GRADE 10_MATHEMATICS GEOMETRY PERMUTATION.pptx

GRADE 10 MATHEMATICS (GEOMETRY,
PERMUTATIONS AND COMBINATIONS PARTS ONLY)
SAMSUDIN N. ABDULLAH, Ph.D.
Master Teacher II
Esperanza National High School
Esperanza, Sultan Kudarat, Region XII, Philippines
Email Address: samsudinabdullah42@yahoo.com

Prepared By:
SAMSUDIN N. ABDULLAH, Ph.D.
Master Teacher I
Esperanza National High School
Poblacion, Esperanza, Sultan Kudarat

Trapezoid
Parallelogram
Triangle
Rectangle
Circle
Oblong
Rhombus
Pentagon
Which of the following is considered to be the simplest and the most
perfect figure? Why?
Square

ACTIVITY
Use your protractor and answer the following questions.
Consider Figure 1 for numbers 1 – 4 and Figure 2 for numbers 5 – 8.
75o
90o
A
B
C
1. m∠A = ______. How is it related to the
measure of its intercepted arc BC which
measures 90o
?
________________________________
2. m∠B = 75o
. Its intercepted arc is AC which
measures ______.
3. The intercepted arc of ∠C is AB. If mAB = 120o
,
then m∠C = ______.
4. The measure of an inscribed angle is _______
(twice, one-half, thrice, one-third) the measure
of its intercepted arc.
5. m∠M = 95o
and m∠R = ______.
6. m∠M + m∠R = ______.
7. If m∠O = 100o
, then m∠E = ______.
8. Opposite angles of an inscribed quadrilateral
are _________________________
(complementary, supplementary).
M
O
R
E
Figure 1
Figure 2
120o

The first angle of an
inscribed triangle is ten less
than thrice the second angle. If
the intercepted arc of the third
angle is one-half the measure
of the circle, find the measure
of the first angle.

The BEST Teaching Strategy is
where the students learn BEST.
Thank you so much for inviting
me as a demonstration teacher.
Hope to see you in the
Department of Education,
equipped with highest teaching
competency.

Use the Pythagorean Formula to find the value of
x. Corresponding points are given for each item.
1.
8 cm
9 cm
x Answer: 𝟏𝟒𝟓 cm
2.
11 m
10 m
x Answer: 𝟐𝟏 m
2 points
2 points

Answer: 𝟒 𝟏𝟏 units
3.
x
15
7
Answer: 8 𝟕 units
4.
6
2
x
2 points
3 points

Answer: 2 𝟔𝟏 cm
6.
5
7
x
Answer: 2 𝟏𝟎𝟓 units
4 points
5.
13 cm
5 cm
15 cm
x
5 points

7.
8
11
x
Answer: 𝟐𝟑𝟓 units
3 points
8.
90o
cm
cm
x
Answer: 4 cm
3 points

Answer: 𝟗𝟑 cm
5 points
Answer: 𝟓𝟕 cm
10.
11 cm
16 cm
x
9.
17 cm
10 cm
12 cm
x
5 points

Lesson4 SegmentsFormedbySecantsandTangents
Iftwosecantsintersectintheexteriorofthecircle,theproductofthelengthofone
secantsegmentandthelengthofitsexternalpartisequaltotheproductofthelengthofthe
othersecantsegmentandthelengthofitsexternalpart.
C
A
D
B
Q
Formula: QA • QB = QC • QD
Example: If QA = 17, QC = 19, QD = 10, find QB.
Solution:
17QB = 19(10)
QB =
QB = 11.18 units

If two chords intersect in the interior of the circle, then the product of the lengths of the
segments of the chord is equal to the product of the lengths of the segments of the other chord.
If a tangent segment and a secant intersect in the exterior of a circle, then the square of
the length of the tangent segment is equal to the product of the length of the secant segment
and its external part.
C
B
D
A
K
Formula: AK • BK = CK • DK
Example: If AK = 3, BK = 5, CK = 6, find DK.
Solution:
3(5) = 6DK = DK or
15 = 6DK DK = 2.5 units
A
P
Q
Formula: PQ2
= AQ • BQ
Example: If PQ = 5, BQ = 3, find AB.
Solution:
52
= 3AQ AQ = 8.33 units
25 = 3AQ AB = 8.33 – 3
= AQ or AB = 5.33 units
B

1. Two chords ST and KP intersect at Q. If the segments of ST have lengths 6 cm and 8 cm,
find the length of the other segment of the chord if the length of one of the segments is
3 cm.
Figure:
S
T
K
P
Q
Solution:
SQ • QT = KQ • QP
6(8) = 3KQ
48 = 3KQ
= KQ or
KQ = 16 cm
6 cm
8 cm
3 cm
?
2
1
Another Examples:
1. In the figure, CU = 6, UT = 9 and CA = 8. What is CB?
Solution:
CU • CT = CA • CB
6(6 + 9) = 8CB
6(15) = 8CB
90 = 8CB
C
U
T
B
A
6
9
8
= CB or
CB = 11.25 units
?

3
1. In the figure, GE is tangent to circle Q at E. If GA = 4 cm and AT = 5 cm, find GE.
G
A
T
E
4 cm
5 cm
?
Solution:
GE2
= GA • GT
GE2
= 4(9)
GE =
GE = 6 cm
• Q
Exercises:
A. Form an equation to solve for x.
1.
9
x
4
3
3x = 4(9)
3x = 36
x = 12 units

2.
x
4
6
4x = 6²
4x = 36
x = 9 units
3.
x
4
6
4(4 + x) = 6²
16 + 4x = 36
4x = 20
x = 5 units

15
8
5
x
15x = 5(13)
15x = 65
x = 4.33 units
15
8
5
x
x(x + 15) = 5(13)
x² +15x = 65
x² + 15x – 65 = 0
x₁ = .51, x₂ = -18.51
(accepted) (discarded)
4.
5.

x² = 3(12)
x² = 36
x = 𝟑𝟔
x = 6 units
3(3 + x) = 4(10)
9 + 3x = 40
3x = 31
x = 10.33 units
2(3x – 2) = 5x
6x – 4 = 5x
x = 4 units
6. 8. 6.
9
3
x
6
4
3
x
2
5
x
3x – 2

A. Use the figure to answer the following:
1. If SA = 6, SI = 15 and SL = 8,
find SN. ___________
2. If SA = 8, AI = 16 and SN = 16,
find SL. ___________
3. If SN = 20, SL = 6 and AI = 27,
find SI. ___________
4. If AN = 25, NY = 18 and LI = 27, find AY and LY. ___________
5. If AY = 6, AN = 17, LY = 8, find YI. ___________
6. If ST = 12 and SI = 24, find SA. ___________
7. If AY = 6, YN = 5 and IY = 10, find LI. ___________
8. If LN = 14, SI = 21 and SN = 22, find SA. ___________
9. If SA = 3 3 and AI = 6 3, find ST. ___________
10. If SN = 12, SL = 2 3 and SA = 3, find SI. ___________
I
A
Y
N
L
T
S
11.25 units
12 units
30.89 units
7 units, 6 units or 7 units, 21 units
8.25 units
6 units
13 units
8.38 units
9 units
13.86 units or 8 𝟑 units

Solutions: Let x = missing segment.
1. 6(15) = 8x
90 = 8x
11.25 = x or x = 11.25 units
2. 8(24) = 16x
192 = 16x
12 = x or x = 12 units
3. 6(20) = x(x + 27)
120 = x² + 27x
0 = x² + 27x – 120
x₁ = .89, x₂ = -30.89 (discarded)
SI = SA + AI = 3.89 + 27 = 30.89 units
I
A
Y
N
L
T
S

4. AY = 25 – 18 = 7 units
7(18) = x(27 – x)
126 = 27x – x²
x² - 27x + 126 = 0
x₁ = units, x₂ = 21 units
5. 6(11) = 8x
66 = 8x
8.25 = x or x = 8.25 units
6. 12² = 24x
144 = 24x
6 = x or x = 6 units
I
A
Y
N
L
T
S

7. 6(5) = 10x
30 = 10x
3 = x
x = 3 + 7 = 10 units
8. 8(22) = 21x
176 = 21x
8.38 = x or x = 8.38 units
9. (3 𝟑 )(9 𝟑 ) = x²
81 = x²
𝟖𝟏 = x or x = 9 units
10. (2 𝟑 )(12) = 3x
41.57 = 3x
13.86 = x or x = 13.86 units
I
A
Y
N
L
T
S

Another Exercise: Find the value of x in each
figure.
1.
7 cm
8 cm x
12 cm
8(x + 8) = 7(19)
8x + 64 = 133
8x = 69
x = 8.63 cm
3x = 7²
3x = 49
x = 16.33 cm
2.
3 cm
7 cm
x

Answer Key (Mastery Test)
1. 40ᵒ
2. 92ᵒ
3. 55ᵒ
4. 5 ᵒ
5. 5ᵒ
6. 48ᵒ
7. 100ᵒ
8. 102.5ᵒ
9. 100ᵒ
10. 140ᵒ
11. 180ᵒ
12. 180ᵒ
1 . 120ᵒ
14. 240ᵒ
15. 40ᵒ
1 . 20ᵒ
17. 70ᵒ
18. 50ᵒ
19. 40ᵒ
20. 50ᵒ
21. 20ᵒ
22. 40ᵒ
2 . 70ᵒ
24. 110ᵒ

25. 180ᵒ
2 . 40ᵒ
27. 50ᵒ
28. 20ᵒ
29. 40ᵒ
0. 50ᵒ
1. 0ᵒ
2. 120ᵒ
33. 110ᵒ
4. 180ᵒ
5. 2 0ᵒ
36. A =
𝟏
𝟐
bh or A =
𝒃𝒉
𝟐
37. A = s²
38. A = lw or A = bh
39. A =
𝟏
𝟐
h(b₁ + b₂)
40. A =
𝟏
𝟐
d₁d₂ or A = bh

41. A = bh
42. A = (
𝒂𝒓𝒄
𝟑𝟔𝟎
)(𝜋r²)
43. C = 𝜋d or C = 2𝜋r
44. P = a + b + c
45. 5x = 3(11)
x = 6.6 cm
46. 5(x – 5) = 3(11)
x = 11.6 cm

47. 8x = 10(24)
x = 30 m
48. 8(x + 8) = 10(24)
x = 22 m
49. x(x + 10) = 12²
x = 8 units
50. x² = 3(21)
x = 7.94 units

Find the area of the shaded region. ENHS is a
square inscribed in ⦿O. Show all your
necessary solutions.
N
E
H
S
•
8 cm

Theorem:
If a line is tangent to the circle, then it is
perpendicular to the radius drawn to the point of
tangency.
•
•

Ashaded region •
= A - A
circle square
= 3.14(8)² - ( 𝟏𝟐𝟖)²
= 200.96 - 128
= 72.96 square cm
Ashaded region
8 cm

Solve each problem. Draw figure for each problem.
Show your solution.
1. The segments of one of two intersecting chords
are x + 2 and x + 6 and x – 4 and x + 18 for the
other. Find the length of each chord.
Figure:
x + 2
x + 6
x – 4
x + 18
Solution:
(x + 2)(x + 6) = (x – 4)(x + 18)
x² + 6x + 2x + 12 = x² + 18x – 4x – 72
8x + 12 = 14x – 72
-6x = -84
x = 14
1st segment –x + 2 = 14 + 2 = 16 units
2nd segment – x + 6 = 14 + 6 = 20 units
3rd segment – x – 4 = 14 – 4 = 10 units
4th segment – x + 18 = 14 + 18 = 32 units
36 units
42 units
The 1st chord is 36 units and the 2nd chord is 42 units.

2. A secant and a tangent segment intersect in the
exterior of a circle. The chord which is a part of the
secant is three times as long as the tangent segment and
the external segment of the secant is 10 cm. Find the
length of the tangent segment.
Solution:
Let x = tangent segment
3x = internal segment
x² = 10(3x + 10)
x² = 30x + 100
x² – 30x – 100 = 0
x₁ = .0 , x₂ = -3.03 (discarded)
The tangent segment measures
33.03 cm.
Figure:
10 cm
3x
x

1. The segments of two intersecting chords are 2x + 2
and 3x – 5 of one and x + 1 and x + 5 of the other.
Find the length of each chord.
Figure:
3x – 5
2x + 2
x + 5
x + 1
Solution:
(2x + 2)(3x – 5) = (x + 5)(x + 1)
6x² – 10x + 6x – 10 = x² + x + 5x + 5
5x² – 10x – 15 = 0 or x² – 2x – 3 = 0
x₁ = -1 (discarded), x₂ = (accepted)
1st chord – 2x + 2 + 3x – 5
= 2(3) + 2 + 3(3) – 5 = 12 units
2nd chord – x + 5 + x + 1
= 3 + 5 + 3 + 1 = 12 units
The measure of each chord is 12 units.

2. The radii of two circles are 3 and 6 while the distance
between their centers is 18. Find the length of their
common external tangent segment.
Solution:
Let x = the common external tangent.
x² = 18² - 3²
x = 𝟑𝟏𝟓
x = 3 𝟑𝟓 or 17.75 units
The measure of the common external tangent
is 3 𝟑𝟓 𝐨𝐫 𝟏𝟕. 𝟕𝟓 𝐮𝐧𝐢𝐭𝐬.
Figure:
• •
3 3
18
3
x =?
x

3. Find the distance between the centers of two
circles if their radii are 5 cm and 17 cm and their
common external tangent is 16 cm long.
Solution:
Let x = distance
between the centers of
two circles.
x² = 16² + 12²
x² = 400
x = 𝟒𝟎𝟎
x = 20 𝒄𝒎
The center of the first circle is 20 𝒄𝒎 away from the center
of the second circle.
Figure:
• •
5 cm
12
16 cm
16 cm
5 cm
x

Find the area of the shaded region.
•
6 cm 60⁰
60⁰
60⁰
30⁰ 30⁰
60⁰
30⁰
6 cm
60⁰ 6 cm
3 cm
3 𝟑 cm

Ashaded region
=A - A
A = 𝜋r² = 3.14(6)² = 113.04 sq. cm
Asmall
=
𝟏
𝟐
(3)(3 𝟑) = 7.79 sq. cm
Aequilateral
= 6(7.79) = 46.74 sq. cm
Ashaded region
equilateral
= 113.04 – 46.74 = 66.30 sq. cm

V. Use the Pythagorean Theorem to solve
for x. Give your answer in a simplest form of
radical (5 points each).
12 𝟔 cm
6 𝟏𝟑 cm
x
1. 2.
x
5 𝟏𝟏 m
10 𝟐 m
90ᵒ

1.
12 𝟔 cm
6 𝟏𝟑 cm
x
Solution:
(12 )² = (6 1 )² + 𝑥²
864 = 468 + x²
396 = x²
11 = 𝑥
6 𝟏𝟏 = 𝒙 𝒐𝒓 𝒙 = 𝟔 𝟏𝟏 𝒄𝒎

2.
10 𝟐 m
5 𝟏𝟏 m
x
Solution:
x² = (5 11)² + (10 2)²
x² = 275 + 200
x² = 475
x = 25 19
𝒙 = 𝟓 𝟏𝟗 𝒎

Proving Theorems is the
most challenging topic in
Geometry. It requires our
critical thinking, reasoning
power and most especially
our patience in applying
TRIAL & ERROR method.

Theorem:
Two inscribed angles that have the same arc
are congruent.
A
B
C
D
Given:∠Cand∠DareinscribedangleswiththesameinterceptedarcAB.
Prove:∠C≌∠D

Proof:
Statements Reasons
1 ∠Cand∠Dareinscribedangleswiththesameinterceptedarc
AB. 1 Given
2
m∠C=
1
2
mAB, m∠D=
1
2
mAB
⌒
⌒
2 The measure of the inscribed angle
is one-half the measure of its
interceptedarc.
3 m∠C=m∠D 3 TransitiveProperty
4 ∴∠C≅∠D 4 Definitionofcongruentangles
A
B
C
D

Theorem:
The measure of an angle formed by a tangent and
a secant drawn to the point of contact is one-half the
measure of the intercepted arc.
•
A
B
•
C
Given: BC is a tangent intersecting the
secant BA at B.
Prove: m∠ABC = mAB
•
⌒
•
D
•
•
• A

Proof:
Statements Reasons
1
BCisatangentintersectingthesecantBAatB. 1 Given
2 TheLinePostulate
3
DrawAD. 3 TheLinePostulate
4
m∠BAD=
1
2
mBD
⌒
interceptedarc.
5 m∠BAD=
1
2
(180o
) 5 Asemicirclemeasures180o
.
2
DrawBDwhichpassesthroughthecenterofthecircle.

6 m∠BAD=90o
7 ∠BADisarightangle. 7 Definitionofarightangle
8 ΔBADisarighttriangle. 8 Definitionofarighttriangle.
9 ∠ADBand∠ABDarecomplementaryangles. 9 The two acute angles of a right
triangleare complementary.
10 m∠ADB+m∠ABD=90o
10 Definitionofcomplementaryangles
11 The line drawn from the point of
tangency passing through the
center of the circle is perpendicular
with the tangent line.
6 Simplifying
11
BD⊥BC

14 m∠ABC+m∠ABD=m∠DBC 14 AngleAdditionPostulate
15 m∠ABC+m∠ABD=90o
15 Substitution
16 m∠ADB+m∠ABD=m∠ABC+m∠ABD 16 TransitivePropertyofEquality
17 m∠ADB=m∠ABC 17 AdditionPropertyofEquality
18
m∠ADB=
1
2
mAB
⌒
interceptedarc.
19
∴m∠ABC=
1
2
mAB
⌒ 19 Substitution
12 ∠DBCisarightangle. 12 Definitionofaperpendicularity
13 m∠DBC=90o
13 Definitionofarightangle
•
O
•
D A
B
•
C
•
•

•
O
•
D A
B
•
C
•
•
Proof:
Statements Reasons
1
BC is a tangent intersecting the secant AB at B.
1 Given
2
Draw BD which passes through the center of the circle.
2 The Line Postulate
3
Draw AD.
3 The Line Postulate
4
m∠BAD =
1
2
mBD
intercepted arc.
5 m∠BAD =
1
2
(180o
) 5 A semicircle measures 180o
.
6 m∠BAD = 90o
6 Simplifying
7 ∠BAD is a right angle. 7 Definition of a right angle
8 ΔBAD is a right triangle. 8 Definition of a right triangle.
9 ∠ADB and ∠ABD are complementary angles. 9 The two acute angles of a right
triangle are complementary.
10 m∠ADB + m∠ABD = 90o
10 Definition of complementary angles
11 BD ⊥ BC 11 The line drawn from the point of
tangency passing through the
center of the circle is perpendicular
with the tangent line.
12 ∠DBC is a right angle. 12 Definition of a perpendicularity
13 m∠DBC = 90o
13 Definition of a right angle
14 m∠ABC + m∠ABD = m∠DBC 14 Angle Addition Postulate
15 m∠ABC + m∠ABD = 90o
15 Substitution
16 m∠ADB + m∠ABD = m∠ABC + m∠ABD 16 Transitive Property of Equality
17 m∠ADB = m∠ABC 17 Addition Property of Equality
18
m∠ADB =
1
2
mAB
intercepted arc.
19
∴ m∠ABC =
1
2
mAB
19 Substitution
⌒
⌒
⌒

How do similar
triangles differ from
congruent triangles?

Similar triangles are two or more triangles
wherein all corresponding angles are congruent and all
corresponding sides are proportional. They have the
same shape but they are different in sizes.
ΔONB ∼ ΔDIG
O
N
B
D
I
G

Congruent triangles are two or more triangles
wherein all corresponding parts are congruent. They
have the same shape and size.
All congruent triangles are similar but not all
similar triangles are congruent.
ΔTOP ≅ ΔMAD
O
T
P A
M
D

Theorem:
If two chords intersect in the interior of the
circle, then the product of the measures of the
segments of the first chord is equal to the product
of the measures of the segments of the second
chord.
Given: AB and CD intersect in the interior of the circle at K.
Prove: AK • BK = CK • DK
K
C B
D
A
C
A
B
D
K
(
(
)
)

Proof:
Statements Reasons
1
ABandCDintersectintheinteriorofthecircle@K. 1 Given
2
DrawADandBC. 2 TheLinePostulate
3 ∠A≅∠C
C
A
B
D
K
(
(
)
)
4 Angles that have the same
interceptedarcarecongruent.

4 ∠B≅∠D 4 Angles inscribed in the same
arcarecongruent.
5 If two corresponding angles of
triangles are congruent, then
the triangles are similar (The
AA Similarity Theorem).
6
AK
CK
=
DK
BK
6 In similar triangles,
corresponding sides are
proportional.
7 ∴AK•BK=CK•DK 7 Multiplication Property of
Equality
C
A
B
D
K
(
(
)
)
5 ΔAKD∼ΔCKD

Theorem:
If a tangent segment and a secant segment
intersect in the exterior of a circle, then the square of
the length of the tangent segment is equal to the
product of the lengths of the secant segment and its
external part.
Given: PQ is a tangent segment and AQ is a secant segment.
Prove: PQ2
= AQ • BQ
A
Q
P
B
A
Q
P
B
)
(
Improvised Figure

Proof:
Statements Reasons
1
PQisatangentsegmentandAQisasecantsegment. 1 Given
2
DrawAPandBP. 2 TheLinePostulate
3 ∠Q≅∠Q 3 ReflexiveProperty
A
Q
P
B
)
(
(
A
B
P
Q
Q
P

4 ∠A≅∠BPQ 4 Angles inscribed in the same
arcarecongruent.
5 ΔBPQ∼ΔPAQ
6
PQ
AQ
=
BQ
PQ
proportional.
7 ∴PQ2
=AQ•BQ 7 Multiplication Property of
Equality
A
Q
P
B
)
(
(
A
B
P
Q
Q
P

Theorem:
If two secants intersect in the exterior of the circle,
the product of the length of one secant segment and the
length of its external part is equal to the product of the
length of the other secant segment and the length of its
external part.

Proof:
Statements Reasons
1
ABandCDintersectoutsidethecircle@Q. 1 Given
2
DrawADandBC. 2 TheLinePostulate
3 ∠A≅∠C 3 Angles inscribed in the same
arcarecongruent.
4 ∠Q≅∠Q 4 ReflexiveProperty
Q
B
A
C
D
Q
C
B
A
Q
D

5ΔDAQ∼ΔBCQ
6
QA
QC
=
QD
QB
proportional.
7∴QA•QB=QC•QD 7 Multiplication Property of
Equality
Q
B
A
C
D
Q
C
B
A
Q
D

TEAM QUIZ
Instruction:
Provide 15 pcs of
𝟏
𝟖
and 1 pc of
𝟏
𝟒
sheet of
paper . Each
𝟏
𝟖
sheet will be used for your answer
in each question and the
𝟏
𝟒
sheet will be for the
list of your group members.
After the allotted time in writing the correct
statement or reason, the team leader will give
the paper to the teacher. Majority of the
questions is worth 3 points and some are 5
points.

Theorem:
The measure of an angle formed by two
secants intersecting within the circle equals one-
half the sum of the measures of the intercepted
arcs.
Given: AC and BD intersect within the circle at X.
Prove: m∠BXA =
1
2
(mAB + mCD)
B
A
C
D
X
⌒ ⌒

Proof:
Statements Reasons
1
ACandBDintersectwithinthecircle@x. 1 Given
2
DrawAD. 2 TheLinePostulate
3 m∠BXA=m∠BDA+m∠CAD
4
m∠BDA=
1
2
mAB,
⌒
4 The measure of the inscribed
ofitsinterceptedarc.
B
A
C
D
X
3 The measure of an exterior
angle is equal to the sum of its
remote interior angles.

5
m∠CAD=
1
2
mCD
⌒
ofitsinterceptedarc.
6
m∠BDA+m∠CAD=
1
2
mAB+
1
2
mCD
⌒ ⌒
6 AdditionPropertyofEquality
7
m∠BDA+m∠CAD=
1
2
(mAB+mCD)
⌒ ⌒
7 Distributive Property of
Equality
8
∴m∠BXA=
1
2
(mAB+mCD)
⌒
⌒
8 SubstitutionProperty
B
A
C
D
X

Theorem:
The measure of an angle formed by two
secants intersecting outside the circle equals one-
half the difference of the measures of the
intercepted arcs.
Given: TP and RP intersect outside the circle at P.
Prove: m∠SPV =
1
2
(mRT – mSV) P
S
V
R
T
⌒
⌒

Proof:
Statements Reasons
1
TPandRPintersectoutsidethecircle@P. 1 Given
2
DrawRS. 2 TheLinePostulate
3 m∠TSR=m∠SRV+m∠SPV
4 m∠SPV=m∠TSR–m∠SRV 4 AdditionPropertyofEquality
P
S
V
R
T
3 The measure of an exterior
angle is equal to the sum of its
remote interior angles.

5
m∠TSR=
1
2
mTR
⌒
6
m∠SRV=
1
2
mSV
⌒
7
m∠SPV=
1
2
mTR–
1
2
mSV
⌒ ⌒
7 LawofSubstitution
8
∴m∠SPV=
1
2
(mTR–mSV)
⌒
⌒
8 DistributiveProperty
P
S
V
R
T

Quiz (
𝟏
𝟐
CW). Find the value of x in each figure (5 points each).
Round off your answer to the nearest hundredths. Show your
solutions.
1. 2.
7 cm
5 cm
11 cm
x
8
10
15
x

3. 4.
7
12
x
17 m
x
13 m
5. 6.
15
8
x
11 cm
7 cm
x
90o

7. 8.
40o
xo
140o
110o
145o
xo

Module V – Plane Coordinate Geometry
Review Lesson
Cartesian Plane or Rectangular Coordinate System
Y
X
QI (+, +)
QII (-, +)
QIII (-, -) QIV (+, -)
Origin
(Analytic Geometry)

Y
X
•
•
•
•
•
•
•
•
• •
•
A
B
C
D
E
(2, 1)
(1, 2)
(-2, 3)
(-3, 2)
(0, 1)
F
H
I
G
K
J
(-2, 0)
(0 -1)
(-3, -2)
(-1, -3)
(2, -1)
(4, -2)
2 4
-2
2
-2
Write the coordinates of the following points.

Lesson I – Distance Between the Two Points
Y
X
P₁(x₁, y₁)
P₂(x₂, y₂)
x₁ x₂
y₁
y₂
d = ? y2 – y1
x2 – x1
Using the Pythagorean Theorem:
c² = a² + b²
d² = (x₂ - x₁)² + (y₂ - y₁)²
d = (x₂ − x₁)² + (y₂ − y₁)² Distance Formula
•
•

Y
X
•
•
•
•
•
•
•
•
•
•
•
•
•
•
1. A
2. B
3. C
4. D
5. E
6. E
7. F
8. G
9. H 10. I
11. J
12. K
13. L
14. M
•
15. N
2 4 6
2
-2
-2
(4, 1)
(4, 3)
(2, 2)
(1, 3)
(0, 2)
(-1, 2)
(-2, 1)
(-1, 0)
(-2, -1) (0, −
𝟑
𝟐
)
(1, 0)
(
𝟕
𝟐
, −
𝟓
𝟐
)
(5, -1)
(6, 0)
(
𝟏𝟏
𝟐
, −
𝟓
𝟐
)
I. Give the coordinates of the following points.

Y
X
2 4 6 8
-2
-4
2
4
-2
-4
•
•
1. d₁ = ?
•
A
B
C
2. d₂ = ?
. d₃ = ?
II. Answer as required. Give your answer in a simplest form of radical
and in approximate value rounded off to nearest hundredths .
4. P = ?
ΔABC
5. W = ?
6. L = ?
7. A = ?
8. diagonal = ?
9. P = ?
10. A = ?
Δ
•
•
•
•
•
•
•

Answer Key (Test II)
1) 2 𝟏𝟎 𝒐𝒓 𝟔. 𝟑𝟐 𝒖𝒏𝒊𝒕𝒔
2) 𝟐 𝟏𝟕 𝒐𝒓 𝟖. 𝟐𝟓 𝒖𝒏𝒊𝒕𝒔
3) 2 𝟓 𝒐𝒓 𝟒. 𝟒𝟕 𝒖𝒏𝒊𝒕𝒔
4) (2 𝟏𝟎 + 𝟐 𝟏𝟕 + 2 𝟓) or 19.04 units
5) 3 units
6) 8 units
7) 24 sq. units
8) 𝟕𝟑 𝒐𝒓 8.54 units
9) 22 units
10) 12 sq. units

Solution:
d₁(A, B) = − 4 + (1 + 1)²
= 1 + 4 = 𝟓 or 2.24 units
d₂(A, C) = −4 − 4 + (−2 + 1)²
= 4 + 1 = 𝟔𝟓 or 8.06 units
d₃(B, C) = −4 − + (−2 − 1)²
= 49 + 9 = 𝟓𝟖 or 7.62 units
P = ( 5 + 5 + 58 ) or 17.92 units
3. Find the perimeter of the triangle with vertices
A(4, -1), B(3, 1) and C(-4, -2). Draw the triangle.

X
Y
2 4
-2
-4
2
-2
A(4, -1)
B(3, 1)
C(-4, -2)
Is ΔABC a right triangle? Let’s verify.
c² = a² + b²
( 𝟔𝟓)² ≟ ( 𝟓 )² + ( 𝟓𝟖)²
5 ≟ 5 + 58
5 ≠ ΔABC is not a right triangle.
A(4, -1), B(3, 1) and C(-4, -2)
•
•
•
Figure:

4. Find the area of a square, the perimeter and
the length of its diagonal whose vertices are (3, 1),
(-1, 5), (-5, 1) and (-1, -3).
Y
X
2
4
2 4
-2
-4
-6
-2
-4
(3, 1)
(-1, 5)
(-5, 1)
(-1, -3)
diagonal = ?
s = ?
•
•
•
•
Figure:

s = −1 − + 5 − 1 = 1 + 1
= 2 = 𝟒 𝟐 units
P = 4s = 4(4 2) = 16 𝟐 units
diagonal = s 2 =(4 2)( 2) = 4(2) = 8 units
A = s² = ( 2)² = 32 square units

5. Show that the points (3, 1), (-1, 3), (-3, 0) and (1, -2) are
vertices of a parallelogram. Sketch the figure.
Y
X
1 2
-1
-2
1
3
-3
-1
-2
(3, 1)
2
3
(-1, 3)
(-3, 0)
d₁ = ?
d₂ = ?
d₃ = ?
d₄ = ?
•
•
•
•
(1, -2)
Show that d₁ = d₂ & d₃ = d₄ and m₁ = m₂ and m₃ = m₄.
Figure:

d₁ = −1 + + − 0 = 4 + 9 = 1 𝑢𝑛𝑖𝑡𝑠
d₂ = − 1 + 1 + 2 = 4 + 9 = 1 𝑢𝑛𝑖𝑡𝑠
d₃ = 1 + + −2 − 0 = 1 + 4
= 20 𝑜𝑟 2 5𝑢𝑛𝑖𝑡𝑠
d₄ = + 1 + 1 − = 1 + 4
= 20 𝑜𝑟 2 5𝑢𝑛𝑖𝑡𝑠
m₁ =
−
− +
= ; m₂ =
+
−
=
m₃ =
−
− −
= − ; m₄ =
+
− −
= −
d₁ & d₂ is the first pair of opposite sides, d₃ & d₄ is the second pair of
opposite sides, m₁ & m₂ are the respective slopes of the 1st pair of opposite
sides, m₃ & m₄ are the respective slopes of the 2nd pair of opposite sides.
Since that d₁ = d₂ & d₃ = d₄; m₁ = m₂ & m₃ = m₄ the figure is a
parallelogram.

Assignment (1 whole)
1. The distance between (-3, -1) and
(3, y) is 𝟑 𝟓. Find y.
2. Show that the triangle with vertices
D(-7, 5), J(-3, -2) and P(4, 2) is an
isosceles right triangle.
3. Show that the points N(5, 4), I(-5, 8),
E(-7, 3) and L(3, -1) are vertices of a
rectangle.

Many geometric properties
can be verified or proven by
using a coordinate plane. A
proof that uses figures on a
Cartesian plane to prove
geometric properties and
relationships is called a
Coordinate Proof.

1. Show that the triangle whose vertices are (-6, 3),
(4, 5) and (-1, -2) is an acute triangle. Draw the
triangle.
Solution:
c = 4 + + (5 − )² = 100 + 4 = 104
a = −1 + + (−2 − )² = 25 + 25 = 50
b = −1 − 4 + (−2 − 5)² = 25 + 49 = 74
To be an acute triangle, the square of the longest side
must be less than the sum of the squares of the other sides.
Show that c² < a² + b².
𝟏𝟎𝟒
𝟐
⩻ 𝟓𝟎
𝟐
+ 𝟕𝟒
𝟐
⃪ 104 < 124
Since that c² < a² + b², the triangle is an acute Δ.

Figure:
X
Y
2 4 6
-2
-4
-6
2
4
6
-2
-4
(-6, 3)
(4, 5)
(-1, -2)
•
•
•

2. Show that the points N(5, 4), I(-5, 8), E(-7, 3) and
L(3, -1) are vertices of a rectangle.
Solution:
m₁(N, I) =
y₂ − y₁
x₂ − x₁
=
𝟖 − 𝟒
−𝟓 − 𝟓
= −
𝟒
𝟏𝟎
or −
𝟐
𝟓
m₂(E, L) =
y₂ − y₁
x₂ − x₁
=
−𝟏 − 𝟑
𝟑 + 𝟕
= −
𝟒
𝟏𝟎
or −
𝟐
𝟓
m₃(I, E) =
y₂ − y₁
x₂ − x₁
=
𝟑 − 𝟖
−𝟕 + 𝟓
=
−𝟓
−𝟐
or
𝟓
𝟐
m₄(N, L) =
y₂ − y₁
x₂ − x₁
=
−𝟏 − 𝟒
𝟑 − 𝟓
=
−𝟓
−𝟐
or
𝟓
𝟐
Dummy Figure:
N(5, 4)
I(-5, 8)
E(-7, 3) L(3, -1)

NI = 𝟓 + 𝟓 𝟐 + (𝟒 − 𝟖)² = 𝟏𝟎𝟎 + 𝟏𝟔 = 𝟏𝟏𝟔 𝒐𝒓 𝟐 𝟐𝟗
EL = −𝟏 − 𝟑 𝟐 + (𝟑 + 𝟕)² = 𝟏𝟔 + 𝟏𝟎𝟎 = 𝟏𝟏𝟔 𝒐𝒓 𝟐 𝟐𝟗
IE = −𝟓 + 𝟕 𝟐 + (𝟖 − 𝟑)² = 𝟒 + 𝟐𝟓 = 𝟐𝟗
NL = 𝟓 − 𝟑 𝟐 + (𝟒 + 𝟏)² = 𝟒 + 𝟐𝟓 = 𝟐𝟗
Since that m₁ = m₂ and m₃ = m₄; NI = EL and IE = NL,
we have shown that there are two pairs of parallel and
congruent sides. m₁m₃ = m₂m₄ = (−
𝟐
𝟓
)(
𝟓
𝟐
) = -1 signifies
that 4 pairs of consecutive sides are perpendicular with
each other. In other words, 4 angles are all 90⁰.
Thus, the given points are vertices of a rectangle
NIEL.
Dummy Figure:
N(5, 4)
I(-5, 8)
E(-7, 3) L(3, -1)

Figure:
X
Y
2 4 6
-2
-4
-6
-8
2
4
6
8
-2
N(5, 4)
I(-5, 8)
E(-7, 3)
L(3, -1)
•
•
•
•

3. Draw the triangle whose vertices are A(3, 3),
B(-3, 1), C(-1, -3). Then find the measure of each
angle.
X
Y
1 2 3
-1
-2
-3
1
2
3
-1
-2
-3
•
•
•
A(3, 3)
B(-3, 1)
C(-1, -3)

Solution:
m₁(A, B) =
𝟑 − 𝟏
𝟑 + 𝟑
=
𝟐
𝟔
or
𝟏
𝟑
m₂(A, C) =
𝟑+𝟑
𝟑+𝟏
=
𝟔
𝟒
or
𝟑
𝟐
m₃(B, C) =
𝟏+𝟑
−𝟑+𝟏
=
𝟒
−𝟐
or -2
tan A =
𝐦₂−𝐦₁
𝟏+𝐦₁𝐦₂
=
𝟑
𝟐
−
𝟏
𝟑
𝟏+(
𝟏
𝟑
)(
𝟑
𝟐
)
tan A =
𝟕
𝟗
A = tan⁻¹(
𝟕
𝟗
) = 37ᵒ52’ 0’’
tan C =
𝐦₃−𝐦₂
𝟏+𝐦₂𝐦₃
=
−𝟐 −
𝟑
𝟐
𝟏+(
𝟑
𝟐
)(−𝟐)
tan C =
𝟕
𝟒
C = tan⁻¹(
𝟕
𝟒
) = 60ᵒ15’18’’
B = 180 – (37ᵒ52’ 0’’ + 60ᵒ15’18’’)
B = 81ᵒ52’12’’)

Lesson II – Midpoint of a Line Segment
X
Y
x₁
y₁
P₁(x₁, y₁)
P₂(x₂, y₂)
x₂
y₂
•
•
•
x
y M(x, y)
•
• x =
x₂ + x₁
𝟐
y =
y₂ + y₁
𝟐
Midpoint Formula

Examples:
1. Draw the line segment with endpoints (4, 3)
and (-2, 1). Then find its midpoint.
X
Y
1 2 3 4 5
-1
-2
1
2
3
(4, 3)
x = 4 − 2
𝟐
= 𝟏
(-2, 1)
y = + 1
𝟐
= 𝟐
The midpoint is (1, 2).
•
(1, 2)
Figure:
•
•

2. The midpoint of a line segment is (2, -1) and its
first endpoint is (-1,-3). Find the coordinates of its
second endpoint.
Solution:
Let(x,y)=(2,-1),(x1,y1)=(-1,-3)and(x2,y2)=?
x=
𝒙₁+𝒙₂
𝟐
4=-1+x₂
5=x₂orx₂=5
y=
𝒚₁+𝒚₂
𝟐
2=
−𝟏+𝒙₂
𝟐 -1=
−𝟑+𝒚₂
𝟐
-2=-3+y2
1=y2 ory2=1
The coordinates of the
2nd endpoint of a line
segment are (5, 1).
-
-
-
-
y
•
•
•
(-1, -3)
(2, -1)
(5, 1)

3. Find the intersection of the main diagonal and
the secondary diagonal of a parallelogram with
vertices (5, 0), (2, 4), (-2, -1) and (1, -5).
X
Y
2 4 6
-2
2
4
-2
-4
(5, 0)
(2, 4)
(-2, -1)
(1, -5)
•
Point of intersection = ?
xm =
𝒙₁+𝒙₂
𝟐
=
𝟐+𝟏
𝟐
=
𝟑
𝟐
ym =
𝒚₁+𝒚₂
𝟐
=
𝟒−𝟓
𝟐
=−
𝟏
𝟐
(
𝟑
𝟐
, −
𝟏
𝟐
)
xs =
𝒙₁+𝒙₂
𝟐
=
𝟓−𝟐
𝟐
=
𝟑
𝟐
ys =
𝒚₁+𝒚₂
𝟐
=
𝟎−𝟏
𝟐
=−
𝟏
𝟐
(
𝟑
𝟐
, −
𝟏
𝟐
)
The point of intersection of the two diagonals is (
𝟑
𝟐
, −
𝟏
𝟐
).
Figure:
•
•
•
•

Assignment (1 whole)
1. The point (0, 1) is the midpoint of a line segment
joining (x, 4) and (-3, y), find x and y.
2. A median of a triangle is a line segment from a vertex
to the midpoint of the opposite side. Find the lengths of
the medians of the triangle with vertices (4, 0), (2, 1)
and (-1, -5).
3. To be a right triangle, the midpoint of the hypotenuse
must be equidistant from the three vertices. Show
that the points (5, 0), (-3, 4) and (3, -2) are the
vertices of a right triangle.
4. Draw the triangle whose vertices are S(5, 2), T(-5, 1)
and E(-2, -11). Then find the measures of the three
angles.

Lesson III – Division of a Line Segment
X
Y
x₁
y₁
P₁(x₁, y₁)
x
y
P(x, y)
x₂
y₂
P₂(x₂, y₂)
•
•
•
P₁P
P₁P₂
= 𝑟
P₁P
P₁P₂
=
𝑥 − 𝑥₁
x − 𝑥₁
𝑥 − 𝑥₁
x − 𝑥₁
= 𝑟
x = x₁ + r(x₂ − x₁)
y = y₁ + r(y₂ − y₁)

Examples:
1. Find the two trisection points of the line
segment joining the points P₁(-2, -1) and P₂(4, 2).
Draw the line segment and indicate the points of
trisection.
Solution:
•
• •
•
3
P1
P2
P4
P3
1
2
(-2, -1)
(4, 2)
x3= x1 + r(x2 – x1) = -2 +
𝟏
𝟑
(4 + 2) = -2 + 2 = 0
y3 = y1 + r(x2 – y1) = -1 +
𝟏
𝟑
(2 + 1) = -1 + 1 = 0
∴P₃(0, 0)
x4= x1 + r(x2 – x1) = -2 +
𝟐
𝟑
(4 + 2) = -2 + 4 = 2
y4 = y1 + r(x2 – y1) = -1 +
𝟐
𝟑
(2 + 1) = -1 + 2 = 1
∴P₄(2, 1)

X
Y
1 2 3 4
-1
-2
1
2
3
-1
P₁(-2, -1)
P₂(4, 2)
•
•
•P₃(0, 0)
1st trisection
•P₄(2, 1)
2nd trisection
Figure:
4

2. The line segment joining the points P₁(-5, 2)
and P₂(3, -5) is extended beyond P₂ so that its
length is doubled. Find the terminal point P.
Solution:
Notethatr=2.
•
•
•
P1
P
P₂
(-5, 2)
(x, y) = ?
(3, -5)
r = 2
x=x1+r(x2–x1)
=-5+2(3+5)
=-5+16
x=11
y=y1+r(y2–y1)
=2+2(-5–2)
=2–14
y=-12

Y
X
2 4 6 8 10
-2
-4
2
-2
-4
-6
-8
-10
-12
P₁(-5, -2)
P₂( -5)
P(11, -12)
•
•
•
Hence, the terminal point is (11, -12).
•
•
•
P1
P
P₂
(-5, 2)
(x, y) = ?
(3, -5)
r = 2

3. Three consecutive vertices of a parallelogram
are (6, 1), (-2, 3) and (-6, 0). Find the coordinates
of the fourth vertex.
Dummy Figure:
•
•
•
•P₁(6, 1)
P₂(-2, 3)
P₃(-6, 0)
P₄(x, y)
Solution:
•
x=
𝒙₁+𝒙₂
𝟐
=
𝟔−𝟔
𝟐
=
𝟎
𝟐
=0
y=
𝒚₁+𝒚₂
𝟐
=
𝟏+𝟎
𝟐
=
𝟏
𝟐
M(0,
𝟏
𝟐
)
x=-2+2(0+2)=-2+4=2
x=x₂+r(x–x₂)
y=y₂+r(y–y₂) y=3+2(
𝟏
𝟐
–3)=3+2(–
𝟓
𝟐
)=3–5=-2

X
Y
2 4 6 8
-2
-4
-6
-8
1
2
3
4
5
-1
-2
-3
P₁( , 1)
P₂(-2, 3)
P₃(-2, 3)
P₄(2, -2)
Figure:
•
•
•
•
•
The 4th vertex of a parallelogram is (2, -2).

Assignment (1 whole).Draw figure for
each problem.
1. Divide the segment joining (5, 1) and (-4, 3) into
four equal parts and find the points of division.
2. The line segment joining (3, 5) and (-3, -5) is
divided into two segments, one of which two-
thirds as long as the other. Find the point of
division.
3. A median of a triangle is a line segment joining
the vertex of a triangle to the midpoint of the
opposite side. Find the point on its median that is
two-thirds of the distance from the vertex to the
midpoint of the opposite side of the triangle
whose vertices are (4, 2), (-4, 3) and (-2, -3).

I. Find the distance between the two points. Write your
answer in a simplest form of radical (3 points each).
1. (-3, 4) and (5, -2)
Answer: 10 units
2. (3, 4) and (0, 0)
Answer: 5 units
3. (5, 2) and (-4, -1)
Answer: 𝟑 𝟏𝟎 units
4. (0, 4) and (-4, -2)
Answer: 𝟐 𝟏𝟑 units
5. (-4, -2) and (4, -2)
Answer: 8 units
Group Quiz

Answer: 10
II. Given the distance between the two points, find
the value of x or y. Quadrant is specified for the point
which has the lacking coordinate (5 points each).
1. d = 5 𝟐 ; (2, -3) and (x, 2) in QII
Choices: -7, -6, -5, -4, -3, -2, -1
2. d = 3 𝟓 ;(5, -1) and (8, y) in QI
Choices: 7, 6, 5, 4, 3, 2, 1
3. d = 𝟓𝟐 ; (1, -8) and (x, -2) in QIII I
Choices: -7, -6, -5, -4, -3, -2, -1
4. d = 𝟏𝟎 ; (-2, -9) and (1, y) in QIV
Choices: -4, -5, -6, -7, -9, -10
5. d = 𝟔𝟐𝟔 ; (-15, 8) and (x, 9) in QI

1. Point F is 5 units from point D whose coordinates
are (6, 2). If the x-coordinate of F is 10 and it lies in
the first quadrant, what is its y-coordinate?
Answer: 5
2. A new transmission tower will be put midway
between two existing towers. On the map drawn on
a coordinate plane, the coordinates of the first
existing tower are (-5, -3) and the coordinates of the
second existing tower are (9, 13). What are the
coordinates of the point where the new tower will be
placed?
Answer: (2, 5)
III. Solve the following problems (10 points each).

Module V – Equation of the Circle
X
Y
•
C(h, k)
x
h
k
•
•
y
r
x – h
y – k
Using the Pythagorean Theorem
c² = a² + b²
r² = (x – h)² + (y – k)² or
(x – h)² + (y – k)² = r² Standard Form of the
Equation of a Circcle
If the center of a circle is at
the origin, the equation is:
x² + y² = r²
Examples:
1. What is the equation of a
circle in standard form
whose center is at the
origin and its radius is 5?
Sketch the graph.
Solution:
x² + y² = r²
x² + y² = 5²
x² + y² = 25
-
-
-
-
•
Figure:

2. Find the equation of the circle in
general form whose center is (3, -4)
and its radius is 6. Sketch the curve.
Solution:
(x – h)² + (y – k)² = r²
(x – 3)² + (y + 4)² = 6²
x² – 6x + 9 + y² + 8y + 16 = 36
x² + y² – 6x + 8y – 11 = 0
Figure:
3. Find the center and the radius of the
circle whose equation is
x² + y² + 10x – 4y + 13 = 0 . Then draw
the circle.
Solution:
x² + y² + 10x – 4y +13 = 0
x² + 10x + 25 + y² – 4y + 4 = -13 + 25 + 4
(x + 5)² + (y – 2)² = 16
(x + 5)² + (y – 2)² = 4²
C(-5, 2) and r = 4
Figure:
- - -
-
•
-
-
C(-5, 2)
-
-
-
-
•
-
-
C( , -4)

4. What is the equation of the circle in general form whose
center is (0, -9) and its radius is 5 𝟑? No need to sketch the
curve.
Solution:
x² + (y + 9)² = (5 𝟑)²
x² + y² + 18y + 81 = 75
x² + y² + 18y + 6 = 0
5. If the center of a circle is (11, 0) and its radius is 6 𝟓 ,
then what is its equation in general form?
Solution:
( x – 11)² + y ² = (6 𝟓)²
x² – 22x + 121 + y² = 180
x² + y² – 22x – 59 = 0

6. What is the center of the circle
x² + y² –14y + 5 = 0? Find its radius.
Solution:
x² + y² – 14y + 5 = 0
x² + (y² –14y + 49) = -5 + 49
x² + (y – 7)² = 44
C(0, 7) & r = 𝟒𝟒 𝒐𝒓 𝟒(𝟏𝟏) 𝒐𝒓 𝟐 𝟏𝟏
7. Determine the center and radius of a circle
x²+ y² – 15 = 8x.
Solution:
x²+ y² – 15 = 8x.
x² – 8x + 16 + y² = 15 + 16
(x – 4)² + y² = 31
C(4, 0) & r = 𝟑𝟏

8. Determine the center and the radius of the circle
whose equation is x² + y² – 3x – 11y +10 = 0. No
need to sketch the circle.
Solution:
x² + y² – 3x – 11y +10 = 0
x² – 3x + y² – 11y = -10
(x² – 3x +
𝟗
𝟒
) + (y² – 11y +
𝟏𝟐𝟏
𝟒
) = -10 +
𝟗
𝟒
+
𝟏𝟐𝟏
𝟒
(x –
𝟑
𝟐
)² + (y –
𝟏𝟏
𝟐
)² =
𝟒𝟓
𝟐
C(
𝟑
𝟐
,
𝟏𝟏
𝟐
) and r =
𝟒𝟓
𝟐
or
𝟗𝟎
𝟐
or
𝟑 𝟏𝟎
𝟐

9. Determine the center and the radius of the circle
whose equation is 2x² + 2y² + 12x – 8y +11 = 0.
No need to sketch the circle.
Solution:
2x² + 2y² + 12x – 8y +11 = 0
2x² + 12x + 2y² – 8y = -11
2(x² + 6x + 9) + 2(y² – 4y + 4) = -11 + 18 + 8
2(x + 3)² + 2(y – 2)² = 15
C(-3, 2) and r =
𝟏𝟓
𝟐
or
𝟑𝟎
𝟐

10. Find the center and the radius of the circle
whose equation is 3x² + 3y² – 15x – 21y – 8 = 0.
No need to sketch the circle.
Solution:
3x² + 3y² – 15x – 21y – 8 = 0
3x² – 15x + 3y² – 21y = 8
3(x² – 5x +
𝟐𝟓
𝟒
) + 3(y² – 7y +
𝟒𝟗
𝟒
) = 8 +
𝟕𝟓
𝟒
+
𝟏𝟒𝟕
𝟒
3(x –
𝟓
𝟐
)² + 3(y –
𝟕
𝟐
)² =
𝟏𝟐𝟕
𝟐
(x –
𝟓
𝟐
)² + (y –
𝟕
𝟐
)² =
𝟏𝟐𝟕
𝟔
C(
𝟓
𝟐
,
𝟕
𝟐
) and r =
𝟏𝟐𝟕
𝟔
or
𝟕𝟔𝟐
𝟔

11. Find the center and the radius of the circle
whose equation is 5x² + 5y² – 4x + 9y – 7 = 0. No
need to sketch the circle.
Solution:
5x² + 5y² – 4x + 9y – 7 = 0
5x² – 4x + 5y² + 9y = 7
5(x² –
𝟒
𝟓
x +
𝟏𝟔
𝟏𝟎𝟎
) + 5(y² +
𝟗
𝟓
y +
𝟖𝟏
𝟏𝟎𝟎
) = 7+
𝟒
𝟓
+
𝟖𝟏
𝟐𝟎
5(x –
𝟒
𝟏𝟎
)² + 5(y +
𝟗
𝟏𝟎
)² =
𝟐𝟑𝟕
𝟐𝟎
(x –
𝟒
𝟏𝟎
)² + (y +
𝟗
𝟏𝟎
)² =
𝟐𝟑𝟕
𝟏𝟎𝟎
C(
𝟐
𝟓
, −
𝟗
𝟏𝟎
) and r =
𝟐𝟑𝟕
𝟏𝟎𝟎
or
𝟐𝟑𝟕
𝟏𝟎

Exercises:
A. Write the equation of each of the following circles in
standard form given the center and the radius.
CENTER RADIUS
1 origin 12
2 (2, 6) 9
3 (-7, 2) 15
4 (-4, -5) 5 𝟐
5 (10, -8) 3 𝟑
6 (-3, -7) 𝟏𝟏
7 (0, -8) 4 𝟕
8 (5, 0) 6 𝟓
9 (3, -7) 7 𝟑
10 (-7, 3) 3 𝟕
STANDARD FORM
1 x² + y² = 144
2 (x – 2)² + (y – 6)² = 81
3 (x + 7)² + (y – 2)² = 225
4 (x + 4)² + (y + 5)² = 50
5 (x – 10)² + (y + 8)² = 27
6 (x + 3)² + (y + 7)² = 11
7 x² + (y + 8)² = 112
8 (x – 5)² + y² = 180
9 (x – 3)2
+ (y + 7)2
= 147
10 (x + 7)2
+ (y – 3)2
= 63

B. Give the center and the radius of each circle. Then,
write it in general form.
STANDARD FORM
1 (x – 2)² + (y – 4)² = 36
2 (x + 4)² + (y – 9)² = 144
3 (x – 6)² + (y – 1)² = 81
4 (x – 8)² + (y + 7)² = 225
5 x² + (y – 5)² = 36
6 (x – 7)² + y² = 64
7 x² + (y + 2)² = 49
8 (x + 2)² + y² = 100
9 (x – 5)2
+ (y – 5)2
= 27
10 (x + 4)2
+ (y + 4)2
= 32
CENTER RADIUS GENERAL FORM
1 (2, 4) 6 x2
+ y2
– 4x – 8y – 16 =0
2 (-4, 9) 12 x2
+ y2
+ 8x – 18y – 47 = 0
3 (6, 1) 9 x2
+ y2
– 12x – 2y – 44 = 0
4 (8, -7) 25 x2
+ y2
– 16x + 14y – 112 = 0
5 (0, 5) 6 x2
+ y2
– 10y – 11 = 0
6 (7, 0) 8 x2
+ y2
– 14x – 15 = 0
7 (0, -2) 7 x2
+ y2
+ 4y – 45 = 0
8 (-2, 0) 10 x2
+ y2
+ 4x – 96 = 0
9 (5, 5) 3 𝟑 x2
+ y2
– 10x – 10y + 23 = 0
10 (-4, -4) 𝟒 𝟐 x2
+ y2
+ 8x + 8y = 0

Exercises:
A. Find the center and the radius of each of the
following circles (2 points each for #s 1 – 5 and 3 points
each for #s 6 – 10).
1. x² + y² – 121 = 0
2. (x – 8)² + (y + 3)² = 100
3. (x + 9)² + (y + 11)² = 13
4. (x – 5)² + (y + 1)² = 99
5. x² + y² – 6x – 40 = 0
6. x² + y² + 4x – 10y = 20
7. x² – 8x + y² – 12y – 29 = 0
8. x² + y² + 6x – 8y + 20 = 0
9. x² + y² – 4x – 2y – 27 = 0
10. 2x² + 2y² + 6x + 4y + 1 = 0
1. C(0, 0) & r = 11
2. C(8, -3) & r = 10
3. C(-9, -11) & r = 𝟏𝟑
4. C(5, -1) & r = 3 𝟏𝟏
5. C(3, 0) & r = 7
6. C(-2, 5) & r = 7
7. C(4, 6) & r = 9
8. C(-3, 4) & r = 𝟓
9. C(2, 1) & r = 𝟒 𝟐
10. C(−
𝟑
𝟐
, -1) & r =
𝟏𝟏
𝟐

B. Write the equation of the circle in general form whose
center and radius are given (3 points each for #s 1 – 5 and
5 points each for #s 6 – 10).
1. C(0, 0) & r = 5
2. C(0, 0) & r =
3. C(0, -5) & r = 7
4. C(2, 0) & r = 3 2
5. C(-3, 5) & r = 6
6. C(-4, -3) & r = 2 5
7. C(−
𝟑
𝟐
, 3) & r = 2
8. C(
𝟓
𝟐
, −
𝟑
𝟐
) & r = 3
9. C(
𝟕
𝟐
, −
𝟗
𝟐
) & r = 5
10. C(
𝟏
𝟐
, −
𝟑
𝟐
) & r =4 11
1. x² + y² – 25 = 0
2. x² + y² – 3 = 0
3. x² + y² + 10y –24 = 0
4. x² + y² – 4x – 14 = 0
5. x² + y² + 6x – 10y – 2 = 0
6. x² + y² + 8x + 6y + 5 = 0
7. 4x² +4y² + 12x –24y +29 = 0
8. 2x² + 2y² – 10x + 6y –1 = 0
9. 2x² + 2y² – 14x + 18y +15 = 0
10. 2x² + 2y² – 2x + 6y – 347 = 0

C. A general form of a circle is given. Transform each
equation to standard form. Then give the coordinates of
the center and radius.
1. SF: (x – 1)²+ (y – 4)² = 64
2. SF: (x +2)²+ (y – 2)² = 36
3. SF: (x +5)²+ (y + 2)² = 32
4. SF: x²+ (y + 4)² = 100
5. SF: (x -
𝟐
𝟑
)²+ (y -
𝟏
𝟑
)² = 4
1. x² + y² – 2x – 8y – 47 = 0
Center: (1, 4) & Radius: 8 units
2. x ² + y² + 4x – 4y – 28 = 0
Center: (-2 , 2) & Radius: 6 units
3. x ² + y² +10x + 4y – 3 = 0
Center: (-5, -2) & Radius: 4 𝟐 units
4. x ² + y² +8y – 84 = 0
Center: (0, -4) & Radius: 10 units
5. 9x ² + 9y² – 12x – 6y – 31 = 0
Center: (
𝟐
𝟑
,
𝟏
𝟑
) & Radius: 2 units

Problem Solving that Applies the Concept of the
Equation of the Circle
x
y
-
-
-
-
-
-
y
•
•
•
-
1. What is the equation of the
circle in general form with a
radius of 6 units and is tangent
to the line y = 1 at (2, 1)?
Figure:
(x – 2)² + (y +4)² = 36
x² – 4x + 4 + y² + 8y + 16 = 36
x² + y² – 4x + 8y – 16 = 0
2. A line passes through the center
of a circle and intersects it at
points (4, -1) and (10, 7). What is
the equation of the circle in
standard form?
Solution:
Solution:
x =
4 + 10
𝟐
= 𝟕
y =
−1 + 7
𝟐
= 𝟑
C(7, 3)
(x – 7)² + (y – 3)² = 25
r = 4 − 7 + (−1 − )²
= 𝟐𝟓
r = 5 units

Figure:
x
y
•
•
•-
Whatis
acircle?

Module VI – Permutations and Combinations
This module deals with the counting techniques,
permutations and combinations. These three topics are also
known as Combinatorics. These topics are widely applied in
many fields of human endeavor such as statistics,
economics and applied sciences.
Lesson I – Fundamental Counting Principle
If one thing can occur in m ways and a second thing
can occur in n ways, and a third thing can occur in r ways,
and so on, then the sequence of things can occur in
m x n x r x … ways
Let us study the following examples.
1. A certain tourist would like to go to Banaue Rice Terraces. There are
three routes from Manila to Baguio and two routes from Baguio to
Banaue. In how many ways the tourist may use to reach Banaue from
Manila?

Diagram:
Manila Baguio Banaue Rice Terraces
R1
R2
R3
R4
R5
Manila to
Banaue
Rice Terraces
R4 R5
R1 (R1, R4) (R1, R5)
R2 (R2, R4) (R2, R5)
R3 (R3, R4) (R3, R5)
∴There are six ways
which the tourist may use
to reach Banaue from
Manila.
2. How many three-digit numbers can be made from the digits 1, 2
and 3?
a) Allowing no repetition of the digits
∴There are six 3-digit
numbers that can be made
from the digits 1, 2 and 3
with no repetition of the
digits.

b) Allowing repetition of the digits
Diagram:
1 111
1 2 112
3 113
1 121
1 2 2 122
3 123
1 131
3 2 132
3 133
1 211
1 2 212
3 213
1 221
2 2 2 222
3 223
1 131
3 2 132
3 133
1 311
1 2 312
3 313
1 321
3 2 2 322
3 323
1 331
3 2 332
3 333
Hundreds
Digit
Tens
Digit
Units
Digit
Number
∴There are 27 three-
digit numbers that can be
made from the digits 1, 2
and 3 allowing the
repetition of the digits.
Whatmethodare
youapplying?

3. How many different 3-digit numbers can be formed
from the digits 1 to 9, if: a) repetitions are allowed? b)
repetitions are not allowed?
Solution:
a) 9(9)(9) = 729 b) 9(8)(7) = 504
4. Manila and Quezon City are connected by 4 roads
while Quezon City and Caloocan City by 3 roads. In how
many ways can one drive from Manila to Quezon City to
Caloocan City?
Solution:
Event 1 – Going from Manila to Quezon City = 4
Event 2 – Going from Quezon to Caloocan City = 3
4(3) = 12 ways

Exercise:
A. Use the Fundamental Counting Principle (FCP)
to answer each question. Draw diagram for each
problem.
1. If there are 4 routes from town A to town B and 5 routes
from town B to town C, how many different routes may
be taken in going from town A to town C?
2. How many different arrangements of X, Y and Z on a
straight if (a) no repetition of the letters is allowed; (b)
repetition is allowed?
3. How many a) two-digit numbers, b) three-digit numbers
can be formed from the digits 1, 3, 5, 7 and 9 if no
repetition of digits is allowed?
4. How many four-digit numbers can be formed from the
digits 1, 2, 3 and 4 if each number is to begin and end
with an odd digit? No repetition of digits is allowed.

B. Use the Fundamental Counting Principle (FCP)
to answer each question. No need to draw diagram.
1. If you buy two pairs of pants, four shirts, and two
pairs of shoes, how many new outfits consisting of a
new pair of pants, one shirt, and one pair of shoes
would you have?
2. Paloma is taking a matching test in which he is
supposed to match four answers with four questions.
In how many different ways can he answer the four
questions?
3. A plate number is made up of two consonants
followed by three nonzero digits followed by a vowel.
How many plate numbers are possible if (a)
repetition of the letters and digits is not allowed?; (b)
repetition of the letters and digits is allowed?

4. Three cards are drawn in succession and without
replacement from a deck of 52 cards.
a. Find in how many ways we can obtain the King
of hearts, the Ace of diamonds and the Queen of clubs in
that order.
b. Find the total number of ways in which the
three cards can be selected.
5. A die is rolled and a coin is tossed. Determine the
number of different possible outcomes by using the
Fundamental Counting Principle. List all the possible
outcomes by constructing a tree diagram.
6. A license plate consists of three letters followed by a
three-digit number. How many plates are available for
distribution if 0 is not allowed to appear as the first digit
in the number.

Another Exercise:
A. How many two-digit numbers may be formed from
each given set of digits if repetitions are not allowed?
1. {1, 2, 3}
2. {1, 2, 3, 4}
3. {1, 2, 3, 4, 5}
4. {1, 2, 3, 4, 5, 6}
5. {1, 2, 3, 4, 5, 6, 7}
B. How many three-digit numbers can be formed from
each set of digits if zero is not to be used as the first digit
and repetitions are allowed?
1. {0, 1, 2, 3}
2. {0, 1, 2, 3, 4}
3. {0, 1, 2, 3, 4, 5}
4. {0, 1, 2, 3, 4, 5, 6}

Quiz (one-half CW)
A. Complete the table with the correct answer.
Number of Objects (n) Number of Objects Taken at
a Time at a Time(r)
Number of Possible
Arrangement
1 2 1
2 2 2
3 3 1
4 3 2
5 3 3
6 4 1
7 4 2
8 4 3
9 4 4
10 10 3
11 15 4
12 20 5
13 7 4
14 25 3
15 21 5
2
2
3
6
6
4
12
24
24
720
32 760
1 860 480
840
13 800
2 441 880

Quiz (one-half CW)
B. Solve for the unknown value in each item.
1. P(6,6)=______ 6.P(8,r)=6720
2. P(7,r)=840 7.P(8,3)=_____
3. P(n,3)=60 8.P(n,4)=3024
4. P(n,3)=504 9.P(12,r)=1320
5. P(10,5)=_____ 10.P(13,r)=156
Answers
1. 720
2. 4
3. 5
4. 9
5. 30 240
6. 5
7. 336
8. 9
9. 3
10. 2

Answer each permutation mentally.
1. A teacher wants to assign 4 different tasks to his 4
students. In how many possible ways can he do it?
2. In a certain general assembly, three major prizes are at
stake. In how many ways can the first, second and third
prizes be drawn from a box containing 120 names?

C. Five different Mathematics books and six different
Grammar books are to be arranged on a shelf. How many
possible arrangements can be made if:
1. the books on the same subjects are to be
arranged together?
2. the books are to be arranged together?

Definition:
If n, in general, is a positive integer, then n
factorial denoted by n! is the product of all positive
integers less than or equal to n.
n! = n(n – 1)(n – 2)(n – 3) … (2)(1)
As a special case, we define
0! = 1.
Examples:
Compute the following:
1. 7! = 7(6)(5)(4)(3)(2)(1) = 5 040
2.
𝟖!
𝟒!
=
𝟖(𝟕)(𝟔)(𝟓)(𝟒!)
𝟒!
= 1 680
3.
𝟗!
𝟒!𝟑!
=
𝟗(𝟖)(𝟕)(𝟔)(𝟓)(𝟒!)
𝟒!𝟑!
=
𝟏𝟓 𝟏𝟐𝟎
𝟑(𝟐)(𝟏)
= 2 520

Lesson II – Permutations
The process of ordering or arranging the elements
of a set such that one is the first, another is the second
and so on is called PERMUTATION.
In symbols, we denote permutation of n objects
taken r at a time as:
P(n, r) =
𝒏!
𝒏 −𝒓 !
= n(n – 1)(n – 2)…(n – r + 1), 0≤ r ≤ n
Examples:
A. Evaluate the following:
1. P(7, 3) =
!
!
=
( )( )( !)
!
2. P(4, 4) =
!
!
=
( )( )( )
= 24
3. P(8, 5) =
!
!
=
( )( )( )( )( !)
!
= 7(6)(5) = 210
= 6 720

4. 9P3 =
!
!
=
( )( )( !)
!
= 9(8)(7) = 504
5. 12P5 =
!
!
=
( )( )( )( )( !)
!
= 95 040
6. P(15, 6) =
!
!
=
( )( )( )( )( )( !)
!
= 3 603 600
B. Solve the following problems by mental
solution only.
1. Ten runners join a race. In how many
possible ways can they be arranged as first,
second and third placers?
Solution:
10(9)(8) = 720

2. If Cardo has 12 t-shirts, 6 pairs of pants,
and 3 pairs of shoes, how many possibilities
can he dress himself up for the day?
Solution:
12(6)(3) = 216
3. In how many ways can Aling Belen
arrange 6 potted plants in a row?
Solution:
6! = 6(5)(4)(3)(2)(1) = 720

4. How many 4-digit numbers can be formed from
the numbers 1, 3, 4, 6, 8 and 9 if repetition of
digits is not allowed?
Solution:
P(6, 4) =
!
!
=
( )( )( )( !)
!
= 360
5. If there are 3 roads from Town A to Town B and
4 roads from Town B to Town C, in how many ways
can one go from Town A to Town C and back to
Town A, through Town B, without passing through
the same road twice?
Solution:
3(2)(4)(3) = 72

6. Suppose that in a certain association, there are 12
elected members of the Board of Directors. In how
many ways can a president, a vice president, a
secretary, and a treasurer be selected from the board?
Solution:
P(12, 4) =
!
!
=
( )( )( )( !)
!
7. You want to order your lunch from the school
canteen which offers student meals consisting of 1 cup
of rice, 1 meat dish, and 1 vegetable dish. How many
choices do you have for your meal if there are 3 choices
of meat dishes and 2 choices of vegetable dishes?
Solution:
1(3)(2) = 6
= 11 880

8. In how many ways can you place 9 different
books on a shelf if there is space enough for only 5
books?
Solution:
P(9, 5) =
!
!
=
( )( )( )( )( !)
!
9. A dress-shop owner has 8 new dresses that she
wants to display in the window. If the display
window has 5 mannequins, in how many ways can
she dress them up?
Solution:
= 15 120
P(8, 5) =
!
!
=
( )( )( )( )( !)
!
= 6 720

Lesson III – Circular Permutations and
Distinguishable Permutations
A special case of permutation where the
arrangement of things is in circular pattern is called
CIRCULAR PERMUTATION. The most common example of
this type is the seating arrangement of people around a
circular table.
C
A
B
B
C
A
C
A
B
Formula:
P = (n – 1)!
Note that 1 circular
permutation corresponds to
3 linear permutations.
C
A
B
Circular Permutation Linear Permutation
A B C
B C A
C A B

The number of permutations of n objects where
a things are alike, b things are alike and c things are
alike, and so forth, is said to be Distinguishable
Permutation. The formula is as follows:
P =
𝑛!
𝑎!𝑏!𝑐!
A. Identify what type of permutation is applied then
solve.
1. There are 12 people in a dinner gathering. In how
many ways can the host (one of the 12) arrange
his guests around a dining table if
a. they can sit on any of the chairs
b. 3 people insist on sitting beside each other?
c. 2 people refuse to sit beside each other?

2. In how many different ways can 9
people be seated at a round table?
3. In how many ways can 7 students be
seated in a round table if 2 particular
students must be seated next to each
other?
4. In how many ways can 9 students be
seated in a round table if 2 particular
students must not be seated next to

B. Find the number of distinguishable
permutations of the letters in eaach
word.
a. HONOR
b. COMMITTEE
c. PHILIPPINES
d. MISSISSIPPI
e. TOMATO
f. BASKETBALL
g. SPOUSE

Lesson IV – Combinations
The number of combinations of n
things taken at r at a time is
nCr =
𝒏!
𝒏 −𝒓 !𝒓!
Examples:
A. Compute the following:
1. 7C2
2. 6C4
3. 12C9
4. 15C2
5. 20C8

B. Solve the following.
1. How many different committees of 5
people can be appointed from a group of
10 people?
2. Twelve members of a club will form
different committees of three members
each. How many committees can be
formed?
3. Using the same conditions as in Problem 2,
the question is: If Clark is one of the
members, in how many committees will he
be included?

4. There are five black balls and four red balls in a
container. How many different groups of four
balls each, so that each group contains that:
a. all balls are of the same color?
b. one ball is black and three are red?
c. two are black and two are red?
5. How many lines are determined by 8 points,
no 3 points of which are collinear?
6. At a party, each guest shook hands with
every other guest exactly once. There were a
total of 190 handshakes. How many guests
were there?

Module VII – Probability
Lesson 1 Sample Space and Events
A sample space denoted by S is the se of all possible outcomes of an experiment. Each
possible outcome or element of the set is called a point or a sample point. In other words, an
element of the set is called a point or a sample point in the sample space.
An event is any subset of a sample space.
Examples:
1. Experiment of Tossing a Coin
S = {h, t}
2. Experiment of Tossing Two Coins
S = {(h, h), (h, t), (t, h), (t, t)}
3. Experiment of Rolling a Die
S = {1, 2, 3, 4, 5, 6}

4. ExperimentofRollingTwoDice(Oneisred,theotherisgreen.)
Thesamplespaceofthisexperimentisillustratedbelow.
R/G 1 2 3 4 5 6
1 {(1,1) (2,1) (3,1) (4,1) (5,1) (6,1)
2 (1,2) (2,2) (3,2) (4,2) (5,2) (6,2)
3 (1,3) (2,3) (3,3) (4,3) (5,3) (6,3)
4 (1,4) (2,4) (3,4) (4,4) (5,4) (6,4)
5 (1,5) (2,5) (3,5) (4,5) (5,5) (6,5)
6 (1,6) (2,6) (3,6) (4,6) (5,6) (6,6)}

Examples:
1. Whatistheprobabilityofgettinganevennumberintheexperimentofrollingadie?
P(A)=
𝒏(𝑨)
𝒏(𝑺)
=
3
6
=
1
2
Solution:
S={1,2,3,4,5,6} n(S)=6
A={2,4,6} n(A)=3

2.Whatistheprobabilitythatthesumofthefacesofthetwodiceis8?
Solution:
F ={(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}
n(F) =5
n(S) =36
P(F)=
5
36
=
5
36

Exercise:
Determine the cardinality of each sample
space.
1. Throwing a coin and a die together
2. Throwing two dice
3. Tossing three coins
a. Determine the sample space S.
b. Determine the event M in which exactly 2
tails appear.
c. Determine the event N in which at least
one tail appears.

B. Answer as required.
1. Let S = {x, y, z}; find all events of this
experiment.
2. Let S = {x, y, z, v}; find all events of this
experiment.
3. Let S be a sample space of an experiment with n
outcomes. Determine the number of events of
this experiment.
4. Let S = {5, 10, 15, 20, 25, 30} be the sample
space of an experiment and let T = {5, 10},
V = {5, 20, 30}, and W = {10, 15, 25} be the
events of the experiment.

Counting Techniques and Probability of
Compound Events
1. A group of 6 men and 10 women must
form a four-person committee. How
many committees are possible if the
committee must consist of the
following?
a. two men and two women
b. any mixture of men and women
c. a majority of women
d. a majority of men

2. A box contains 7 red balls, 5 white balls,
and 3 green balls. In how many ways can
we select 3 balls such that
a. they are all red?
b. they are all white?
c. they are all green?
d. they are of different colors?
e. two are red and one is white?
f. exactly 2 are white?
g. exactly 1 is green?
h. none is green?

3. A box of 10 articles contains one that is
defective. In how many ways can Marivic select 4
articles such that
a. the defective is included?
b. the defective is not included?
4. A box contains 4 white balls, 3 red balls, and 3
green balls. If three balls are drawn
simultaneously at random, what is the probability
that
a. they are all white?
b. two are red and one is green?
c. exactly two are green?
d. none is white?
e. they are of different colors?

5. There are 7 blocks in a box, each having one of
the colors of the rainbow. You are asked to draw
one block from the box at random.
a. Write the sample space for this
experiment. Represent each block by the initial of
each color.
b. Write out an event A such that n(A) = 3.
c. Write out an event B such that B is a
subset of A.
d. What is the probability of event A? event
B?
e. P(A and B) = ?

6. A three-digit number with repetition of digit is
made from the digits 3 to 8. What is the
probability that the number is
a. odd?
b. even?
c. between 200 and 500?
d. a multiple of 5?
e. between 500 and 750?
7. In Gabriel’s pocket there are 6 different coins: 5
cents, 10 cents, 25 cents, 50 cents, 1 peso and 2
peso. If he draws two coins, what is the
probability that the total amount is less than 1
peso?

That ends our lesson for
TODAY. Thank you for
LISTENING.

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