Proof :
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$% (& $( %&
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Solution:
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DUHD
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Proved
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DUHD VLGHV
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'$2&
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Proof :
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DUHD
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Proved
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5.2 Construction
Introduction
We have already learnt to construct triangles, squares, rectangles, rhombus, parallelograms, trapezium
and quadrilateral, in previous classes and also have learned about the following theorems:
¾ Area of parallelograms standing on the same base or equal bases and between the same
parallel lines are equal.
¾ Area of triangle is equal to the half of the area of parallelogram standing on the same base or
equal bases between same parallel lines.
¾ Area of triangles standing on the same base or equal bases and between same parallel lines
are equal.
Here we are going to construct equivalent figures based on these theorems with given
measurements. We follow the following steps before drawing the actual figure.
x Draw a rough sketch of the figure from the given condition.
x Write the given measurements in rough sketch.
x Analyze the figure and construct the actual figure.
Construction of a parallelogram whose area is equal to the area of a given triangle.
Example 1: Construct a triangle ABC in which AB = 6.6cm, BC = 5.3cm and AC = 7.1cm.
a) Construct a parallelogram having an angle 60o and equal in area to
the triangle.
b) Construct a parallelogram with one side 6cm equal in area to the
triangle.
a) Steps
i. Draw a 'ABC with AB = F E C
6.6cm, BC = 5.3cm and AC = M N
7.1cm.
ii. Draw MN parallel to AB
through the point C.
[Making BAC = ACM]
iii. Locate the point D as midpoint
of AB. AD B
iv. Draw an angle ADE = 60o at D,
such that DE meets MN at E. (a)
v. From the point E, draw an arc
with radius EF = AD to cut MN
at F.
vi. Join A and F.
Now, ADEF is the required parallelogram which is equal in area to the triangle
ABC.
Geometry Mathematics - 10|201
b) Steps
i. (i), (ii) and (iii) are same MF E CN
steps of (a)
ii. From the point D draw an arc
with radius 6cm to cut MN at
E.
iii. From the points E, draw an
arc with radius EF = AD to
cut MN at F.
iv. Join A and F. AD B
Now, ADEF is the required (b)
parallelogram which is equal
in area to the triangle ABC.
Construction of a triangle whose area is equal to the area of a given parallelogram.
Example 2: Construct a parallelogram ABCD in which AB = 5.7cm, AC = 6.6cm and
BD = 8.4cm and construct
a) a triangle with one side 7.7cm and equal in area to the parallelogram
ABCD.
b) a triangle having an angle 45o and equal in area to the parallelogram
ABCD.
a) Steps CF
B
i. Draw a parallelogram ABCD D
having AB = 5.7cm, AvC = 6.6cm
and BD = 8.4cm.
ii. Produce AB to E such that AB = BE.
iii. From A, draw an arc with radius A E
AF=7.7cm to cut DC (or DC
produced) at F.
iv. Join A, F and E, F.
Now, 'AEF is the required
triangle, which is equal to
ABCD in area.
b) Steps CF
i. (i) and (ii) steps are same as (a). D
ii. At point E, draw an angle AEF = 45o
BE
45o, such that EF meets DF at F.
iii. Join A and F. Geometry
Now, 'AEF is the required A
triangle, which is equal to
ABCD in area.
202|Mathematics - 10
Example 3: Construct a rhombus ABCD with diagonal AC = 8cm. and diagonal BD =
6cm. Also construct a triangle AEF equal in Area to the rhombus having
one side 9cm.
Steps
1. Draw BD = 6cm. and draw perpendicular bisector of BD which intersect BD at O.
2. From O take OA = 4cm. and OC = 4cm. along the bisector. Thus rhombus ABCD is
constructed.
3. AB is produced to E such that AB = BE.
4. Take AF = 9cm. to cut DC produced at F.
5. Join E and F. Thus 'AEF = rhombus ABCD is constructed.
Construction of a triangle whose area is equal to the area of a given quadrilateral.
Example 4: Construct a quadrilateral ABCD in which AB = 5.4cm, BC = 5cm,
CD = 4.9cm, AD = 6.1cm and BAD = 60o. Construct a triangle equal in
area to the quadrilateral.
Steps D C
i. Construct a quadrilateral ABCD in
which AB = 5.4cm, BC = 5.1cm,
CD = 4.9cm, AD = 6.1cm and
BAD = 60o.
ii. From C draw CE//BD.
iii. Join D and E. A B E
iv. 'ADE is the required triangle
whose area is equal to the area of
quadrilateral ABCD.
Verification :
Here,
BD || EC (By construction)
i) Area of 'BDC = Area of 'BDE
[They are standing on same base BD and betn same parallel BD and CE.
ii) Area of 'ADE = 'ABD + 'BDE
iii) Area of Quad ABCD = 'ABD + 'BDC
? Area of Quad ABCD = Area of 'ADE
Proved.
Geometry Mathematics - 10|203
Exercise 5.2
1. a) Construct a Parallelogram ABCD in which AB = 6.2cm, BC = 5.2cm and ABC = 60o.
Also construct a parallelogram BCMN with a side 7.3cm and equal area with the
parallelogram ABCD.
b) Construct a parallelogram PQRS in which PQ = 6cm, QR = 7cm, PR = 8cm. Also
construct a parallelogram YZQR with an angle ZQR = 75q and equal area with the
parallelogram PQRS.
c) Construct a parallelogram MNOP in which MO = 8cm. NP = 7cm and NO = 5.6cm.
Also construct a rectangle XNOY having equal area with the parallelogram MNOP.
2. a) Construct a parallelogram XYMN in which XY = 7.3cm, YM = 6.2cm XYM = 75q.
Also construct a triangle XAB with a side 8cm and equal area with the parallelogram
XYMN.
b) Construct a parallelogram PQRS in which PQ = 5.8cm PS = 4.9cm and angle QPR = 30o
and construct a triangle QSZ having equal area with the parallelogram PQRS.
c) Construct a parallelogram ABCD in which AB = 5.9cm AC = 8.6cm and
BD = 6.8cm. And construct a triangle with the side 9cm. having the area half of the area
parallelogram ABCD.
3. a) Construct a Parallelogram ABCD in which AC = 7cm., BC = 5.2cm and ACD = 30q.
Also construct a triangle CPQ having an angle CPQ = 75q, and equal area with the
parallelogram ABCD.
b) Construct a parallelogram PQRS in which PQ = 5.2cm, PR = 7.6cm, QS = 8.2cm and
construct a triangle BAR with an angle ARQ = 45o equal area with the parallelogram
PQRS.
c) Construct a Parallelogram ABCD in which AB = 5.2cm, BC = 6.1cm and ABC = 120 o
and construct a triangle AEF with an angle AFE = 30o equal area with the parallelogram
ABCD.
4. a) Construct a triangle PQR in which p = 7.3cm, Q = 75 o and R = 45o and construct a
triangle SQR having SQ = 6.7cm. equal area to the triangle PQR.
b) Construct a triangle EFG in which EF = 4.8cm EFG = 45 o and FG = 5.9cm. Also
construct an isosceles triangle AFG, whose area equal to the area triangle EFG.
c) Construct a triangle ABC in which B = 60 o, AB = 7cm. AC = 6cm. Also construct a
triangle MBA equal area to the triangle ABC.
5. a) Construct a triangle ABC in which AB = 7.1cm, BC = 5.9cm and AC = 6.3 cm, Also
construct a rectangle BDCF equal an area to the triangle ABC.
b) Construct a triangle XYZ in which YZ = 6cm YXZ = 75o, XYZ = 60o. Also
construct a parallelogram YABC with a side YA = 7.6cm equal area to the triangle XYZ.
c) Construct a triangle WXY in which XY = 7.4cm X = 45 o, W = 60 o. Also construct a
parallelogram YPQR where QPY = 135 o equal area to the triangle WXY.
204|Mathematics - 10 Geometry
d) Construct a parallelogram having an angle 60o, equal area to the bigger triangle ABC
having AB = 5cm, B = 45 o, AC = 4cm.
6. a) Construct a quadrilateral ABCD in which AB = BC = 5.3cm, CD = AD = 4.8cm and
BAD = 60o and construct a triangle ADE equal in area to the quadrilateral ABCD.
b) Construct a quadrilateral PQRS in which PQ = 5.5cm, QR = 4.8cm, RS = 4.2cm,
PS = 5.9cm, QS = 5.1cm and construct a triangle PST equal in area to the quadrilateral
PQRS.
7. a) Construct a quadrilateral DEFG in which DE = 4.8cm, EF = 5.7cm, FG = 5.1cm,
DG = 6.2cm, DF = 4.9cm and construct a triangle CEF equal in area to the quadrilateral
DEFG.
b) Construct a quadrilateral ABCD in which AB = BC = 5.9cm, CD = AD = 4.9cm,
ABC = 45o and construct a triangle BCE equal in area to the quadrilateral ABCD.
8. a) Construct a rhombus ABCD in which diagonals AC = 7cm BD = 6.5cm. And also
construct a ' ABZ = having equal area with the rhombus ABCD.
b) Construct a rhombus PQRS in which a side PQ = 5cm. diagonal PR = 8cm. And also
construct a 'QRW having equal area with the rhombus PQRS.
9. a) Construct a square ABCD in which AB = 6cm. and construct a 'ABM having equal area
with the square ABCD.
b) Construct a rectangle PQRS in which PQ = 5cm. and QR = 7cm. and construct a 'QXY
with QX = 7.5cm having equal area with the rectangle PQRS.
c) Construct a rectangle WXYZ in which the length of diagonal WY = 8cm. and angle
between the diagonals is 60. Also construct a 'WMN with a side 7.2cm and having
equal area with the rectangle WXYZ.
Geometry Mathematics - 10|205
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TXDGULODWHUDO ZKHUH DOO WKH YHUWLFHV RI WKH TXDGULODWHUDO DUH RQ WKH 2
FLUFXPIHUHQFH RI WKH FLUFOH 7KXV WKH TXDGULODWHUDO $%&' LV FDOOHG D
F\FOLF TXDGULODWHUDO
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ͮDĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬ *HRPHWU\
7KHRUHP
The arcs subtending equal angles at the centre of a circle are equal.
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Experiment: L 'UDZ WKUHH FLUFOHV RI GLIIHUHQW UDGLL ZLWK FHQWUH 2
& $ & &
' 2
2 $
2
% ' %'
LL LLL
$%
L
LL 'UDZ WZR HTXDO FHQWUDO DQJOHV $2% DQG &2' DW WKH FHQWUH RI HDFK FLUFOH
To verify: ܣܦ ܤ ܥ
Verification: 0HDVXUH WKH OHQJWK RI ܣܤDQG ܦ ܥDQG FRPSOHWH WKH IROORZLQJ WDEOH 0HDVXUH WKH OHQJWK
RI DUF ZLWK WKH KHOS RI D WKUHDG DQG D UXOHU
)LJ ۯ۰ ۱۲ 5HVXOW
L
LL
LLL
Conclusion: 7KH DERYH H[SHULPHQW VKRZV WKDW WKH DUFV VXEWHQGLQJ HTXDO DQJOHV DW WKH FHQWUH RI D
FLUFOH DUH HTXDO
&RQYHUVH RI WKHRUHP
Equal arcs of a circle subtend equal angles at the centre.
([SHULPHQWDO YHULILFDWLRQ
Experiment: L 'UDZ WKUHH FLUFOHV RI GLIIHUHQW UDGLL ZLWK FHQWUH 2
& ' &
$ 2
2
$% % '
L LL
LL 'UDZ HTXDO DUFV $% DQG '& ZLWK WKH KHOS RI SHQFLO FRPSDVV
*HRPHWU\ DĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬͮ
To verify: $2% &2'
Verification: 0HDVXUH WKH FHQWUDO DQJOHV $2% DQG &2' DQG WDEXODWH
)LJ $2% &2' 5HVXOW
L $2% &2'
LL $2% &2'
LLL $2% &2'
Conclusion: 7KH DERYH H[SHULPHQW VKRZV WKDW WKH HTXDO DUFV RI D FLUFOH VXEWHQG HTXDO DQJOHV DW WKH
FHQWUH
7KHRUHP
Equal chords of a circle cut off equal arcs in the circle.
([SHULPHQWDO YHULILFDWLRQ
Experiment: 'UDZ WKUHH FLUFOHV RI GLIIHUHQW UDGLL ZLWK FHQWUH 2 DQG GUDZ HTXDO FKRUGV $% DQG &' LQ
HDFK FLUFOH
% $& $&
2 2
$
%'
LL
2
& '
L % '
LLL
To verify: ൌ ࡰ
Verification: 0HDVXUH WKH OHQJWK RI DQG ZLWK WKH KHOS RI WKUHDG DQG UXOHU DQG WDEXODWH
)LJ /HQJWK RI ۯ۰ /HQJWK RI ۱۲ 5HVXOW
L
LL
LLL
Conclusion: 7KH DERYH H[SHULPHQW VKRZV WKDW WKH HTXDO FKRUGV RI D FLUFOH FXW RII HTXDO DUFV LQ WKH
FLUFOH
*HRPHWU\
ͮDĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬ
&RQYHUVH RI WKHRUHP
Equal arcs subtend equal chords in the circle.
Experiment: 'UDZ WKUHH FLUFOHV RI GLIIHUHQW UDGLL ZLWK FHQWUH 2 DQG GUDZ HTXDO DUFV $% DQG &' LQ
HDFK FLUFOH ZLWK WKH KHOS RI SHQFLO FRPSDVV
$& & &
'$
2
2 2
$ '
' %
LL
% % LLL
L
To verify: с
Verification: 0HDVXUH WKH OHQJWKV RI FKRUGV $% DQG &' ZLWK WKH KHOS RI D UXOHU DQG WDEXODWH
)LJ /HQJWK RI ۯ۰ /HQJWK RI ۱۲ 5HVXOW
L
LL
LLL
Conclusion: 7KH DERYH H[SHULPHQW VKRZV WKDW HTXDO DUFV VXEWHQG HTXDO FKRUGV LQ WKH FLUFOH
7KHRUHP
The angle at the centre of a circle is twice the angle at the circumference standing on the
same arc.
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Experiment: L 'UDZ WKUHH FLUFOHV RI GLIIHUHQW UDGLL ZLWK FHQWUH 2
& &
2 2 2&
% % $ %
LLL
$ $
LL
L
*HRPHWU\ DĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬͮ
LL 'UDZ WKH FHQWUDO DQJOH $2% DQG DQJOH DW WKH FLUFXPIHUHQFH $&% VWDQGLQJ RQ WKH VDPH
DUF $% LQ HDFK FLUFOH
To verify: $2% $&%
Verification: 0HDVXUH $2% DQG $&% E\ SURWUDFWRU LQ HDFK ILJXUH DQG WDEXODWH
)LJ $&% $2% 5HVXOW
L $&% $2%
LL $&% $2%
LLL $&% $2%
Conclusion: 7KH DERYH H[SHULPHQW VKRZV WKDW WKH DQJOH DW WKH FHQWUH RI D FLUFOH LV WZLFH WKH DQJOH DW
FLUFXPIHUHQFH
7KHRUHWLFDO 3URRI
*LYHQ L 2 LV WKH FHQWUH RI WKH FLUFOH &
LL $2% LV WKH FHQWUDO DQJOH DQG $&% LV WKH 2
DQJOH DW WKH FLUFXPIHUHQFH RI WKH FLUFOH
VWDQGLQJ RQ WKH VDPH DUF $%
7R SURYH $2% $&% $' %
&RQVWUXFWLRQ -RLQ &2 DQG SURGXFH &2 WR SRLQW '
Proof :
6WDWHPHQWV 5HDVRQV
,Q '$2&
L $2 &2
LL 2$& 2&$ L 5DGLL RI WKH VDPH FLUFOH
LL %DVH DQJOHV RI DQ LVRVFHOHV WULDQJOH
LLL $2' 2$& 2&$ LLL ([WHULRU DQJOH RI D WULDQJOH LV HTXDO W
WKH VXP RI WZR RSSRVLWH LQWHULRU DQJOHV
LY $2' 2&$ 2&$ LY %HLQJ 2$& 2&$
Y $2' 2&$
6LPLODUO\ LQ '%2& Y )URP LY
%2' 2&% 6DPH DV DERYH IDFWV DQG UHDVRQV
$2' %2' 2&$ 2&%
$2% 2&$ 2&% )URP Y RI DQG
$&% :KROH SDUW D[LRP
Proved
ͮDĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬ *HRPHWU\
$OWHUQDWLYH PHWKRG %
2 &
$
Proof :
6WDWHPHQW 5HDVRQ
$2& R 5HODWLRQ EHWZHHQ FHQWUDO DQJOH DQG LWV
RSSRVLWH DUF RI WKH FLUFOH
$%& R
$%& R 5HODWLRQ EHWZHHQ LQVFULEHG DQJOH DQG
?$2& $%& LWV RSSRVLWH DUF RI WKH FLUFOH
)URP
)URP DQG
Proved
7KHRUHP
The angles at the circumference of a circle standing on the same arc are equal.
Or
Angles standing at the same segment of a circle are equal.
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Experiment: L 'UDZ WKUHH FLUFOHV LQ GLIIHUHQW UDGLL ZLWK FHQWUH 2
LL 'UDZ LQVFULEHG DQJOHV $%& DQG $'& RQ WKH VDPH DUF $&
% ' %%
'
'
2 $2 &
2
&
$ $ &
L LL LLL
To verify: $%& $'&
Verification: 0HDVXUH $%& DQG $'& ZLWK WKH KHOS RI SURWUDFWRU LQ HDFK ILJXUH DQG WDEXODWH
)LJ $%& $'& 5HVXOW
$%& $'&
$%& $'&
$%& $'&
Conclusion: 7KH DERYH H[SHULPHQW VKRZV WKDW WKH DQJOHV DW WKH FLUFXPIHUHQFH RI D FLUFOH VWDQGLQJ RQ
WKH VDPH DUF DUH HTXDO
DĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬͮ
*HRPHWU\
7KHRUHWLFDO SURRI %
'
*LYHQ L 2 LV WKH FHQWUH RI D FLUFOH
LL $%& DQG $'& DUH WKH LQVFULEHG DQJOHV 2
VWDQGLQJ RQ WKH VDPH DUF $& $
7R SURYH $%& $'& &
&RQVWUXFWLRQ -RLQ $ DQG & ZLWK FHQWUH 2
Proof : 6WDWHPHQW 5HDVRQ
&HQWUDO DQJOH LV GRXEOH WKH DQJOH DW WKH
$2& $%&
FLUFXPIHUHQFH VWDQGLQJ RQ WKH VDPH DUF
$2& $'& 6DPH UHDVRQLQJ DV QR
$%& $'&
$%& $'& )URP DQG
'LYLGLQJ ERWK VLGHV E\
Proved
%
$OWHUQDWLYH PHWKRG '
2&
$
Proof
6WDWHPHQW 5HDVRQ
$%& R 5HODWLRQ EHWZHHQ LQVFULEHG DQJOH DQG LWV
RSSRVLWH DUF RI WKH FLUFOH
$'& R 6DPH UHDVRQLQJ DV
?$%& $'& )URP DQG
7KHRUHP Proved
The angle at semi circle is a right angle. % &
2
7KHRUHWLFDO SURRI $
*LYHQ L 2 LV WKH FHQWUH RI D FLUFOH
ͮDĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬ *HRPHWU\
LL $2& LV WKH GLDPHWHU RI WKH FLUFOH DQG $%& LV
DQ DQJOH LQ VHPL FLUFOH
7R SURYH $%& R
Proof :
6WDWHPHQWV 5HDVRQV
$2& R
$%& $2& %HLQJ VWUDLJKW DQJOH
$%& î R ,QVFULEHG DQJOH LV KDOI RI WKH FHQWUDO DQJOH
? $%& R VWDQGLQJ RQ VDPH DUF RI WKH FLUFOH
)URP VWDWHPHQW DQG
)URP VWDWHPHQWV
Proved
7KHRUHP
The opposite angles of a cyclic quadrilateral are supplementary.
Or
Sum of two opposite angles of a cyclic quadrilateral is 180o (two right angles).
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Experiment: L 'UDZ WKUHH FLUFOHV RI GLIIHUHQW UDGLL ZLWK FHQWUH 2
LL 'UDZ D F\FOLF TXDGULODWHUDO $%&' LQ HDFK FLUFOH
$
$ '
$ %'
'
% &% & &
L LL LLL
To verify: $%& $'& R %$' %&' R
Verification: 0HDVXUH WKH RSSRVLWH DQJOHV RI TXDGULODWHUDO $%&' ZLWK WKH KHOS RI D SURWUDFWRU LQ HDFK
ILJXUH DQG WDEXODWH
)LJ $ & $ & % ' % ' 5HVXOW
, $ & R
% ' R
LL $ & R
% ' R
LLL $ & R
% ' R
Conclusion: 7KH DERYH H[SHULPHQW VKRZV WKDW WKH RSSRVLWH DQJOHV RI D F\FOLF TXDGULODWHUDO DUH
VXSSOHPHQWDU\
*HRPHWU\ DĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬͮ
7KHRUHWLFDO SURRI
*LYHQ L 2 LV WKH FHQWUH RI WKH FLUFOH
LL $%&' LV D F\FOLF TXDGULODWHUDO
7R SURYH %$' %&' R
$%& $'& R
&RQVWUXFWLRQ -RLQ %2 DQG '2 6XSSRVH REWXVH %2' [ DQG UHIOH[ %2' \
Proof
6WDWHPHQW 5HDVRQ
L %$' %2' [ ,QVFULEHG DQJOH LV KDOI RI WKH DQJOH DW WKH
L FHQWUH RI D FLUFOH VWDQGLQJ RQ WKH VDPH DUF
%&'
LL %&' %2' \ LL 6DPH UHDVRQ DV LQ L VWDQGLQJ RQ WKH
VDPH DUF %$'
LLL %$' %&' %2' [ LLL )URP L DQG LL RI
%2' \
LY %$' %&' [ \ LY )URP LLL RI
Y %$' %&' î R Y $QJOH LQ D FRPSOHWH WXUQ LV R
LH [ \ R
YL %$' %&' R YL )URP Y RI
6LPLODUO\ $%& $'& R %\ MRLQLQJ $2 DQG &2 VDPH DV DERYH
IDFWV DQG UHDVRQV
Proved
$
$OWHUQDWLYH PHWKRG
2
%'
Proof : &
L 6WDWHPHQW 5HDVRQ
LL
%$' R L 5HODWLRQ EHWZHHQ LQVFULEHG DQJOH DQG
LWV RSSRVLWH DUF
%&' R
LL 6DPH DV UHDVRQ L
LLL %$' %&' R LLL )URP L DQG LL
LY %$' %&' R FLUFXPIHUHQFH LY :KROH SDUW D[LRP
Y %$' %&' î R Y &LUFXPIHUHQFH { R
YL %$' %&' R
6LPLODUO\ $%& $'& R YL )URP Y
6DPH DV DERYH IDFWV DQG UHDVRQV
Proved
ͮDĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬ *HRPHWU\
&RUROODU\
7KH H[WHULRU DQJOH RI D F\FOLF TXDGULODWHUDO LV HTXDO WR WKH VXP RI LWV RSSRVLWH LQWHULRU DQJOHV YHULI\
H[SHULPHQWDOO\
([SHULPHQWDO YHULILFDWLRQ
L 'UDZ WKUHH FLUFOHV LQ GLIIHUHQW UDGLL ZLWK FHQWUH 2
LL 'UDZ D F\FOLF TXDGULODWHUDO $%&' DQG SURGXFH D VLGH %& XS WR (
L LL LLL
7R YHULI\ %$' '&(
0HDVXUH WKH %$' DQG '&( ZLWK WKH KHOS RI SURWUDFWRU WKHQ WDEXODWH WKHP
ILJXUH %$' '&( 5HVXOW
L
LL %$' '&(
LL %$' '&(
%$' '&(
&RQFOXVLRQ 7KH DERYH H[SHULPHQW VKRZV WKDW WKH H[WHULRU RU DQJOH RI WKH F\FOLF TXDGULODWHUDO LV HTXDO
WR WKH VXP RI LWV RSSRVLWH LQWHULRU DQJOHV
7KHRUHWLFDO SURRI
*LYHQ R LV WKH FHQWUH RI FLUFOH
$%&' LV D F\FOLF TXDGULODWHUDO VLGH LV %& LV
SURGXFHG XSWR (
7R SURYH %$' '&(
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3URRI 6WDWHPHQW 5HDVRQV
%$' %&'
EHLQJ RSSRVLWH DQJOHV RI D F\FOLF TXDGULODWHUDO
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3URYHG
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([DPSOH )LQG WKH YDOXHV RI x DQG y IURP WKH IROORZLQJ ILJXUHV
D % E % F
\ $ 2 [& &\ 2
\ R [
2 '
%
[
& $ R '
R (
$
Solution:
D L 2$& 2&$ R
LL $2& R 2$& >EHLQJ 2$ 2& EDVH DQJOHV@
R î R >VXP RI DQJOHV RI D WULDQJOH LV R@
R R
, QVFULEHG DQJOH LV KDOI RI WKH DQJOH DW WKH FHQWUH
?$2& [ R
RI D FLUFOH VWDQGLQJ RQ WKH VDPH DUF
LLL $%& $2&
>WKH\ DUH VWDQGLQJ RQ WKH VDPH DUF %&@
? $%& \ î R
>VWDQGLQJ RQ WKH VDPH DUF @
R 7KH H[WHULRU DQJOH RI D F\FOLF TXDGULODWHUDO
E L %'& %$& R
LL $&% R %$& $%&' LV HTXDO WR WKH LQWHULRU RSSRVLWH DQJOH
R ± R
7KH DQJOH DW WKH FHQWUH RI D FLUFOH LV WZLFH WKH
? [ R
LLL $&% $'% R LQVFULEHG DQJOH VWDQGLQJ RQ WKH VDPH DUF
? \ R
F L $'( $&% R
? \ R
LL $2% $&%
? [ R
([DPSOH ,Q WKH JLYHQ ILJXUH 234 R DQG 254 R )LQG 4
WKH PHDVXUHV RI 325 DQG 365
Solution: 2
-RLQ 4 DQG 2 35
L 243 234 R DQG 245 254 R
6
LL 345 243 245 %D VH DQJOHV RI DQ LVRVFHOHV WULDQJOH
R R R
ͮDĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬ *HRPHWU\
LLL 325 345
î R 7KH DQJOH DW WKH FHQWUH RI D FLUFOH LV WZLFH WKH
R
LQVFULEHG DQJOH VWDQGLQJ RQ WKH VDPH DUF 365
LY 365 R 345 6XP RI WKH RSSRVLWH DQJOHV RI D F\FOLF
R ± R TXDGULODWHUDO LV R &
%
R $
([DPSOH ,Q WKH DGMRLQLQJ ILJXUH 2 LV WKH FHQWUH
RI D FLUFOH ,I 2$% R ILQG $&%
2
Solution:
L 2$% 2%$ R >%HLQJ 2$ 2%@
LL $2% R 2$% >6XP RI DQJOHV RI D WULDQJOH LV R@
R ± î R
R ± R R
LLL 5HIOH[ $2% R ± R >2QH FRPSOHWH WXUQ LV R@
LY $&% UHIOH[ $2% ,QVFULEHG DQJOH LV KDOI WKH DQJOH DW WKH FHQWUH RI
î R R D FLUFOH VWDQGLQJ RQ VDPH DUF
([DPSOH ,Q WKH JLYHQ ILJXUH 2 LV WKH FHQWUH RI D FLUFOH LQ $ (
ZKLFK $2% LV D GLDPHWHU DQG $%&' LV D F\FOLF 2%
TXDGULODWHUDO ,I $(' R ILQG WKH VL]H RI %&'
Solution:
-RLQ % DQG (
L $(% R >DQJOH DW VHPL FLUFOH@ ' &
LL %(' R $('
R ± R R >EHLQJ FRPSOHPHQWDU\ DQJOHV@
LLL %&' R %(' >VXP RI WKH RSSRVLWH DQJOHV RI D F\FOLF TXDGULODWHUDO@
R ± R R (' 2 &
([DPSOH ,Q WKH DGMRLQLQJ ILJXUH $%&' LV D F\FOLF TXDGULODWHUDO
$
,I $(' R DQG $)% ILQG WKH %&' [
Solution: /HW %$) [
1RZ %$) '$( [ >YHUWLFDOO\ RSSRVLWH DQJOHV DUH HTXDO@ %
)
$'& [ >VLQFH H[WHULRU DQJOH LV HTXDO WR WKH VXP RI WZR LQWHULRU RSSRVLWH DQJOHV @
$%& [ >VLQFH H[WHULRU DQJOH LV HTXDO WR WKH VXP RI WZR LQWHULRU RSSRVLWH DQJOHV @
?$%& $'& >VXP RI WKH RSSRVLWH DQJOHV RI WKH F\FOLF TXDGULODWHUDO@
[ [
RU [
RU [
*HRPHWU\ DĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬͮ
? %$) %&' [ ([ WHULRU DQJOH RI D F\FOLF TXDGULODWHUDO LV HTXDO
? %&' WR LQWHULRU RSSRVLWH DQJOH
([DPSOH ,Q WKH JLYHQ ILJXUH 3456 LV D F\FOLF 3
TXDGULODWHUDO 36 LV SURGXFHG WR 7 LI 65 LV DQ 6
DQJOH ELVHFWRU RI 467 SURYH WKDW 35 45 7
Solution:
*LYHQ L 3456 LV D F\FOLF TXDGULODWHUDO 45
LL 65 LV DQ DQJOH ELVHFWRU RI 467 465 765
7R SURYH 35 45
Proof :
6WDWHPHQWV 5HDVRQV
465 765 *LYHQ
465 435 7KH\ DUH VWDQGLQJ RQ WKH VDPH DUF 45
765 435 )URP VWDWHPHQW DQG
765 345 ([WHULRU DQJOH RI WKH F\FOLF TXDGULODWHUDO LV
HTXDO WR WKH LQWHULRU RSSRVLWH DQJOHV
435 345 )URP VWDWHPHQW DQG
?35 45 %DVH DQJOHV EHLQJ HTXDO
Proved
([DPSOH ,Q WKH ILJXUH DORQJ VLGH $% &' 3URYH WKDW %0& $1'
Solution:
*LYHQ L $% &' $ %
'
LL %0& DQG '1$ DUH VWDQGLQJ RQ WKH DUFV %$& &
DQG $%' UHVSHFWLYHO\
7R SURYH %0& $1' 0
&RQVWUXFWLRQ -RLQ $ DQG ' 1
Proof :
6WDWHPHQW 5HDVRQ
%$' $'& %HLQJ DOWHUQDWH DQJOHV VLQFH $% &'
5HODWLRQ EHWZHHQ WKH DQJOH DW WKH FLUFXPIHUHQFH RI
WKH FLUFOH DQG WKH RSSRVLWH DUF
)URP VWDWHPHQW
$GGLQJ RQ ERWK VLGHV
:KROH SDUW D[LRP
'LYLGLQJ ERWK VLGHV E\
$1' %0& )URP 5HODWLRQ EHWZHHQ WKH DQJOH DW WKH
FLUFXPIHUHQFH RI WKH FLUFOH DQG WKH RSSRVLWH DUF
Proved
ͮDĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬ *HRPHWU\
([DPSOH ,Q WKH JLYHQ ILJXUH $34& DQG '34% $3 '
LQWHUVHFWHG DW 3 DQG 4 $% DQG &' SDVV %
WKURXJK WKH SRLQW 3 & $ ' DQG % DUH MRLQHG WR &
WKH SRLQW 4 3URYH WKDW $4& %4'
Solution:
*LYHQ L $34& DQG '34% DUH WZR LQWHUVHFWLQJ FLUFOHV ZKLFK 4
LQWHUVHFW DW 3 DQG 4
LL $% DQG &' SDVV WKURXJK WKH SRLQW 3
7R SURYH $4& %4'
Proof
6WDWHPHQWV 5HDVRQV
%4' %3' ,QVFULEHG DQJOHV VWDQGLQJ RQ WKH VDPH
DUF %'
%3' $3& 9HUWLFDOO\ RSSRVLWH DQJOHV DUF HTXDO
%4' $3& )URP VWDWHPHQW DQG
$3& $4& 6WDQGLQJ RQ WKH VDPH DUF $&
%4' $4& )URP VWDWHPHQW DQG
Proved
([DPSOH ,Q WKH DGMRLQLQJ ILJXUH '$%& LV DQ LQVFULEHG $
WULDQJOH ( LV WKH PLG SRLQW RI DUF %& ZKHUH ( '
DQG $ DUH MRLQHG 3URYH WKDW 01 %&
Solution: 01
*LYHQ L '$%& LV DQ LQVFULEHG WULDQJOH %( &
LL с
7R SURYH 01 %&
&RQVWUXFWLRQ -RLQ $ DQG '
Proof:
6WDWHPHQW 5HDVRQ
L ෞ L *LYHQ
LL %$( &$( LL $QJOHV DW WKH FLUFXPIHUHQFH RI FLUFOH VWDQGLQJ RQ
WKH HTXDO DUF DUH HTXDO
LLL %$( %'( LLL 6WDQGLQJ RQ WKH VDPH DUF %(
LY &$( %'( LY )URP LL DQG LLL RI
L 0$1 1'0 L )URP LY RI
LL ?0 $ ' DQG 1 DUH LL %HLQJ 0$1 0'1
FRLQ F\FOLF SRLQWV
LLL '$1 '01 LLL %RWK DQJOHV DUH VWDQGLQJ RQ WKH VDPH DUF '1
6LQFH 0 $ ' DQG 1 DUH FRQF\FOLF SRLQWV
L '$& '%& L $JDLQ WKH\ DUH VWDQGLQJ RQ WKH VDPH DUF '& RI
WKH FLUFOH
LL '01 '%& LL )URP VWDWHPHQWV LLL RI DQG L RI
LLL ?01 %& LLL )URP L FRUUHVSRQGLQJ DQJOHV EHLQJ HTXDO
VLQFH '01 '%&
Proved
*HRPHWU\ DĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬͮ
([HUFLVH
*URXS
$
)LQG WKH VL]H RI XQNQRZQ DQJOHV LQ WKH IROORZLQJ ILJXUHV ZKHUH 2 LV WKH FHQWUH RI WKH FLUFOH
& 2 4 ; '
R 2[ [
\
\ % R = 2
2 \
$ 35
[ R
< R ()
L LL LLL LY
& 3'
$ R 2 $ 2 ' 4 \ 6
[ \ ( R 2 [ *
\ [
R 2 \
[ %& R )
%
5
Y YL YLL YLLL
& ; : 4
< R $ R [7
$ [ = 5
2
2 \' \ 2 2
$ ;[ R
[
% = < 6
[L [LL
L[ [
$
% & '[ * \% 4 6
[ [ 2 [ [
$ R R
R \ 2
2 ) 2 3
\ ( ' \
]
R 5
' [LY & [YL
[Y
[LLL
$ ' ) $ '$ '
[ (% 2
% 2 [
' \ R [ & \
&
$ 2 [ &%
%
&
[YLL [YLLL [L[ [[
ͮDĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬ *HRPHWU\
4
D ,Q WKH JLYHQ ILJXUH 2 LV WKH FHQWUH RI WKH FLUFOH ,I 2 5
234 254 R ILQG WKH 345 325 DQG 3 7
567
6 %
E ,Q WKH DGMRLQLQJ ILJXUH 2 LV WKH FHQWUH RI WKH FLUFOH ,I &
2$% R ILQG WKH $&% $
2
F ,Q WKH JLYHQ ILJXUH 2 LV WKH FHQWUH RI WKH FLUFOH LQ ZKLFK 2
2345 LV D SDUDOOHORJUDP )LQG 234 35
G ,Q WKH JLYHQ ILJXUH 2 LV WKH FHQWUH RI WKH FLUFOH 2$ DQG 4
2& DUH DQJOH ELVHFWRUV RI %$& DQG %&$
UHVSHFWLYHO\ ,I $% $& ILQG $2& DQG 2$% $&
2
D ,Q WKH DGMRLQLQJ ILJXUH 2 LV WKH FHQWUH DQG %2( LV D
GLDPHWHU RI WKH FLUFOH ,I '%( [R DQG %(' [R %
ILQG WKH %$' DQG %&' $'
E ,Q WKH ILJXUH DORQJVLGH 2 LV WKH FHQWUH RI WKH FLUFOH ,I % [ [ (
%'& R DQG %$' R ILQG WKH 2$' 2
F ,Q WKH JLYHQ ILJXUH 2 LV WKH FHQWUH RI WKH FLUFOH DQG &
45A36 ,I 345 R ILQG WKH DQJOH 526
2&
'
$%
356
2
G ,Q WKH DGMRLQLQJ ILJXUH 3& %& DQG 3$' R 4%
$
)LQG WKH $3& 3
2
&
D ,Q WKH ILJXUH DORQJVLGH 2 LV WKH FHQWUH RI WKH FLUFOH ,I '
$
%' &' DQG $%' R ILQG WKH %$& %
2
'&
*HRPHWU\ DĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬͮ
$
%
E ,Q WKH JLYHQ ILJXUH $%&' LV D F\FOLF TXDGULODWHUDO ,I
'
R R$' &' DQG $%& R ILQG WKH &$'
F ,Q WKH DGMRLQLQJ ILJXUH $%&' LV D F\FOLF TXDGULODWHUDO &
,I $3% R DQG %4& R ILQG WKH $'&
(
3$
%2 '
&
4
G ,Q WKH DGMRLQLQJ ILJXUH 3456 LV D F\FOLF TXDGULODWHUDO ,I $3
365 R DQG 3$4 ILQG WKH 4%5
4 26
%5
*URXS
%
&
$
D ,Q WKH JLYHQ ILJXUH $% &' VKRZ WKDW (
%$' $(&
%
'
0
E ,Q WKH JLYHQ ILJXUH 34 56 3URYH WKDW 3 24
406 315 56
1
F ,Q WKH JLYHQ ILJXUH LI 1$6 0%5
SURYH WKDW 01 56 $
01
G ,Q WKH JLYHQ ILJXUH '( )* 3URYH 5 6
% 4
WKDW '4) *3(
3 (
D ,Q WKH JLYHQ ILJXUH $% &' 3URYH WKDW '
$(' %)&
)*
E ,Q WKH JLYHQ ILJXUH LI *;) (<+ ()
SURYH WKDW () *+
$ %
& '
( )
* +
;<
ͮDĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬ *HRPHWU\
F ,Q WKH DGMRLQLQJ ILJXUH 01 DQG 56 DUH WZR FKRUGV 06
ZKLFK LQWHUVHFW DW 3 ,I 0$A56 DQG 6%A01 SURYH
WKDW $% 51 3
G ,Q WKH JLYHQ ILJXUH 2 LV WKH FHQWUH RI WKH FLUFOH DQG $%
$% LV D GLDPHWHU ,I с SURYH WKDW $0 21 51
$
D ,Q WKH DGMRLQLQJ ILJXUH 34 DQG 56 DUH WZR FKRUGV RI 02
WKH FLUFOH ZKLFK LQWHUVHFW DW $ 3URYH WKDW
1
L 3$5 сŽ %
LL 4$5 сŽ
6
E $% DQG &' DUH WZR FKRUGV RI WKH FLUFOH WKDW 3 4
$ $
LQWHUVHFW H[WHUQDOO\ DW 3 3URYH WKDW
5 &
%3' сŽ
%
3
'
F ,Q WKH JLYHQ ILJXUH WZR FKRUGV $% DQG &' LQWHUVHFW DW $
ULJKW DQJOH DW ( RI WKH FLUFOH 3URYH WKDW ( '
L с & &
LL с
%
G ,Q WKH JLYHQ ILJXUH 021 LV D GLDPHWHU ZLWK FHQWUH 2 ,I 1
͕ SURYH WKDW $01 $%& $&% $
2
%
0
D ,Q WKH JLYHQ ILJXUH 3 DQG 4 DUH WKH FHQWUH RI WKH FLUFOHV %
ZKLFK LQWHUVHFW DW $ DQG & 3$ DQG 3& DUH SURGXFHG WR $
PHHW WKH RWKHU FLUFOH DW % DQG ' UHVSHFWLYHO\ 3URYH
WKDW $% &' 34
E ,Q WKH JLYHQ ILJXUH 2 LV WKH FHQWUH RI WKH FLUFOH ' DQG ) &
DUH WZR SRLQWV RQ WKH FKRUG $% VXFK WKDW $' %) '
&(
,I с SURYH WKDW $'& %)( 2
$' ) %
*HRPHWU\ DĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬͮ
$
F ,Q WKH JLYHQ ILJXUH WZR FKRUGV %$ DQG &$ PHHW DW $
0 DQG 1 DUH PLG SRLQWV RI DQG UHVSHFWLYHO\ 01 0 ( 2 ) 1
FXWV $% DQG $& DW ( DQG ) UHVSHFWLYHO\ 3URYH WKDW
$( $) %&
G
ǡ
A α
%'
()
Ȍ ǡ
&
$
Ƭ ǡ ǡ
Ǥ 0
$ %
) '
&( 2
D ,Q WKH ILJXUH DORQJVLGH () $& 3URYH WKDW
L % ( ) DQG ' DUH FRQF\FOLF SRLQWV
LL
E ,Q WKH DGMRLQLQJ ILJXUH 2 LV WKH FHQWUH RI D FLUFOH ,I
$%& &$( SURYH WKDW % ' ( DQG ) DUH
FRQFO\FOLF SRLQWV
$ )
F ,Q WKH JLYHQ ILJXUH 2 LV WKH FHQWUH RI WKH FLUFOH DQG &3 2 '
$%A&' DW 3 3URYH WKDW $2() LV D F\FOLF (
TXDGULODWHUDO %
$
54
G ,Q WKH DGMRLQLQJ ILJXUH 3 4 DQG 5 DUH PLG SRLQWV RI % 63 &
(
%& $& DQG $% UHVSHFWLYHO\ ,I $6 A %& SURYH
WKDW 3456 LV D F\FOLF TXDGULODWHUDO $
&
D ,Q WKH JLYHQ ILJXUH 2 LV WKH FHQWUH RI D FLUFOH 2 '
,I $% $& SURYH WKDW %' &' %
*HRPHWU\
ͮDĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬ
E ,Q WKH ILJXUH DORQJVLGH $%&' LV D F\FOLF '
TXDGULODWHUDO %& LV SURGXFHG WR ( VXFK WKDW $
$% &( DQG %' LV DQ DQJOH ELVHFWRU RI $%&
3URYH WKDW '% '( % &(
$
'
% &
F ,Q WKH ILJXUH $% '( DQG $%' &%' 3URYH ( 6
WKDW
3
2
/
45
G 3URYH WKDW 324 526 3/4
H ,Q WKH JLYHQ ILJXUH LV WKH FHQWUH RI FLUFOH &KRUGV
$& DQG %' LQWHUVHFWV DW ( 3URYH WKDW $(%
$2% &2%
)'
,Q WKH DGMRLQLQJ ILJXUH 2 LV WKH FHQWUH RI D FLUFOH $ 2& %
(
LQ ZKLFK $% LV D GLDPHWHU ,I &' &( SURYH
WKDW 2)& 2'&
ʹǤ
ܻܲ α ܯܻǡ
ܻܺԡ
&
͵Ǥ ,Q WKH DGMRLQLQJ ILJXUH FKRUGV $% DQG &' RI D $
03 1
FLUFOH LQWHUVHFW DW ULJKW DQJOH DW 3 1 LV WKH PLG ' %
SRLQW RI %& DQG 13 LV SURGXFHG WR PHHW $' DW
0 3URYH WKDW 30 A $'
ͶǤ
ǡ
ܰܯԡ Ǥ
ܺܯԡ
*HRPHWU\ DĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬͮ
ͷǤ
ǡ α ǡ
͵ ܶܵ α ܴܳ
Ǥ
ǡ
Ǥ
α
Ǥ ǡ
Ǥ
ǡ
Ǥ
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Ǥ
Ǥ
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Ǥ
A Ǥ
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Ǥ
ͳͳǤ ǡ
α ܳܯԡ ǡ
Ǥ
ͳʹǤ ǡ
ǡ Ϊ α ͳͺͲι
*HRPHWU\
ͮDĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬ
7DQJHQW $ 2
7%
$ WDQJHQW WR D FLUFOH LV D VWUDLJKW OLQH WKDW PHHWV DW RQH DQG RQO\ RQH SRLQW
VLQJOH SRLQW RQ WKH FLUFXPIHUHQFH RI WKH FLUFOH ,Q WKH DGMRLQLQJ ILJXUH 2 '%
$% LV D WDQJHQW DQG 7 LV D SRLQW RI FRQWDFW RI WKH FLUFOH :H FDQ GUDZ
QXPHURXV WDQJHQWV LQ D FLUFOH %XW ZH FDQ GUDZ RQO\ WZR WDQJHQWV IURP
WKH H[WHUQDO VLQJOH SRLQW WR D FLUFOH
6HFDQW $ VHFDQW LV D VWUDLJKW OLQH WKDW LQWHUVHFWV D FLUFOH ,W FXWV WKH
FLUFXPIHUHQFH DW DQ\ WZR SRLQWV ,Q WKH ILJXUH DORQJVLGH $&'% LV D
VHFDQW $ VWUDLJKW OLQH $% LQWHUVHFWV WKH FLUFOH DW WZR SRLQWV & DQG ' :H
FDQ GUDZ QXPHURXV VHFDQWV LQ D FLUFOH
&RPPRQ WDQJHQWV $&
,I D VWUDLJKW OLQH LV WDQJHQW WR WZR FLUFOHV WKHQ LW LV FDOOHG FRPPRQ WDQJHQW ,Q WKH DGMRLQLQJ ILJXUHV
$% LV WKH FRPPRQ WDQJHQW WR WKH FLUFOHV
$ $ $&
&
'
& %
% %
7KHRUHP
The tangent to the circle at any point is perpendicular to the radius at the point of contact.
([SHULPHQWDO YHULILFDWLRQ
Experiment: L 'UDZ WKUHH FLUFOHV RI GLIIHUHQW UDGLL ZLWK FHQWUH
2
LL 'UDZ WDQJHQWV $&% DW WKH SRLQW RI FRQWDFW & DQG MRLQ 2 DQG & LQ HDFK FLUFOH
22 2
&%
&% $&% $
$
To verify: 2& A $%
Verification: 0HDVXUH 2&$ RU 2&% DQG WDEXODWH
)LJXUH 2&$ RU 2&% 5HVXOWV
L 2&$ 2&% R
LL
LLL
Conclusion: 7KH DERYH H[SHULPHQW VKRZV WKDW WKH WDQJHQW WR WKH FLUFOH DW DQ\ SRLQW LV SHUSHQGLFXODU WR
WKH UDGLXV DW WKH SRLQW RI FRQWDFW
*HRPHWU\ DĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬͮ
7KHRUHP
The lengths of tangents to a circle drawn from an external point are equal to each other.
([SHULPHQWDO YHULILFDWLRQ
Experiment: L 'UDZ WKUHH FLUFOHV RI GLIIHUHQW UDGLL ZLWK FHQWUH
2
LL 'UDZ WDQJHQWV 3$ DQG 3% IURP DQ H[WHUQDO SRLQW 3 ZKHUH $ DQG % DUH SRLQW RI FRQWDFW
UHVSHFWLYHO\
$3
2
$ 3 2
%
%$
%2
LL LLL
3
L
To verify: 3$ 3%
Verification: 0HDVXUH WKH OHQJWK RI 3$ DQG 3% DQG WDEXODWH
)LJXUH 3$ 3% 5HVXOWV
L 3$ 3%
LL 3$ 3%
LLL 3$ 3%
Conclusion: 7KH OHQJWK RI WDQJHQWV WR D FLUFOH GUDZQ IURP DQ H[WHUQDO SRLQW DUH HTXDO
7KHRUHP
The angles made by a tangent of a circle with a chord at the point of contact are
respectively equal to the angles in the alternate segments of the circle.
([SHULPHQWDO YHULILFDWLRQ
Experiment: L 'UDZ WKUHH FLUFOHV RI GLIIHUHQW UDGLL ZLWK FHQWUH
2
LL 'UDZ WDQJHQWV 3$4 ZKHUH $ LV WKH SRLQW RI FRQWDFW
& & %
2
&%
$4
' 2% 2'
LLL
'
3 $ 43 $ 43
L LL
To verify: &$4 $'& DQG &$3 $%&
ͮDĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬ *HRPHWU\
Verification: 0HDVXUH WKH DQJOHV &$4 $'& &$3 DQG $%& DQG WDEXODWH
)LJXUH &$4 $'& 5HVXOWV &$3 $%& 5HVXOWV
L &$4 $'& &$3 $%&
LL &$4 $'& &$3 $%&
LLL &$4 $'& &$3 $%&
Conclusion: 7KH DERYH H[SHULPHQW VKRZV WKDW WKH DQJOH PDGH E\ D WDQJHQW RI D FLUFOH ZLWK D FKRUG DW
WKH SRLQW RI FRQWDFW DUH UHVSHFWLYHO\ HTXDO WR WKH DQJOHV LQ WKH DOWHUQDWH VHJPHQWV RI WKH
FLUFOH
:RUNHG 2XW ([DPSOHV
([DPSOH )LQG WKH VL]H RI DQJOHV PDUNHG IURP WKH IROORZLQJ FLUFOHV ZLWK FHQWUH 2
DQG SRLQW RI FRQWDFW ZLWK WDQJHQWV DUH DOVR JLYHQ DV LQ WKH ILJXUHV
D 6 E $ F
4
2
2[ 2 R
R 6[
3 45 R [ 357
% &'
D L 245 R >5DGLXV RI WKH FLUFOH LV SHUSHQGLFXODU WR WKH WDQJHQW DW WKH SRLQW RI
FRQWDFW
LL 246 264 R ± R R > 26 24 DQG FRPSOHPHQWDU\ DQJOHV @
LLL 426 [ R u R R R R >6XP RI WKH DQJOHV RI D WULDQJOH@
E L %$& R ± R R >5HPDLQLQJ DQJOH RI ULJKW DQJOHG WULDQJOH@
LL %2' u R R >&HQWUDO DQJOH LV GRXEOH WKH DQJOH DW WKH FLUFXPIHUHQFH @
LLL 2'% [ R R R >5HPDLQLQJ DQJOHV RI D ULJKW DQJOHG WULDQJOH@
F L 534 345 R > 35 45@
LL 754 u 345 u R R >([WHULRU DQJOH LV HTXDO WR WKH VXP RI WZR
LQWHULRU RSSRVLWH DQJOHV@
LLL 754 564 [ R >%HLQJ DQJOHV RI WKH DOWHUQDWH VHJPHQWV RI WKH FLUFOH @
([DPSOH ,Q WKH DGMRLQLQJ ILJXUH 2 LV WKH FHQWUH RI D 3 &
FLUFOH 3& LV D WDQJHQW DQG & LV WKH SRLQW RI 2
FRQWDFW 3$% LV D VHFDQW ZKHUH 3$ FP DQG $
$% FP )LQG WKH OHQJWK RI 3&
Solution: L 2 LV WKH FHQWUH RI D FLUFOH %
*LYHQ
LL 3& LV D WDQJHQW DQG & LV WKH SRLQW RI FRQWDFW
LLL 3$% LV VHFDQW
7R ILQG /HQJWK RI 3&
*HRPHWU\ DĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬͮ
&RQVWUXFWLRQ -RLQ $ DQG % ZLWK WKH SRLQW RI FRQWDFW &
D ,Q '3%& DQG '$3&
L %3& $3& >&RPPRQ DQJOH @
LL 3%& $&3 >$QJOH DW WKH DOWHUQDWH VHJPHQWV RI WKH FLUFOH DUH HTXDO @
LLL 3&% 3$& >5HPDLQLQJ DQJOHV RI D WULDQJOH @
?'3%& a'$3& >E\ $ $ $@
E $3&3 3%&3 >&RUUHVSRQGLQJ VLGHV DUH SURSRUWLRQDO@
F 3& %3 $3 FP u FP RU 3& FP FP
([DPSOH ,Q WKH DGMRLQLQJ ILJXUH 2 LV WKH FHQWUH RI D FLUFOH $ %
$% DQG $& DUH WZR WDQJHQWV ZKLFK DUH GUDZQ 2
IURP WKH H[WHUQDO SRLQW $ WR WKH FLUFOH ZKHUH %
DQG & DUH SRLQWV RI FRQWDFW UHVSHFWLYHO\ &
3URYH WKDW 2%& %$&
Solution:
*LYHQ L 2 LV WKH FHQWUH RI D FLUFOH
LL $% DQG $& DUH WDQJHQWV ZKHUH % DQG & DUH SRLQWV RI FRQWDFW
7R SURYH 2%& %$&
Proof:
6WDWHPHQWV 5HDVRQV
$%& $&% %HLQJ $% $&
%$& R $%&
%$& R $%& 6XP RI WKH DQJOHV RI D WULDQJOH LV
R
'LYLGLQJ ERWK VLGHV E\
2%$ R 5DGLXV RI WKH FLUFOH LV SHUSHQGLFXODU
WR WKH WDQJHQW DW WKH SRLQW RI FRQWDFW
2%& R $%&
2%& %$& %HLQJ FRPSOHPHQWDU\ DGMDFHQW
DQJOHV
)URP DQG
Proved
ͮDĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬ *HRPHWU\
([HUFLVH
)LQG WKH VL]H RI DQJOHV PDUNHG IURP WKH IROORZLQJ FLUFOHV ZLWK FHQWUH 2 DQG SRLQW RI FRQWDFW
ZLWK WDQJHQWV DV JLYHQ LQ WKH ILJXUHV
% '
\
2 $[ 2 5 2 6[ 2 R &
R $
\
'
[ R
[\
$ % && % '3 7 4
L LL LLL LY
0 3 $
' $ 2 [%
\
2%
$ 2 \% ' \ 2 R )
[
R R [
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1 4% '& (
Y YL YLL YLLL
'& 4 5
R \ 2
R
$2 [ %&
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$
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3
$ (& % ' % \&[ '
\
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2 2[ D
2 2 D
[
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% &' 4 56
$ $
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DĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬͮ
*HRPHWU\
3 3 5 33 5
R \ $ \\
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[ R
2
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& V [L[
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[YLL [YLLL
D ,Q WKH DGMRLQLQJ ILJXUH 2 LV WKH FHQWUH RI WKH FLUFOH $
'%( LV D WDQJHQW ZKHUH % LV WKH SRLQW RI FRQWDFW ,I 2 &
$%' R DQG %2& R ILQG $%2 DQG R
$&2 R
'% (
E ,Q WKH JLYHQ ILJXUH 2 LV WKH FHQWUH RI WKH FLUFOH $% $ 3
LV D WDQJHQW ZKHUH 4 LV WKH SRLQW RI FRQWDFW ,I D
435 354 ILQG WKH VL]H RI 465 DQG $43 4 2
D 6
%5
$
F ,Q WKH JLYHQ ILJXUH 2 LV WKH FHQWUH RI WKH FLUFOH '&(
LV D WDQJHQW ZKHUH & LV WKH SRLQW RI FRQWDFW ,I R )
&' &) DQG %$& R ILQG WKH $%)
2
%
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G ,Q WKH DGMRLQLQJ ILJXUH 2 LV WKH FHQWUH RI WKH FLUFOH $ '
$%&' LV D F\FOLF TXDGULODWHUDO 3%4 LV D WDQJHQW ZLWK 2
SRLQW RI FRQWDFW % ,I %& &' DQG '%& R ILQG &
WKH 3%'
3 % 4
%
D ,Q WKH DGMRLQLQJ ILJXUH 37 LV D WDQJHQW DQG 3$% LV $2
VHFDQW ,I $% 3$ FP ILQG WKH OHQJWK RI 37
37
E ,Q WKH JLYHQ ILJXUH 2 LV WKH FHQWUH RI WKH FLUFOH ,I 2
4
25 45 FP DQG 35 FP ILQG WKH 345 DQG 5
OHQJWK RI 34
3
ͮDĂƚŚĞŵĂƚŝĐƐ Ͳ ϭϬ *HRPHWU\
F ,Q WKH JLYHQ ILJXUH 2 LV WKH FHQWUH RI WKH FLUFOH %$ 2
LV WDQJHQW ZKHUH $ LV SRLQW RI FRQWDFW ,I 2& FP &
DQG %& FP ILQG WKH OHQJWK RI $%
% $
$
G ,Q WKH JLYHQ ILJXUH ;< LV D GLDPHWHU ZLWK FHQWUH 2 3 <
3$ DQG 3% DUH WDQJHQWV GUDZQ IURP WKH H[WHUQDO 2
SRLQW 3 RI WKH FLUFOH ,I 3; FP DQG ;< FP ;
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Exercise: 5.1.1 (Area of Triangle and Quadrilateral)
1. 54cm2 2. 8 2 cm 3. 5cm 4. 10cm 5. 13cm 6. 5 2 cm 7. 4cm 8. 5cm
9. 9cm2 10. 27cm2 11. 48cm2 12. 35cm2 13. 12cm 14. 3cm
Exercise: 5.1.2
Group ‘A’ 2. 10cm2 3. 9 3 cm2 4. 30cm2 5. 6cm2, 12cm2 6. 4cm
1. 18cm2 8. 5cm 9. 7cm2 10. 40cm2
7. 6 6 cm2
Exercise: 5.3 (Circle)
Group ‘A’
1. i) 54° ii) 43° iii) 112°, 56° iv) 48°, 96° v) 26°, 26° vi) 43°, 43° vii) 102°, 129°
x) 27°, 54° xiv) 28°, 96°
viii) 35° ix) 45°, 45° xvii) 90° xi) 88° xii) 82° xiii) 132°, 48°
b) 129° d) 30°
xv) 102°, 78° xvi) 43°, 86°, 47° d) 28° xviii) 44°, 44° xix) 90°, 45°
xx) 90° 2. a) 54°, 108°, 54° c) 60° d) 30°, 120° 3. a) 126°, 54°
b) 22° c) 64° 4. a) 92° b) 48° c) 45°
Exercise: 5.4 (Tangent)
Group ‘A’ ii) 48°, 84° iii) 45°, 45° iv) 32°, 58° v) 62°, 28° vi) 58°, 32° vii) 52°, 52°
1. i) 27° ix) 21°, 21° x) 52°, 76° xi) 84° xii) 114° xiii) 48°, 66° xiv) 68°, 22°
xvi) 30°, 60° xvii) 118°, 59° xviii) 92°, 88° xix) 136°, 44° xx) 54°, 72° 2. a) 16°, 37°
viii) 52°, 52°
xv) 29°, 61° c) 24cm d) 12cm, 12cm
b) 30°, 60° c) 34° d) 132° 3. a) 6 6 cm b) 30°, 12cm
236|Mathema s - 10 Geometry
Chapter
6 Trigonometry
Objectives
At the end of this chapter, the Students will be able to
• calculate the area of triangle and quadrilateral by using trigonometrical
formulae.
• Solve the problems related to height and distance by using trigonometrical
ratios.
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chart of trigonometrical ratios and their values with standard angles, angle
of elevation and angle of depression by using local materials if necessary,
model of clinometer, picture of sextant and theodilite.
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Prime Mathematics 10
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