Vardaan
Class 7 Maths • Chapter 10

Practical Geometry

Vardaan Learning Institute • Detailed Notes with Practice Questions

📐 1. Tools Used in Construction

Tool Used For
Ruler (Scale) Drawing straight lines of known length
Compass Drawing arcs and circles; copying lengths
Protractor Measuring and drawing angles
Set-square Drawing parallel lines and right angles
Divider Comparing lengths

🔺 2. Construction of Triangles

A unique triangle can be constructed if we are given enough information. The following conditions are sufficient:

Conditions for unique triangle construction: Note: AAA does not give unique triangle (infinite triangles possible). SSA is ambiguous (0, 1 or 2 triangles possible).

Construction 1 — SSS (All three sides given)

📐 Construct △ABC with AB=5cm, BC=6cm, AC=4cm
  1. Draw base BC = 6 cm using ruler.
  2. With B as centre and radius = 5 cm (=AB), draw an arc above BC.
  3. With C as centre and radius = 4 cm (=AC), draw another arc cutting the previous arc at A.
  4. Join AB and AC. △ABC is ready.
BC = 6 cm A B C AB=5 AC=4

Construction 2 — SAS (Two sides + included angle)

📐 Construct △PQR with PQ=4cm, ∠Q=60°, QR=5cm
  1. Draw base QR = 5 cm.
  2. At Q, draw ∠Q = 60° using protractor. Draw ray QX.
  3. On ray QX, mark P at 4 cm from Q (PQ = 4cm).
  4. Join PR. △PQR is ready.

Construction 3 — ASA (Two angles + included side)

📐 Construct △LMN with ∠L=60°, LM=5cm, ∠M=75°
  1. Draw base LM = 5 cm.
  2. At L, draw angle = 60° (draw ray LX).
  3. At M, draw angle = 75° (draw ray MY).
  4. The two rays LX and MY intersect at N. △LMN is ready.
  5. Note: ∠N = 180° − 60° − 75° = 45°.

Construction 4 — RHS (Right angle, hypotenuse, side)

📐 Construct right △ABC with right angle at B, hypotenuse AC=5cm, BC=3cm
  1. Draw BC = 3 cm.
  2. At B, draw a ray BX ⊥ BC (90° angle).
  3. With A as centre (first draw A tentatively on BX), use Pythagoras: AB = √(AC²−BC²) = √(25−9) = 4 cm. Mark A at 4 cm on BX.
  4. Join AC. Verify AC = 5 cm. △ABC is ready.

📏 3. Construction of Special Lines

Construction Steps Summary
Perpendicular Bisector of a segment AB Take compass with radius > half AB. Draw arcs from both A and B (same radius). Join the two intersection points → perpendicular bisector
Angle Bisector of ∠AOB Draw arc from O cutting OA at P and OB at Q. With same radius from P and Q, draw arcs intersecting at R. Join OR → angle bisector
Parallel Line through a point P to line AB Using corresponding angles (or using ruler + set square). Set square method: place set square edge on AB, slide ruler along, draw line through P
✏️ Practice Questions — Practical Geometry (20 Questions)

Section A — Identify & Plan Easy

Q1. Can a △ be constructed with sides 3cm, 4cm, 8cm? Why?
Q2. Which condition (SSS/SAS/ASA/AAS/RHS) applies: AB=5cm, BC=6cm, AC=7cm?
Q3. Which condition applies: ∠P=40°, PQ=6cm, ∠Q=70°?
Q4. Which condition applies: right ∠ at B, AB=5cm, AC (hypotenuse)=13cm?
Q5. Why can't we construct a unique triangle given only 3 angles?
Q6. Two sides of a triangle are 5cm and 12cm and angle between them is 90°. What is the hypotenuse? Which condition to use?

Section B — Construction Steps Medium

  1. Q7. Construct △ABC: AB=4cm, BC=5cm, CA=6cm. Write all steps.
  2. Q8. Construct △PQR: PQ=5cm, ∠P=60°, ∠Q=45°. Also find ∠R.
  3. Q9. Construct △XYZ: XY=6cm, ∠X=30°, XZ=5cm (SAS).
  4. Q10. Construct a right-angled triangle with hypotenuse=10cm and one leg=6cm.
  5. Q11. Construct the perpendicular bisector of a line segment of length 8cm.
  6. Q12. Draw any angle of 60°. Then bisect it to get 30°.
  7. Q13. Construct an equilateral triangle with side 5cm. Hint: What are all angles?
  8. Q14. Construct isosceles △ABC with AB=AC=5cm and BC=4cm.

Section C — Reasoning / Challenge Hard

  1. Q15. Construct △DEF where DE=4cm, EF=3cm, DF=5cm. Check if it's a right triangle. At which vertex is the right angle?
  2. Q16. In △ABC, ∠A=50°, ∠B=70°. Construct the triangle if BC=6cm. (Use AAS/ASA — figure out which.)
  3. Q17. Given 3 angles of a triangle as 40°, 60°, 80° — can you construct distinct triangles? List 3 possible side combinations.
  4. Q18. Explain why SSA is NOT a reliable construction criterion using a specific example (ambiguous case).
  5. Q19. Construct △PQR where ∠Q=90°, hypotenuse PR=7.5cm, PQ=4.5cm. Verify QR using Pythagoras.
  6. Q20. Construct a line through point P(outside a line l) parallel to l using a ruler and compass only (not set square). Write all steps.
✅ Key Answers: Q1: No (3+4=7, not > 8) | Q2: SSS | Q3: ASA | Q4: RHS | Q5: Infinite similar triangles possible | Q6: 13cm, RHS | Q8: ∠R=75° | Q15: Right angle at E (3²+4²=5²)

📝 Quick Revision

  1. Tools: Ruler (length), Compass (arcs), Protractor (angles), Set-square (right angles/parallels)
  2. 5 construction criteria: SSS, SAS, ASA, AAS, RHS → each gives a unique triangle
  3. AAA → NOT unique (similar triangles). SSA → ambiguous (0, 1, or 2 triangles possible)
  4. Perpendicular bisector: equidistant from both endpoints. Angle bisector: equidistant from both sides
  5. Triangle possible only if sum of any two sides > third side