Class 7 Mathematics Federal Board of Intermediate and Secondary Education (FBISE) (FBISE) 📚 Model Textbook of Mathematics Grade 7 (FBISE / NBF)

Class 7 Mathematics Ch 10 Mastery Guide: Triangle Construction, SSS/SAS/ASA/RHS & Perpendicular Distance (FBISE)

📖 Chapter 10: Practical Geometry 📅 Updated: Sep 13, 2026

🗺️ Teacher & Parent Roadmap: Unit 10 (Practical Geometry)

Practical geometry bridges the abstract beauty of geometric theory with concrete hands-on construction. While earlier grades focused on freehand sketches and visual identification, Grade 7 students master rigorous Euclidean compass-and-straightedge constructions. They learn why certain side combinations form valid triangles while others fail, how to construct acute, right, and obtuse triangles across four distinct criteria (SSS, SAS, ASA, RHS), and mathematically prove why the perpendicular path is the shortest distance between a point and a line.

🎯 Core Learning Objectives:

Apply the Triangle Inequality Theorem (a + b > c) to verify if a triangle can exist before attempting construction.
Construct equilateral, isosceles, and scalene triangles using compass arcs.
Execute constructions across all four geometric criteria: SSS, SAS, ASA/AAS, and RHS.
Construct a perpendicular from an external point to a line segment and prove the shortest distance property.

⚠️ Common Misconceptions & Pitfalls:

Error: Trying to build a triangle where two sides sum to less than or equal to the third (e.g. 2 + 5 = 7 or 2 + 4.2 < 7.3). The arcs will never intersect in 2D space!
Pitfall: Confusing SAS (included angle between the two given sides) with ASS (angle not included, which can cause ambiguous cases).
Misconception: Assuming the shortest route to a line is an angled oblique path. (Fact: The perpendicular line at exactly 90° is always strictly the shortest distance).

📐 Core Concepts: Triangles & Geometric Foundations

A triangle is a closed two-dimensional rectilinear plane figure bounded by three straight line segments joined at three non-collinear vertices. It is the fundamental building block of all polygons and architectural trusses because it is the only polygon that is rigid and cannot deform without changing the lengths of its sides.

1. Taxonomy & Classification of Triangles

Triangles are classified along two independent dimensions: by their side lengths and by their interior angle measures.

Equilateral Triangle

3 Equal Sides; 3 Equal Angles (60°)

Isosceles Triangle

2 Equal Sides; 2 Equal Base Angles

Scalene Triangle

All 3 sides & angles are unequal

Right-Angled Triangle

One 90° Right Angle; Longest side is Hypotenuse

Obtuse-Angled Triangle

One angle is greater than 90°

Acute-Angled Triangle

All three angles are strictly less than 90°

2. The Triangle Inequality Theorem

The Law of Triangle Existence:
A triangle with side lengths a, b, and c can exist if and only if the sum of the lengths of any two sides is strictly greater than the length of the third side:

                (1) a + b > c      (2) b + c > a      (3) c + a > b
            

Practical shortcut: It is sufficient to check that the sum of the two smallest sides is strictly greater than the largest side.

✓ Triangle Possible (3 + 4 = 7 > 5)

Arcs intersect cleanly at a distinct 3rd vertex C

✗ Triangle Impossible (2 + 4.2 = 6.2 < 7.3)

Sum of radii is less than base; arcs will NEVER touch

3. The 4 Standard Triangle Construction Protocols

To construct a unique triangle, we need exactly three independent pieces of geometric data. The four standard cases are:

1. SSS (Side-Side-Side)

Base drawn with ruler; other 2 sides located by compass arcs

2. SAS (Side-Angle-Side)

Base drawn; included angle constructed; ray cut to length

3. ASA / AAS

Base drawn; two angles constructed at ends; rays intersect

4. RHS (Right-Hypotenuse-Side)

90° ray raised; arc of hypotenuse radius cuts vertical ray

4. Perpendicular Distance & The Shortest Path Theorem

When measuring the distance between an external point and a straight line, which path should we take? Geometry proves that the perpendicular segment dropped from the point to the line is the unique shortest path.

Theorem: Perpendicular Distance is the Shortest Distance

In right triangle ΔARS, the oblique line AS is the hypotenuse, so by Pythagoras: AS² = AR² + RS² > AR² ⇒ AS > AR.

📝 100% Complete Textbook Exercise Solutions

Exercise 10.1: Construction of Triangles & Perpendiculars

Question 1: Construct the triangles where possible.

(a) 3 cm, 4 cm, 5 cm:
Check: 3 + 4 = 7 > 5 (✓ Possible).
Steps:
1. Draw base AB = 5 cm.
2. From A, draw arc with r = 3 cm.
3. From B, draw arc with r = 4 cm intersecting at C.
4. Join AC and BC.
Type: Scalene Right-Angled Triangle (3² + 4² = 5²).

(b) 5 cm, 7 cm, 2 cm:
Check: 5 + 2 = 7 (NOT > 7).
Result: NOT POSSIBLE.
Reason: Sum of two sides equals the third side; the arcs meet on the line, collapsing into a flat line segment without area.

(c) 2 cm, 4.2 cm, 7.3 cm:
Check: 2 + 4.2 = 6.2 < 7.3.
Result: NOT POSSIBLE.
Reason: Sum of the two smaller sides is less than the third side. The arcs cannot reach each other.

(d) 4.6 cm, 4.6 cm, 4.6 cm:
Check: 4.6 + 4.6 = 9.2 > 4.6 (✓ Possible).
Steps:
1. Draw base AB = 4.6 cm.
2. With radius 4.6 cm, draw arcs from A and B intersecting at C.
3. Join AC and BC.
Type: Equilateral Triangle (all angles = 60°).

Question 2: Construct triangle ABC: AB = 6 cm, BC = AC = 5 cm. Name with respect to sides.
Steps: Draw base AB = 6 cm. From A and B, draw arcs of radius 5 cm intersecting at C. Join AC and BC.
Classification: Isosceles Triangle (since two sides AC = BC = 5 cm).

Question 3: Construct triangles with given sides and included angle:
• (a) Sides 4 cm and 5.2 cm, included angle 75°: Draw base 5.2 cm, construct 75° angle ray at endpoint, cut off 4 cm, join. Type: Acute-angled Triangle.
• (b) Sides 6 cm and 3.8 cm, included angle 120°: Draw base 6 cm, construct 120° obtuse angle ray, cut off 3.8 cm, join. Type: Obtuse-angled Triangle.

Question 4: Construct ΔABC with ∠A = 45°, ∠B = 90°, AB = 6.5 cm. Measure third angle.
Third Angle: ∠C = 180° - (90° + 45°) = 45°.
Type: Right-Angled Isosceles Triangle (since ∠A = ∠C = 45°, sides AB = BC = 6.5 cm).

Question 5: Construct ΔPQR with ∠P = 60°, ∠Q = 60°, PQ = 5.2 cm. Measure third angle & name triangle.
Third Angle: ∠R = 180° - (60° + 60°) = 60°.
Classification: Equilateral Triangle (w.r.t sides) and Acute-Angled / Equiangular Triangle (w.r.t angles).

Question 6: Construct right triangle with hypotenuse 7 cm and side 5.5 cm (RHS Case).
Steps: Draw base 5.5 cm, raise perpendicular ray (90°), swing arc of radius 7 cm from other endpoint cutting ray at vertex.
• (a) Length of 3rd side: √(7² - 5.5²) = √(49 - 30.25) = √18.75 ≈ 4.33 cm (measured ~4.3 cm).
• (b) Name w.r.t. sides: Scalene Triangle (sides: 5.5 cm, 4.33 cm, 7 cm).

Question 7: Construct ΔABC with AB = 5.2 cm, ∠B = 90°, AC = 6.5 cm. Type w.r.t. angles.
Calculated other leg: BC = √(6.5² - 5.2²) = √(42.25 - 27.04) = √15.21 = 3.9 cm.
Type: Right-angled Triangle.

Question 8: Construct ΔABC (AB = 6 cm, BC = 5.3 cm, AC = 5.6 cm). Draw perpendicular from C on AB.
Steps: Construct ΔABC using SSS. With center C, draw arc cutting AB at two points. Draw intersecting arcs below to locate perpendicular line. Join and measure.
Perpendicular Altitude: h ≈ 4.6 cm (exact ~4.55 cm).

Question 9: Construct ΔXYZ with XY = 5.4 cm, ∠X = 75°, ∠Y = 60°. Measure YZ and XZ.
Third Angle: ∠Z = 180° - 135° = 45°.
Measured lengths: YZ ≈ 7.4 cm (exact: 7.38 cm) and XZ ≈ 6.6 cm (exact: 6.61 cm).

Question 10: Construct triangle with each side 6 cm. Measure three angles. What did you notice?
Angle Measures: ∠A = 60°, ∠B = 60°, ∠C = 60°.
Key Discovery: Whenever all three sides of a triangle are equal, all three interior angles are guaranteed to be equal to 60° (Equilateral = Equiangular).

Review Exercise 10: Master Review & MCQs

Question 1: Multiple Choice Questions (MCQs)

(i) In ΔABC, AB = AC. The triangle is: (d) isosceles

(ii) Choose correct statement: (b) A triangle can have only one right angle

(iii) Sum of two sides of triangle is ... third side: (b) greater than

(iv) Sum of measures of three angles: (c) 180°

(v) Triangle having all sides equal: (a) equilateral

(vi) Angle present in all types of triangles: (a) acute (at least 2 acute angles)

(vii) a and b opposite equal sides: (c) a = b

(viii) Interior angle of equilateral triangle: (d) 60°

(ix) Side opposite right angle: (b) hypotenuse

(x) If AB = 5 cm, BC = 8 cm, AC cannot be: (d) 2.9 cm (5 + 2.9 = 7.9 < 8)

Question 2: Construct the following triangles:
• (a) Sides 4.3 cm, 6 cm, 3.8 cm: Draw base 6 cm, draw arcs 4.3 cm and 3.8 cm (Scalene triangle).
• (b) XY = 5.6 cm, YZ = 4 cm, ∠Y = 75°: SAS construction with included angle 75°.
• (c) RQ = 5.2 cm, ∠R = 30°, ∠Q = 120°: Third angle ∠P = 30° (Obtuse Isosceles triangle with PQ = 5.2 cm).
• (d) Hypotenuse 5.7 cm and side 4.5 cm: RHS right triangle; other leg = √(5.7² - 4.5²) ≈ 3.5 cm.

Question 3: Construct ΔLMN of suitable measurement and draw perpendicular from N to LM.
Example: Equilateral Δ with side 6 cm. Dropped perpendicular altitude measures h = √(6² - 3²) = √27 ≈ 5.2 cm.

Question 4: Draw AB = 5 cm. Point P outside. Draw perpendicular PM and prove it is shortest.
Proof: For any other point Q on line AB (Q ≠ M), triangle ΔPMQ is right-angled at M. In ΔPMQ, PQ is the hypotenuse. By Pythagoras theorem, PQ² = PM² + MQ² > PM² ⇒ PQ > PM. Thus PM is strictly the shortest distance.

📌 Chapter 10 Quick Revision & Mastery Formulas

Triangle Inequality:
a + b > c (Sum of two smaller sides must exceed longest side).

Angle Sum Property:
∠A + ∠B + ∠C = 180°
Equilateral angle: 180° / 3 = 60°

Pythagoras Theorem:
Hypotenuse² = Base² + Perpendicular²
c² = a² + b² (Right triangles).

Shortest Distance:
The perpendicular line segment from a point to a line is strictly the shortest distance.

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