chord of half the arc by (CD) the height of the arc, and the quotient is (CF) the diameter of the circle.

(3) To find (AB) the chord, when (CF) the diameter and (CD) the height of the arc are given.

Multiply together the two parts (CD and DF) into which the diameter (CF) is cut by the chord (AB); and the square root of this product gives (AD or BD) half the chord of the arc. The double of this

will be (AB) the chord of the arc.

(4) To find (AC) the chord of half the arc, when (CD) the height of the arc and (CF) the diameter are given.

Multiply together (CF) the diameter and (CD) the height of the arc; and the square root of the product is (AC) the chord of half the arc.

(5) To find (CD) the height of the arc, when (AC) the chord of half the arc, and (CF) the diameter, are given.

Divide the square of (AC) the chord of half the arc by (CF) the diameter of the circle; and the quotient is (CD) the height or versed sine of the arc.

Note 1.-The formula AD2=DFX CD can be thus proved: In the figure given above, because ADE is a right-angled triangle, we have (see Rule 2, Prob. I) :

A D2=AE2-DE2-EF2-DE2 (since EF AE, being radii of the circle)

=(EF+DE) (EF-DE)=FD(EC-DE), since EF=EC;

therefore A D2=FDXCD.

Note 2.-The formula Ac2=CDX CF can be proved in the following manner :—

Since ADC (see figure given) is a right-angled triangle, we have

AC2=AD2+CD2.

But in Note 1, it has been proved that AD2=FDX CD.
Substituting this, we have

A C2= (FD × CD) +CD2=CD(FD+CD)=CD×CF.

Note 3.-The arch of a bridge is frequently in the shape of the arc of a circle, in which case

CD is the rise, or versed sine, or height above the piers.

AB is the span or chord of the arch.

c, the middle point of

the arch, is the crown. AC is the distance of

B

D

the spring of the arch to the crown c, or the chord of half the arch.

oc is the radius with which the arch was drawn.

Example 1.-The chord of an arc is 30 inches, and its height is 10 inches; find the diameter of the circle.

Now, by Rule 1, (FD) part of the diameter wanting chord2-height.

=302÷10=900÷10=90 inches.

Then diameter = (FD+CD) = 90+10=100 in. = 8 ft. 4 in.

and

Example 2.-The chord of half an arc is 12 ft. 8 in., the height of the chord is 1 ft. 7 in.; find the diameter of the circle.

Now, 12 ft. 8 in.=152 in.; and 1 ft. 7 in.=19 in.

Then, by Rule 2, diameter = square of chord of half the arc divided by height

=1522-19=-23104-19-1216 in.=101 ft. 4 in.

Example 3.-The height of an arc is 16 ft., and the diameter of the circle is 65 ft.; find the chord of the arc. The two sections (CD, DF) into which the diameter is divided are 16 and (65–16), or 16 and 49 ft.

Then by Rule 3, AD, half the chord of the arc

= √/CDXDF=√16 × 49=√784=28 ft.

Therefore AB, the chord of the arc,=2 × 28=56 ft.

Example 4.-The height of an arc is 36 ft., and the diameter of the circle is 196 ft.; find the chord of half the arc. By Rule 4: AC, the chord of half the arc, = square root of the product of the diameter and height.

= √196 × 36=√7056=84 ft.

Example 5.-The chord of half an arc is 36 ft., and the diameter of the circle is 108 ft.; find the height of the arc.

By Rule 5: The height of the arc=chord of half the arc2 diameter

=362 108=1296÷108=12 ft.

(1) The chord of half an arc is 25 ft., and the height of the arc is 15 ft.; find the diameter of the circle.

(2) The chord of half an arc is 17 ft., and the height of the arc is 7 ft.; find the diameter of the circle.

(3) The chord of half an arc is 17 ft. 9 in., and the height of the arc is 6 ft. 3 in.; find the diameter of the circle.

(4) The chord of half an arc is 25.5 ft., and the versed sine of the arc is 5.1 ft.; find the diameter of the circle. (5) The chord of half an arc is 15 ft., and the diameter of the circle is 40 ft.; find the height of the arc.

(6) The chord of half an arc is 12 ft., and the diameter of the circle is 36 ft.; find the height of the arc.

(7) The chord of half an arc is 28 ft., and the diameter of the circle is 56 ft.; find the height of the arc.

(8) The chord of half an arc is 60 ft., and the diameter of the circle is 125 ft.; find the height of the arc.

(9) The height of an arc is 9 in., and the diameter of the circle is 25 in.; find the chord of half the arc.

(10) The height of an arc is 2 ft. 1 in., and the diameter of the circle is 14 ft. 1 in.; find the chord of half the arc. (11) The height of an arc is 49 ft., and the diameter of the circle is 12.1 ft.; find the chord of half the arc.

(12) The chord of half an arc is 17.5 ft., and the diameter of the circle is 175 ft.; find the height of the arc.