Chapter 5 · Exercise 2Textbook pages 91–92

Exercise 2 – Animated Solutions ഹരണം

Five questions that use division of square roots, rationalising a denominator and the identity (x + y)(x − y) = x² − y² – plus one beautiful geometry puzzle. Try each question in your notebook first, use the hint if you are stuck, then press Next ▶ to watch the solution grow.

📖 Tools you will use

Question 1 · (√2 + 1)(√2 − 1) = 1 and what it is good for

Q1Prove that (√2 + 1)(√2 − 1) = 1. Using this, find (i) 1/(√2 − 1) and (ii) 1/(√2 + 1), each correct to two decimal places.

Use (x + y)(x − y) = x² − y² with x = √2, y = 1. For (i), multiply the top and bottom of the fraction by (√2 + 1) – the denominator becomes 1.

🎬 Solution animation: the identity as a picture, then the two reciprocals
Answer: (√2 + 1)(√2 − 1) = (√2)² − 1² = 2 − 1 = 1. (i) 1/(√2 − 1) = √2 + 1 ≈ 2.41. (ii) 1/(√2 + 1) = √2 − 1 ≈ 0.41.
💡 Why this trick matters

Dividing by 0.414… by hand is painful. Dividing by 1 is trivial. Multiplying top and bottom by the "partner" (√2 + 1) turns the awkward denominator √2 − 1 into the whole number 1. This is called rationalising the denominator – you will use it again and again in Class 10.

Question 2 · An equilateral triangle with height 4 cm

Q2In the figure the height of an equilateral triangle is 4 cm. Find the length of its sides, correct to the millimetre.

Call the side a. The height cuts the triangle into two right triangles with base a/2. Use Pythagoras, or the shortcut height = √3 × (a/2).

🎬 Solution animation
Answer: ¾a² = 16 ⇒ a = 8/√3 = 8√3/3 ≈ 4.619 cm ≈ 46 mm.

Question 3 · The pinwheel of equilateral triangles

Q3In the figure all the red triangles are equilateral. What is the ratio between the sides of the outer square and the inner square?

The outer square is four identical rectangles arranged around the small square. In one rectangle, the diagonals meet at its centre, so a red triangle has base = short side and height = half the long side. Equilateral ⇒ height = (√3/2) × base.

🎬 Solution animation
Answer: outer : inner = (√3 + 1)h : (√3 − 1)h = (√3 + 1) : (√3 − 1) = (2 + √3) : 1 ≈ 3.73 : 1.
🎚️ Play: only one shape of rectangle makes the red triangles equilateral

Drag the slider. The red triangles are equilateral only when the long side is √3 ≈ 1.732 times the short side – then the apex angle is exactly 60° and all three sides of the red triangle are equal. The ratio outer : inner = (r + 1)/(r − 1) changes with the shape.

Question 4 · Square roots of mixed numbers

Q4Prove that √2⅔ = 2√⅔ and √3⅜ = 3√⅜. Can you find other numbers like these?

Turn the mixed number into an improper fraction (2⅔ = 8/3) and look for a perfect-square factor: 8/3 = 4 × 2/3.

🎬 Solution animation
Answer: 2⅔ = 8/3 = 4 × ⅔ ⇒ √(2⅔) = 2√⅔. 3⅜ = 27/8 = 9 × ⅜ ⇒ √(3⅜) = 3√⅜. In general n + n/(n² − 1) = n³/(n² − 1) = n² × n/(n² − 1), so √(n + n/(n²−1)) = n√(n/(n²−1)): √(4⁴⁄₁₅) = 4√(4/15), √(5⁵⁄₂₄) = 5√(5/24), …
🎚️ Make your own examples

For every whole number n ≥ 2, the mixed number "n and n over (n² − 1)" has a square root that is n times the square root of its fraction part. The decimal check in the last line shows both sides agree.

Question 5 · Which quotients are whole numbers or fractions?

Q5In the pairs below, dividing the first number by the second sometimes gives a natural number or a fraction. Find them: (i) √72, √2 (ii) √27, √3 (iii) √125, √50 (iv) √10, √2 (v) √20, √5 (vi) √18, √8.

√a ÷ √b = √(a ÷ b). Simplify the fraction a/b first; the answer is a whole number or fraction exactly when both the numerator and the denominator of the simplified fraction are perfect squares.

Answer: (i) 6 (ii) 3 (v) 2 are natural numbers; (vi) 3/2 is a fraction; (iii) √(5/2) and (iv) √5 are irrational.
🎚️ Divide any two square roots

Try a = 125, b = 50 (irrational), a = 18, b = 8 (fraction 3/2), a = 20, b = 5 (whole number).

Check yourself

🧭 What Exercise 2 taught us

Next: Section 3 – Areas of triangles and Heron's formula ▶