This is the Grade 10 Mathematics Quarter 2 Module 7 self-learning module, “Solving Problems on Circles” (competency M10GE-IIf-2). It is an Alternative Delivery Mode (ADM) module from the Department of Education – Cordillera Administrative Region, written for learners studying at home with help from a teacher or parent. After working through it, learners can use what they know about chords, arcs, angles, tangents, and secants to solve real-life word problems about circles.
On this page 7 sections
What learners will learn
The module applies earlier lessons on chords, arcs, central angles, inscribed angles, secants, tangents, segments, and sectors of a circle. By the end, learners should be able to:
- Solve problems involving chords, arcs, central angles, and inscribed angles of circles.
- Solve problems involving tangents and secants of circles.
Key rules used in the module
The module reviews these facts in its “What I Have Learned” section and uses them in every word problem:
| Situation | Rule |
|---|---|
| Central angles | Central angles of a circle with no common interior points add up to 360°. |
| Inscribed angle | Its measure is one-half the measure of its intercepted arc. |
| Tangent line | It is perpendicular to the radius drawn to the point of tangency. |
| Area of a sector | Area of the circle × (measure of the sector’s arc ÷ 360°). |
| Arc length | (degree measure of the arc × 2πr) ÷ 360°. |
| Two secants from an outside point | One secant segment × its external part = the other secant segment × its external part. |
| Two intersecting chords | The product of the segments of one chord = the product of the segments of the other chord. |
Worked examples in the module
Each example follows four steps: Understand the Problem, Devising a Plan, Carrying out the Plan, and Looking Back (checking the answer by working backwards).
Chords, arcs, central and inscribed angles
- Ferris wheel: Each central angle between spokes is 15°. Since 360° ÷ 15° = 24 spokes, and each spoke holds two seats, the wheel seats 24 × 2 = 48.
- Star-shaped garden path: A regular five-pointed star in a circle cuts the circle into five arcs of 360° ÷ 5 = 72° each. Each inscribed angle at a star point is half of that, or 36°.
- Dart board: A 40 cm board with 20 equal sectors has arcs of 18° and a radius of 20 cm. Each sector has an area of (1/20) × 400π = 20π cm², about 62.8 cm².
- Iron arch: Two concentric semicircles with inner radius 2 m and outer radius 2.8 m have arc lengths of 2π m (about 6.28 m) and 2.8π m (about 8.79 m). The total iron needed is about 15.07 m.
Tangents and secants
- Elastic band on a circular board: With SY = 15 cm, TY = 6 cm, and LY = 7 cm, the rule SY · TY = EY · LY gives 90 = 7(EY), so EY is about 12.86 cm.
- Repaired round table: Two wooden strips form intersecting chords with FS = 6 ft, SA = 4 ft, and LS = 3 ft. From 6 × 4 = 3 × SG, SG = 8 ft, so the second strip LG = 3 + 8 = 11 ft.
Parts of the module
- What I Know: a 10-item multiple-choice pretest. Its items include clock angles, pie charts, wheel spokes, a dart board, a circular garden with tangent pathways, and ropes across a pool.
- What’s In: review drills. “Half, equal, or twice as?” covers arcs, central angles, and inscribed angles. “Sum or Difference?” covers angles formed by tangents and secants.
- What’s More: the six worked word problems, with two practice sets. Assessment 1, “It’s Your Turn!”, has problems on a motorcycle wheel, a clock, a fruit pie, and a garage arch. Assessment 2, “Solve my length!”, has problems on pool ropes and a circular frame.
- What I Have Learned: six fill-in-the-blank statements of the key rules.
- What I Can Do: “The Huge Ceiling Ball!” Learners find the radius of a hanging light ball from a 50 cm string and a 30 cm tangent.
- Assessment: a 10-item multiple-choice posttest.
- Additional Activity: “Guess What!” Learners find the major and minor arcs when two tangent pathways meet at 50°.
- Answer Key: answers for all tests and activities.
How to use this module
- Learners: Start with the pretest to see what you already know. Then work through the parts in order. Write all answers on a separate sheet of paper, not on the module. For each word problem, follow the four steps and check your answer at the end. Use the answer key honestly, only after you have tried each task.
- Parents and facilitators: The module includes Notes to the Teacher with tips for supporting home-based learning. Encourage your child to show every step of the solution, because the rubrics reward complete, accurate steps.
- Teachers: The practice problems, “What I Can Do”, and the Additional Activity come with scoring rubrics. Assessments 1 and 2 are scored 5, 3, or 1 per item. “What I Can Do” and the Additional Activity are scored up to 20 points. These rubrics can guide how you check learners’ submitted solutions.
Frequently asked questions
What grade and quarter is this circles module for?
It is for Grade 10 Mathematics, Quarter 2, Module 7. It covers competency M10GE-IIf-2, solving problems on circles.
Does the module have an answer key?
Yes. The answer key covers the pretest, the review drills, Assessments 1 and 2, What I Have Learned, What I Can Do, the posttest, and the Additional Activity.
What problem-solving method does the module teach?
Every word problem uses four steps: Understand the Problem, Devising a Plan, Carrying out the Plan, and Looking Back. In the last step, learners check the answer by working backwards.
How do you solve intersecting chords problems in this module?
Multiply the two segments of one chord and set the result equal to the product of the two segments of the other chord. In the table example, 6 × 4 = 3 × SG, so SG = 8 ft.
What is the radius of the ceiling ball in the What I Can Do task?
The answer key gives 40 cm. The tangent from the ceiling (30 cm) is perpendicular to the radius, and the 50 cm string reaches the center, so this forms a right triangle with the string as its longest side.
Full text of the module
The text below is a transcription of the PDF above, as printed.
Mathematics Quarter 2 – Module 7: Solving Problems on Circles
10
CO_Q1_Mathematics 10_ Module 7
[illustration: learners studying outdoors and at a table, with a teacher, a globe, and a lamp]
Mathematics – Grade 10 Alternative Delivery Mode Quarter 2 – Module 7: Solving Problems on Circles (M10GE-IIf-2) First Edition, 2020
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Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this module are owned by their respective copyright holders. Every effort has been exerted to locate and seek permission to use these materials from their respective copyright owners. The publisher and authors do not represent nor claim ownership over them.
Published by the Department of Education Secretary: Leonor Magtolis Briones Undersecretary: Diosdado M. San Antonio
| Development Team of the Module | |
|---|---|
| Writer’s Name: | Esther B. Guitobon |
| Editor’s Name: | Heather G. Banagui |
| Reviewer’s Name: | Bryan A. Hidalgo |
| Layout Artist: | Jecson L. Oafallas |
| Management Team: | May B. Eclar Benedicta B. Gamatero Carmel F. Meris Ethielyn E. Taqued Edgar H. Madlaing Marciana M. Aydinan Lydia I. Belingon |
Printed in the Philippines by:
Department of Education – Cordillera Administrative Region
Office Address: Wangal, La Trinidad, Benguet Telefax: (074) 422-4074 E-mail Address: car@deped.gov.ph
Mathematics Quarter 2 – Module 7: Solving Problems on Circles
10
Introductory Message
This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson.
Each SLM is composed of different parts. Each part shall guide you step-by-step as you discover and understand the lesson prepared for you.
Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self-check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these.
In addition to the material in the main text, Notes to the Teacher are also provided to our facilitators and parents for strategies and reminders on how they can best help you on your home-based learning.
Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task.
If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator.
Thank you.
What I Need to Know
Concepts on circles are basic necessity in many fields such as construction, landscaping and engineering. Aside from these concepts, the properties of lines and its subsets and their relationship to circles are important. These are the lines that either touch or intersect circles.
In this module, we will be applying previously learned concepts on chords, arcs, central angles, inscribed angles, secants, tangents, segments and sectors of a circle to solve problems. Specifically, you will be able to
- Solve problems involving chords, arcs, central angles, and inscribed angles of circles; and
- Solve problems involving tangents and secants of circles.
What I Know
Description: This section will test how much you already know about problems involving circles and its chords, arcs, central angles, inscribed angles, tangent lines and secant lines.
Instruction: Read each item carefully. Choose the letter of your answer and write it on your answer sheet.
Which of the following is the measure of the angle determined by the hands of a clock at five o’clock?
- a. 90°
- b. 120°
- c. 150°
- d. 180°
A circle is divided into 6 equal arcs. Which of the following is the degree measure of each arc?
- a. 30°
- b. 60°
- c. 90°
- d. 120°
A wheel has 30 spokes which are evenly spread. Which of the following is the measure of each central angle formed by the spokes?
- a. 12°
- b. 15°
- c. 24°
- d. 30°
In a pie chart showing the expenditures of the Lopez family, the measure of the central angle corresponding to their food expenses is 120°. If their monthly income is Php 35,000.00, how much of it is spent for food?
- a. Php 11,666.67
- b. Php 17,500.00
- c. Php 5,666.67
- d. Php 12,000.00
Catherine designed a pendant. It is a regular hexagon set in a circle. Suppose the vertices are connected by line segments and meet at the center of the circle as shown in the figure. Which of the following is the measure of each of the smallest angle formed at the center?
- a. 22.5°
- b. 45°
- c. 60°
- d. 72°
[illustration: regular hexagon set in a circle, with its vertices joined by line segments to the center]
A dart board has a diameter of 40 cm and is divided into 20 congruent sectors. Which of the following is the area of one of the sectors?
- a. 20π cm2
- b. 40π cm2
- c. 80π cm2
- d. 800π cm2
Mr. Alex designed a semicircle arch made of bent iron for their gate’s entrance. Suppose the diameter of the semicircle is 3 meters, which of the following should be the length of the iron before bending to avoid wastage?
- a. 1.5π m
- b. 3π m
- c. 6π m
- d. 9π m
A circular garden has pathways which are tangent to the garden whose center is 10 meters away from the entrance as shown in the figure below. If the radius of the circular garden is 5 meters, how long is each pathway?
[illustration: circle labeled “Garden” with a “Pathway” tangent to it from “Entrance”; a dashed line runs from the entrance through the center]
- a. 3√5 m
- b. 5√3 m
- c. 25 m
- d. 75 m
A family tied a pair of ropes to their circular pool, to serve as a guide for their children who are practicing how to swim. The ropes intersect at the interior forming segment as shown in the figure below. Find the missing segment measure of the rope in meters.
[chart: diagram of a circle labeled “Circular Pool” with two intersecting chords; segment values 6 and 4 on one chord, 3 on the other, and the unknown segment labeled x]
- a. 5 m
- b. 6 m
- c. 7 m
- d. 8 m
A circular medallion is hung such that the strings are tangent to the circle as shown in the figure. If the major arc measures 230°, find the degree measure of the angle formed by the strings.
[illustration: circular medallion hung by two strings tangent to the circle; labels “String” and “Medallion”; a small diamond shape inside the circle]
- a. 40°
- b. 50°
- c. 130°
- d. 230°
What’s In
Let us have a review of what you have already learned about arcs, central angles, inscribed angles, tangents and secants of circles by doing the following activities.
Half, equal, or twice as?
In the figure that follows, A, R, O, L and P are points on circle ʘC. Fill in the blanks with the word half, equal or twice to show the relationship of the degree measures of arcs, central angles and inscribed angles formed.
[illustration: circle ʘC with points P, L, A, O and R on it; lines drawn from P, L and A through C, and chords connecting the points]
- The measure of central angle ∠OCR is _____ to the degree measure of $\widehat{OR}$.
- The degree measure of $\widehat{LO}$ is _____ to the measure of central angle ∠LCO.
- The measure of inscribed angle ∠OPR is _____ the measure of central angle ∠OCR.
- The measure of central angle ∠PCA is _____ the measure of inscribed angle ∠POA.
- The measure of inscribed angle ∠PRL is _____ the measure of central angle ∠PCL.
Sum or Difference?
In the figure that follows, K, M, P, and N are points on ʘQ. Fill in the blanks with the word sum, difference or one-half to show the relationship of angles formed by intersecting tangents and secants.
[transcriber’s note: the printed footer on these pages reads “CO_Q1_Mathematics 10_ Module 7”, which differs from the file name CO_Math10_Q2_Module7.pdf. The pages are transcribed as printed.]
What’s More
Word Problem 1:
Each central angle formed by consecutive spokes of a Ferris wheel measures 15°. What is the seating capacity if two seats are anchored to each spoke in the Ferris wheel?
Step 1: Understand the Problem
In the problem, the angles formed in between consecutive spokes can be considered as central angles of circles. We are asked to determine the seating capacity of the Ferris wheel if a central angle formed by consecutive spokes measures 15°.
To solve, remember that the sum of the central angles of a circle with no common interior points is 360˚.
Step 2: Devising a Plan
We need to find how many spokes are there in the Ferris Wheel. Since the central angle formed by consecutive spokes measures 15° and a whole revolution is 360°, then get the quotient between 360° and 15°. After which, multiply the quotient by two since there are two seats in each spoke.
Step 3: Carrying out the Plan
360°÷15°= 24 Getting the quotient of 360° and 15° to find the number of spokes
24 × 2= 48 Multiplying to two to get the number of seating capacity.
Step 4: Looking Back
To check whether there are really 48 seating capacity, solve the problem backwards. Verify first if the product of 24 and 2 is really equivalent to 48. Next, verify the product of 24 and 15° if it is really equivalent to 360°. The two steps are easily verified; thus, we can conclude that our answer is correct.
Therefore, there are 48 seating capacity.
Word Problem 2:
There are circular gardens having paths in the shape of a regular star pentagon like the one shown in the figure.
[illustration: circular garden with a regular five-pointed star whose five points touch the circle]
a. Determine the measure of an arc intercepted by an inscribed angle formed when the vertices of the star are connected.
b. What is the measure of an inscribed angle in a garden with a five-pointed star?
Step 1: Understand the Problem
In the garden having paths in the shape of a regular star pentagon, connect each vertex to the other vertices as shown in the figure at the right. The figure shows the intercepted arcs of the inscribed angle formed by two consecutive dotted lines. For example, inscribed angle, ∠A is intercepted by C̅B̅̂. In the problem, we need to find the measure of the inscribed angles and intercepted arcs.
[illustration: five-pointed star inscribed in a circle, with dotted lines joining the vertices; vertices labeled A (top), B (lower left) and C (lower right)]
Step 2: Devising a Plan
Remember that from the given, the star pentagon formed is regular and the degree measure of one complete circle is 360˚. Thus, to find the measure of each intercepted arc, get the quotient of 360˚ and the number of intercepted arcs.
To find the measure of an inscribed angle, use the theorem: “If an angle is inscribed in a circle, then the measure of the angle is equal to one-half the measure of its intercepted arc.”
Step 3: Carrying out the Plan
360°÷5= 72˚ Getting the quotient of 360° and 5 to find the degree measure of an intercepted arc. Thus, the measure of each intercepted arc of the 5 angles of the regular star pentagon is 72˚.
72˚ ÷ 2= 36˚ Getting the half of the degree measure of an intercepted arc since inscribed angle is one-half of its intercepted arc. This results to one inscribed angle equal to 36°.
Step 4: Looking Back
Check if twice the measure of an inscribed angle is equal to the measure of an intercepted arc and five times the degree measure of an intercepted arc is 360˚. In the problem, the measure of an arc intercepted by an inscribed angle formed when the vertices of the star are connected should be 72˚. In addition, the sum of the measure of all intercepted arc should be 360°.
Word Problem 3:
A dart board has a diameter of 40 cm and is divided into 20 congruent sectors. What is the area of one of the sectors?
[photo: dart board with 20 numbered sectors and three darts]
Step 1: Understand the Problem
Considering the dart board at the right, the darkest region is a sector of the circular dart board. The area of one sector is what the problem intends to find. To find its area, get the product of the area of the circle and the (measure of the corresponding arc of the sector) / 360.
[illustration: dart board with one sector shaded black]
Step 2: Devising a Plan
From the given, to find the area of a sector of the circle, find first the measure of the arc which corresponds to a sector. This can be done by dividing the degree measure of a whole circle (360°) by 20 sectors.
The area (A) of the circle can be computed using A = πr². To get the radius, divide the diameter by two.
Finally, to determine the area of each sector of the dart board, get the product of (measure of the corresponding arc of the sector) / 360 and the area of the circle.
Step 3: Carrying out the Plan
From the plan, the following steps is followed to find the area of each sector of the dart board
a. Find the measure of each arc by getting the quotient of 360˚ and 20.
360˚ ÷ 20 = 18˚
b. Find the radius of the dart board.
40 ÷ 2 = 20 cm. is the length of the radius
c. Find the ratio of the measure of an arc to 360˚,
Ratio= (measure of the arc) / 360 = 18 / 360 = 1 / 20
d. Find the area (A) of the circle using the equation, A = πr²
A = π(20)² Substitute the radius from letter b
A = 400 cm² Simplify Exponents
e. Finally, find the area of each sector by getting the product of the ratio in step c and the area in step d.
Area of each sector = 1/20 · 400π
= 20π Substitute 3.14 to π
= 62.83 cm²
Step 4: Looking Back
To check the computed area of each sector
a. Start with 20 π. Multiply it by the number of sectors which is 20.
b. Multiply the result by 1/20
c. Divide the answer in b by π. After which, extract the square root.
d. Multiply the result of c by 2. The product should be 40. If you obtain a different answer, go back to step a.
e. Finally get the product of 18 and 20. The obtained answer should be equivalent to 360.
Word Problem 4:
Mr. Celso designed an arch made of bent iron for the top of a school’s main entrance as shown in the figure at the right. Each of the 12 segments between the two concentric semicircles are 0.8 meter long. Suppose the diameter of the inner semicircle is 4 meters, what is the total length of the bent iron used to make this arch?
[illustration: arch of two concentric semicircles joined by 12 segments; labels 0.8m (width of one segment band) and 4m (diameter of the inner semicircle)]
Step 1: Understand the Problem
The problem asked to determine the total length of the bent iron used to make the arch. Each of the 12 segments joining the inner and outer circle measures 0.8 meter and the diameter of the inner circle is 4 meters
Step 2: Devising a Plan
To solve the total bent iron used to make the arch, get the sum of the inner arc length and outer arc length. In solving for each arc length, remember that the length of an arc can be determined by using the proportion (degree measure of the arc) / 360° = (arc length) / 2πr where r is the radius of the circle. Multiply 2πr both sides of the proportion then simplify. We obtain
arc length = (degree measure of the arc)(2πr) / 360°
In the given, the radius of the inner arc length is 2m and the radius of the outer arc length is the sum of 2 m and 0.8 m. While the inner arc length and outer arc length measures 180° since they are both semicircles.
Step 3: Carrying out the Plan
Solve separately the inner and outer arc length then get their sum. Therefore, the solution is as follows:
a. Finding the Arc length of the inner arc:
arc length = (degree measure of the arc)(2πr) / 360°
= (180°)(2π)(2) / 360° Substituting the arc length of a semicircle and the radius of the inner arc
= 2π Simplifying
<!– page 17 is transcribed from attachment page 1 –>
Substituting 3.14 for π
b. Finding the Arc length of outer arc:
arc length = (degree measure of the arc)(2πr) / 360°
= (180°)(2π)(2.8) / 360° Substituting the arc length of a semicircle and the radius of the outer arc which is the sum of 2 and 0.8
= 5.6π / 2 Simplifying
= 8.79m Substituting 3.14 for π and further simplifying
After obtaining the length of the inner and outer arcs, get the sum to finally obtain the total length of the bent iron used to make the arch.
6.28 + 8.79 = 15.07 m
Step 4: Looking Back
Solve backwards to check if the solution is correct. Start with the measure of the length of each arc then solve backwards until the radii is obtained. The solution could be as follows:
a. Radius of the outer arc, starting from 5.6π / 2
Multiply 5.6π / 2 by 360° / (180°)(2π). Obtained value should be 2.8. If not, check what went wrong in our solution.
b. Radius of the inner arc starting from 2π
Multiply 2π by 360° / (180°)(2π). Obtained value should be 2. If not, check our solution.
Assessment 1: It’s Your Turn!
Apply the concepts learned in the previous activities to be able to solve the following problems.
Rowel is designing a motorcycle wheel. He decided to put 6 spokes which divide the rim into 6 equal parts. What is the degree measure of each angle formed by the spokes at the hub?
Find the degree measure of the angle formed by the hands of a clock at 5:00 in the afternoon.
[illustration: clock face showing hands at 5:00]
- A fruit pie is divided into 3 equal parts forming central angles. Find the area of each sector formed if the fruit pie has a radius of 6 inches.
[illustration: circle divided into three sectors by radii]
- Mr. Alex designed a semicircle arch made of bent iron for the door of his garage. Suppose the diameter of the semicircle is 3 meters, how long should the iron be before bending to avoid wastage?
[illustration: semicircle arch labeled 3m]
Rubrics: To guide you with your solution, please refer to the following:
| Score per item | Descriptors |
|---|---|
| 5 | All the steps to answer a word problem are followed and each step is presented accurately. |
| 3 | The steps to answer a word problem are followed but the answer provided is incomplete. |
| 1 | The solution is erroneous. |
Word Problem 5:
In a circular board, an elastic band is tied to different points of the circle namely points S and E. The band is then stretched to a certain point outside the circle forming secants SY and EY. If SY= 15cm, TY= 6cm, and LY= 7cm, what is the length of secant EY? Refer to the figure at the right.
[chart: diagram; untitled figure beside Word Problem 5] Labels: S, T, L, E, Y.
Step 1: Understand the Problem
The word problem deals with secant segments. The problem asked to find the length of secant EY. The given are: length of secant SY, length of external secant TY and the length of external secant LY.
Step 2: Devising a Plan
To solve the problem, use the following theorem: If two secant segments are drawn to a circle from an exterior point, then the product of the lengths of one secant segment and its external secant segment is equal to the product of the lengths of the other secant segment and its external secant segment. Applying the theorem in the figure, we have,
SY ∙ TY = EY ∙ LY
Step 3: Carrying out the Plan
a. Review the given measurements.
SY = 15 cm TY = 6 cm LY = 7 cm
b. To solve the unknown segment EY, substitute the given in the theorem SY ∙ TY = EY ∙ LY
(15)(6) = (EY)(7) Substituting the given measurements
90 = 7(EY) Simplifying
90 / 7 = 7(EY) / 7 Dividing both sides of the equation by 7
EY = 12.86 cm. Simplify to obtain 12.86 cm., the length of EY
Step 4: Looking Back
To check if the solution is correct, substitute the given and computed value in the theorem SY ∙ TY = EY ∙ LY as follows
(15)(6) = (12.86)(7) Substituting the values.
90 = 90 Simplify. Since we obtain 90 on both sides of the equation, thus our solution is correct.
Word Problem 6:
An old circular table was repaired by nailing two wooden strips on its back, forming intersecting chords. Find the length of the second piece of wood LG if the dimensions of the chords are as follows:
FS = 6ft SA = 4ft LS = 3ft
[chart: diagram; untitled figure beside Word Problem 6] Labels: L, 3ft, F, 6ft, S, 4ft, A, G.
Step 1: Understand the Problem
The given word problem involves intersecting chords of a circle. We are given the length of segment LS, FS and segment SA. Segments FS and SA form chord FA. Using the given and concepts of secant segments, we are asked to find for the length of segment LG.
Step 2: Devising a Plan
Since we are dealing with intersecting chords of a circle, use the following theorem: If two chords of a circle intersect, then the product of the measures of the segments of one chord is equal to the product of the measures of the segments of the other chord. In symbols, FS ∙ SA = LS ∙ SG
Step 3: Carrying out the Plan
a. From the problem, the following are the given measurements:
FS = 6 ft
SA = 4 ft LS = 3 ft LG is unknown
b. To solve for LG, start with the theorem.
FS ∙ SA = LS ∙ SG Substituting the given values
(6)(4) = 3(SG) Getting the product
24 = 3(SG) Divide both sides by 3
24 / 3 = 3(SG) / 3
SG = 8 ft Simplifying. Therefore, SG is equal to 8 feet.
c. To finally solve for LG,
LG = LS + SG
LG = 3 + 8 Substituting the given value of LS and the obtained value of SG
LG = 11 Simplifying. Therefore, LG is equal to 11 feet
Step 4: Looking Back
To check if we had a correct answer, we don’t need to start checking the obtained value of LG since we can easily identify that 3 + 8 is really equivalent to 11. Thus, to check our answer, we start with the value of SG. We can substitute the obtained value and the given to the theorem FS ∙ SA = LS ∙ SG
(6)(4) = (3)(8) Substituting the values
24 = 24 Obtaining equal values on both sides of the equation, thus we obtain a correct value of SG
Assessment 2: Solve my length!
Apply the concepts learned to solve for the given word problems. Use the following guide questions to help you in presenting your solution.
Guide Questions: ✓ What are the problems all about? ✓ What theorems or concepts can you use to solve the problems? ✓ How can you formulate equations based from the theorems or concepts? ✓ How will you use the equations to solve for the unknowns? ✓ Did you check your answer?
- A family tied a pair of ropes to their circular pool to serve as guide for their children who are practicing how to swim. The ropes intersect at the interior of the pool with three segments formed measuring 5m, 6m, and 12 m as illustrated in the figure at the right. Find the missing segment measure of the rope in meters.
[illustration: circular pool with two intersecting ropes; labels: Circular Pool, 5m, 6m, 12m, x]
- A circular frame is supported with sticks on its backside as shown in the figure at the right. Find the unknown measure of the external secant segment x.
[illustration: circular frame with two sticks meeting at an exterior point; labels: 7in, x, 8in, 21in, Circular Frame]
Rubrics: To guide you with your solution, please refer to the following
| Score per item | Descriptors |
|---|---|
| 5 | All steps to answer word problem are followed and each step is presented accurately. |
| 3 | The steps to answer word problem are followed but the answer provided is incomplete. |
| 1 | The solution is erroneous. |
What I Have Learned
Fill in the blanks with your learnings regarding solving problems on circles
If a line is tangent to a circle, then it is _____ to the radius drawn to the point of tangency.
The sum of the central angles of a circle with no common interior points is _____.
If an angle is inscribed in a circle, then the measure of the angles is _____ the measure of the intercepted arc.
The area of a sector of a circle is the product of the area of the circle and the _____.
If two secant segments are drawn to a circle from an exterior point, then the product of the lengths of one secant segment and its external secant segment is _____ to the product of the lengths of the other secant segment and its external secant segment.
If two chords of a circle intersect, the product of the measures of the segments of one chord is _____ to the product of the measures of the segments of the other chord.
What I Can Do
The Huge Ceiling Ball!
Ronnie used a golden string to hang a huge circular light ball in the middle of their living room. The string measures 50 centimeters and it reaches until the center of the circular light ball. Suppose the tangent line from the ceiling where the string is hoisted to the circular light ball is 30 cm, what is the radius of the circular light ball?
Rubrics: Your output will be rated as follows
| Score | Descriptors |
|---|---|
| 20 | The diagram was well illustrated and with correct label. All computations are accurate and the final answer is expressed correctly in a complete sentence. |
| 15 | The diagram is illustrated. The computations are correct but the final answer is not clearly expressed. |
| 10 | Some parts of the diagram are missing. Some computations are not accurate. |
| 5 | A diagram is present and an answer is identified though incorrect. |
Assessment
Instruction: Read carefully each item. Choose the letter of your answer and write it on your answer sheet.
Which of the following is the measure of the angle determined by the hands of a clock at four o’clock?
a. 90°
b. 120°
c. 150°
d. 180°Which of the following is the degree measure of an arc formed when a circle is divided into 5 equal arcs?
a. 30°
b. 60°
c. 72°
d. 144°A wheel has 12 spokes which are evenly spread. Which of the following is the measure of each central angle formed?
a. 12°
b. 15°
c. 24°
d. 30°In a pie chart showing the expenditures of the Lopez family, the measure of the central angle corresponding to their food expenses is 120°. If their monthly income is Php 35,000.00, which of the following cost corresponds to their food expenses?
a. Php 11,666.67
b. Php 17,500.00
c. Php 5,666.67
d. Php 12,000.00Angel wants to draw a pie chart that shows how she spent her 30-day vacation. If she used six days of it practicing her piano recital, which of the following degree measure of a central angle does corresponds to its sector?
a. 22.5°
b. 45°
c. 72°
d. 90°A dart board has a diameter of 30 cm. and is divided into 15 congruent sectors. Which of the following is the area of one of the sectors?
a. 15π cm²
b. 20 cm²
c. 80 cm²
d. 800π cm²Mr. Alex designed a semicircle arc made of bent iron for their gate’s entrance. Suppose the diameter of the semicircle is 3 meters, how long should the iron be before bending to avoid wastage? (Round off answer to two decimal places)
a. 4.71 m
b. 5.71 m
c. 9.42 m
d. 10.55 mA circular garden has pathways which are both tangent to the garden whose center is 10 meters away from the entrance as shown in the figure below. If the radius of the circular garden is 5 meters, which of the following is the length of each pathway?
[illustration: circular garden with two tangent pathways meeting at an entrance; labels: Garden, Pathway, Entrance]
a. 3√5 m b. 5√3 m c. 25 m d. 75 m
- A circular weaved winnower is supported by a pair of sturdy rattan that intersects at its back as shown in the figure. Which of the following choices is the missing length?
[illustration: circle with two intersecting rattan supports; labels: 5 in, 8 in, 16 in, x]
a. 5 in b. 10 in c. 12 in d. 15 in
- A circular mirror is hung such that the strings are tangent to the circle as shown in the figure. If the major arc measures 240°, which of the following is the degree measure of the angle formed by the strings.
[illustration: circular mirror with two tangent strings meeting above it; label: mirror]
a. 60° b. 80° c. 130° d. 230°
Additional Activity
Guess What!
Mario is looking at a circular garden having pathways tangent to the circle from an exterior point as shown in the sketch below. If the angle formed by the tangent pathways in the exterior of the circle is 50°, what are the degree measures of the major arc and the minor arc intercepted by the tangents in the circle? Explain how you arrived at your answer.
[illustration: circle with two tangent pathways meeting at an exterior point; label: 50 degrees]
[transcriber’s note: the printed page numbers on this sheet are 21 to 23, and the last page has no number; the markers follow the requested sequence]
Rubrics:
| Score | Descriptors |
|---|---|
| 20 | The answer is accurate and explanation is well expressed supported by solution. |
| 15 | The answer is accurate but the explanation is not well supported by a solution. |
| 10 | The answer is inaccurate though the explanation contains valid points. |
| 5 | The explanation is not valid resulting to an inaccurate answer. |
You can do it!
Answer Key
[transcriber’s note: the answer key page is printed rotated 180 degrees, with its boxes in mixed orientation; the answers below are listed in reading order]
What I Know
- c
- b
- a
- a
- c
- a
- a
- b
- d
- b
What’s In
- difference 1. equal
- difference 2. equal
- sum 3. half
- sum 4. twice
- One-half 5. twice
What I Can Do
[illustration: circle with a sector marked x, a line segment labeled 50cm and a segment labeled 30cm]
The radius of the ball is 40 cm.
What’s More :
Assessment 1
- 60 degrees
- 150 degrees
- 37.68 square inches
- 4.71 meters
Assessment 2
- 10 meters
- 5 inches
Assessment
- b
- c
- d
- a
- c
- a
- a
- b
- b
- a
What I Have Learned
- perpendicular
- 360°
- one-half
- measure of corresponding arc of the sector / 360°
- Equal
- equal
Additional Activity
The major arc is equal to 230 degrees and the minor arc is 130 degrees.
References
Callanta, Melvin M., Canonigo, Allan M., Chua, Arnaldo I., Cruz, Jerry D., Esparrago, Mirla S., Garcia, Elino S., Magnaye, Aries N., Orines, Fernando B., Perez, Rowena S., Ternida, Conception S.2015. Mathematics-Grade 10 Learner’s Module, Meralco Avenue, Pasig City: REX Book Store, Inc.
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