✦ NCF 2023 • COMPETENCY-FOCUSED QUESTION BANK ✦
Chapter 6: Triangles
Class 10 Standard Mathematics (Case Study Bank)
Curated By Akash Srivastva
✨ Interactive TLMs, Worksheets & Digital Study Notes:
www.akashmaths.online
Case Study 1
Drone Photogrammetry
4 Marks
Drone Mapping & Photogrammetry
An autonomous Agri-Drone maps farm crop health from vertex A, projecting field-of-view triangle ABC. Software superimposes a virtual grid calibration line DE parallel to ground boundary BC (DE ∥ BC).
Q1. Conceptual Understanding
1 Mark
State the mathematical theorem establishing AD/DB = AE/EC when DE ∥ BC in △ABC.
(A) Pythagoras Theorem
(B) Basic Proportionality Theorem (Thales Theorem)
(C) Angle Bisector Theorem
(D) Converse of Pythagoras Theorem
Q2. Application
1 Mark
If AB = 12 m, AD = 3 m, and AE = 4.5 m, calculate the exact length of boundary segment EC.
(A) 9.0 m
(B) 12.0 m
(C) 13.5 m
(D) 15.0 m
Q3. Analytical Reasoning
2 Marks
If AD = x, DB = x + 2, AE = x + 3, and EC = x + 7 with DE ∥ BC, find the value of x.
(A) x = 3
(B) x = 4
(C) x = 5
(D) x = 2
OR (Alternative Q3)
If AD/DB = 2/3 and total boundary AC = 15 m with DE ∥ BC, find the individual lengths of AE and EC.
(A) AE = 5 m, EC = 10 m
(B) AE = 6 m, EC = 9 m
(C) AE = 4 m, EC = 11 m
(D) AE = 7 m, EC = 8 m
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | Basic Proportionality Theorem (BPT) / Thales Theorem: "If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio." | [0.5 Mark] for naming BPT. [0.5 Mark] for theorem statement. |
| Q2 | DB = AB − AD = 12 − 3 = 9 m. By BPT: AD/DB = AE/EC ⇒ 3/9 = 4.5/EC ⇒ 1/3 = 4.5/EC ⇒ EC = 13.5 m. |
[0.5 Mark] for proportionality ratio. [0.5 Mark] for EC = 13.5 m. |
| Q3 | x / (x + 2) = (x + 3) / (x + 7) ⇒ x(x + 7) = (x + 2)(x + 3). x² + 7x = x² + 5x + 6 ⇒ 2x = 6 ⇒ x = 3. |
[1 Mark] for cross-multiplication. [1 Mark] for solving x = 3. |
| Q3 (OR) | AE/EC = 2/3. Let AE = 2k, EC = 3k ⇒ 5k = 15 ⇒ k = 3. AE = 2(3) = 6 m, EC = 3(3) = 9 m. |
[1 Mark] for ratio formulation. [1 Mark] for AE = 6 m and EC = 9 m. |
Case Study 2
Structural Engineering
4 Marks
Sustainable Truss Bridge Support
A steel truss bridge lateral support is reinforced with triangular frame PQR. Two horizontal steel tie-rods AB and CD are welded across members PQ and PR parallel to base span QR.
Q1. Proportional Parallelism
1 Mark
If PC = 6 m, CQ = 18 m, PD = 8 m, and DR = 24 m, prove mathematically if tie-rod CD is parallel to QR.
(A) Yes, PC/CQ = PD/DR = 1/3, so CD ∥ QR by Converse of BPT
(B) No, ratios are unequal
(C) Parallel by RHS Congruence
(D) Cannot be determined
Q2. Similarity Criterion
1 Mark
Identify the similarity criterion proving △PCD ∼ △PQR when CD ∥ QR.
(A) SSS Similarity
(B) SAS Similarity
(C) AA Similarity Criterion
(D) RHS Similarity
Q3. Analytical Proportions
2 Marks
If AB ∥ CD with PA = 3 m, AC = 3 m, and CD = 9 m, calculate the length of upper tie-rod AB.
(A) 4.0 m
(B) 4.5 m
(C) 5.0 m
(D) 6.0 m
OR (Alternative Q3)
If PQ = 24 m and PR = 32 m, find PY on PR such that PX = 9 m creates tie-rod XY ∥ QR.
(A) PY = 10 m
(B) PY = 14 m
(C) PY = 15 m
(D) PY = 12 m
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | PC/CQ = 6/18 = 1/3; PD/DR = 8/24 = 1/3. Since PC/CQ = PD/DR, by the Converse of BPT, CD ∥ QR. |
[0.5 Mark] for evaluating ratios. [0.5 Mark] for Converse of BPT conclusion. |
| Q2 | ∠CPD = ∠QPR (common), ∠PCD = ∠PQR (corresponding). Hence △PCD ∼ △PQR by AA Similarity Criterion. |
[0.5 Mark] for angle reasoning. [0.5 Mark] for AA Similarity. |
| Q3 | PC = PA + AC = 3 + 3 = 6 m. △PAB ∼ △PCD ⇒ PA/PC = AB/CD ⇒ 3/6 = AB/9 ⇒ 1/2 = AB/9 ⇒ AB = 4.5 m. |
[1 Mark] for side ratio setup. [1 Mark] for AB = 4.5 m. |
| Q3 (OR) | PX/PQ = PY/PR ⇒ 9/24 = PY/32 ⇒ 3/8 = PY/32 ⇒ PY = (3 × 32)/8 = 12 m. | [1 Mark] for proportionality equation. [1 Mark] for PY = 12 m. |
Case Study 3
Height Estimation
4 Marks
Shadow-Casting & Height Estimation
A giant Deodar tree AB casts ground shadow BC = 24 m at 11:00 AM. At the same moment, vertical calibration pole DE = 3 m casts shadow EF = 4 m:
• Tree shadow (BC): 24 m
• Pole height (DE): 3 m
• Pole shadow (EF): 4 m
Q1. Conceptual Understanding
1 Mark
Explain why △ABC and △DEF are similar and specify the similarity criterion.
(A) SAS Similarity (equal sides ratio)
(B) AA Similarity (∠B = ∠E = 90° and equal sun elevation angle ∠C = ∠F)
(C) SSS Similarity
(D) RHS Congruence
Q2. Application
1 Mark
Using the properties of similar triangles, calculate the exact height of tree AB.
(A) 18 meters
(B) 16 meters
(C) 20 meters
(D) 21 meters
Q3. Analytical Shadows
2 Marks
When tree shadow becomes 36 m, a young sapling casts shadow of 1.6 m. Find the sapling's height.
(A) 0.6 m
(B) 1.0 m
(C) 0.8 m (80 cm)
(D) 1.2 m
OR (Alternative Q3)
Find line-of-sight distance AC from shadow tip C to tree top A and find hypotenuse ratio AC : DF.
(A) AC = 30 m; Ratio AC : DF = 6 : 1
(B) AC = 32 m; Ratio AC : DF = 5 : 1
(C) AC = 28 m; Ratio AC : DF = 6 : 1
(D) AC = 30 m; Ratio AC : DF = 4 : 1
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | ∠B = ∠E = 90° and ∠C = ∠F (same solar elevation angle). △ABC ∼ △DEF by AA Similarity Criterion. |
[0.5 Mark] for angle equalities. [0.5 Mark] for AA criterion. |
| Q2 | AB/DE = BC/EF ⇒ AB/3 = 24/4 ⇒ AB/3 = 6 ⇒ AB = 18 m. | [0.5 Mark] for proportion. [0.5 Mark] for AB = 18 m. |
| Q3 | H_sapling / 18 = 1.6 / 36 ⇒ H_sapling = 18 × (1.6 / 36) = 1.6 / 2 = 0.8 m (80 cm). | [1 Mark] for setup. [1 Mark] for 0.8 m. |
| Q3 (OR) | AC = √(18² + 24²) = √(324 + 576) = √900 = 30 m. DF = √(3² + 4²) = 5 m ⇒ AC : DF = 30 : 5 = 6 : 1. |
[1 Mark] for AC = 30 m. [1 Mark] for ratio 6 : 1. |
Case Study 4
Maritime Navigation
4 Marks
Autonomous Ship Guidance Corridor
An autonomous harbor vision system projects similar triangular corridors for a pilot boat (△ABC) and cargo ship (△PQR) with AB/PQ = 3/5. Altitudes AD and PM represent safety stopping depths.
Q1. Altitudes Ratio Property
1 Mark
Prove why AD/PM = AB/PQ for similar triangles △ABC ∼ △PQR with altitudes AD and PM.
(A) By SSS Similarity of whole triangles
(B) △ABD ∼ △PQM by AA similarity (∠B = ∠Q and ∠ADB = ∠PMQ = 90°)
(C) By equal areas theorem
(D) By BPT Corollary
Q2. Altitude Calculation
1 Mark
If safety altitude PM for the cargo ship is 25 m, calculate stopping altitude AD for pilot boat.
(A) 15 meters
(B) 12 meters
(C) 18 meters
(D) 20 meters
Q3. Perimeter & Median Scaling
2 Marks
If perimeter of larger corridor △PQR is 75 m, calculate total perimeter of pilot boat corridor △ABC.
(A) 40 meters
(B) 50 meters
(C) 45 meters
(D) 35 meters
OR (Alternative Q3)
If AX and PY are corresponding medians of △ABC and △PQR, and PY = 30 m, calculate median AX.
(A) 16 meters
(B) 18 meters
(C) 20 meters
(D) 15 meters
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | In △ABD and △PQM: ∠B = ∠Q and ∠ADB = ∠PMQ = 90° ⇒ △ABD ∼ △PQM (AA criterion) ⇒ AD/PM = AB/PQ. | [0.5 Mark] for sub-triangles similarity. [0.5 Mark] for ratio proof. |
| Q2 | AD/25 = 3/5 ⇒ AD = 25 × (3/5) = 15 m. | [1 Mark] for AD = 15 m. |
| Q3 | Perimeter(ABC) / Perimeter(PQR) = AB/PQ ⇒ P_ABC / 75 = 3/5 ⇒ P_ABC = 75 × (3/5) = 45 m. | [1 Mark] for perimeter ratio theorem. [1 Mark] for 45 m. |
| Q3 (OR) | AX/PY = AB/PQ ⇒ AX/30 = 3/5 ⇒ AX = 30 × (3/5) = 18 m. | [1 Mark] for median ratio property. [1 Mark] for AX = 18 m. |
Case Study 5
Sports Biomechanics
4 Marks
Archer Posture Frame
An Olympic archery biomechanics tracking system models posture as △ABC (shoulder A, elbow B, wrist C). Sensor line DE connects D on AB and E on AC with AD = 12 cm, DB = 18 cm, AE = 16 cm, EC = 24 cm, DE = 14 cm.
Q1. Converse of BPT
1 Mark
Prove why sensor line segment DE is parallel to forearm line segment BC.
(A) By equal angles sum
(B) AD/DB = 12/18 = 2/3 and AE/EC = 16/24 = 2/3; parallel by Converse of BPT
(C) By Pythagoras theorem
(D) Parallel by mid-point theorem
Q2. Perimeter Ratio
1 Mark
Find the ratio of perimeter of △ADE to perimeter of full arm frame △ABC.
(A) 2 : 5
(B) 2 : 3
(C) 3 : 5
(D) 4 : 9
Q3. Forearm Length & Angles
2 Marks
Calculate the exact length of forearm segment BC.
(A) 30 cm
(B) 32 cm
(C) 35 cm
(D) 42 cm
OR (Alternative Q3)
If alignment ∠ADE = 65° and wrist ∠ACB = 45°, calculate the archer's shoulder angle ∠BAC.
(A) 60°
(B) 70°
(C) 75°
(D) 80°
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | AD/DB = 12/18 = 2/3; AE/EC = 16/24 = 2/3. Since AD/DB = AE/EC, by Converse of BPT, DE ∥ BC. |
[0.5 Mark] for evaluating ratios. [0.5 Mark] for Converse of BPT conclusion. |
| Q2 | AB = AD + DB = 12 + 18 = 30 cm. Perimeter ratio = AD/AB = 12/30 = 2/5 = 2 : 5. |
[0.5 Mark] for ratio relation. [0.5 Mark] for 2 : 5. |
| Q3 | DE/BC = AD/AB ⇒ 14/BC = 12/30 = 2/5 ⇒ BC = (14 × 5)/2 = 35 cm. | [1 Mark] for side proportion. [1 Mark] for BC = 35 cm. |
| Q3 (OR) | Since DE ∥ BC, ∠ABC = ∠ADE = 65°. In △ABC: ∠BAC = 180° − (65° + 45°) = 180° − 110° = 70°. |
[1 Mark] for ∠B = 65°. [1 Mark] for ∠BAC = 70°. |
Case Study 6
Telecom Infrastructure
4 Marks
5G Network Mast & Signal Shadow
A vertical 5G mast PQ = 30 m and building AB = 18 m stand upright on horizontal ground. Collinear sight lines connect sensor S, roof corner A, and mast tip P with SB = 27 m.
Q1. Triangle Similarity
1 Mark
Prove why △SAB ∼ △SPQ and identify the criterion.
(A) SAS Criterion
(B) AA Similarity (∠S common, ∠ABS = ∠PQS = 90°)
(C) RHS Congruence
(D) SSS Similarity
Q2. Separation Distance
1 Mark
Find horizontal distance BQ between building base and 5G mast base.
(A) 15 meters
(B) 20 meters
(C) 18 meters
(D) 27 meters
Q3. Aerial Paths & Pole Heights
2 Marks
If sensor-to-building distance SA = 22.5 m, calculate drone aerial flight distance AP from roof to mast tip.
(A) 15.0 meters
(B) 12.5 meters
(C) 17.5 meters
(D) 20.0 meters
OR (Alternative Q3)
Point M on SQ divides SM:MQ = 2:1. Find the required height of vertical calibration pole MN.
(A) 16 meters
(B) 18 meters
(C) 24 meters
(D) 20 meters
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | ∠ABS = ∠PQS = 90° and ∠S is common ⇒ △SAB ∼ △SPQ by AA Similarity Criterion. | [0.5 Mark] for angle setup. [0.5 Mark] for AA criterion. |
| Q2 | SB/SQ = AB/PQ ⇒ 27/SQ = 18/30 = 3/5 ⇒ SQ = (27 × 5)/3 = 45 m. BQ = SQ − SB = 45 − 27 = 18 m. |
[0.5 Mark] for finding SQ = 45 m. [0.5 Mark] for BQ = 18 m. |
| Q3 | SA/SP = AB/PQ ⇒ 22.5/SP = 3/5 ⇒ SP = (22.5 × 5)/3 = 37.5 m. AP = SP − SA = 37.5 − 22.5 = 15 m. |
[1 Mark] for SP = 37.5 m. [1 Mark] for AP = 15 m. |
| Q3 (OR) | SM = (2/3) × 45 = 30 m. MN/PQ = SM/SQ ⇒ MN/30 = 30/45 = 2/3 ⇒ MN = (2 × 30)/3 = 20 m. |
[1 Mark] for SM calculation. [1 Mark] for pole height 20 m. |
Case Study 7
Autonomous Vehicles
4 Marks
LiDAR Distance Sensor Calibration
An autonomous car LiDAR bumper sensor O projects beam cone to parallel target boards AB and CD (AB ∥ CD). Central axis altitudes are OP = 4 m (to board AB = 1.6 m) and OQ = 10 m (to board CD).
Q1. Triangle Similarity
1 Mark
Prove why △OAB ∼ △OCD for parallel target boards AB ∥ CD.
(A) SSS Similarity
(B) AA Similarity (corresponding angles ∠OAB = ∠OCD and ∠OBA = ∠ODC)
(C) SAS Similarity
(D) RHS Similarity
Q2. Altitudes Property
1 Mark
State the relationship between the board widths ratio AB/CD and altitudes ratio OP/OQ.
(A) AB/CD = OP/OQ
(B) AB/CD = (OP/OQ)²
(C) AB/CD = √(OP/OQ)
(D) AB/CD = 2(OP/OQ)
Q3. Target Width & Ray Segment
2 Marks
Calculate the exact width of the second calibration target board CD.
(A) 3.6 meters
(B) 4.0 meters
(C) 4.5 meters
(D) 5.0 meters
OR (Alternative Q3)
If diagonal range reading OA = 4.5 m, find the length of ray segment AC between boards.
(A) 6.00 meters
(B) 6.25 meters
(C) 7.00 meters
(D) 6.75 meters
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | Since AB ∥ CD: ∠OAB = ∠OCD and ∠OBA = ∠ODC (corresponding angles). △OAB ∼ △OCD by AA Similarity Criterion. |
[0.5 Mark] for corresponding angles. [0.5 Mark] for AA criterion. |
| Q2 | The ratio of corresponding sides of similar triangles equals the ratio of their corresponding altitudes: AB/CD = OP/OQ. | [1 Mark] for stating AB/CD = OP/OQ. |
| Q3 | 1.6 / CD = 4 / 10 ⇒ 4 × CD = 16 ⇒ CD = 4 m. | [1 Mark] for proportion setup. [1 Mark] for CD = 4 m. |
| Q3 (OR) | OA/OC = OP/OQ ⇒ 4.5/OC = 4/10 ⇒ OC = (4.5 × 10)/4 = 11.25 m. AC = OC − OA = 11.25 − 4.5 = 6.75 m. |
[1 Mark] for OC = 11.25 m. [1 Mark] for AC = 6.75 m. |
Case Study 8
Sustainable Architecture
4 Marks
A-Frame Sustainable Solar Cabin
An eco-cabin front elevation forms isosceles triangle ABC with AB = AC = 5 m and base BC = 6 m. Loft floor beam DE is built parallel to base BC, attached 2 m down from peak A (AD = AE = 2 m).
Q1. Ratio of Slopes
1 Mark
Find the exact ratio of the upper sloped roof segment to total roof length (AD/AB).
(A) 1 : 2
(B) 2 : 5 (or 0.4)
(C) 2 : 3
(D) 3 : 5
Q2. Division Ratio
1 Mark
Calculate DB and find the ratio in which loft floor DE divides sloped side AB.
(A) DB = 3 m; Ratio AD : DB = 2 : 3
(B) DB = 2.5 m; Ratio AD : DB = 1 : 1
(C) DB = 3 m; Ratio AD : DB = 3 : 2
(D) DB = 4 m; Ratio AD : DB = 1 : 2
Q3. Loft Dimensions
2 Marks
Calculate the exact structural width of wooden loft floor DE.
(A) 2.0 meters
(B) 2.2 meters
(C) 2.4 meters
(D) 2.8 meters
OR (Alternative Q3)
Calculate the exact perimeter of upper triangular attic space △ADE above the loft floor.
(A) 6.4 meters
(B) 6.0 meters
(C) 7.2 meters
(D) 5.8 meters
📋 View Step-by-Step Marking Scheme & Solution▼
| Question | Expected Answer / Step-by-Step Solution | CBSE Marks Allocation |
|---|---|---|
| Q1 | AD/AB = 2/5 = 2 : 5 (or 0.4). | [1 Mark] for ratio 2 : 5. |
| Q2 | DB = AB − AD = 5 − 2 = 3 m. Division ratio AD/DB = 2/3 = 2 : 3. |
[0.5 Mark] for DB = 3 m. [0.5 Mark] for ratio 2 : 3. |
| Q3 | DE ∥ BC ⇒ △ADE ∼ △ABC ⇒ DE/BC = AD/AB ⇒ DE/6 = 2/5 ⇒ DE = (2 × 6)/5 = 2.4 m. | [1 Mark] for similarity proportion. [1 Mark] for DE = 2.4 m. |
| Q3 (OR) | Perimeter(ADE) = AD + AE + DE = 2 + 2 + 2.4 = 6.4 m. | [1 Mark] for side addition setup. [1 Mark] for 6.4 m. |
Live Practice: Chapter 6 Triangles
60:00