Login / Register
Follow Us
0
0
Register Free!
Register Free!
Online Mock Tests & Exam Preparation | Target Exams
Select Medium:
🇬🇧 English Medium Active 🇮🇳 বাংলা মাধ্যম 🇮🇳 हिन्दी माध्यम
West Bengal Board (WBBSE) Class 10 (Madhyamik) Mathematics Set 02 (Advanced Level) English Medium 100% Free Practice Test

West Bengal Board (WBBSE) Class 10 (Madhyamik) Mathematics: Theorems Related to Tangent to a Circle Online Test (Set 02 (Advanced Level))

Theorems Related to Tangent to a Circle — Set 02 (Advanced Level)

Practise free online CBT mock test for West Bengal Board (WBBSE) Class 10 (Madhyamik) Mathematics Chapter 15 'Theorems Related to Tangent to a Circle'. 20 MCQs, 20-minute timer, instant scorecard, and step-by-step solutions aligned with official board syllabus.

📝
20 MCQs
Questions
⏱️
20 Mins
Duration
🎯
20 Marks
Total Marks
➕
+1.0
Per Correct
🛡️
0.0
Negative Mark
🌐
English
Language
Start Test Now (English Medium)
Instant Scorecard Step-by-Step Solutions WBBSE Madhyamik Blueprint

Chapter Study Guide & In-Depth Notes

Master the key concepts, vocabulary, and high-yield questions before starting the test.

3 Sections

WBBSE Class 10 Mathematics: Theorems related to Circle (Chapter 3)

WBBSE Madhyamik Ganit Prakash Ch 3

In the WBBSE Class 10 Madhyamik Mathematics curriculum (Ganit Prakash / গণিত প্রকাশ), Theorems related to Circle (Chapter 3) forms the foundation of secondary Euclidean geometry (বৃত্ত সম্পর্কিত উপপাদ্য). A circle is defined as the geometric locus of all points in a plane equidistant from a fixed point called the centre ($O$), with the constant distance denoted as the radius ($r$). A straight line segment connecting any two points on the circumference is known as a chord ($AB$). The longest chord passing through the centre is the diameter ($d = 2r$).

The fundamental geometric pillars of this chapter are established by Theorems 32 and 33: "The perpendicular drawn from the centre of a circle to a chord which is not a diameter, bisects the chord." Conversely, "The line segment joining the centre of a circle and the midpoint of a chord (not a diameter) is perpendicular to the chord." If $OM \perp AB$, then $AM = MB = \frac{1}{2}AB$, and inside right-angled triangle $\triangle OMA$, the Pythagorean relation governs all metric dimensions: $$r^2 = d^2 + \left(\frac{L}{2}\right)^2$$ where $r = OA$ is the radius, $d = OM$ is the perpendicular distance from the centre to the chord, and $L = AB$ is the chord length.

Furthermore, chords of a circle equidistant from the centre are equal in length ($d_1 = d_2 \iff L_1 = L_2$), and conversely, equal chords are strictly equidistant from the centre. In concentric circles (circles sharing the same centre with radii $r < R$), a chord of the outer circle tangent to the inner circle is bisected at the point of tangency by the radius of the inner circle. Thorough conceptual and algebraic mastery of these theorems guarantees 100% marks in both geometric riders (proofs) and objective numerical problems in the WBBSE Madhyamik board examination.

Core Moral & Key Takeaway:

Mastering centre-chord perpendicularity $OM \perp AB$, chord bisection $AM = \frac{L}{2}$, and the Pythagorean metric $r^2 = d^2 + (L/2)^2$ guarantees complete success in WBBSE Madhyamik circle geometry.

Core Theorems, Formulas & Geometric Terminology

Theorems & Formula Sheet
Term / Formula Type Definition / Meaning Explanation / Notes Example / Usage
Theorem 32 Theorem Line joining centre to chord midpoint is perpendicular: OM ⊥ AB. Line segment joining circle centre to chord midpoint is perpendicular to the chord. "If M is midpoint of chord AB, then ∠OMA = 90°."
Theorem 33 Theorem Perpendicular from centre bisects chord: AM = MB = L/2. Perpendicular drawn from centre of circle to a chord bisects the chord. "If OM ⊥ AB, then AM = MB = AB/2."
Chord-Radius Pythagorean Metric Master Formula r² = d² + (L/2)² Radius squared equals perpendicular distance squared plus half-chord squared. "If r = 10 cm, d = 6 cm, then half-chord = √(100 - 36) = 8 cm, chord = 16 cm."
Perpendicular Distance (d) Derived Formula d = √(r² - (L/2)²) Perpendicular distance from centre = √(radius² - (chord/2)²). "For r = 13 cm and chord = 24 cm, d = √(169 - 144) = 5 cm."
Chord Length (L) Derived Formula L = 2 · √(r² - d²) Total chord length = 2 × √(radius² - distance²). "For r = 5 cm and d = 3 cm, L = 2 × √(25 - 9) = 8 cm."
Equidistant Chords Property Property L₁ = L₂ ⇔ d₁ = d₂ Equal chords are equidistant from the centre and vice versa. "Two chords of 14 cm each are at identical distances from centre."
Parallel Chords on Opposite Sides Distance Rule Total distance D = d₁ + d₂ Distance between parallel chords on opposite sides of centre is the sum of distances. "If d₁ = 4 cm and d₂ = 3 cm, then distance between them is 4 + 3 = 7 cm."
Concentric Circles Tangent Chord Identity L = 2√(R² - r²) Length of outer chord touching inner circle of radius r. "For R = 5 cm, r = 3 cm, chord length L = 2√(25 - 9) = 8 cm."

Key Features & Devices Explored:

Right-Triangle Construction Method
Always drop perpendicular OM from centre O to chord AB and join radius OA to formulate right-triangle OMA.
Half-Chord Caution Rule
Always apply Pythagorean metric using half-chord length (L/2), and double the result to get full chord length.
Parallel Chords Position Test
Verify whether chords lie on opposite sides (d₁ + d₂) or the same side (|d₁ - d₂|) of the circle centre.
Concentric Circle Tangent Property
Radius to tangent forms 90° angle at contact point, making inner radius r the perpendicular distance d.

High-Yield Revision Points & Solved Madhyamik Board Exam Q&A

WBBSE Exam Revision Notes
⚡ High-Yield Exam Takeaways:
  • Common Pythagorean Triplets: Questions in Madhyamik examinations frequently utilize standard integer triplets: (3, 4, 5), (6, 8, 10), (5, 12, 13), and (8, 15, 17).
  • Chord Bisection: The perpendicular from centre bisects the chord ($AM = MB$). This geometric property is frequently tested in 2-mark short questions.
  • Diameter Property: The diameter is the maximum possible chord ($d = 2r$), and its perpendicular distance from the centre is zero ($d = 0$).
  • Distance vs. Chord Length: Chords closer to the centre are longer; as perpendicular distance increases, chord length decreases.
  • Rider Presentation: When writing geometric proofs for Madhyamik riders, format your answer strictly into four sections: Given, To Prove, Construction, and Proof.
Frequently Asked Questions & Model Answers:
Q1 In a circle of radius 10 cm, find the perpendicular distance from the centre to a chord of length 16 cm.
Answer: Here radius $r = 10\text{ cm}$ and chord $AB = 16\text{ cm}$. Draw perpendicular $OM \perp AB$. By Theorem 33, $OM$ bisects $AB$: $$AM = \frac{1}{2}AB = \frac{16}{2} = 8\text{ cm}$$ In right-angled $\triangle OMA$, by Pythagoras theorem: $$OM = \sqrt{OA^2 - AM^2} = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6\text{ cm}$$ Hence, the perpendicular distance from the centre is 6 cm.
Q2 In a circle of radius 13 cm, a chord is at a distance of 5 cm from the centre. Find the total length of the chord.
Answer: Given radius $r = 13\text{ cm}$ and distance $d = OM = 5\text{ cm}$. In right $\triangle OMA$: $$AM = \sqrt{r^2 - d^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ cm}$$ Since $OM$ bisects the chord $AB$, the total chord length is: $$AB = 2 \times AM = 2 \times 12 = 24\text{ cm}$$
Q3 In a circle of radius 5 cm, two parallel chords of lengths 8 cm and 6 cm are drawn on opposite sides of the centre. Find the distance between the two chords.
Answer: Here radius $r = 5\text{ cm}$. For first chord $AB = 8\text{ cm}$: $$d_1 = \sqrt{5^2 - (8/2)^2} = \sqrt{25 - 16} = \sqrt{9} = 3\text{ cm}$$ For second chord $CD = 6\text{ cm}$: $$d_2 = \sqrt{5^2 - (6/2)^2} = \sqrt{25 - 9} = \sqrt{16} = 4\text{ cm}$$ Since the chords are on opposite sides of the centre, total distance between them is: $$D = d_1 + d_2 = 3 + 4 = 7\text{ cm}$$
Q4 Two concentric circles have radii 5 cm and 3 cm. Find the length of the chord of the outer circle which touches the inner circle.
Answer: Let radius of outer circle $R = 5\text{ cm}$ and inner circle $r = 3\text{ cm}$. The chord $AB$ of the outer circle touches the inner circle at $M$, so $OM \perp AB$ and $OM = r = 3\text{ cm}$. In right $\triangle OMA$: $$AM = \sqrt{R^2 - r^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\text{ cm}$$ Since $OM$ bisects chord $AB$, total length is: $$AB = 2 \times AM = 2 \times 4 = 8\text{ cm}$$

Sample Questions Preview

Free Sneak Peek

Here is a sneak peek of the questions included in this test set. Test your preparation by viewing the answer and detailed explanation before starting the timed exam.

Q1
In a circle with center O, tangents PA and PB are drawn from P. If ∠APB = 60° and OP = 10 cm, find the radius of the circle.
O কেন্দ্রবিশিষ্ট একটি বৃত্তে, স্পর্শক PA এবং PB P থেকে আঁকা হয়। ∠APB = 60° এবং OP = 10 সেমি হলে, বৃত্তের ব্যাসার্ধ নির্ণয় করুন।
केंद्र O वाले एक वृत्त में, P से स्पर्शरेखाएँ PA और PB खींची जाती हैं। यदि ∠APB = 60° और OP = 10 सेमी है, तो वृत्त की त्रिज्या ज्ञात कीजिए।
A
10 cm
B
5 cm
C
5√3 cm
D
2.5 cm
Correct Answer: Option B
Explanation:
OP bisects ∠APB => ∠APO = 30°. In right ΔOAP: sin 30° = OA
OP দ্বিখণ্ডিত ∠APB => ∠APO = 30°। ডানদিকে ΔOAP: sin 30° = OA
OP, ∠APB => ∠APO = 30° को समद्विभाजित करता है। दाएं ΔOAP में: पाप 30° = OA
Q2
According to the Alternate Segment Theorem, the angle between a tangent and a chord through the point of contact is equal to:
অল্টারনেট সেগমেন্ট থিওরেম অনুসারে, যোগাযোগের বিন্দুর মধ্য দিয়ে একটি স্পর্শক এবং একটি জ্যার মধ্যে কোণ সমান:
वैकल्पिक खंड प्रमेय के अनुसार, संपर्क बिंदु के माध्यम से स्पर्श रेखा और जीवा के बीच का कोण बराबर होता है:
A
The central angleকেন্দ্রীয় কোণकेन्द्रीय कोण
B
Double the alternate angleবিকল্প কোণ দ্বিগুণ করুনवैकल्पिक कोण को दोगुना करें
C
The angle subtended by the chord in the alternate segmentবিকল্প সেগমেন্টে জ্যা দ্বারা সাবটেনড কোণएकांतर खंड में जीवा द्वारा बनाया गया कोण
D
90°
Correct Answer: Option C
Explanation:
The Alternate Segment Theorem states that the angle between a tangent and a chord equals the angle in the alternate segment.
বিকল্প সেগমেন্ট থিওরেম বলে যে একটি স্পর্শক এবং একটি জ্যার মধ্যবর্তী কোণটি বিকল্প অংশের কোণের সমান।
वैकल्पिक खंड प्रमेय बताता है कि स्पर्शरेखा और जीवा के बीच का कोण वैकल्पिक खंड में कोण के बराबर होता है।

Ready to attempt all 20 questions in exam environment?

Get real-time countdown timer, question palette navigator, and instant analytics.

How Chapter-Wise Online CBT Tests Build Winning Habits for Future Competitive Exams

Early practice in high school builds the mental muscle, speed, and analytical discipline required to ace NTSE, Olympiads, JNVST, CUET, and Foundation JEE/NEET.

Competitive Edge

Real-Time Clock Awareness & Pacing

समय प्रबंधन एवं गति नियंत्रण

In school examinations, students write answers without active pacing. In national entrance tests, students have barely 45-60 seconds per MCQ. Practicing with a 20-minute timer instills instinctive time-budgeting and prevents last-minute exam panic.

Mastery of the Elimination Technique

विकल्प निष्कासन विधि में दक्षता

Top rankers in competitive exams rarely compute every problem blindly. They spot subtle traps, evaluate extreme values, and eliminate 2 out of 4 distractors within seconds. Our 4-option sets teach critical distractor analysis.

CBT Screen Stamina & Digital Comfort

कंप्यूटर आधारित परीक्षा का सहज अभ्यास

Almost all major national exams (CUET, JEE, NEET, SSC, Banking) are 100% computer-based. Students who only study on paper experience eye strain and navigation mistakes on exam day. Early CBT practice creates effortless digital fluency.

Immunity to Negative Marking

सटीकता और नकारात्मक अंकन से सुरक्षा

Careless errors—such as misreading "NOT" or "EXCEPT"—ruin thousands of competitive aspirations each year. Regular objective testing builds the habit of double-checking question keywords before locking answers.

Instant Diagnostic Feedback Loop

तत्काल त्रुटि निवारण एवं आत्म-मूल्यांकन

Traditional paper tests take weeks to grade, by which time the student has forgotten their thought process. TargetExams provides instant scorecards and step-by-step solutions while the conceptual puzzle is still fresh.

Start Building Your Competitive Edge from Class

Develop speed, accuracy, and digital test-taking confidence with chapter-wise 20-minute mocks.

Syllabus Concepts Evaluated in this Chapter

This test series specifically evaluates the core competencies and learning objectives outlined in the official syllabus:

  • Definitions and fundamental properties of circle, centre, radius, diameter, chord, and arc.
  • Theorem 32: Formal proof that the line joining the centre to the midpoint of a chord is perpendicular to the chord.
  • Theorem 33: Formal proof and applications of perpendicular from centre bisecting the chord.
  • Application of Pythagorean metric $r^2 = d^2 + (L/2)^2$ to solve numerical board exam problems.
  • Calculating distance between parallel chords lying on opposite sides or the same side of the centre.
  • Determining lengths of tangent chords in systems of concentric circles.

Exam Rules & Test Taking Guidelines

  • Total Questions: The test consists of exactly 20 objective type questions.
  • Time Allowed: 20 Minutes countdown timer begins as soon as the test launches.
  • Marking Scheme: +1.0 mark is awarded for each correct answer.
  • Zero Penalty: There is 0.0 negative marking for incorrect or unattempted questions.
  • Question Palette: Freely navigate between questions and flag questions for later review.
  • Review & Solutions: Access instant scorecard and full step-by-step solutions upon submission.

Frequently Asked Questions (FAQs)

Yes! Foundation Practice Set 01 for every chapter in West Bengal Board (WBBSE) Class 10 (Madhyamik) Mathematics is 100% free to attempt. Students can test their preparation without any subscription or payment.
This test strictly follows the official West Bengal Board (WBBSE) syllabus and curriculum for Class 10 (Madhyamik), covering textbook concepts, formulas, and board-standard MCQs for Chapter 15.
This dedicated test page is conducted in English Medium. If you prefer taking the test in Bengali or Hindi, you can switch using the language switcher bar at the top or toggle languages during the exam.
The test comprises 20 MCQs to be completed in 20 minutes. Each question carries +1.0 mark for a correct response, and there is 0.0 negative marking for incorrect or unattempted questions.
Yes! Instantly after submitting your test, you receive a comprehensive digital scorecard with accuracy analysis, time breakdown, and step-by-step verified explanations for all 20 questions.

Practice Sets in this Chapter

Set 01 (Foundation)

20 MCQs • 20 Mins
View

Set 02 (Advanced)

20 MCQs • 20 Mins
Current

More Chapters in this Subject

🎯 Start Preparation

Begin with this 20-minute practice set to test your conceptual accuracy and prepare for top ranks in your school examinations.

Theorems Related to Tangent to a Circle (Set 02 (Advanced Level))
20 MCQs • 20 Mins • 100% Free Practice Test

Select Test Language Medium

Choose the language medium you want to attempt this online test in:

বাংলা মাধ্যম (Bengali Medium) WBBSE Official
পশ্চিমবঙ্গ মধ্যশিক্ষা পর্ষদ (WBBSE) পাঠ্যক্রম অনুযায়ী সম্পূর্ণ বাংলা মাধ্যমে পরীক্ষা দিন।
English Medium
Displays questions, options, and explanations in English medium.
हिन्दी माध्यम (Hindi Medium)
प्रश्नों और विकल्पों को हिन्दी माध्यम में प्रदर्शित करेगा।
English + বাংলা (Bilingual Bengali)
Displays questions and options in English & Bengali side-by-side.
English + हिन्दी (Bilingual Hindi)
Displays questions and options in English & Hindi side-by-side.