Chapter 5 · Exercise 3Textbook pages 96–98

Exercise 3 – Animated Solutions ത്രികോണപ്പരപ്പുകൾ

Six questions on areas of equilateral triangles, hexagons, isosceles triangles and Heron's formula – and then the textbook's appendix (അനുബന്ധം): the full proof of Heron's formula, one algebraic step at a time.

📖 Tools you will use

Question 1 · Equilateral triangles with sides 10, 5 and √3

Q1The side of an equilateral triangle is given. Find its area: (i) 10 cm (ii) 5 cm (iii) √3 cm.

Area = √3 × (half the side)². For (iii), half the side is √3/2 and its square is 3/4.

🎬 Solution animation
Answer: (i) √3 × 5² = 25√3 ≈ 43.30 sq cm. (ii) √3 × (5/2)² = 25√3/4 ≈ 10.83 sq cm. (iii) √3 × (√3/2)² = √3 × ¾ = 3√3/4 ≈ 1.30 sq cm.

Question 2 · A regular hexagon of side 6 cm

Q2Find the area of a regular hexagon (സമഷഡ്ഭുജം) with all sides 6 cm.

Join the centre to the six vertices. The six triangles are all equilateral (each has a 60° angle at the centre and two equal sides).

🎬 Solution animation: the hexagon falls apart into six equilateral triangles
Answer: One triangle: √3 × 3² = 9√3. Hexagon: 6 × 9√3 = 54√3 sq cm ≈ 93.53 sq cm.
🐝 Real life

Honeycomb cells are regular hexagons – they tile a surface with the least wax for the most storage. A cell of side 3 mm holds 6 × √3 × 1.5² ≈ 23.4 sq mm of floor area. Hexagonal floor tiles, bolt heads and pencils are all measured the same way.

Question 3 · Equilateral triangle with height 12 cm

Q3The height of an equilateral triangle is 12 cm. Find its perimeter and area.

height = √3 × (a/2), so a/2 = 12/√3. Rationalise: 12/√3 = 12√3/3 = 4√3.

🎬 Solution animation
Answer: half side = 12/√3 = 4√3 ⇒ side = 8√3 cm; perimeter = 24√3 cm ≈ 41.57 cm; area = ½ × 8√3 × 12 = 48√3 sq cm ≈ 83.14 sq cm (or √3 × (4√3)² = 48√3).

Question 4 · Hexagon with opposite sides 6 cm apart

Q4In a regular hexagon the distance between two parallel (opposite) sides is 6 cm. Find its perimeter and area.

The distance between opposite sides is twice the height of one of the six equilateral triangles: 2 × (√3/2)a = √3·a.

🎬 Solution animation
Answer: √3·a = 6 ⇒ a = 6/√3 = 2√3 cm; perimeter = 12√3 cm ≈ 20.78 cm; area = 6 × √3 × (√3)² = 6 × 3√3 = 18√3 sq cm ≈ 31.18 sq cm.

Question 5 · Triangle with sides 8, 6, 6

Q5Find the height and area of a triangle with sides 8 cm, 6 cm and 6 cm.

It is isosceles with base 8. The height bisects the base into 4 and 4; Pythagoras gives h² = 6² − 4².

🎬 Solution animation
Answer: h² = 36 − 16 = 20 ⇒ h = √20 = 2√5 cm ≈ 4.47 cm; area = ½ × 8 × 2√5 = 8√5 sq cm ≈ 17.89 sq cm. Heron check: s = 10, √(10 × 2 × 4 × 4) = √320 = 8√5 ✓.

Question 6 · Heron's formula three times

Q6Find the area of each triangle: (i) 4 cm, 5 cm, 7 cm (ii) 4 cm, 13 cm, 15 cm (iii) 5 cm, 12 cm, 13 cm.

s = half the perimeter. Then multiply s, s − a, s − b, s − c and take the square root; simplify by pulling out perfect squares.

🎬 Solution animation
Answer: (i) s = 8: √(8 × 4 × 3 × 1) = √96 = 4√6 ≈ 9.80 sq cm. (ii) s = 16: √(16 × 12 × 3 × 1) = √576 = 24 sq cm. (iii) s = 15: √(15 × 10 × 3 × 2) = √900 = 30 sq cm (a right triangle: ½ × 5 × 12 = 30 ✓).
🧮 Heron calculator

Appendix · Why Heron's formula works അനുബന്ധം

This is the textbook's proof (pages 96–98). It is exactly the 8-7-3 calculation from Section 3, done with letters instead of numbers. Take a triangle with base a, other sides b and c, height h from the top vertex, and let the foot of the height split the base into x (under side c) and a − x (under side b).

🎬 The proof, one step at a time
💡 How to read the proof

Every step uses only two tools: Pythagoras and the identity p² − q² = (p + q)(p − q). The clever part is introducing s = half the perimeter at the very end, which turns four long brackets into the neat product s(s − a)(s − b)(s − c). You do not need to memorise the proof – but following it once shows you where the formula comes from.

Check yourself

🧭 What Exercise 3 taught us

Next: Real-world applications ▶