School Geometry Iii Iv Hall 2e6j2o

=

to inscribe a polygon of

draw an angle AOB

/'

_/

/

And if gruent isosceles triangles. the polygon has n sides, each of the .L'AOB, BOC,

circle.

^^—^^^ ^^D

Let AB, BC, CD, ... be consecutive sides of a regular polygon inscribed in

n

sides in a given circle,

at the centre equal to

This gives

.

n

the length of a side AB ; and chords equal to AB may now be The resulting figure will set oflf round the circumference. clearly be equilateral and equiangular. (ii) To circumscribe a polygon of n sides about the circle, the points A, B, C, D, ... must be determined as before, and tangents drawn to the circle at these points. The resulting figure may readily be proved equilateral and equiangular.

This method gives a

Note. the angle

360"

n

strict

geometrical construction only

when

can be drawn with ruler and comes.

EXERCISES. 1. (i)

Give

strict constructions for inscribing in

a regular hexagon 2.

About a

;

(ii)

a regular octagon

;

a circle (radius 4 cm.) a regular dodecagon.

(iii)

circle of radius 1 '5" circumscribe

a regular hexagon ; (ii) a regular octagon. Test the constructions by measurement, and justify them by proof. (i)

3. An equilateral triangle and a regular hexagon are inscribed in a given circle, and a and b denote the lengths of their sides prove that :

(i)

4.

area of triangle = i (area of hexagon)

By means

(ii)

a^=3b^.

of your protractor inscribe a regular heptagon in a Calculate and measure one of its angles ; and the length of a side.

circle of radius 2".

measure

;

.

problems on circles and polygons.

Problem To draw a

circle (i)

in

(ii)

31.

about a regular polygan.

Let AB, BC, CD, DE, ... be consecutive sides of a reeular polyeron of n sides. Bisect the ^'ABC, CO meeting at O.

Then O

BCD by

Outline of Proof.

conclude that All the hisectoi's of

(ii)

circle.

/<^^^^^^~^^^^vXv

OD

^\\

//

\\

//

N ^^/^^ \\_ V] \Sv / xy// ^\>^ ^-^^Kl/

a(

jj^D

f

I

OD; and from

A'OCB, OCD, shew that

(i)

BO,

the centre both of the

is

and circumscribed

inscribed

201

the congruent

Hence we

bisects the Z.CDE.

the angles of thejpolygon

meet at O.

Prove that OB = OC = OD=...; from Theorem Hence O is the circum-centre.

Draw

OP, OQ, OR,

Prove that

...

perp. to AB, BC, CD,

OP = OQ = OR=

*

Hence O

'

is

...

;

6.

...

from the congruent A'OBP,

the in-centre.

EXERCISES. a regular hexagon on a side of 2*0". Draw the inscribed and circumscribed circles. Calculate and measure their diameters to the nearest hundredth of an inch. 1.

Draw

2. Shew that the area of a regular hexagon inscribed in a circle three-fourths of that of the circumscribed hexagon.

is

Find the area of a hexagon inscribed in a circle of radius 10 cm. to the nearest tenth of a sq. cm.

ABC

is an isosceles triangle inscribed in a circle, 3. If of the angles B and double of the angle A ; shew that a regular pentagon inscribed in the circle.

C

4.

On

having each is a side of

BC

a side of 4 cm. construct (without protractor)

(i) a regular hexagon ; (ii) a regular octagon. In each case find the approximate area of the figure.

' ;

GEOMETRY.

THE CIRCUMFERENCE OF A CIRCLE.

By

experiment and measurement it is found that the length roughly 3| times the length of diameter that is to say

of the circumference of a circle is its

:

circumference

_

diameter

and

can be proved that this

it

is

^^

^

,

^

the same for

all circles.

A more

correct value of this ratio is found by theory to be 3-1416 ; while correct to 7 places of decimals it is 3-14:15926o Thus the value 31 (or 3* 14^5) is too great, and correct to 2 places only.

The

ratio

diameter

is

which the circumference of any circle bears to denoted by the Greek letter tt ; so that circumference

Or,

if

its

= diameter x tt.

r denotes the radius of the circle, circumference

where to

= 2r x tt = 27rr

we

are to give one of the values 3|, 3*1416, oi 3-1415926, according to the degree of accuracy required in the final result. tt

Note. The theoretical metliods by which ir is evaluated to any required degree of accuracy cannot be explained at this stage, but its value may be easily verified by experiment to two decimal places.

For example round a cylinder ends overlap. At any point in through both folds. Unwrap and the distance between the pin holes

wrap a

strip of paper so that the the overlapping area prick a pin straighten the strip, then measure this gives the length of the circumMeasure the diameter, and divide the first result by the second. :

:

ference.

Ex. find of TT.

1.

From

CiRCUHFERENCR.

these data

and record the value

Find the three results.

mean

of

the

Ex. 2. A fine thread is wound evenly round a cylinder, and it is found that the length required for 20 complete turns is 75*4". The diameter of the cylinder is 1 -2" find roughly the value of t. :

A

bicycle wheel, 28" in diameter, makes 400 revolutions in travelling over 977 yards. From this result estimate the value of ir.

Ex.

3.

;

;

CIRCUMFERENCE AND AREA OF A CIRCLE.

203

THE AREA OF A CIRCLE. Let AB be a side of a polygon of n sides circumscribed about a circle whose centre is O and radius r. Then we have

Area of polygon

= 71. AAOB = 7^. jABxOD = J nAB xr = J (perimeter of polygon) x r .

and

this is true

Now

however many

number

sides the

polygon

may

have.

the perimeter and area of the polygon may be made to differ from the circumference and area of the circle by quantities smaller than any that can be named hence ultimately if

the

of sides is increased

without

limit,

;

Area

of

circle

=J =1

.

circumference x r

.

27rr

Xr

AI.TERNATIVE METHOD.

Suppose the circle divided into any even number of sectors having equal central angles denote the number of sectors bj"^ 7i. Let the sectors be placed side by side as represented in the diagram ; the area of the circle = the area of the fig. ABCD ; then :

and

however great n may be. number of sectors is increased, each arc is decreased the outlines AB, CD tend to become straight, and the angles at D and B tend to become rt. angles.

this is true

Now so that

as the (i)

(ii)

;

;

;

GEOMETRY.

204

Thus when n is increased without limit, the fig. ABCD ultimately becomes a rectangle, whose length is the semi-circumference of the circle, and whose breadth is its radius. :. Area of circle = ^ circumference x radius .

= J.27rr xr=7rr2. THE AREA OF A SECTOR.

If

make an angle of 1°, they cut off an arc whose length = ^^ of the circumference (ii) a sector whose area = ^|^^ of the circle .'.if the angle AOB contains D degrees, then

two

radii of a circle

(i)

and

(i)

(ii)

the arc

AS

the sector

= -^^

AOB = k^t:

= ^^

=J

.

of the circwmference of the area of the

circle

of (J circumference x radius)

arc

AB X radius.

THE AREA OF A SEGMENT.

The area of a minor segment is found by subtracting from the corresponding sector the area of the triangle formed by the chord and the radii. Thus Area of segment ABC = sector OACB -

triangle

AOB.

area of a major segment is most simply found by subtracting the area of the corresponding minor segment from the area of the circle.

The

;

CIRCUMFERENCE AND AREA OF A CIRCLE.

205

EXERCISES. [In each case choose the value of degree of accuracy.

v

so as to give

a

result

of the assigned

'\

1.

whose

Find to the nearest millimetre the circumferences of the circles (ii) 100 cm. radii are (i) 4'5 cm.

2.

circles

Find to the nearest hundredth of a square inch the areas of the whose radii are (i) 2*3". (ii) 10*6". Find to two places of decimals the circumference and area a square whose side is 3*6 cm.

3.

of a

circle inscribed in

4. In a circle of radius 7*0 cm. a square is described find to the nearest square centimetre the difference between the areas of the circle and the square. :

Find to the nearest hundredth of a square inch the area of the 5. circular ring formed by two concentric circles whose radii are 5 '7" and 4-3".

Shew that the area of a ring lying between the circumferences of 6. circles is equal to the area of a circle whose radius is the length of a tangent to the inner circle from any point on the outer.

two concentric

A rectangle whose sides are 8*0 cm.

and 6*0 cm. is inscribed in a Calculate to the nearest tenth of a square centimetre the total area of the four segments outside the rectangle. 7.

circle.

to the nearest tenth of an inch the side of a square whose equal to that of a circle of radius 5".

Find

8.

area

is

9.

A circular ring is formed by the circumference of

taking

two concentric

of the ring is 22 square inches, and its width is 1 '0" as ^^, find approximately the radii of the two circles.

The area

circles.

t

Find to the nearest hundredth of a square inch the difierence between the areas of the circumscribed and inscribed circles of an equilateral triangle each of whose sides is 4", 10.

11. Draw on squared paper two circles whose centres are at the points (1'5", 0) and (0, "8"), and whose radii are respectively '7' and 1 "O". Prove that the circles touch one another, and find approximately their circumferences and areas.

Draw a

circle of radius I'O" having the point (TG", 1*2") as Also draw two circles with the origin as centre and of radii 1*0^' and 3'0" respectively. Shew that each of the last two circles touches the first. 12.

centre.

GEOMETRY.

206

EXERCISES.

On the

Inscribed, Circumscribed,

and Escribed Circles of a

Triangle. (Theoretical.)

Describe a circle to touch two parallel straight lines and a third straight line which meets them. Shew that two such circles can be drawn, and that they are equal. 1.

Triangles which have equal bases and eqtud vertical angles have 2. tqual circumscribed angles.

ABC is a triangle, and 3. circumscribed circles ; if A, I,

I,

S

S are the centres of the inscribed are collinear, shew that AB = AC.

and

The sum

of the diameters of the inscribed and circumscribed a right-angled triangle is equal to the sum of the sides containing the right angle. 4.

circles of

If the circle inscribed in the triangle ABC touches the sides D, E, F ; shew that the angles of the triangle DEF are respectively

5.

-at

90-4, 9»-|' 90-16.

,and I, B,

is the centre of the circle inscribed in the triangle ABC, the centre of the escribed circle which touches BC shew that Ij, C are concyclic.

If

I

Ij

;

In any triangle the difference of two sides is equal to the 7. difference of the segments into which the third side is divided at the point of of the inscribed circle. In the triangle ABC, and S are the centres of the inscribed and 8. circumscribed circles shew that IS subtends at A an angle equal to half the difference of the angles at the base of the triangle. I

:

Hence shew that if AD is bisector of the angle DAS.

drawn perpendicular

to

BC, then

Al

is

the

The diagonals of a quadrilateral ABCD intersect at O shew 9. that the centres of tlie circles circumscribed about the four triangles AOB, BOC, COD, DOA are at the angular points of a parallelogram. :

In any triangle ABC,

10.

and

O

is

if

I

is

the centre of the inscribed circle,

produced to meet the circumscribed circle at O shew that the centre of the circle circumscribed about the triangle BIC. Al

if

11. circle

;

is

;

Given the base, altitude, and the radius of the oiroumscribed construct the triangle.

12. Three circles whose centres are A, B, C touch one another shew that the inscribed circle ol externally two by two at D, E, F the triangle ABC is the circumscribed circle of the triangle DEF. :

207

THE ORTHOCENTRE OF A TRIANGLE.

THEOREMS AND EXAMPLES ON CIECLES AND TRIANGLES. THE ORTHOCENTRE OF A TRIANGLE. The perpendiculars drawn from

I.

the vertices

of a triangle

to the

opposite sides are concurrent.

the A ABC, let AD, BE be the drawn from A and B to the opposite sides and let them intersect at O. CO and produce it to meet AB at

In

perp"

;

;

F. It IS required to to

shew that

CF

is

perp.

AB. DE.

Then, because the

^"^

OEC,

ODC

are

rt.

angles,

the points O, E, C,

.'.

the

.'.

Z-

DEC = the A DOC, = the

Again, because the the

.'.

/.

vert. Z."

are concyclic

:

same segment; opp. L FOA.

AEB,

the points A, E, D,

.•.

D

Z.DEB = the LDAB,

in the

ADB B

are

rt.

angles,

are concyclic

in the

:

same segment.

FAO = the sum of the L" DEC, DEB = a rt. angle Theor. the remaining Z.AFO = art. angle: that is, CF is perp. to AB.

the sum of the L" FOA,

:

.•.

Hence the three perp» AD, BE, CF meet

at the point

16.

O. Q.E.D.

Definitions. (i)

The

vertices

intersection of the perpendiculars drawn from the a triangle to the opposite sides is called its

of

ortliocentre. (ii)

The

diculars

is

triangle

formed by ing the feet

of the perpen-

called the pedal or orthocentric triangle.

GEOMETRY.

208

II. In an acute-angled triangle the perpendictUars draxon from the vertices to the opposite aides bisect the angles of the pedal triangle through

which they .

In the acute-angled A ABC, let AD, BE, be the perp» drawn from the vertices to the opposite sides, meeting at the ortho-

CF

centre

O

;

and

let

DEF be the pedal' triangle.

It is required to "prove that

AD, BE, CF

bisect respectively

FDE, DEF, EFD.

the L*

It may be shewn, as in the last theorem, that the points O, D, C, E are concyclic ; :.

the

Z.

ODE = the

Z.OCE,

Similarly the points O, D, B, .'.

F are

in the

same segment.

concyclic

Z.ODF=the Z.OBF,

the

°

in the

;

same segment.

But the Z.OCE = the Z.OBF, each being the comp' .-.

Similarly

BE

it

may

the

Z.

of the

Z.BAO.

ODE = the ÔDF.

be shewn that the L*

DEF, EFD

by

are bisected

and CF.

q.e.d.

Corollary, (i) Every tioo sides of the pedal triangle are equally inclined to that side of the original triangle in which they meet. For the

ODE

L EDC = the comp* of

the

Z.

= the comp* of = the Z.BAC.

the

Z.OCE

Similarly

it

may be shewn the

.-.

Z.

that the

Z.FDB = the Z.BAC,

EDO = the Z.FDB = the

Z.A.

may be proved that Z.DEC = the Z.FEA = the LB, and the Z.DFB = the Z.EFA = the Z.C.

In like manner

it

the

Corollary.' one another

and

(ii)

The

DEC, AEF, ABC.

triangles

to the triangle

DBF

are equiangular to

Note. If the angle BAC is obtuse, then the perpendiculars bisect externally the corresponding angles of the peaal triangle.

BE,

CF

THE ORTHOOENTRE OF A TRIANGLE.

209

EXERCISES.

I/O

1.

AD

is

ABC

orthocentre of the triangle and if the perpendicular = DG. to meet the circum-cirde in G, prove that

is the

produced

0D

2. In an acute-angled triangle the three aides are the external bisectors of the angles of the pedal triangle : and in an obtuse-angled triangle the sides containing the obtuse angle are the internal bisectors of the corresponding angles of the pedal triangle. 3.

I/O

BOC, BAC 4.

I/O

is the

orthocentre o/ the triangle

ABC, shew

that the angles

are supplementary. is the

orthocentre o/ the triangle ABC, then any one o/ the C is the orthocentre o/ the triangle whose vertices are

four points O, A, B, the other three.

The three circles which through two vertices o/ a triangle orthocentre are each equal to the circum-circle o/ the triangle.

5. its

and

D, E are taken on the circumference of a semi-circle described 6. on a given straight line AB the chords AD, BE and AE, BD intersect (produced if necessary) at F and G shew that FG is perpendicular :

:

toAB. 7.

ABC

is

circum-circle

:

a triangle, O is its orthocentre, and AK a diameter of the shew that BOCK is a parallelogram.

The orthocentre of a triangle is ed to the middle point of the and the ing line is produced to meet the circum-circle prove that it will meet it at the same point as the diameter which es 8.

base,

:

through the vertex,

The perpendicular from the vertex of a triangle on the base, and 9. the straight line ing the orthocentre to the middle point of the base, are produced to meet the circum-circle at P and shew that PQ is parallel to the base.

Q

:

The distance o/ each vertex o/ a triangle /ram the orthocentre is 10. double o/ the perpendicular draum /rom the centre of the circum-circle to the opposite side. 11.

Three

circles are described

each ing through the orthocentre

of a triangle and two of its vertices : shew that the triangle formed ing their centres is equal in all respects to the original triangle. 12.

by

Construct a triangle, having given a vertex, the orthocentre,

and the centre of the

circum-circle.

;

:

;

GEOMETRY.

210

LOCI. Given the base and vertical angle of a triangUy find the locus oj

III. its

orthocentre.

Let BC be the given base, and X the given angle ; and let BAG be any triangle on the base BC, having its vertical Z.A equal to the L X. Draw the perp» BE, CF, intersecting at the orthocentre O. It is required to find the locus

Since the

Proof.

Z."

OFA,

of O.

OEA are rt.

fuigles, /.

E

the points O, F, A, .•.

the

Z.

the vert. opp.

.*.

But the

Z.

A

is

are concyclie

FOE L

;

the supplement of the Z. A BOC is the supplement of the is

:

constant, being always equal to the .•.

its

supplement

is

LX

Z.

A.

;

constant

A

that is, the BOC has a fixed base, and constant vertical angle ; hence the locus of its vertex O is the arc of a segment of which BC the chord.

Given the base and vertical angle of a triangle, find the locus qf

IV.

the in-centre.

Let BAC be any triangle on the given base BC, having its vertical angle equal to the given Z.X; and let Al, Bl, CI be the Then is the inbisectors of its angles. I

eentre. It is required to find the locus

Proof. by A, B,

by

is

of

I.

Denote the angles of the A ABC C ; and let the L BIC be denoted

I.

Then from the (i)

I

and from the (ii)

A BIC,

so that .'. ,

+ iB + iC = two

A + B + C = two rt. JA + ^B + ^C = one rt.

angles angles

;

angle.

taking the differences of the equals in - iA = one rt. angle = one rt. angle + Â.

(i)

and

(ii),

:

I

or,

rt.

A ABC,

I

constant, being always equal to the L X ; .'. is constant the loous of is the arc of a segment on the fixed chord

But A

is

I

.*.

I

80

:

EXERCISES ON LOCI.

211

EXERCISES ON LOCI. 1. Given the base BC and the vertical angle the locus of the ex-centre opposite A.

A

of

a triangle

;

find

Through the extremities of a given straight line AB any two 2. are drawn ; find the locus of the interparallel straight lines AP, section of the bisectors of the angles PAB, QBA.

BQ

Find the locus of the middle points of chords of a circle drawn 3. through a fixed point. Distinguish between the cases when the given point is within, on, or without the circumference. Find the locus of the points of of tangents 4. a fixed point to a system of concentric circles.

drawn from

5. Find the locus of the intersection of straight lines which through two fixed points on a circle and intercept on its circumference an arc of constant length.

6.

A and B

and PQ and QB. 7.

is

BAG

are two fixed points on the circumference of a circle, any diameter: find the locus of the intersection of PA

is

any triangle described on the fixed base BC and having and BA is produced to P, so that BP is

a constant vertical angle equal to the

sum

;

of thd sides containing the vertical angle

:

find the

locus of P. 8. AB is a fixed chord of a circle, and AC is a moveable chord ing through A if the parallelogram CB is completed, find the locus of the intersection of its diagonals. :

A

9. straight rod PQ slides between two rulers placed at right angles to one another, and from its extremities PX, are drawn perpendicular to the rulers find the locus of X.

QX

:

Two circles intersect at A and B, and through P, any point on 10. the circumference of one of them, two straight lines PA, PB are drawn, and produced if necessary, to cut the other circle at X and Y find the locus of the intersection of AY and BX. 11. Two circles intersect at A and B HAK is a fixed straight line drawn through A and terminated by the circumferences, and PAQ is any other straight line similarly drawn find the locus of the intersection of HP and QK. ;

:

GEOMETRY.

212

simson's line. 2%€ feet of the perpendicvlara drawn to the three triangle from any point on its circum-circle are collinear. V.

sides

<^ o

Let P be any point on the circuni -circle of the A ABC and let PD, PE, PR be the perps. drawn from P to the sides. ;

It is required to prove that the points D, E, are collinear.

FE and ED

FE and ED

then

same straight

will be

F

:

shewn

to be in the

line.

PA, PC.

Because the

Proof.

.*.

.'.

the

Z.«

PEA, PFA are

^PEF = the

since the points A, P, C,

Again because the .-.

L?

the

:

B

L PAB

are concyclic.

PEC, PDC

C

the points P, E, D,

are

rt.

angles,

are concyclic.

PED = the supp* of the Z. PCD = the supp* of the L PEF. FE and ED are in one st. line.

the .•.

Ohs. triangle

angles,

Z-PAF, in the same segment

= the suppt of = the Z.PCD,

.•,

rt.

the points P, E, A, F are concyclic

Z-

FED

known

The

line

ABC

for the point P.

is

as the Pedal or Slmson's Line of the

EXERCISES.

From any

P on

the circum-circle of the triangle ABC, perpendiculars PD, PF are drawn to BC and AB if FD, or FD produced, cuts AC at E, shew that PE is perpendicular to AC. 1.

point

:

Find the locus of a point which moves so that

2.

are

drawn from

it

if

perpendiculars

to the sides of a given triangle, their feet are

collinear. 3.

ABC

and AB'C are two triangles with a common angle, and meet again at P shew that the feet of perpenP to the lines AB, AC, BC, B'C are collinear.

their oircum-circles diculars drawn from

A

;

triangle is inscribed in a circle, and any point P on the circum4. ference is ed to the orthocentre of the triangle: shew that thii ing line is bisected by the pedal of the point P.

THE TRIANGLE AND

ITS CIRCLES.

THE TRIANGLE AND

ITS CIRCLES.

213

D, E, F are the, points of of the inscribed circle of the triangle ABC, and D^, Ej, Fj the points of of the escribed circle, which touches BC and the other sides produced : a, b, o denote the length qf the sides BC, CA, AB ; s the semi-perimeter of the triangle, and r, Tj me radii of the inscribed and escribed circles.

VI.

Prove thefdlomng

equalities :

(i)

AE =AF =8 -a,

BD=BF=8-6, CD =CE =s-c (ii)

(iii)

(iv)

(V) (vi)

AEi=AFi=s. CDi=:CEi=5-&, BDi = BFi=s-c.

CD =BDi, and BD=CD,. EE, = FF, = a. The area

of the

A ABC=rs =r^{8-a).

(vii)

Draw

prove that

the above figure in the case

r=s-c;

r,

when C

= 8-b.

is

a right angle, and

214

GEOMETRY.

VII. |i

>

'a*

's

BC, CA,

Prove (i)

(ii)

In the triangle ABC, is the centre of the inscribed circle, and ^he centres of the escribed circles touching resvectively the sides and the other sides produced. I

AB

the following properties

The points A,

I,

The points

A,

(iii)

The

(iv)

77ie

1.^,

:

\ are collinear 1^

so are B,

:

are collinear

;

so are

I3,

I,

Ig ;

B,

Ij

aiid C, ;

and

I,

Ij,

Is.

C,

Ig.

triangles BIjC, CI2A, AI^B are equiangular to one another.

triangle

Ijljls

is

equiangular

ing the points of of the inscribed

to

the triangle

formed by

circle.

(v) Of the four points I, Ij, Ig, I3, ea^h is the orthocentre of the triangle whose vertices are the other three. (vi)

points

I,

The four Ij,

I2, I3,

circles, each of which es through three of the are all equal.

THE TRIANGLE AND

215

ITS CIRCLES.

EXERCISES.

With

the figure given on page 214 shew that if the circles whose centres are I, Ij, Ig, I3 touch BC at D, Dj, Dg, D3, then 1.

(i)

(iii)

Shew

2.

DD2=DiD3=&.D2D3=6 + c.

that the orthocentre

of the inscribed and escribed

(ii)

(iv)

DD3=DiD2=c. DDi = 6-'C.

and veHicea of a

circles

triangle are the centres

of the pedal triangle.

Given the base and vertical angle of a triangle, find the locus of the 3. centre of the escribed circle which touches the base. Given the base and vertical angle of a 4. of the circum-circle is fxed.

t7'iangle,

shew that the centre

Given the base BC, and the vertical angle A of the triangle, find 5. the locus of the centre of the escribed circle which touches AC. 6. Given the base, the vertical angle, and the point of with the base of the in-circle ; construct the triangle.

Given the base, the vertical angle, and the point of with' 7. the base, or base produced, of an escribed circle ; construct the triangle. 8. I is the centre of the circle inscribed in centres of the escribed circles; shew that llj, circumference of the circum-circle.

a

and Ij, Ig, I3 the are bisected by the

triangle,

llg,

II3

ABC is a triangle, and Ig, I3 the centres of the escribed circles 9. which touch AC, and AB respectively shew that the points B, C, Ig, l| lie upon a circle whope centre is on the circumference of the circum:

circle of the triangle

ABC.

With three given points as centres describe three circles touch10. ing one another two by two. How many solutions will there be ? Given the centres of the

11.

threfe escribed circles;

construct the

triangle.

Given the centre of the inscribed circle, and the centres of 12. escribed circles ; construct the triangle. 13.

circle

;

two

Given the vertical angle, perimeter, and radius of the inscribed construct the triangle.

Given the vertical angle, the radius of the inscribed 14. the length of the perpendicular from the vertex to the base the triangle.

circle, ;

and

construct

In a triangle ABC, is the centre of the inscribed circle ; shew 15. that the centres of the circles circumscribed about the triangles BIC, CIA, AIB lie on the circumference of the circle circumscribed about the given triangle. I

;

GEOMETRY.

216

THE NINE-POINTS CIRCLE. VIII. In any triangle the middle points of the sides, the feet of the perpendiculars from the vertices to the opposite sides, and the middle points of the lines ing the orthocentre to the vertices are concyclic.

In the A ABC, let X, Y, Z be the middle points of the sides BC, CA,

AB;

let D, E, F be the feet of the perp' to these sides from A, B, C ; let be the orthocentre, and a, p, y the middle points ot OA, OB, OC.

O

It is required to

prove that

the nine points X, Y, Z, D, E, F, a, are concyclic.

/S,

y

XY, XZ, Xa, Ya, Za. since

from the A ABO, AZ = ZB, and Aa = aO,

Za

par» to

Now .-.

is

BO.

And from

the

Ex.

A ABC, ZX

.'.

But

BO

.•.

is,

a

and Xa

is

that

the

BX = XC,

and

L XZa

is

a

rt. rt.

AC

angle with

;

angle.

Similarly, the L XYa is a rt. angle. the points X, Z, a, Y are concyclic

C

:

of the circle which es through X, Y, Z on the a diameter of this circle. lies

Similarly

it

may

be shewn that

Again, since .*.

Similarly .'.

since BZ = ZA, par* to AC.

is

produced makes a .•.

DC

X

2, p. 64.

the circle on

Xa

and y

lie

on the

O" of this oirole.

a rt. angle, as diameter es through D. is

be shewn that E and F the points X, Y, Z, D, E, F, a, it

may

/S

aDX

lie /3,

on the

7 are

O"

of this oirole Q.B.D. concyclic.

;

From this property the circle which es through the middl« Obs. points of the sides of a triangle is called the Nine-Points Circle ; many of its properties may be deriv^ from the fact of its being the oiroum« oirole of the pedal triangle.

THE NINE-POINTS CIRCLE. To prove

217

that

of the nine-points circle is the middle point of the straight line which s the orthocentre to the circum-centre. (i)

the centre

(ii) the radius of the nine-points circle is half the radius of the circum-circle. (iii)

and

centre,

the centroid is collinear with the circum-centre, the nine-points the orthocentre.

In the A ABC, let X, Y, Z be the D, E, F middle points of the sides ;

the feet of the perp* O the orthocentre ; S and N the centres of the circumscribed and nine-points circles ;

respectively.

To prove that N is the middle SO. It may be shewn that the pern, to XD from its middle point bisects SO (i)

point of

;

Theor. 22. Similarly the perp. to EY at its middle point bisects SO that is, these perp» intersect at the middle point of :

.*.

SO

:

XD

since and EY are chords of the nine-points circle, and EY at rt. angles the intersection of the lines which bisect Theor. 31, Cor. its centre : .•. q.e.d. the centre N is the middle point of SO.

And

XD

is

1.

To prove that the radius of the nine-points circle is half the (ii) radium of the circum-circle. By the last Proposition, Xa is a diameter of the nine-points circle. .*. the middle point of Xa is its centre but the middle point of SO is also the centre of the nine-points circle. {Proved.) Hence Xo and SO bisect one another at N. :

Then from the A» SNX, ONa,

SN = ON,

C

because

-(

and

NX = Na,

land the Z.SNX = the Z.ONa; .-.

SX = Oa =Ao.

And SX .-.

is

also par^ to Aa,

SA = Xa.

But SA is a radius of the circum-circle ; and Xo is a diameter of the nine-points circle .*.

;

the radius of the nine-points circle is half the radius of the ciroum< circle. fSee also p. 267, Examples 2 and 3.] q.e.d.

218

GEOMETRY. (iii)

To prove

that the centroid is collinear with points S,

N, O.

AX and draw ag par' to SO. Let AX meet SO at G.

Then from the A AGO, since Aa=oO, and ag is par' to OG, kg=gQ. Ex. 1, p. 64. .-.

And from

the

and

A Xap,

NG

is

since

aN = NX,

par' to ag,

gQ^GX. AG = §of AX;

:. .-.

.*.

G is the centroid of the triangle ABC. Theor. III., Cor., p. 97.

That

is,

the centroid is collinear with the points S, N, O. q.e.d.

EXERCISES. Given the hose and vertical angle of a triangle, find the locus of the centre of the nine-points circle. 1.

2.

O,

is

The nine-points circle of any triangle ABC, whose orthocentre is also the nine-points circle of each of the triangles AOB, BOO.

COA. If I, Ij, 1.2, U are the centres of the inscribed and escribed circles 3. of a triangle ABu, then the circle circumscribed about ABC is the nine-points circle of each of the four triangles formed by ing three of the points I, l^, I2, I3. 4. All triangles which have the same orthocentre and the same circumscribed circle, have also the same nine-points circle.

Given the base and vertical angle of a triangle, shew that one 5. angle and one side of the pedal triangle are constant. Given the base and vertical angle of a triangle, find the locus of 6. the centre of the circle which es through the three escribed centres.

Note.

For some other important properties of the Nine-pointa page 310.

Circle see Ex. 54,

;

PART

IV,

ON SQUAKES AND EECTANGLES IN CONNECTION WITH THE SEGMENTS OF A STRAIGHT LINE. THE GEOMETRICAL EQUIVALENTS OF CERTAIN ALGEBRAICAL FORMULA Definitions.

A

rectangle ABCD is said to be i. contained by two adjacent sides AB, AD for these sides fix its size

A

and shape.

rectangle whose adjacent sides are AB, AD is denoted AB, AD ; this is equivalent to the product AB AD.

the red.

by

.

Similarly a square on AB, or AB^.

drawn on the

side

AB

is

denoted by

the sq.

2. If a point X is taken in a straight line AB, or in AB produced, then X is said to divide AB into the

two segments AX, XB

;

being in either case

the distances of

the dividing point

X from

a

x ~

Fiff- i-

the segments the extremities

w,

B

^

1,

In Fig.

2,

AB AB

is

X

^

^^S-

of the given line AB.

In Fig.

B

'

2.

said to be divided internally at X.

divided externally at X.

Obs. In internal division the given line the segments AX, XB.

In external division the given segments AX, XB.

line

AB

is

AB

is

the sum of

the different of the

;

;

GEOMETRY.

220

Theorem If of two

50.

[Euclid

11. 1.]

straight lines, one is divided into

any number of parts,

the rectangle contained by the two lines is equal to the

rectangles contained by the v/ndivided line

and

sum of

the

the several parts of

the divided line.

be

a k

k

k

Let AB and K be the two given st. lines, and let AB be divided into any number of parts AX, XY, YB, which contain respectively a, b, and c units of length; so that AB contains a + b + c units. Let the line K contain k units of length. It

is

the

required rect.

to

prove that

AB, K = rect. AX, K

+ rect.

XY, K

+ rect.

YB, K;

namely that

+ b + c)k=

{a

ak

+

+

bk

ek.

Draw AD perp. to AB and equal to K. Through D draw DC par^ to AB. Through X, Y, B draw XE, YF, BC par^ to AD.

Construction.

The fig. AC = the fig. Proof. of these, by construction,

AE + the

fig.

XF + the

fig.

YC;

and

fig.

AC = rect. AB, K {fig. fig. fig.

AE = rect. XF = rect. YC = rect.

;

and contains

(a

+ b + c)k

units of area

AX, K ; and contains ak units of area bk XY, K ; YB, K; ck

;

Hence the rect. AB, K or,

(o

= rect.

+ 6 + c)^=

AX, K

ak

+ rect. XY, K + rect. YB, K; ck. bk + + Q.E.D.

::

SQUARES AND RECTANGLES.

Corollaries.

[Euclid

II.

2

221

and

3.]

Two special cases of this Theorem deserve attention. (i) When AB is divided only at one point X, and when undivided line

AD

X

A

Then the That

sq.

the

equal to AB.

is

D on AB = the

B

EC rect. AB,

AX + the

rect.

AB, XB.

is,

The square on

the given

contained by the whole lin^

lirie

and

is

sum of the rectangles of the segments.

equal to the

eojch

Or thus AB2 = AB.AB

= AB(AX + XB) = AB.AX + AB.XB. (ii)

When AB

divided at one point X, and equal to one segment AX.

is

undivided line AD

is

That

rect.

AB,

AX = the

the

EC

D

Then the

when

sq.

on AX + the

rect.

AX, XB.

is,

The rectangle contained by the whole to tJie square on that segment with

equal

line

and one segment

is

the rectangle contained hp

the two segments.

Or thus

AB.AX = (AX + XB)AX = AX2 + AX.XB.

GEOMETRY.

222

Theorem If a

51.

[Euclid

II. 4.]

straight line is divided internally at

any

point, the square

on the given line is equal to the sum of the squares on the two segments together -with twice the rectangle contained by the segments.

(aV


Theorem If a

52.

straight line is divided

on

[Euclid

7.]

any point, the square sum of the squares on the two

externally at

the given line is equal to the segments diminislied by t%dce the

A-*-

II.

223

rectangle

a- B --^-X

containsd

by the

GEOMETRY.

224

Theorem The

rectangle contained by their

Let the given let

11.

5 and 6.]

on two straight

sum and

lines is equal to the

difference.

g-C

A<.

and

[Euclid

53.

difference of the squares

AC be placed

lines AB,

them contain a and

b units of

in the same st. length respectively.

line^

It is required to prove that

AB2 - AC2 = (AB + AC) (AB - AC) namely that

a^

-

b^

;

={a + b){a-b).

On AB and AC draw the squares ABDE, and produce CF to meet ED at H.

Construction.

ACFG

;

Then GE = CB = a-& Proof.

Now

AB2 - AC^ = the

sq.

units.

AD - the

sq.

AF

= the rect. CD + the rect. QH = DB.BC +GF.GE = AB.CB +AC.CB = (AB + AC)CB = (AB + AC)(AB-AC). That

is,

a2

-

62

=(a + b)(a-b). Q.E.IX

;

:


225

Corollary. If a straight line is bisected, arid also divided (internally or externally) into two unequal segments, the rectangle contained by these segments is equal to the difference of the squares on half the line and on the line between the ts of

Y

X

A

That

Fig.

I.

AB

is

is, if

nally in Fig.

1,

bisected at

X and in Fig.

1,

in Fig.

2,

Y

B

Fig.

and externally

in Fig.

For in the

X

A

B

section.

2.

also divided at Y, inter2,

then

AY YB = AX2 - XY^ AY YB = XY^ - AX2. .

.

first case,

AY YB = (AX + XY) (XB - XY) .

= (AX + XY)(AX-XY) = AX2-XY2. The second

case

may

be similarly proved.

EXERCISES.

Draw diagrams on squared paper

1.

straight line

to

(i)

(ii)

four-times the square on half the line nine- times the square

;

on one-third of the

Draw diagrams on squared paper

2.

shew that the square on a

is

line.

to illustrate the following

algebraical formulae (i)

(ii) (iii)

(iv)

In Theor.

3.

(x + 7)2=a;2+14a; + 49.

(a + 6

+ c)2=a2 + 62 + c2 + 26c + 2ca + 2a6. + bd. (a; + 7){a; + 9) = a;2+16x + 63. {a + b){c-{-d)=ac-\-ad-{-bc

50, Cor.

find the area of the 4.

find

In Theor. AB. In Theor.

5.

= 24

sq.

6.

sq. in.

fig.

50, Cor.

AB = 4

cm., and the

fig.

AE = 9-6

sq. cm.,

(ii), if

51, if the fig.

AX = 2-1", and

AG = 36

sq.

the

fig.

XC = 3-36

sq. in.,

cm., and the rect. AX,

XB

cm., find AB.

In Theorem 52, ,

(i), if

XC.

find

if

the

fig.

AG = 9 -61

sq. in.

,

and the

[For further Examples on Theorems 50-53 see H.S.O.

fig.

DG = 6 '51

AB.

P

p. 230.]

GEOMETRY.

Theorem In an oUuse-an^led the

obtuse angle

is

II. 12.]

on

triangle, the square

the side svhtending

sum of tlie squares on the sides together with twice the rectangle con-

equal to the

containing the obtuse angle tained by one of those sides

upon

[Euclid

64.

and

the projection of

the other side

it.

Let ABC be a triangle obtuse-angled at C; and let AD be drawn perp. to BC produced, so that CD is the projection of the side CA on BC. [See Def. p. 63.] It

is

required to prove tlmt

AB2 = BC2 + CA2 + 2BC.CD. Because

Proof.

.-.

BD

is

BD2

= BC2 + CD2 + 2 BC

To each

^

the sfum of the lines BC, CD,

of these equals

.

add

CD. DA^.

\

Then BD2 + DA2 = BC2 + (CD2 + DA2) + 2 BC. But BD2+ DA2 = AB21 „ >, and CD2 + DA2 = CA2J' Hence

,

.,

for the

AB2 = BC2 + CA2

+ 2 BC

,

z.

.

r^ D

\

Theak^ 51.

.

IS

CD.

„

a

.

CD. ,

rt. z..

/ <}.]C.D.


Theorem In

55.

[Euclid

II.

22?

13.]

every triangle the square on the side subtending

an

acute angle

sum of the squares on the sides containing diminislied by twice the rectangle contained hy one of and the projection of the other side upon it. is

that angle

equal to the

those sides

Fig:. 2.

Let ABC be a triangle in whicb the z. C is acute be drawn perp. to BC, or BC produced ; so that projection of the side CA on BC.

;

AD

It

is

and

CD

is

let

the

required to prove that

--

^

AB2 = BC2 + CA2 - 2BC CD. .

Since in both figures

Proof.

BD

is

the

difference of

the lines

BC, CD, .-.

BD2 = BC2 + CD2-2BC.CD.

To each

Thear. 52.

of these equals

add

DA2.

Then BD2+ DA2 = BC2 + (CD2+

DA2)

-2BC CD

But BD2 + DA2 = ABn f , i, for the and CD2+DA2 = CA2j'

r^

4..

z.

(i)

.

•

D

IS

a

rt.

L.

Hence AB2 = BC2 + CA2 - 2BC CD. .

Q.E.D.

^Q^

'

^ ^^

iV

;

:

GEOMETRY.

228

Summary of Theorems

C

D (i)

If

D

the Z.ACB

B

the

Z- ACB

If the

Z.ACB

If

DC

B

C(D)

is obtuse,

= BC2 + CA2 + 2BC

AB2 (ii)

54 and 55.

29,

is

.

CD.

The&r.

AB2 = BC2 + CA2. (iii)

54

a right anghj The&r. 29.

is acute,

AB2

= BC2 + CA2 - 2BC

.

CD.

TJieoi'.

55.

Observe that in (ii), when the aACB is right, AD coincides with AC, so that CD (the projection of CA) vanishes hence, in this case,

Thus the enunciation

three

The square on a less

than the

may

results

side of

sum of

a

.

be

collected

in

a

single

triangle is greater than, equal

the squares

angle contained by those sides

2BC CD = 0.'

mi

to,

or

the other sides, according as the

is obtuse,

a right angle, or acute

;

the

difference in cases of inequality being twice the rectangle contained by one of the two sides and the projection on it of the other.

EXERCISES. In a triangle ABC, a = 21 cm., 6 = 17 cm., c = 10 cm. By how square centimetres does c^ fall short of a^ + b^^ Hence or otherwise calculate the projection of AC on BC. 1.

many 2.

ABC

is

an isosceles triangle in which

Shew

perpendicular to AC. 3.

In the

A ABC,

AB = AC and BE is drawn

that BC2 = 2AC

:

.

C£.

shew that

(i)

if

the

(ii)

if

the

LC = G0\ ^C = 120%

then c'^=a^ + b'^-a^;

then c^=a'^ + V^+db.

squares and rectangles.

Theorem

229

56.

In any triangle the sum of the squares on two sides is equal to twice the square on half the third side together with twice the square on

the

median which

Let ABC be a

bisects the third side.

triangle,

and AX the median which

bisects the

base BC. It is required to prove that

AB2 + AC2 = 2BX2

Draw AD

+ 2AX2.

BC and consider the case in which AB and AC are unequal, and AD falls within the triangle. Then of the z.*AXB, AXC, one is obtuse, and the other acute. Let the ÂXB be obtuse. perp. to

;

Then from the A AX B, AB2

And from

the

= BX2 + AX2 + 2BX XD. .

AC2 = XC2

Adding these

Thear. 54.

A AXC, results,

+ AX2 - 2XC

.

XD.

Thecyr. 55.

and ing that XC = BX,

we have AB2 + AC2 = 2BX2 + 2AX2. Q.E.D.

Note.

The proof may

perpendicular

AD

falls

easily be

adapted to the case in which the

outside the triangle.

EXERCISE. In any triangle the difference of the squares on two sides is equal to twice the rectangle contained by the base and the intercept between the middle point of the base and the foot of the perpendicular draion from the vertical angle to the base.

GEOMETRY.

230

EXERCISES ON THEOREMS 50-53. 1.

AB

is

Use the

CJorollaries of

Theorem 50

to

shew that

if

a straight line

divided internally at X, then

AB2=AX2 + XB2 + 2AX.XB. 2.

If

a straight line

AB

is

bisected at

X and

produced to Y, and

if

AY. YB = 8AX2, shew that AY = 2AB. The sum of the squares on two stinight lines is never less than 3. twice the rectangle contained by the straight lines. Explain this statement by reference to the diagram of Theorem 52. it from the formula (a - h)^ = a^-\-lfi- 2ah.

Also deduce 4.

In the formula

(a

+ &) (a

and enunciate verbally the

-

&)

= a^ - i^,

substitute a =

^ —^, b = —n^* DC

"4"

iK

7/

t/

resulting theorem.

If a straight line is divided internally at Y, shew that the 5. rectangle AY, YB continually diminishes as Y moves from X, the midpoint of AB.

Deduce

this

(i)

(ii)

6.

nally, case

from the Corollary of Theorem 53 from the formula a6= f

^

—

)

~ (

;

o

)

*

line AB is bisected at X, and also divided (i) interexternally into two unequal segments at Y, sheio that in either

If a straight (ii)

AY2 + YB2=2(AX2 + XY2).

[Proof of case

[Euclid

II. 9, 10.]

(i).

AY2 + YB2 = AB2 - 2AY YB = 4AX2 - 2 (AX + X Y) AX = 4AX2 - 2 AX2 - X Y2)

TJieor. 51.

.

(

(

X Y) Theor. 53.

=2AX2 + 2XY2. Case 7.

(ii)

If

may

AB

be derived from Theorem 52 in a similar way.]

is

divided internallj' at Y, use the result of the last in the value of AY^ + YB'^, as Y moves

example to trace the changes from A to B.

8. In a right-angled triangle, if a perpendicular is drawn from the right angle to the hypotenuse, the square on this perpendicular is equal to the rectangle contained by the segments of the hypotenuse. 9.

ABC

is

an

and AY is drawn to cut the base Prove that YC, for internal section ; YC, for external section.

isosceles triangle,

internally or externally at Y.

AY2 = AC2 - BY

AY2= AC2+ BY

.

.

80

EXERCISES.

EXERCISES ON THEOREMS 54-56.

AB is a straight line 8 cm. in length, and from its middle point 1. as centre with radius 5 cm. a circle is drawn ; if P is any point on shew that circumference, the

O

AP2+BP2=82sq. cm. 2.

BC is bisected at X. If a =17 cm., cm., calculate the length of the median AX, and

In a triangle ABC, the base

6 = 15 cm.,

and

c

=S

deduce the Z.A. 3.

of a triangle = 10 cm., and the sum of the squares sq. cm. ; find the locus of the vertex.

The base

on

theother sides = 122 is

\4) Prove that the sum of the squares on the sides of a parallelogram equal to the sum of the squares on its diagonals. of a

^_^_^^

rhombus and

^35ESoft^r diagonal

its

to within

shorter ndiagonat isach measure

-S-" ;

'Or'.

In any quadrilateral the squares on the diagonals are together twice the sum of the squares on the straight lines ing the middle points of opposite sides. [See Ex. 7, p. 64.] eqHgiJ to

l^

A BCD

is

O any point within it shew that OA^ + O02=OB2 + OD2. BCi^s^-o5_and OA,2±OC2-.2U sq. in., find the distance a rectangle, and

:

section of tha diagonals.

The sum of the squares on the sides of a quadrilateral is greater than the sum of the squares on its diagonals by four times the square 7.

on the straight

line

which s the middle points

of the diagonals.

In a triangle ABC, the angles at B and C are acute are drawn perpendicular to AC, AB respectively', prove that 8.

if

;

BE,

CF

BC2=AB.BF + AC.CE. Three times the sum of the squares on the sides of a triangle equal to four times the sum of the squares on the medians. 9.

10.

ABC

medians

is

a triangle, and

O

the point of intersection of

is

its

shew that

:

AB2+BC2 + CA2=3(OA2 + OB2 + OC2). 11.

If

a straight line

AB

bisected at X, and also divided (inter-

is

nally or externally) at Y, then

AY2 + YB2 = 2 (AX2 + X Y2). Prove

this

from Theorem

limiting position

if

[See p. 230 Ex.

by considering a

when the vertex C

In a triangle ABC, «iBX=nXC, shew that 12.

56,

falls

the base

at

Y

BC

in the base is

CAB

triangle

AB.

divided at

7wAB2+wAC2=wBX2 + nXC2+(w+w)AX2.

6. ]

in the

X

so that

GEOMETRY.

232

KECTANGLES IN CONNECTION WITH CIRCLES. THE0REai5jJ

[Euclid III. 35.]

If two chords of a circle cut at a point within contained by their segments are egimh

it,

the rectangles

D ABC,

In the point

CD

let AB,

be chords cutting at the interna]

X

It

is

required to prove that the

Let

O

Supposing bisecting

red AX, XB = the

recL CX,

XD.

be the centre, and r the radius, of the given

OE drawn

perp. to the chord AB,

circle.

and therefore

it.

OA, OX.

The

Proof.

rect.

AX,

XB = (AE + EX) (E B - EX) = (AE + EX)(AE-EX) = AE2 - EX2 Thear. = (AE2 + OE2)-(EX2 + OE2)

= the L' at E are

Similarly

it

0X2,

since

rt. l'.

may

be shewn that

the rect. CX, ,-.

-

r2

53.

the rect. AX,

XD = r2-OX2. XB = the rect.

CX, XD. Q.E.D.

Corollary. thord which

JEach rectangle

is bisected at the

is

equal

given point X.

to the

square on half the

RECTANGLES IN CONNECTION WITH CIRCLES.

Theorem If two chm'ds of a

circle,

233

[Euclid III. 36.]

58.

when produced,

a point outside

cut at

the rectangles containêd hy their segments are

equal.

rectangle is equal to the square on the tangent

from

And

it,

each

the point oj

intersecti&n.

\ In fê 0ABC,^4et AB, CD be chords cuttin^when produced, at the externaLôint X ; and let XT be a tangent drawn from "^ that point. _ It is required to prove that the rect.

Let

O

XB = ^ê

reel CX,

XD = the

sq.

on XT.

be the centre, and r the radius of the given

Suppose bisecting

AX,

OE drawn

perp. to the chord AB,

circle.

and therefore

it.

OA, OX, OT.

The

Proof.

rect.

AX,

XB = (EX + AE) (EX - EB) = (EX + AE)(EX-AE) = EX2 - AE2 Them-. = (EX2 + OE2)-(AE2 + OE2)

= the

z."

at

E are

rt.

_

0X2

it

.

.-.

gince

j-2^

^^

may be shewn that the rect. CX, XD = 0X2 _ ^2. And since the radius OT is perp. to the Similarly

53.

• .

the rect. AX,

tangent XT,

XT2 = 0X2 - ^2^

XB = the rect.

The&r. 2 9.

CX,

XD = the

sq.

on XT.

Q.E.D.

;;

;

;

GEOMETRY.

234

Theorem If from a point

[Euclid III 37.]

59.

a

two straight lines are drawn, one of which cuts the circle, and the other meets it ; and if the rectangle contained by the whole line which cuts the circle and the part of it outside the circle is equal to the square on the line which meets the circle, then the lin£ whicJi meets the circle is a tangent to it. outside

circle

From X a point outside the ©ABC, let two straight lines XC be drawn, of which XA cuts the circle at A and B, and XC meets it at C and let the rect. XA, XB = the sq. on XC.

XA,

It is required to prove that

Proof.

XC

touches the circle at C.

Suppose XC meets the circle again at D then XA XB = XC XD. .

.

Thear. 58.

XA XB = XC^

But by hypothesis,

.

.-.

XC.XD = XC2; XD = XC.

Hence XC cannot meet the

circle

again unless the points of

section coincide

that

is,

XC

is

a tangent to the

circle.

Q.E.D.

NoTK ON Theorems

57, 58.

ing that the segments into which the chord AB is divided at X, internally in Theorem 57, and externally in Theorem 58, are in each case AX, XB, we may include both Theorems in a single enunciation.

If any number of chords of a within or vnthout a chords are equal.

circle

are draion through a given point contained by the segments of the

circle, the rectangles


235

EXERCISES ON THEOREMS 57-59. {Numerical and Graphical.) a circle of radius 5 cm., and within it take a point X 3 cm. from the centre O. Through X draw any two chords AB, CD. (i) Measure the segments of AB and CD hence find approximately the areas of the rectangles AX.XB and CX.XD, and compare the

Draw

1.

;

results.

M N which OXM calculate

(ii) Draw the chord right-angled triangle

(iii)

differs

is

X

bisected at

;

and from the

the value of XM^.

Find by how much per cent, your estimate of the from its true value.

rect.

AX,

XB

Draw a circle of radius 3 cm., and take an external point X 2. 5 cm. from the centre O. Through X draw any two secants XAB, XCD. (i)

Measure XA, XB and XC, XD; hence find approximately the XA XB and XC XD, and compare the results. Draw the tangent XT ; and from the right-angled triangle XTO

rectangles (ii)

.

.

calculate the value of XT^. (iii)

differs

Find by how much per from

its

AB,

CD

XB = l-2",

and

3.

of

cent,

your estimate of the

rect.

AX,

XB

true value. are

two

CX = 2-7".

straight lines intersecting at X. AX = 1-8'', If A, C, B, D are concyclic, find the length

XD.

Draw a

circle

through A, C, B, and check your result by measure-

ment.

A

4. secant XAB and a tangent external point X.

A

5.

(i)

If

(ii)

If

XT

are

drawn

to a circle from an

XA=0-6", and XB=2-4", find XT. XT=7-5 cm., and XA = 4-5 cm., find XB.

serai-circle is

drawn on a given line AB and from X, any XM is drawn to AB cutting the circum;

point in AB, a perpendicular ference at M shew that :

AX

=2-5", If of the semi-circle. (i)

(ii)

find

If

AX.XB = IVIX2. find XB

and MX=2-(y',

;

hence find the diameter

the radius of the semi-circle = 3 '7 cm., and

AX = 4-9

cm.,

MX.

6. A point X moves within a circle of radius 4 cm, and PQ is any chord ing through X ; if in all positions PX.XQ=12 sq. cm,, find the locus of A. ,

What will

the locus be

PX.XQ = 20sq.

cm.?

if

X

moves outside the same

circle,

so that

GEOMETRY.

236

EXERCISES ON THEOREMS 57-59. {Theoretical.)

ABC

a triangle right-angled at C and from drawn to the hypotenuse shew that

/\. AjIO

is

is

;

C a perpendicular

:

AD.DB = CD2. two

If

2.

intersect,

circles

common chord two

and through any point

CD are drawn, AX.XB=CX.XD.

chords AB,

that

one in each

3. Deduce from Theorem 58 that the tangents drawn rom any external point are equal.

If

4.

in their

two

If a

5.

and

common tangent

shew that

B,

AB

the points A, B, C,

and B

to

shew

to a circle

them from any point

is

drawn

to

two

circles

which cut at A

produced bisects PQ,

lines AB, CD intersect at X so that AX XB deduce from Theorem 57 (by redrictio ad absurdum) that

= CX.XD, 7.

PQ

drawn

in their

are equal.

two straight

If

6.

circles intersect, tangents

common chord produced

X

circle,

D

.

are concyclic.

In the triangle ABC, perpendiculars AP, BQ are drawn from shew that to the opposite sides, and intersect at O

A

:

AO.OP=BO.OQ. 8.

CD

;

is

ABC

a triangle right-angled at C, and from drawn to the hypotenuse shew that is

C

a perpendicular

:

AB.AD=AC2. Through A, a point of intersection of two circles, two straight lines CAE, DAF are drawn, each ing through a centre and terminated by the circumferences shew that 9.

:

CA.AE = DA.AF.

^1^

If

from any external point P two tangents are drawn to a O and radius r ; and if OP meets the chord

given circle whose centre is shew that of at ;

Q

11.

AB at

(or

0P.0Q=r3.

AB is a fixed AB produced)

P and

the circle

diameter of a

circle,

any straight at Q, shew that ;

if

and CD is perpendicular to drawn from A to cut CD

line is

AP AQ = constant. .

a fixed point, and CD a fixed straight line ; AP is anv straight line drawn from A to meet CD at P; if in AP a point Q is taken so that AP AQ is constant, find the locus of Q. 12.

A

is

.

)

:


237

EXERCISES ON THEOREMS 57-59. {Miacdlaneoua.

The chord of an arc of a circle = 2c, the height Shew by Theorem 57 that

1.

of the

arc=^, the

radius = r.

h{2r-h)=c'^.

Hence

find the diameter of a circle in a segment 8" in height.

which a chord 24" long outs

The radius of a circular arch is 25 feet, 2. find the span of the arch. If the height is

how much

reduced by 8

feet,

will the span be reduced

is

height

is

18 feet

the radius remaining the same, by

Employ the equation h{2r-h) = c^

3.

its

?

hy a diagram

Check your calculated results graphically represents 10 feet. whose chord

and

off

in

which

to find the height of

1"

an arc

16 cm., and radius 17 cm.

Explain the double result geometrically.

d denotes the

If

4.

and t the length Theorem 58 that

circle,

Hence

shortest distance from an external point to a' of the tangent from the same point, shew by

^ (^ ^ 2r) = t^. when

find the diameter of the circle your result graphically.

cZ=l'2",

and t=2'4"; and

If the horizon visible to an observer on a cliff 330 feet above the 5. sea-level is 22J miles distant, find roughly the diameter of the earth.

Hence find the approximate distance at which a bright light raised 66 feet above the sea is visible at the sea-level. 6.

arc,

If h is the height of an arc of radius prove that b^=2rh.

and

h the

chord of half the

A semi-circle is described on AB as diameter, and any two chords

7.

AC,

r,

BD

are

drawn

intersecting at

P

:

shew that

AB2=AC.AP-fBD.BP.

Two circles intersect at B and C, and the two direct common 8. tangents AE and DF are drawn if the common chord is produced to meet the tangents at G and H, shew that :

GH2=AE2-fBC2. 9.

and

If

PM

from an external point P a secant PCD is drawn to a is perpendicular to a diameter AB, shew that

PIV|2=PC.PD-fAM.MB.

circle,

GEOMETRY.

238

PROBLEMS.

C

Problem

a square eqwil in

32.

a/rea to

^ a

X Let ABCD be the given rectangle. Produce AB to E, making BE equal to BC. Construction. a semi-circle; and produce CB to meet the circumference at F.

On AE draw

Then BF

is

a side of the required square.

Let X be the mid-point XF.

Proof.

of AE,

and r the radius

of

the semi-circle.

Then the

rect.

AC = AB BE .

= (r-f-XB)(r-XB) = r2-XB2 = FB2, from the rt. Corollary.

To

describe

angled

a square equal in area

A FBX. to

any given

rectilineal figure.

Reduce the given

Draw a Apply

figure to a triangle of equal area.

rectangle equivalent to this triangle.

Prob. 18. Proh. 17

to the rectangle the construction given above.

:

PROBLEMS ON CIRCLES AND RECTANGLES.

239

EXERCISES. 1.

area.

Draw a

What

rectangle 8 cm. by 2 cm., the length of each side ?

and construct a square of equal

is

2. Find graphically the side of a square equal in area to a rectangle whose length and breadth are 3*0" and 1'5". Test your work oy measurement and calculation. 3.

Draw any

is 3 -75 sq. in. and construct a Find by measurement and calculation the length

rectangle whose area

square of equal area. of each side.

;

Draw an equilateral triangle on a side of 3", and construct a 4. rectangle of equal area [Problem 17]. Hence find by construction and measurement the side of an equal square.

ABCD from the following data: A = 65°; BC=CD=5 cm. Reduce this figure to a triangle and hence to a rectangle of equal area. Construct an equal square, and measure the length of its side. 5.

Draw

a quadrilateral

AB = AD = 9cm. [Problem

'

6.

;

18],

Divide AB, a line 9 cm. in length, internally at X, so that square on a side of 4 cm.

AX XB = the .

Hence give a graphical solution, correct to the the simultaneous equations

first

decimal place, of

:

x + i/=9,

xi/=lQ.

your unit of length, solve the following equations by a graphical construction, correct to one decimal figure 7.

Taking yu'

a-s

:

x + y = 40,

xy = lQ9.

8. The area of a rectangle is 25 sq. cm. , and the length of one side 7*2 cm. ; find graphically the length of the other side to the nearest millimetre, and test your drawing by calculation. is

9.

Divide AB, a line 8 cm. in length, externally at X, so that square on a side of 6 cm. [See p. 245.]

AX XB = lhe .

Hence

find a graphical solution, correct to the first decimal place, of

the equations

x-y=%,

xy=^Q.

10. On a straight line AB draw a semi -circle, and from any point P on the circumference draw PX perpendicular to AB. AP, PB, and denote these lines by x and y. Noticing that (i) x"^ + y"^ = kE^ (ii) a:y=2AAPB=AB. PX ; devise a -,

graphical solution of the equations a;2

:

+ y2=100;

a;y=25.

;

GEOMETRY.

S40

Problem To

and one fart may

the whole

he equal to the square

Let AB be the

st.

line to

that

on

the other part.

3r---vX a-x R

A^

way

33.

divide a given straight line so that the rectangle contained hy

be divided at a point X in such a

AB.BXÂX^.

Draw Construction. AC.

BC

perp. to AB,

and make BC equal

half AB.

From CA From AB Then AB Proof.

is

cut off

CD

cut off

AX equal

equal to CB. to AD.

divided as required at X.

Let AB = a units

of length,

and

let

AX = «.

ThenBX = a-ic; AD = a;; BC = CD = |.

Now that

or, is,

angled

A ABC;

each of these equals take ax

a^-ax = x'^; a{a-x) = x'^,

then that

rt.

= (AC-BC)(AC + BC); a^ = x{x-\-a) = x^ + ax.

is,

From

AB2 = AC2- BC2, from the

AB.BX = AX2. EXERCISE.

Let AB be divided as above at X. On AB, AX, and on opposite sides of AB, draw the squares ABEF, AXGH and produce GX to meet FE at K. In this diagram name rectangular figures equivalent to a^, a;', x{x + a), ax, and a{a-x). Hence illustrate the above proof graphically. ;

H

X G

to

;

MEDIAL SECTION.

241

A

straight line is said to be divided in Medial Section when Note. the rectangle contained by the given line and one segment is equal to the square on the other segment. This division may be internal or external ; that is to say, AB may be divided internally at X, and externally at X', so that

AB.BX=AXa, AB.BX' = AX'2.

(i) (ii)

To

obtain X', the construction of p. 240 must be modified thus

CD

is

AC produced from BA produced,

:

to be cut off from

AX'

in the negative sense.

Algebraical Illustration. If a

St.

line

AB

is

divided at X, internally or externally, so that

AB.BX = AX2, and

if

AB=a, AX = a;, and

consequently

BX = a-x,

then

a{a-x) = x^, a? + ax-a'^=0,

or>

and the roots

of this quadratic,

the lengths of

AX and

namely, —^

—

« ^^^ ~

(

"^"^^

)'

AX'.

EXERCISES. 1.

Divide a straight line 4" long internally in medial section. its length algebraically.

Measure the greater segment, and find

2. Divide AB, a line 2" long, externally in medial section at X'. Measure AX', and obtain its length algebraically, explaining the geometrical meaning of the negative sign. 3.

In the figure of Problem 33, shew that

Hence prove

(i)AX = -2--2;

(n)

AC = ^.

AX'= - -2- + 2 (

[Theor. 29.]

)•

4. If a straight line is divided internally in medial section, and from the greater segment a part is taken equal to the less, shew that the greater segment is also divided in medial section.

H.S.G. Ill -IV.

Q

;

;

;

GEOMETRY.

842

Problem To draw an

isosceles triangle

34.

having each of the angles at the base

d&hble of the vertical angle.

B Construction.

Take any so that

(This construction

With

is

line AB,

c and divide

at X, Prob.

.

shewn separately on the

centre A, and radius AB,

and

it

AB BX = AX2.

in it place the

draw the

33

left.)

BCD

chord BC equal to AX.

AC.

Then ABC

the triangle required.

XC, and suppose a circle drawn through

Proof.

and

is

A,

X

C.

Now, by

construction,

.-.

.*.

BA BX = AX^ .

BC touches

the z.BCX

= the

= BC2; AXC

the

Z.XAC, in the

at

C

alt.

To each add the l XCA

Theor. 59.

segment.

;

then the l BCA = the l XAC + the l XCA = the ext. Z.CXB. *

And the L BCA = the L CBA, for AB = AC. .-. the z. CBX = the z. CXB CX = CB = AX, .'. the ^ XAC = the z. XCA .*. the L XAC + the z. XCA = twice the z. A. But the L ABC = the L ACB = the z. XAC 4- the z. XCA = twice the l A. .-.

;

Pr&ved,

MEDIAL SECTION.

243

EXERCISES.

How many

degrees are there in the vertical angle of an isosceles triangle in which each angle at the base is double of the vertical angle ? 1.

Shew how a

2.

means 3.

angle

of

Problem

right angle

be divided into five equal parts by

In the figure of Problem 34 point out a triangle whose vertical three times either angle at the base.

is

Shew how such a 4.

may

34.

triangle

If in the triangle

may be

constructed.

ABC, the I.B = the Z.C = twice the

*^^^

AB~ 5.

Z.A,

shew

BC_n/5-1 2

In the figure of Problem 34,

•

the two circles intersect at F,

if

shew that (i)

BC = CF;

(ii)

the circle

CF

(iii)

BC,

(iv)

AX, XC,

circle

AXC = the

circum-circle of the triangle

ABC

;

are sides of a regular decagon inscribed in the

BCD CF

the circle

;

are sides of a regular pentagon inscribed in

AXC.

In the figiire of Problem 34, shew that the centre of the circle 6. circumscribed al)out the triangle CBX is the middle point of the arc XC. In the figure of Problem 34. if is the in-centre of the triangle I', S' the in-centre and circum-centre of the triangle CBX, shew that S' = S' I'. 7.

I

ABC, and

I

If a straight line is divided in medial section, the rectangle con8. tained by the sum and difference of the segments is equal to the rectangle contained by the segments. 9.

If a straight line

AB

shew that Also this result

H.S.G. III.-Iv.

is

divided internally in medial section at X,

AB2 + BX^ = 3AX2. by substituting the values given on page

q2

241.

,

GEOMETRY.

244

THE GRAPHICAL SOLXJTION OF QUADRATIC EQUATIONS. From the following constructions, which depend on Problem 32, a graphical solution of easy quadratic equations may be obtained. I.

To

divide a straight line internally so that the rectangle contained may be equal to a given square.

by the segments

Let

AB

be the

line to be divided,

st.

and

DE

a side of the givea

square.

On AB draw

Ck>nstruction.

perp. to

AB and

a semicircle

;

and from

B draw 6F

equal to DE.

From F draw

FCC par^ to AB,

From Then AB

is

C,

cutting the O"* at

C draw CX,

C and C.

C'X' perp. to AB.

divided as required at X, and also at X'.

AX XB = CX2

Proof.

Prob. 32.

.

= BF2 = DE2. Similarly AX'. Application.

X'B = DE2.

The purpose of

Now

two straight AB, and their product, viz.

this construction is to find

AX, XB, having given their sum, the square on DE.

lines

viz.

to solve the equation a;^- 13x + 36 = 0,

numbers whose sum

is 13,

and whose product

we have

is 36,

or

to find

two

6".

To do this graphically, perform the above construction, making AB equal to 13 cm., and DE equal to \/36 or 6 cm. The segments AX, XB represent the roots of the equation, and their values may be obtained by measurement. term of the equation is not a perfect square, as 11=0, v^l must be first got by the arithmetical rule, or graphically by nit'ans of Problem 32.

Note.

in ar*- 7a;

If tlie last

4-

245

GRAPHICAL SOLUTION OF QUADRATICS.

II.

To

divide

X'

Let

a

may

hy the segments

straight line externally so that the rectangle contaitied be equal to a given square.

B X

A

AB be the st.

line to

be divided externally, and

DE

the side of

the given square. Construction. Bisect AB at O.

With

From B draw BF

centre O, and radius

to

perp.

OF draw

AB, and equal to DE.

a semicircle to cut

X and X'. Then AB is divided

externally as required at X,

Proof

AX.XB=X'B.BX,

AB

pro-

duced at

since

and

also at X'.

AX = X'B,

= BF2 = DE2 Application. difference, viz.

Now

Here we find two lines AX, XB, having given their AB, and i\\e\r product, viz. the square on DE.

to solve the equation x'^- 6a;- 16=0,

numbers whose numerical or

Proh. 32.

difference is 6,

we have

to find

and whose product

is

two 16,

42.

To do this graphically, perform the above construction, making AB equal to 6 cm., and DE equal to sfl^ or 4 cm. The segments AX, XB represent the roots of the equation, and their values, as before, may be obtained by measurement. EXERCISES. of

Obtain approximately the roots of the following quadratics by means graphical constructions ; and test your results algebraically. 10a;

+ 16=0.

5a;-

36=0.

a;2- 14a; +

49=0.

a;2_7a;-49=0.

a;2- 12a; + 25=0.

a:2_i0a; + 20=0.

;

GEOMETRY.

246

EXERCISES FOR SQUARED PAPER.

A

1.

through the points (0, 4), (0, 9) touches the Calculate and measure the length of OP,

circle ing

X-axis at P.

With centre at the point (9, 6) a circle is drawn to touch the 2. Find the rectangle of the segments of any chord through O. y-axis. Also find the rectangle of the segments of any chord through the point

(9, 12).

Draw a circle (shewing all lines of construction) through the Find the length of the otlier intercept on points f6, 0), (24, 0), (0, 9). Also find the length of a the y-axis, and by measurement. tangent to the circle from the origin. 3.

i

Draw a circle through the points (10, 0), 4. prove by Theorem 59 that it touches the aj-axis. Find

(i)

If

5.

a

the coordinates of the centre,

5), (0,

20); and

the length of the radius.

(ii)

through the points (16, 0), also es through (0, 24).

circle es

by Theorem 58 that it Find (i) the coordinates of the centre, from the origin.

(0,

(18, 0), (0, 12), sheiv

the length of the tangent

(ii)

Plot the points A, B, C, D from the coordinates (12, 0), f -6, 0), - 8) and prove by Theorem 57 that they are concyclic.

6.

(0, 9), (0,

If r

;

denotes the radius of the

that

circle, shev/

0A2 + 0B2 + 0C2 + 0D2 = 4r». 7.

Draw

a circle (shewing

all lines of

construction) to touch the

and to cut the x-axis at (3, 0). Prove that the circle must cut the tc-axis again at the point (27, 0) and find its radius. your results by nteasurement. Shew that two circles of radius 13 may be drawn through the 8. point (0. 8) to touch the x-axis and by means of Theorem 68 find the y-axis at the point

(0, 9),

;

length of their 9.

how and

common

chord.

Given a circle of radius 15, the centre being at the origin, shew to draw a second circle of the same radius touching the given circle also touching the x-axis.

How many

circles can

centre of that in the

first

be so drawn? quadrant.

Measure the coordinates of the

A, B, C. D are four points on the x-axis at distances 6, 9, 15, 25 10. from the origin O. Draw two intersecting circles, one through A, B, and the other through C, D, and hence determine a point P in the X-axis such that

PA.PB=PC.PD. Calculate and measure OP. If the distances of A, B, C,

D from O

are a,

prove that

OP=(o6-cd)/(crf6 -c-d).

6,

c,

d

respectively,

1.

GEOMETRY. Exercises.

Page

181.

Exercises.

Page

187.

108°.

108%

1.

72°,

2.

1-6".

2.

2-3 cm., 4-6 em., 6-9 cm.

4.

6-9 cm.

1.

2-12"; 4-50 sq.

4.

128f ;

1.

3-46"; 4-Or.

4.

(i)

41-57 sq. cm.;

28-3 cm.

1-7".

3.

4.

Exercises.

;

1.

(i)

11-31 cm.;

7.

30-5 sq. cm.

;

200.

Exercises.

Page

201.

(ii)

77*25

sq.

6-28-3

(ii)

sq.

cm.

cm.

8-9".

8.

6-3".

4.

cm.

5-20".

90°.

Page

205.

(i)

4.

56

sq.

9.

4";

37".

;

(ii)

6.

3".

Page 6.

in.

cm.

Areas, 1-54 sq.

10. in.,

352-99 sq.

4398

225.

10cm.

Page

228.

Page

231.

6.

1".

3.

A circle of rad.

6.

0-25".

6 cm.

in.

in.

sq. in.

12-57 sq.

3*14 sq.

15 cm.

Exercises. 8-5 cm.

2-0*

cm.

2.

êrcises.

4.

5.

259-8 sq. cm.

16-62 sq.

Exercises.

2.

199.

8 5 cm.

Page

2.

1018

6-4sq. cm.

sq.

1-.39".

3-2 cm.

Exercises.

Circumferences, 4 '4",

630

3.

7.

1-73".

3.

;1.

Page 4.

in.

Exercises.

8.

198.

20-78 sq. cm.

Exercises.

11.

Page

1-98", 1-6".

in.

m

ANSWERS. Exercises. 16 sq. cm.

1.

(i)

3.

0-8".

6.

Two

4.

(i)

1-2".

(ii)

2.

12-5 em.

235.

(i)

5.

16 sq. cm. (i)

Page

237.

(ii)

1-6", 4-1".

concentric circles, radii 2 cm. and 6 cm.

Exercises. 1.

Page

16 sq. cm.

(ii)

16 sq. cm.

(ii)3-ocm.

—

:

WORKS BY H. S. HALL, M. A., and F. H. STEVENS, M.A. A SCHOOL GEOMETRY,

based on the recommendations of

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