Weeks Secants, Tangents, Segments, and
5–6 Sectors of a Circle
INTRODUCTION
After going through this lesson, you are expected to illustrate secants, tangents, segments,
and sector of a circle, prove theorems on secant and tangent segments of a circle; and solve
problems on circles.
This lesson will add more terminologies to what you have previously learned terms related to
circles and explore the theorems on intersecting chords, tangent lines and secant lines.
DISCUSSION
Learning Task 1
Use the figure below to match the definition in Column A with its description in Column B. Copy
and answer this in your answer sheet:
Column A Column B
1. A line segment, a line, or a ray that
intersects a circle at ⊙B. a. ´
BC
2. The point of tangency between a tangent
and the circle. b. ´
DC
3. A line segment, a line, or a ray that
intersects a circle at exactly two points. c. ´
EC
4. A secant segment with one endpoint on
the circle and one endpoint outside the d. ´
BC
circle.
5. An external secant segment is the part of
a secant segment that is outside a circle. e. point B
6. The tangent segment is a segment whose
endpoints are the point of tangency and f. ´
DC
the fixed point outside the circle.
Sector of a Circle
A sector of a circle is the region bounded by an arc of the circle and the two radii to the
endpoints of the arc. To find the area of a sector of a circle, get the product of the ratio
measure of the arc
and the area of the circle.
360
π r2 c
Formula: Area of sector =
360
where: c is the central angle in degrees
r is the radius of the circle
π is Pi, approximately 3.14
Illustrative Example 1:
Find the area of the sector of radius 6 cm and its central angle is 70°. Note that the measure of
the central angle is equal to its intercepted arc.
π r2 c
Formula: Area of sector =
360
(3.14) ( 6 )2 (70)
= sq. cm.
360
= 21.99 sq. cm.
Segment of a Circle
A segment of a circle is the region bounded by an
arc and its chord.
To find the area of the shaded segment, subtract the
area of triangle from the area of a sector.
Illustrative Example 2:
Solve the area of the shaded region.
Solution:
Area of the segment = Area of the sector – Area of the Rectangle
π r2 c 1 2
Area of the segment = – r sin C
360 2
(8¿¿ 2)60 1
Area of the segment = (3.14) ¿ – ( 8¿¿ 2) ¿sin 60
360 2
Area of the segment = 5.78 sq. m.
Arc Length
The length of an arc can be determined by using the proportion
measure of the central angle arc length A l
= or = , where A is the degree measure of the
360 2π r 360 2π r
arc, r is the radius of the circle, and l is the arc length.
In the given proportion, 360 is the degree measure of the whole circle, while 2 πr is the
circumference.
Illustrative Example 3:
If ∠ BEA=90 ° and the radius is 6 cm, what is the length of arc intercepted
by the angle?
Solution:
A l
360
= 2π r
90 l
360
= 2(3.14)(6 cm)
360l = 2(90)(3.14)(6 cm)
360l 3391.2
360
= 360
= 9.42 cm
EVALUATION
Learning Task 1
In⊙A below, m∠LAM = 42, m∠HAG = 30, and ∠KAH is a right angle. Find the following measure
of an angle or an arc.
1. m∠LAK = _____
2. m∠JAK = _____
3. m∠LAJ = _____
4. m∠JAH = _____
5. m∠KAM = _____
6. m ^
LK = _____
7. m ^
JK = _____
8. m ^
LMG = _____
9. m ^
JH = _____
10. m ^
KLM = _____
Learning Task 2
The radius of ⊙O below is 5 units. Find the length of each of the following arcs given the degree
measure. Answer the questions that follow.
1. m^PV = 45; length of ^
PV = _______
2. m PQ = 60; length of PQ
^ ^ = _______
3. mQR = 90; length of QR
^ ^ = _______
4. m RTS = 120;
^ length of RTS
^ = _______
5. mQRT = 95;
^ length of QRT
^ = _______
Learning Task 3
Find the area of the shaded region of each circle. Show your solutions on the separate paper.
1. 2.
3. 4.
TEACHER’S REMARKS: _______________________________________________________
