✦ NCF 2023 • COMPETENCY-FOCUSED QUESTION BANK ✦
Chapter 10: Circles
Class 10 Standard Mathematics (Case Study Bank)
Curated By Akash Srivastva
✨ Interactive TLMs, Worksheets & Digital Study Notes:
www.akashmaths.online
Case Study 1
Space Science
4 Marks
Satellite Communication & Earth Horizon
A satellite S orbits Earth (center O, radius R = 6400 km) at altitude h = 1600 km. A radio line-of-sight signal ST touches Earth tangentially at horizon point T.
Q1. Tangent-Radius Perpendicularity
1 Mark
State the geometric theorem relating radius OT and line of sight ST, and state angle ∠OTS.
(A) 60° (Theorem on chords)
(B) 90° (Theorem 10.1: Tangent is perpendicular to radius through point of contact)
(C) 45° (Angle in semicircle)
(D) 30° (Tangent ratio property)
Q2. Horizon Distance
1 Mark
Calculate the exact horizon transmission distance ST (in km) when satellite altitude h = 1600 km.
(A) 4500 km
(B) 4800 km
(C) 5200 km
(D) 6000 km
Q3. Subtended Center Angle
2 Marks
Find the cosine of coverage angle cos(∠TOS) subtended at Earth's center for this configuration.
(A) 3/5
(B) 1/2
(C) 4/5 (or 0.8)
(D) √3/2
OR (Alternative Q3)
If upgraded orbit makes angle ∠TOS = 60°, calculate the new transmission distance ST.
(A) 6400√3 km
(B) 3200√3 km
(C) 6400 km
(D) 12800 km
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | Theorem 10.1 states that the tangent at any point of a circle is perpendicular to the radius through point of contact. Hence ∠OTS = 90°. | [0.5 Mark] for Theorem 10.1. [0.5 Mark] for ∠OTS = 90°. |
| Q2 | OS = 6400 + 1600 = 8000 km. In right △OTS: ST = √(OS² − OT²) = √(8000² − 6400²) = 4800 km. | [1 Mark] for ST = 4800 km. |
| Q3 | cos(∠TOS) = Adjacent / Hypotenuse = OT / OS = 6400 / 8000 = 4/5 = 0.8. | [1 Mark] for ratio setup. [1 Mark] for 4/5 (0.8). |
| Q3 (OR) | tan 60° = ST / OT ⇒ √3 = ST / 6400 ⇒ ST = 6400√3 km. | [1 Mark] for tan 60° relation. [1 Mark] for ST = 6400√3 km. |
Case Study 2
Automotive Design
4 Marks
Vehicle Wheel & Mudguard Design
An EV wheel (center O, radius r = 15 cm) is protected by a mudguard bracket anchored at chassis point P (OP = 25 cm). Supporting struts PA and PB are tangent at A and B.
Q1. Tangent Length Equality
1 Mark
If strut PA = 20 cm, write the length of PB using Theorem 10.2.
(A) 15 cm
(B) 20 cm
(C) 25 cm
(D) 10 cm
Q2. Quadrilateral Perimeter
1 Mark
Calculate the perimeter of quadrilateral OAPB formed by radii and struts.
(A) 60 cm
(B) 80 cm
(C) 70 cm
(D) 75 cm
Q3. Geometric Chord Analysis
2 Marks
If installation angle ∠APB = 60°, prove △PAB is equilateral and find chord length AB.
(A) AB = 20 cm (equilateral triangle)
(B) AB = 15 cm
(C) AB = 25 cm
(D) AB = 10√3 cm
OR (Alternative Q3)
Prove quadrilateral OAPB is cyclic and determine the angle sum ∠AOB + ∠APB.
(A) 180° (Opposite angles sum to 180°)
(B) 360°
(C) 90°
(D) 270°
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | Theorem 10.2: Tangents drawn from an external point to a circle are equal in length. Hence PB = PA = 20 cm. | [1 Mark] for PB = 20 cm. |
| Q2 | Perimeter = OA + AP + PB + BO = 15 + 20 + 20 + 15 = 70 cm. | [1 Mark] for perimeter = 70 cm. |
| Q3 | PA = PB ⇒ ∠PAB = ∠PBA. Since ∠APB = 60°, 2∠PAB = 120° ⇒ ∠PAB = 60°. All angles are 60°, so △PAB is equilateral, hence AB = PA = 20 cm. |
[1 Mark] for proving equilateral. [1 Mark] for AB = 20 cm. |
| Q3 (OR) | ∠OAP = ∠OBP = 90° (Theorem 10.1). Sum of opposite angles = 90° + 90° = 180° ⇒ OAPB is cyclic, and ∠AOB + ∠APB = 180°. |
[1 Mark] for cyclic proof. [1 Mark] for angle sum 180°. |
Case Study 3
Robotic Packaging
4 Marks
Belt-and-Pulley Mechanical System
A conveyor pulley (center O, radius r = 7 cm) is driven by tangent belt segments TA and TB from tensioner drive T, with TA = 24 cm.
Q1. Center Distance
1 Mark
Calculate straight-line distance OT from pulley center O to tensioner drive T.
(A) 20 cm
(B) 25 cm
(C) 28 cm
(D) 31 cm
Q2. Central Angle
1 Mark
If angle between belt segments at drive ∠ATB = 120°, find central angle ∠AOB.
(A) 90°
(B) 45°
(C) 60°
(D) 30°
Q3. Chord Length & Adjustment
2 Marks
Calculate the exact length of contact chord AB (rounded to two decimal places).
(A) 12.50 cm
(B) 14.20 cm
(C) 13.44 cm
(D) 15.00 cm
OR (Alternative Q3)
If belt angle ∠ATB is adjusted to 60°, calculate the new distance OT.
(A) 14 cm
(B) 12 cm
(C) 16 cm
(D) 21 cm
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | In right △OAT: OT = √(OA² + TA²) = √(7² + 24²) = √625 = 25 cm. | [1 Mark] for OT = 25 cm. |
| Q2 | ∠AOB + ∠ATB = 180° ⇒ ∠AOB = 180° − 120° = 60°. | [1 Mark] for ∠AOB = 60°. |
| Q3 | Area(OAT) = 1/2 × 7 × 24 = 1/2 × 25 × AM ⇒ AM = 168 / 25 = 6.72 cm. AB = 2 × AM = 2 × 6.72 = 13.44 cm. |
[1 Mark] for AM = 6.72 cm. [1 Mark] for AB = 13.44 cm. |
| Q3 (OR) | ∠ATO = 60° / 2 = 30°. sin 30° = OA / OT ⇒ 1/2 = 7 / OT ⇒ OT = 14 cm. | [1 Mark] for sin 30° relation. [1 Mark] for OT = 14 cm. |
Case Study 4
Urban Architecture
4 Marks
Concentric Circular Pathways
A public plaza has concentric circular pathways of radii R = 10 m and r = 6 m centered at O. Concrete walkway AB is a chord of the outer circle and tangent to the inner circle at P.
Q1. Geometric Relation
1 Mark
State the relationship between OP and AB at contact point P, and explain why P bisects AB.
(A) OP ⊥ AB (tangent perpendicular to radius); perpendicular from center bisects chord
(B) OP ∥ AB
(C) OP = AB/4
(D) P divides AB in ratio 2:1
Q2. Walkway Length
1 Mark
Calculate the exact length of the glass walkway AB using R = 10 m and r = 6 m.
(A) 12 m
(B) 16 m
(C) 18 m
(D) 20 m
Q3. Equilateral Corridor & Annulus
2 Marks
If outer radius is twice inner radius (R = 2r), prove △ABC formed by tangents must be equilateral.
(A) Scalene triangle
(B) Right-angled isosceles triangle
(C) Equilateral triangle (all interior angles = 60°)
(D) Cannot be determined
OR (Alternative Q3)
Calculate the exact area of the ring-shaped region (annulus) between pathways in terms of π.
(A) 36π m²
(B) 64π m²
(C) 100π m²
(D) 16π m²
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | OP ⊥ AB by Theorem 10.1. The perpendicular from the center of a circle to a chord bisects the chord, so P is the midpoint of AB. | [0.5 Mark] for OP ⊥ AB. [0.5 Mark] for bisector property. |
| Q2 | In right △OPA: AP = √(10² − 6²) = √(100 − 36) = √64 = 8 m. AB = 2 × AP = 2 × 8 = 16 m. |
[0.5 Mark] for AP = 8 m. [0.5 Mark] for AB = 16 m. |
| Q3 | sin(∠OAP) = OP / OA = r / (2r) = 1/2 ⇒ ∠OAP = 30°. Full angle ∠A = 2(30°) = 60°. By symmetry, ∠B = 60° and ∠C = 60°, proving △ABC is equilateral. |
[1 Mark] for angle derivation. [1 Mark] for equilateral conclusion. |
| Q3 (OR) | Area = πR² − πr² = π(10² − 6²) = π(100 − 36) = 64π m². | [1 Mark] for formula. [1 Mark] for 64π m². |
Case Study 5
Astrophysics
4 Marks
Transit of Venus Geometry
Telescopic imaging tracks Venus (center O, radius r = 3000 km) from station P with tangents PA and PB where ∠APB = 40°.
Q1. Central Angle
1 Mark
Find the exact angle subtended by the points of contact at the center (∠AOB).
(A) 120°
(B) 130°
(C) 140°
(D) 150°
Q2. Perpendicular Angle
1 Mark
State angle ∠OAP formed between radius OA and tangent line PA with theorem.
(A) 90° (Theorem 10.1: Tangent is perpendicular to radius)
(B) 40°
(C) 60°
(D) 70°
Q3. Base Angles & Path Length
2 Marks
Calculate exact measures of base angles ∠OAB and ∠OBA in △OAB.
(A) 30° each
(B) 20° each
(C) 25° each
(D) 40° each
OR (Alternative Q3)
If OP = 6000 km and OA = 3000 km, calculate length of tangent signal path PA.
(A) 3000√3 km
(B) 3000√2 km
(C) 4500 km
(D) 6000 km
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | ∠AOB + ∠APB = 180° ⇒ ∠AOB = 180° − 40° = 140°. | [1 Mark] for ∠AOB = 140°. |
| Q2 | ∠OAP = 90° by Theorem 10.1 (tangent is perpendicular to radius through point of contact). | [1 Mark] for 90° and theorem. |
| Q3 | OA = OB ⇒ ∠OAB = ∠OBA. In △OAB: 140° + 2∠OAB = 180° ⇒ 2∠OAB = 40° ⇒ ∠OAB = ∠OBA = 20°. |
[1 Mark] for angle property. [1 Mark] for 20° each. |
| Q3 (OR) | In right △OAP: PA = √(OP² − OA²) = √(6000² − 3000²) = √(27 × 10⁶) = 3000√3 km. | [1 Mark] for Pythagoras theorem. [1 Mark] for 3000√3 km. |
Case Study 6
Landscape Architecture
4 Marks
Inscribed Botanical Fountain
A circular fountain is inscribed in triangular park △ABC touching AB = 12 m, BC = 14 m, and CA = 10 m at contact points D, E, and F.
Q1. Tangents Parameterization
1 Mark
If AD = x, express lengths BE and CF in terms of x using Theorem 10.2.
(A) BE = 12 − x, CF = 10 − x
(B) BE = 14 − x, CF = 12 − x
(C) BE = x − 2, CF = x + 2
(D) BE = 10 − x, CF = 12 − x
Q2. Linear Formulation
1 Mark
Formulate and solve equation for x using BC = BE + CE = 14 m.
(A) x = 3 m
(B) x = 5 m
(C) x = 4 m
(D) x = 6 m
Q3. Tangent Lengths & Inradius
2 Marks
Calculate individual lengths of boundary segments AD, BE, and CF.
(A) AD = 4 m, BE = 8 m, CF = 6 m
(B) AD = 5 m, BE = 7 m, CF = 6 m
(C) AD = 3 m, BE = 9 m, CF = 6 m
(D) AD = 4 m, BE = 6 m, CF = 8 m
OR (Alternative Q3)
If area of △ABC is 15√7 m², determine the inradius r of the circular fountain.
(A) 5√7 / 6 m
(B) 3√7 / 4 m
(C) √7 / 2 m
(D) 7√5 / 6 m
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | By Theorem 10.2: AF = AD = x ⇒ BE = BD = 12 − x; CF = CE = 10 − x. | [1 Mark] for BE = 12 − x, CF = 10 − x. |
| Q2 | BC = BE + CE ⇒ 14 = (12 − x) + (10 − x) ⇒ 2x = 8 ⇒ x = 4 m. | [1 Mark] for solving x = 4 m. |
| Q3 | AD = 4 m, BE = 12 − 4 = 8 m, CF = 10 − 4 = 6 m. | [1 Mark] for AD and BE. [1 Mark] for CF = 6 m. |
| Q3 (OR) | Semi-perimeter s = (12 + 14 + 10)/2 = 18 m. Area = r × s ⇒ 15√7 = r × 18 ⇒ r = 15√7 / 18 = 5√7 / 6 m. |
[1 Mark] for s = 18 m. [1 Mark] for r = 5√7 / 6 m. |
Case Study 7
E-Mobility Engineering
4 Marks
E-Bike Sprocket Assembly
A circular e-bike sprocket (radius r = 4 cm) is circumscribed by triangular frame △ABC. Contact point D on BC divides it into CD = 6 cm and BD = 8 cm.
Q1. Tangent Segments
1 Mark
Find the exact lengths of contact segments CE and BF.
(A) CE = 8 cm, BF = 6 cm
(B) CE = 6 cm, BF = 8 cm
(C) CE = 4 cm, BF = 4 cm
(D) CE = 7 cm, BF = 7 cm
Q2. Semi-perimeter Expression
1 Mark
If AE = AF = x, express semi-perimeter s of △ABC in terms of x.
(A) x + 12 cm
(B) x + 16 cm
(C) x + 14 cm
(D) 2x + 28 cm
Q3. Frame Dimensions & Total Area
2 Marks
Equate Heron's area to r × s to find x, and determine lengths of sides AB and AC.
(A) x = 7 cm; AB = 15 cm, AC = 13 cm
(B) x = 6 cm; AB = 14 cm, AC = 12 cm
(C) x = 8 cm; AB = 16 cm, AC = 14 cm
(D) x = 5 cm; AB = 13 cm, AC = 11 cm
OR (Alternative Q3)
Calculate the total area of triangular frame △ABC using solved value x = 7 cm.
(A) 72 cm²
(B) 80 cm²
(C) 84 cm²
(D) 96 cm²
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | Tangents from an external point are equal: CE = CD = 6 cm; BF = BD = 8 cm. | [1 Mark] for CE = 6 cm, BF = 8 cm. |
| Q2 | Sides are 14, x + 6, and x + 8. s = (14 + x + 6 + x + 8)/2 = (2x + 28)/2 = x + 14 cm. | [1 Mark] for s = x + 14 cm. |
| Q3 | Area = √[48x(x + 14)] = 4(x + 14) ⇒ 48x(x + 14) = 16(x + 14)² ⇒ 3x = x + 14 ⇒ x = 7 cm. AB = 7 + 8 = 15 cm; AC = 7 + 6 = 13 cm. |
[1 Mark] for solving x = 7 cm. [1 Mark] for AB = 15 cm and AC = 13 cm. |
| Q3 (OR) | s = 7 + 14 = 21 cm. Area = r × s = 4 × 21 = 84 cm². | [1 Mark] for s = 21 cm. [1 Mark] for Area = 84 cm². |
Case Study 8
Aero-Structural
4 Marks
Fuselage Reinforcement Bracket
A quadrilateral bracket ABCD circumscribes a circular drone fuselage (center O), touching sides AB, BC, CD, DA at contact points P, Q, R, S.
Q1. Tangent Pairs
1 Mark
Write relationships between tangent segments from vertex A and vertex C.
(A) AP = AS and CQ = CR
(B) AP = BQ and CQ = DS
(C) AP = CR and AS = CQ
(D) AP = 2AS and CQ = 2CR
Q2. Load-Balancing Property
1 Mark
Prove the structural balance relation for a quadrilateral circumscribing a circle.
(A) AB + CD = AD + BC
(B) AB × CD = AD × BC
(C) AB − CD = AD − BC
(D) AB + BC = CD + DA
Q3. Side Length & Parallelogram
2 Marks
If AB = 18 cm, BC = 21 cm, and CD = 15 cm, calculate the exact length of section AD.
(A) 10 cm
(B) 14 cm
(C) 12 cm
(D) 16 cm
OR (Alternative Q3)
If bracket ABCD is designed as a parallelogram, prove that ABCD must be what shape?
(A) Rhombus (adjacent sides are proven equal)
(B) Rectangle
(C) Square
(D) Kite
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | By Theorem 10.2: AP = AS and CQ = CR. | [1 Mark] for AP = AS, CQ = CR. |
| Q2 | AB + CD = (AP + PB) + (CR + RD) = (AS + BQ) + (CQ + DS) = (AS + DS) + (BQ + CQ) = AD + BC. | [1 Mark] for proving AB + CD = AD + BC. |
| Q3 | AB + CD = AD + BC ⇒ 18 + 15 = AD + 21 ⇒ 33 = AD + 21 ⇒ AD = 12 cm. | [1 Mark] for setup. [1 Mark] for AD = 12 cm. |
| Q3 (OR) | In a parallelogram, AB = CD and AD = BC. Using AB + CD = AD + BC: 2AB = 2AD ⇒ AB = AD. Since adjacent sides are equal, parallelogram ABCD is a rhombus. |
[1 Mark] for showing AB = AD. [1 Mark] for concluding it is a rhombus. |
Live Practice: Chapter 10 Circles
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