7.1 Introduction
7.2 Lines Of A Circle
7.3 Arcs
7.4 Inscribed Angles
7.5 Some Properties Of Tangents, Secants And Chords
7.6 Chords And Their Arcs
7.7 Segments Of Chords, Secants And Tangents
7.8 Lengths of Arcs And Areas Of Sectors
This powerpoint presentation is an introduction for the topic TRIANGLE CONGRUENCE. This topic is in Grade 8 Mathematics. I hope that you will learn something from this sides.
This powerpoint presentation is an introduction for the topic TRIANGLE CONGRUENCE. This topic is in Grade 8 Mathematics. I hope that you will learn something from this sides.
Grade 9 (Alternate) Mathematics III - Learning Modules for EASE Program of DepEdR Borres
Grade 9 (Alternate) Mathematics III - Learning Modules for EASE Program of DepEd
A compilation of Math III learning modules for EASE which can be alternate for Grade 9 Mathematics.
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Learn about the properties of tangents, chords and arcs of the circle. Learn to find measure of the inscribed angle and the property of cyclic quadrilateral
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2. Chapter 7 Circle
7.1 Introduction
7.2 Lines Of A Circle
7.3 Arcs
7.4 Inscribed Angles
7.5 Some Properties Of Tangents, Secants And Chords
7.6 Chords And Their Arcs
7.7 Segments Of Chords, Secants And Tangents
7.8 Lengths of Arcs And Areas Of Sectors
3. Chapter 7 Circle
› 7.1 Introduction
› A circle is defined as a set of all such points in a given
plane which lie at a fixed distance from a fixed point in the
plane. This fixed point is called the center of the circle and
the fixed distance is called the radius of the circle (see
figure 7.1).
Figure 7.1
4. 7.1 Introduction
› Figure 7.1 shows a circle where point P is
the center of the circle and segment PQ
is known as the radius. The radius is the
distance between all points on the circle
and P. It follows that if a point R exists
such that (seg.PQ) > (seg.PR) the R is
inside the circle. On the other hand for a
point T if (seg.PT) > (seg.PQ) T lies
outside the circle. In figure 8.1 since l
(seg.PS) = (seg.PQ) it can be said that
point S lies on the circle.
Figure 7.1
5. 7.2 Lines of a Circle
› The lines in the plane of the circle are classified into three
categories (figure 7.2).
› a) Lines like l which do not intersect the circle.
› b) Lines like m which intersect the circle at only one point.
› c) Lines like n which intersect the circle at two points.
6. 7.2 Lines of a Circle
› Lines like m are called tangents. A
tangent is a line that has one of its
points on a circle and the rest outside
the circle.
› Line n is called a secant of the circle. A
secant is defined as any line that
intersects a circle in two distinct points.
› A segment whose end points lie on a
circle is called a Chord . In figure 7.2
AB is a chord of the circle. Thus a
chord is always a part of secant.
Figure 7.2
7. 7.2 Lines of a Circle
› A circle can have an infinite number of
chords of different lengths (figure 7.3)
› The longest chord of the circle passes
through its center and is called as
the diameter. In figure 7.3 chord CD is the
diameter. It can be noticed immediately that
the diameter is twice the radius of the circle.
The center of the circle is the midpoint of the
diameter. A circle has infinite diameters and
all have the same length.
8. 7.2 Lines of a Circle
› Example 1: A, B, C & D lie on a circle with center P. Classify the
following segments as radii and chords.
› Example 2: Name the secant and the tangent in the following
figure
› Example 3
› P is the center of a circle with radius 5 cm. Find the length of the
longest chord of the circle.
9. 7.3 Arcs
› The angle described by any two radii of a circle is called
the central angle. Its vertex is the center of the circle. In
figure 7.4 ∠APB is a central angle. The part of the circle
that is cut by the arms of the central angle is called an arc.
AB is an arc and so is AOB . They are represented as 𝐴𝐵
and 𝐴𝑂𝐵.
10. 7.3 Arcs
› 𝐴𝐵 is called the minor arc and 𝐴𝑂𝐵 is the major arc. The
minor arc is always represented by using the two end
points of the arc on the circle. However it is customary to
denote the major arc using three points. The two end
points of the major arc and a third point also on the arc. If
a circle is cut into two arcs such that there is no minor or
major arc but both the arcs are equal then each arc is
called a semicircle.
› An arc is measured as an angle in degrees and
also in units of length. The measure of the angle
of an arc is its central angle and the length of
the arc is the length of the portion of the
circumference that it describes.
11. 7.3 Arcs
› Since the measure of the angle of an arc is its central
angle, if two central angles have equal measure then the
corresponding minor arcs are equal.
› Conversely if two minor arcs have equal measure then
their corresponding central angles are equal.
12. 7.4 Inscribed angles
› Whereas central angles are formed by radii, inscribed
angles are formed by chords. As shown in figure 7.5 the
vertex o of the inscribed angle AOB is on the circle. The
minor arc 𝐴𝐵 cut on the circle by an inscribed angle is
called as the intercepted arc.
Figure 7.5
13. 7.4 Inscribed angles
› Theorem: The measure of an inscribed angle is half the
measure of its intercepted arc.
› Proof: For a circle with center O, ∠BAC is the inscribed
angle and arc BXC is the intercepted arc. To prove that m
∠ BAC = ½ m(arc BXC). There arise three cases as shown
in figure 7.6(a), 7.6(b) and 7.6(c).
14. 7.4 Inscribed angles
Figure 7.6 (a)
Case 1: The center is on the angle figure 7.6 (a) join c to O.
D OAC is an isosceles triangle as seg.OA = seg.OC.
Assume m OAC = m OCA = P
m COA = 180 - 2P as sum of all the angles of a triangle is 1800.
COA & COB form a linear pair. Therefore they are supplementary.
m COB = 1800 - m COA
= 1800 - ( 1800 - 2P)
= 2P
= 2 m BAC
But m BAC = m (arc BXC) m BAC = ½ m (arc BXC)
15. 7.4 Inscribed angles
› Theorem: If two inscribed angles intercept the same arc or
arcs of equal measure then the inscribed angles have equal
measure.
› In figure 7.7, ∠CAB and ∠CDB intercept
the same arc CXB.
› Prove that m ∠CAB = ∠ CDB.
x
16. 7.4 Inscribed angles
› From the previous theorem it is known that
› m ∠ CAB = ½ m(arc CXB) and also
› m ∠ CDB = ½ m(arc CXB)
› Therefore m ∠ CAB = m ∠ CDB
Therefore if two inscribed angles intercept the same arc or
arcs of equal measure the two inscribed angles are equal in
measure.
17. 7.4 Inscribed angles
› Theorem: If the inscribed angle intercepts a semicircle the
inscribed angle measures 900.
› The inscribed angle ∠ACB intercepts
a semicircle arc AXB (figure 7.8). We
have to prove that m ∠ ACB = 900.
figure 7.8
18. 7.4 Inscribed angles
› m ∠ACB =1/2 m(arc AXB)
=1/2(180o)
=90o
Therefore if an inscribed angle intercepts
a semicircle the inscribed angle is a right
angle.
figure 7.8
19. Exercises
› Example 1
› a) In the above figure name the central
angle of arc AB.
› b) In the above figure what is the measure
of arc AB.
› c) Name the major arc in the above figure.
20. Exercises
› Example 2
› a) In the above figure name the inscribed
angle and the intercepted arc.
› b) What is m (arc PQ)
21. Exercises
› Example 3
› ∠PAQ and ∠PBQ intercept the same arc PQ what is the
m∠PBQ and m(arc PQ) ?
22. 7.5 Some properties of tangents, secants and
chords
› Theorem: If the tangent to a circle and the radius of the
circle intersect they do so at right angles:
Figure 7.9 (a) Figure 7.9 (b)
In figure 7.9 (a) l is a tangent to the circle at A and PA is the radius.
23. 7.5 Some properties of tangents, secants and
chords
› To prove that PA is perpendicular to l , assume that it is
not.
› Now, with reference to figure 7.9 (b) drop a perpendicular
from P onto l at say B. Let D be a point on l such that B is
the midpoint of AD.
figure 7.9 (b)
24. 7.5 Some properties of tangents, secants and
chords
› In figure 7.9 (b) consider DPDB and DPAB
› seg.BD ≅ seg.BA ( B is the midpoint of AD)
› ∠PBD ≅ ∠PBA ( PB is perpendicular to l )
and seg.PB = seg.PB (same segment)
› DPBD ≅ DPAB (SAS)
› seg.PD ≅ seg.PA corresponding sides of congruent
triangles are congruent.
25. 7.5 Some properties of tangents, secants and
chords
› D is definitely a point on the circle
because (seg.PD) = radius.
› D is also on l which is the tangent. Thus l
intersects the circle at two distinct points A and D. This
contradicts the definition of a tangent.
› Hence the assumption that PA is not perpendicular to l is
false. Therefore PA is perpendicular to l.
26. Angles formed by intersecting chords, tangent
and chord and two secants:
› If two chords intersect in a circle, the angle they form is
half the sum of the intercepted arcs.
› In the figure 7.10 two chords AB and CD intersect at E to
form ∠1 and ∠2.
figure 7.10
27. Angles formed by intersecting two chords
› m∠1 =1/2 (m seg.AD + m seg.BC) and
› m∠2 =1/2 (m seg.BD + m seg.AC)
28. Angles formed by intersecting tangent and
chord
› Tangent Secant Theorem: If a chord intersects the tangent
at the point of tangency, the angle it forms is half the
measure of the intercepted arc.
› In the figure 7.11, l is tangent to the circle. Seg.AB which is
a chord, intersects it at B which is the point of tangency.
Figure 7.11
m∠1 =1/2 (m arc ACB) and
m∠2 =1/2 (m arc AB), also
m∠𝐴𝐵𝐶= ½(m arc AXB) =90o
Figure 7.12
29. Angles formed by intersecting two secants
› If two secants intersect outside a circle half the difference in
the measures of the intercepted arcs gives the angle formed
by the two secants.
› In figure 7.13, l and m are secants. l and m intersect at O
outside the circle. The intercepted arcs are 𝐴𝐵 and 𝐶𝐷.
figure 7.13
mCOD = 1/2 ( m 𝐴𝐵 - m 𝐶𝐷)
30. 7.5 Some properties of tangents, secants and
chords
› Conclusion :
› (a) If two chords intersect in a circle the angle formed is
half the sum of the measures of the intercepted arcs.
› (b) Angle formed by a tangent and a chord intersecting at
the point of tangency is half the measure of the
intercepted arcs.
› (c) Angle formed by two secants intersecting outside the
circle is half the difference of the measures of the
intercepted arcs.
31. Exercises
› Example 1
› In the above figure seg.AB and
seg.CD are two chords
intersecting at X such that
mAXD = 1150 and m (arc CB)
= 450 . Find m arc APD.
Solution: m AXD =1/2( m (arc APD) + m (arc CB) )
m (arc APD) = 2 m AXD - m (arc CB) )
m (arc APD) = 2(115o) – 45o
m (arc APD) = 185o.
32. Exercises
› Example 2
› l is a tangent to the circle at B. Seg. AB
is a chord such that m ∠ABC = 500.
Find the m (arc AB).
Example 3
l and m are secants to the circle
intersecting each other at A.
The intercepted arcs are arc PQ
and arc RS if m∠PAQ = 250 and
m ∠ROS = 800 find m (arc PQ).
33. 7.6 Chords and their arcs
› Theorem: If in any circle two chords are equal in length
then the measures of their corresponding minor arcs are
same.
› As shown in figure 7.14 AB and CD are congruent chords.
Therefore according to the theorem stated above m(arc
AB) = m(arc CD) or mAOB = mCOD.
Figure 7.14
34. 7.6 Chords and their arcs
To prove this, join A, B, C & D with O.
Consider D AOB and D COD
seg.AO @ seg.CO and seg.OB @ seg.OD (radii of a circle are
always congruent).
seg.AB @ seg.CD (given), D AOB @ D COD ( S S S )
AOB @ COD (corresponding angles of congruent triangles are
congruent). Hence arc AB @ arc CD.
Conversely it can also be proved that in the same circle or
congruent circles, congruent arcs have their chords congruent.
35. 7.6 Chords and their arcs
› Theorem: The perpendicular from the
center of a circle to a chord of the circle
bisects the chord.
› In figure 7.15, XY is the chord of a circle with
center O. Seg.OP is the perpendicular from
the center to the chord. According to the
theorem given above seg.XP = seg.YP.
Figure 7.15
O
P
36. 7.6 Chords and their arcs
To prove this, join OX and OY
Consider D OXP and D OXY
Both are right triangles.
hypotenuse seg. OX @ hypotenuse seg. OY
( both are radii of the circle )
seg. OP @ seg OP (same side)
D OXP @ D OYP (H.S.)
seg XP @ seg. YP (corresponding sides of congruent triangles are congruent).
P is the midpoint of seg.XY.
Hence seg.OP which is the perpendicular from the center to the chord
seg.XY bisects the chord seg.XY.
Figure 7.15
O
P
37. 7.6 Chords and their arcs
› Now consider two chords of equal length in
the same circle. Their distance from the
center of the circle is same.
› In figure 7.16 seg. PQ and seg. RS are two
chords in the circle with center O such that
(seg.PQ) = (seg.RS).
› seg.OX and seg. OY are the perpendicular
distances from O to seg.PQ and seg.RS
respectively.
Figure 7.16
38. 7.6 Chords and their arcs
To prove that l (seg.OX) = l (seg.OY) join O to P and R.
Since perpendicular from the center to the chord bisects
the chord l ( seg.PX) = l(seg.RY).
Now consider D POX and D ROY
seg. PX @ seg.RY
seg.OP @ seg.OR. Both being radii onto the circle.
D POX @ D ROY (by H.S.)
seg.OX @ seg.OY (corresponding sides of congruent
triangles are congruent).
Thus if two chords are equal in measure they are equidistant from the center
of the circle.
39. 7.6 Chords and their arcs
› The converse of this theorem is that if two
chords are equidistant from the center of the
circle, they are equal in measure.
› As shown in figure 7.17 if seg. HI and seg. JK
are two chords equidistant from the center of
the circle, they are equal in length.
Figure 7.17
40. Exercises
› Example 1
› AB and CD are chords in a circle with center
O. (seg.AB) = (seg.CD) = 3.5cm & m∠COD = 950.
Find m arcAB.
› Example 2
› PQ is a chord of a circle with center O. Seg.OR is a
radius intersecting PQ at right angles at point T. If PT
= 1.5cm and m arcPQ = 800, find (PQ) and m arcPR.
› Example 3
› Seg HI and seg. JK are chords of equal measure in a
circle with center O. If the distance between O and
seg. HI is 10 cm find the length of the perpendicular
from O onto seg.JK.
41. 7.7 Segments Of Chords, Secants And
Tangents
› Theorem : If two chords, seg.AB and seg.CD intersect
inside a circle at P then
(seg.PA)(seg.PB)=(seg.PC)(seg. PD)
Figure 7.18 (a)
42. 7.7 Segments Of Chords, Secants And
Tangents
› Theorem : If two chords, seg.AB and seg.CD intersect
outside a circle at P then
(seg. PA)(seg. PB) = (seg. PC)(seg. PD)
Figure 7.18 (b)
43. 7.7 Segments Of Chords, Secants And
Tangents
› Consider a secant PAB to a circle, (figure 7.22) intersecting
the circle at A and B and line PT is a tangent then
(seg. PA)(seg. PB) = (seg. PT)2.
Figure 7.19
44. 7.7 Segments of chords secants and tangents
› Theorem: The lengths of two tangent segments from an
external point to a circle are equal.
› As shown in figure 7.20 seg. QR and seg. QS are two
tangents on a circle with P as its center.
Figure 7.19
45. 7.7 Segments of chords secants and tangents
› Example 1
› Two chords seg AB and seg. CD intersect in
the circle at P. Given
that (seg.PC)=(seg.PB)=1.5cm and (seg.PD)
= 3 cm. Find (seg.AP).
› Example 2
› Seg.PA and seg.PC are two secants
where (seg.PB)=3.5 cm, (seg.PD) = 4 cm and
(seg.DC) = 3 cm. Find (seg.AB).
› Example 3
› PT is a tangent intersecting the secant
through AB at P. Given l (seg. PA) = 2.5 cm.
and l (seg.AB) = 4.5 cm., find l (seg PT).
46. 7.8 Lengths of arcs and areas of sectors
› An arc is a part of the circumference of the circle; a part
proportional to the central angle.
› If 3600 corresponds to the full circumference. i.e. 2 p r
then for a central angle of x0(figure 7.20) the
corresponding arc length will be l such that
Figure 7.20
360 𝑜
2𝜋𝑟
=
𝑚 𝑜
𝑙
47. 7.8 Lengths of arcs and areas of sectors
› Analogically consider the area of a sector. This too is
proportional to the central angle. 3600 corresponds to
area of the circle 𝜋𝑟2. Therefore for a central angle m0 the
area of the sector will be in the ratio :
𝜋𝑟2
360 𝑜
=
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑠𝑒𝑐𝑡𝑜𝑟
𝑚 𝑜
48. 7.8 Lengths of arcs and areas of sectors
› Example 1
› In a circle with the radius of 2 cm, the central angle for an
arc AB is 750. Find (seg.AB). Also find the area of the
sector AOB having a central angle of 750
› Solution:
› (seg.AB) =
(75 𝑜)(2𝜋)(2)
360 𝑜 = 2.6 cm.
› Area of sector AOB=
(𝜋)(22)(75 𝑜)
360 𝑜 =2.6cm2