Class 8 : Maths – Lesson 9. The Baudhayana-Pythagoras Theorem
EXPLANATION AND ANALYSIS
🌍 INTRODUCTION — AN ANCIENT IDEA THAT SHAPED GEOMETRY
📐 Geometry is not only about shapes we see today; it is deeply connected with ideas developed thousands of years ago.
🧠 One such powerful idea is the relationship between the sides of a right-angled triangle.
📘 Long before modern mathematics, Indian scholars understood this relationship.
🔍 In ancient Indian texts called the Śulba Sūtras, this idea was clearly stated.
📌 This relationship is known as the Baudhāyana–Pythagoras Theorem.
🎯 In this lesson, we will:
understand right-angled triangles
learn the statement of the theorem
explore its meaning step by step
see how it is useful in daily life
🔷 WHO WAS BAUDHĀYANA?
📜 Baudhāyana was an ancient Indian mathematician and scholar.
🧠 He lived around 800 BCE, much earlier than Greek mathematician Pythagoras.
📘 His work appears in the Baudhāyana Śulba Sūtra, which dealt with:
geometry
constructions
altar designs
🔵 These texts were practical in nature and used for construction and measurement.
🟡 The theorem shows India’s early contribution to mathematical thinking.
📌 Therefore, this result is rightly called the Baudhāyana–Pythagoras Theorem.
📐 UNDERSTANDING A RIGHT-ANGLED TRIANGLE
🧠 A triangle is a closed figure with three sides and three angles.
📘 A right-angled triangle is a triangle in which:
one angle is exactly 90°
📍 Important parts of a right-angled triangle:
Hypotenuse → the longest side, opposite the right angle
Other two sides → the sides forming the right angle
🔵 The hypotenuse is always opposite the right angle.
🟣 The other two sides are shorter than the hypotenuse.
📌 This theorem applies only to right-angled triangles.
📜 STATEMENT OF THE BAUDHĀYANA–PYTHAGORAS THEOREM
📘 The theorem states:
👉 In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
📐 If:
one side = a
another side = b
hypotenuse = c
🧮 Then:
c² = a² + b²
🔵 This statement connects length and area ideas.
🟡 It is a relationship that always holds true for right-angled triangles.
🧠 MEANING OF THE THEOREM (INTUITIVE IDEA)
🧠 The theorem is not just a formula; it explains a geometric balance.
📘 Imagine drawing:
a square on each side of a right-angled triangle
🔵 The area of the square on the hypotenuse
🟡 equals the combined areas of the squares on the other two sides
📌 This idea shows how areas and lengths are connected.
🟣 That is why the theorem involves squares of sides, not just sides.
🔢 WHY SQUARES ARE USED IN THE THEOREM
🧠 The word “square” in the theorem means area, not just multiplication.
📘 Squaring a number represents the area of a square with that side length.
🔵 a² means area of square with side a
🟡 b² means area of square with side b
🟣 c² means area of square with side c
📌 The theorem compares areas, not just numbers.
📐 CONDITIONS FOR APPLYING THE THEOREM
🧠 The Baudhāyana–Pythagoras Theorem works only when:
🔵 the triangle is a triangle
🟡 one angle is exactly 90°
🔴 the hypotenuse is correctly identified
🚫 It does not apply to:
acute-angled triangles
obtuse-angled triangles
📌 Correct identification of the right angle is very important.
🔍 IDENTIFYING A RIGHT-ANGLED TRIANGLE
🧠 Sometimes, we are not told directly whether a triangle is right-angled.
📘 The theorem helps us check.
🔵 If for sides a, b, c:
c² = a² + b²
🟡 then the triangle is right-angled
🟣 and c is the hypotenuse
📌 This makes the theorem a tool for verification.
📊 PRACTICAL UNDERSTANDING (WITHOUT CALCULATIONS)
🧠 Even without heavy calculations, the idea can be understood clearly.
📘 Suppose two sides of a triangle are fixed.
🔵 The third side must adjust so that the area relationship holds.
🟡 This ensures the triangle maintains a right angle.
🟣 Builders and surveyors used this idea practically in ancient times.
📌 Geometry here is deeply connected to real construction.
🏗️ USES IN DAILY LIFE
🧠 The theorem is widely used in real-world situations.
🔵 measuring height of buildings
🟡 checking squareness of rooms
🔴 designing roads and bridges
🟣 navigation and mapping
🟠 computer graphics and design
📘 Even modern technology relies on this ancient idea.
🌍 HISTORICAL IMPORTANCE
📜 The Baudhāyana–Pythagoras Theorem shows that:
advanced geometry existed in India very early
🧠 It reflects:
logical thinking
observation
practical mathematics
📌 This lesson highlights India’s contribution to global knowledge.
⚠️ COMMON MISTAKES TO AVOID
🚫 using the theorem for non-right triangles
🚫 confusing hypotenuse with other sides
🚫 squaring incorrectly
🚫 assuming any triangle follows this rule
✔️ Always check the right angle first.
🌟 IMPORTANCE OF THIS LESSON
🏆 builds strong geometric foundation
🧠 develops logical reasoning
⚡ prepares for higher geometry
📘 connects ancient knowledge with modern maths
🌱 improves problem-solving skills
Understanding this theorem gives confidence in geometry.
🧾 SUMMARY
🔵 right-angled triangle has one 90° angle
🟡 hypotenuse is the longest side
🔴 square of hypotenuse equals sum of squares of other sides
🟣 theorem works only for right-angled triangles
🟠 idea comes from Baudhāyana Śulba Sūtra
🟢 theorem has wide real-life applications
🔁 QUICK RECAP
🔵 right angle is essential
🟡 identify hypotenuse correctly
🟣 use squares of sides
🟠 applies only to right-angled triangles
🔴 ancient Indian origin
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TEXTBOOK QUESTIONS
🔒 ❓ 1. Find the diagonal of a square with sidelength 5 cm.
📌 ✅ Answer:
🟢 Step 1: Understand the geometry
⬥ A square has two equal sides meeting at a right angle (90°).
⬥ Let the side length a = 5 cm.
⬥ The diagonal d forms the hypotenuse of a right-angled triangle with sides 5 and 5.
🔵 Step 2: Apply Baudhāyana-Pythagoras Theorem
⬥ Formula: d² = a² + a²
⬥ d² = 5² + 5²
⬥ d² = 25 + 25 = 50
🟡 Step 3: Solve for diagonal
⬥ d = √50
⬥ We can simplify √50 as √(25 × 2) = 5√2.
⬥ Diagonal = 5√2 cm (approx 7.07 cm).
🔒 ❓ 2. Find the missing sidelengths in the following right triangles:
🔒 ❓ (i) Triangle with legs 7 and 9
📌 ✅ Answer:
🟢 Step 1: Identify sides
⬥ Legs: a = 7, b = 9. Hypotenuse c is unknown.
🔵 Step 2: Apply Theorem
⬥ c² = 7² + 9²
⬥ c² = 49 + 81 = 130
🟡 Step 3: Final Calculation
⬥ Hypotenuse = √130
🔒 ❓ (ii) Triangle with leg 4 and hypotenuse 10
📌 ✅ Answer:
🟢 Step 1: Identify sides
⬥ Hypotenuse c = 10, Leg a = 4. Missing leg b.
🔵 Step 2: Apply Theorem
⬥ b² = c² – a²
⬥ b² = 10² – 4²
⬥ b² = 100 – 16 = 84
🟡 Step 3: Final Calculation
⬥ b = √84 = √(4 × 21) = 2√21
⬥ Missing side = 2√21
🔒 ❓ (iii) Triangle with leg 40 and hypotenuse 41
📌 ✅ Answer:
🟢 Step 1: Identify sides
⬥ Hypotenuse c = 41, Leg a = 40. Missing leg b.
🔵 Step 2: Apply Theorem
⬥ b² = 41² – 40²
⬥ Using difference of squares: (41 – 40)(41 + 40) = 1 × 81 = 81.
🟡 Step 3: Final Calculation
⬥ b = √81 = 9
⬥ Missing side = 9
🔒 ❓ (iv) Triangle with leg 10 and hypotenuse √200
📌 ✅ Answer:
🟢 Step 1: Identify sides
⬥ Hypotenuse c = √200, Leg a = 10. Missing leg b.
🔵 Step 2: Apply Theorem
⬥ b² = (√200)² – 10²
⬥ b² = 200 – 100 = 100
🟡 Step 3: Final Calculation
⬥ b = √100 = 10
⬥ Missing side = 10
🔒 ❓ (v) Triangle with legs 10 and √150
📌 ✅ Answer:
🟢 Step 1: Identify sides
⬥ Leg a = 10, Leg b = √150. Hypotenuse c is unknown.
🔵 Step 2: Apply Theorem
⬥ c² = 10² + (√150)²
⬥ c² = 100 + 150 = 250
🟡 Step 3: Final Calculation
⬥ c = √250 = √(25 × 10) = 5√10
⬥ Hypotenuse = 5√10
🔒 ❓ (vi) Triangle with leg 27 and hypotenuse 45
📌 ✅ Answer:
🟢 Step 1: Identify sides
⬥ Hypotenuse c = 45, Leg a = 27. Missing leg b.
🔵 Step 2: Apply Theorem
⬥ b² = 45² – 27²
⬥ Simplify using common factor 9: Sides are 9 × 5 and 9 × 3.
⬥ This is a scaled 3-4-5 triangle (3 × 9, 4 × 9, 5 × 9).
🟡 Step 3: Final Calculation
⬥ Missing side = 4 × 9 = 36.
⬥ Verification: 36² = 1296. 2025 – 729 = 1296. Correct.
⬥ Missing side = 36
🔒 ❓ 3. Find the sidelength of a rhombus whose diagonals are of length 24 units and 70 units.
📌 ✅ Answer:
🟢 Step 1: Properties of a Rhombus
⬥ The diagonals of a rhombus bisect each other at right angles (90°).
⬥ Half of diagonal 1 (d1) = 24 ÷ 2 = 12.
⬥ Half of diagonal 2 (d2) = 70 ÷ 2 = 35.
🔵 Step 2: Form a right triangle
⬥ The side of the rhombus (s) is the hypotenuse of the triangle formed by these half-diagonals.
⬥ s² = 12² + 35²
🟡 Step 3: Calculate
⬥ s² = 144 + 1225
⬥ s² = 1369
🔴 Step 4: Find square root
⬥ √1369 = 37
⬥ Sidelength = 37 units
🔒 ❓ 4. Is the hypotenuse the longest side of a right triangle? Justify your answer.
📌 ✅ Answer:
🟢 Step 1: Direct Answer
⬥ Yes, the hypotenuse is always the longest side.
🔵 Step 2: Justification
⬥ In any triangle, the side opposite the largest angle is the longest.
⬥ In a right-angled triangle, one angle is 90°.
⬥ The sum of all angles is 180°, so the other two angles must sum to 90°. This means each of them is less than 90° (acute).
🟡 Conclusion
⬥ Since 90° is the largest angle, the side opposite to it (the hypotenuse) must be the longest side.
🔒 ❓ 5. True or False — Every Baudhāyana triple is either a primitive triple or a scaled version of a primitive triple.
📌 ✅ Answer:
🟢 Step 1: Definition
⬥ A “Baudhāyana triple” (Pythagorean triple) is a set of three integers (a, b, c) such that a² + b² = c².
⬥ A “Primitive triple” has no common factor among a, b, c.
🔵 Step 2: Analysis
⬥ Any triple (ka, kb, kc) can be reduced by dividing by the greatest common divisor k to get (a, b, c), which is primitive.
⬥ Conversely, any triple is formed by multiplying a primitive triple by a scale factor k.
🟡 Conclusion
⬥ True.
🔒 ❓ 6. Give 5 examples of rectangles whose sidelengths and diagonals are all integers.
📌 ✅ Answer:
🟢 Step 1: Understand the requirement
⬥ A rectangle with sides L and B and diagonal D forms a right triangle where L² + B² = D².
⬥ We need Pythagorean triples (L, B, D).
🔵 Step 2: List 5 common triples
- Sides: 3, 4, Diagonal: 5 (since 3² + 4² = 5²)
- Sides: 5, 12, Diagonal: 13 (since 5² + 12² = 13²)
- Sides: 8, 15, Diagonal: 17 (since 8² + 15² = 17²)
- Sides: 6, 8, Diagonal: 10 (Scaled version of 3-4-5)
- Sides: 20, 21, Diagonal: 29 (since 20² + 21² = 29²)
🔒 ❓ 7. Construct a square whose area is equal to the difference of the areas of squares of sidelengths 5 units and 7 units.
📌 ✅ Answer:
🟢 Step 1: Analyze Area Requirement
⬥ Area of larger square = 7² = 49.
⬥ Area of smaller square = 5² = 25.
⬥ Required Area = 49 – 25 = 24 sq units.
🔵 Step 2: Determine Side Length
⬥ Let the side of the new square be x.
⬥ x² = 24
⬥ x = √24 = √(4 × 6) = 2√6.
🟡 Step 3: Geometric Construction Logic
⬥ We need to construct a line segment of length √24.
⬥ Use the Baudhāyana theorem: x² + 5² = 7².
⬥ This corresponds to a right triangle where the hypotenuse is 7 and one leg is 5. The other leg will be √24.
🔴 Step 4: Construction Steps- Draw a line. Mark a point A.
- Erect a perpendicular at A. Mark point B such that AB = 5 units.
- From B, draw an arc of radius 7 units to cut the base line at C.
- The length AC is the required side (√24).
- Construct a square using side length AC.
🔒 ❓ 8. (i) Using the dots of a grid as the vertices, can you create a square that has an area of (a) 2 sq. units, (b) 3 sq. units, (c) 4 sq. units, and (d) 5 sq. units?
📌 ✅ Answer:
🟢 (a) Area 2 sq. units
⬥ Yes. Construct a square with side length √2.
⬥ Connect points (0,0) to (1,1). This diagonal of a 1 × 1 unit square has length √(1² + 1²) = √2.
⬥ A square built on this tilted line has Area (√2)² = 2.
🔵 (b) Area 3 sq. units
⬥ No. On an integer grid, the area of a square formed by vertices is always the sum of two squares (a² + b²).
⬥ The number 3 cannot be written as the sum of two perfect squares (integers).
🟡 (c) Area 4 sq. units
⬥ Yes. This is a standard 2 × 2 square aligned with the grid lines.
⬥ Side length = 2. Area = 2² = 4.
🔴 (d) Area 5 sq. units
⬥ Yes. We need a side length of √5.
⬥ Notice 1² + 2² = 5.
⬥ Draw a tilted line connecting (0,0) to (2,1). This length is √5.
⬥ Build a square on this side. Area = 5.
🔒 ❓ (ii) Suppose the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner?
📌 ✅ Answer:
🟢 Step 1: General Rule
⬥ Any square on a grid has a side length that is the hypotenuse of a right triangle with integer legs a and b.
⬥ Side squared (Area) = a² + b².
🔵 Step 2: Conclusion
⬥ The possible integer areas are exactly those integers that can be expressed as the sum of two squares of integers (including 0).
⬥ Examples: 1 (1² + 0²), 2 (1² + 1²), 4 (2² + 0²), 5 (2² + 1²), 8 (2² + 2²), 9, 10, 13, etc.
⬥ Impossible integers: 3, 6, 7, 11, etc.
🔒 ❓ 9. Find the area of an equilateral triangle with sidelength 6 units. [Hint: Show that an altitude bisects the opposite side. Use this to find the height.]
📌 ✅ Answer:
🟢 Step 1: Find the Altitude (Height)
⬥ An altitude in an equilateral triangle divides the base into two equal halves.
⬥ Base = 6 units ➔ Half Base = 3 units.
⬥ The altitude (h) forms a right triangle with the Half Base (3) and the Side (6).
🔵 Step 2: Calculate Height
⬥ h² + 3² = 6²
⬥ h² + 9 = 36
⬥ h² = 27
⬥ h = √27 = √(9 × 3) = 3√3
🟡 Step 3: Calculate Area
⬥ Area = ½ × base × height
⬥ Area = ½ × 6 × 3√3
⬥ Area = 3 × 3√3 = 9√3
🔴 Final Answer
⬥ Area = 9√3 sq units (approx 15.59 sq units).
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OTHER IMPORTANT QUESTIONS
Section A — MCQ (Q1–Q10)
🔒 ❓ Q1. In a right triangle, if the legs are 6 cm and 8 cm, the hypotenuse is:
🟢1️⃣ 10 cm
🔵2️⃣ 12 cm
🟡3️⃣ 14 cm
🟣4️⃣ 16 cm
✔️ Answer: 🟢1️⃣
📌 ✅ Answer:
🔹 c² = 6² + 8²
🔹 c² = 36 + 64
🔹 c² = 100
🔹 c = √100 = 10 cm
🔒 ❓ Q2. Which statement is always true for a right triangle?
🟢1️⃣ a² + b² = c² where c is the longest side
🔵2️⃣ a² = b² + c² where a is the longest side
🟡3️⃣ a + b = c always
🟣4️⃣ a² + b² = c² for any triangle
✔️ Answer: 🟢1️⃣
📌 ✅ Answer:
🔹 Baudhāyana–Pythagoras theorem holds only for right triangles.
🔹 The hypotenuse (longest side) is the side opposite the right angle.
🔒 ❓ Q3. A triangle has sides 9, 12, 15. It is:
🟢1️⃣ right-angled
🔵2️⃣ equilateral
🟡3️⃣ isosceles
🟣4️⃣ not a triangle
✔️ Answer: 🟢1️⃣
📌 ✅ Answer:
🔹 Check: 9² + 12² = 81 + 144 = 225
🔹 15² = 225
🔹 Since 9² + 12² = 15², it is right-angled.
🔒 ❓ Q4. If a² + b² > c² for a triangle with largest side c, then the triangle is:
🟢1️⃣ acute-angled
🔵2️⃣ right-angled
🟡3️⃣ obtuse-angled
🟣4️⃣ impossible
✔️ Answer: 🟢1️⃣
📌 ✅ Answer:
🔹 For largest side c:
🔹 If a² + b² > c² ➡️ acute triangle.
🔒 ❓ Q5. Which of these is a Baudhāyana triple?
🟢1️⃣ (4, 5, 6)
🔵2️⃣ (5, 12, 13)
🟡3️⃣ (6, 7, 8)
🟣4️⃣ (7, 8, 9)
✔️ Answer: 🔵2️⃣
📌 ✅ Answer:
🔹 5² + 12² = 25 + 144 = 169
🔹 13² = 169
🔹 So (5, 12, 13) is a triple.
🔒 ❓ Q6. A right triangle has hypotenuse 17 and one leg 8. The other leg is:
🟢1️⃣ 9
🔵2️⃣ 15
🟡3️⃣ √225
🟣4️⃣ 18
✔️ Answer: 🔵2️⃣
📌 ✅ Answer:
🔹 Other leg² = 17² − 8²
🔹 Other leg² = 289 − 64
🔹 Other leg² = 225
🔹 Other leg = √225 = 15
🔒 ❓ Q7. The hypotenuse is the longest side because:
🟢1️⃣ it is opposite the smallest angle
🔵2️⃣ it is opposite the right angle (largest angle)
🟡3️⃣ it touches both legs
🟣4️⃣ it is always horizontal
✔️ Answer: 🔵2️⃣
📌 ✅ Answer:
🔹 The side opposite the largest angle is the longest side.
🔹 In a right triangle, 90° is the largest angle.
🔒 ❓ Q8. If the legs are a and b, the area of the square on the hypotenuse equals:
🟢1️⃣ a + b
🔵2️⃣ a² + b²
🟡3️⃣ (a + b)²
🟣4️⃣ a² − b²
✔️ Answer: 🔵2️⃣
📌 ✅ Answer:
🔹 Area on hypotenuse = c²
🔹 c² = a² + b²
🔒 ❓ Q9. A triangle has sides 7, 24, 25. The right angle is opposite:
🟢1️⃣ 7
🔵2️⃣ 24
🟡3️⃣ 25
🟣4️⃣ cannot decide
✔️ Answer: 🟡3️⃣
📌 ✅ Answer:
🔹 7² + 24² = 49 + 576 = 625
🔹 25² = 625
🔹 Hypotenuse is 25, right angle is opposite 25.
🔒 ❓ Q10. If (3, 4, 5) is a Baudhāyana triple, then (6, 8, 10) is:
🟢1️⃣ not a triple
🔵2️⃣ also a triple
🟡3️⃣ only sometimes a triple
🟣4️⃣ not a triangle
✔️ Answer: 🔵2️⃣
📌 ✅ Answer:
🔹 Scaling a triple by the same factor keeps the relation true.
🔹 (6, 8, 10) is 2×(3, 4, 5).
Section B — Short Answer (Q11–Q20)
🔒 ❓ Q11. Find the diagonal of a square of side 7 cm.
📌 ✅ Answer:
🔹 Diagonal² = 7² + 7²
🔹 Diagonal² = 49 + 49
🔹 Diagonal² = 98
🔹 Diagonal = √98 = √(49×2) = 7√2 cm
🔒 ❓ Q12. A right triangle has legs 9 cm and 40 cm. Find the hypotenuse.
📌 ✅ Answer:
🔹 c² = 9² + 40²
🔹 c² = 81 + 1600
🔹 c² = 1681
🔹 c = √1681 = 41 cm
🔒 ❓ Q13. A right triangle has hypotenuse 26 cm and one leg 10 cm. Find the other leg.
📌 ✅ Answer:
🔹 Other leg² = 26² − 10²
🔹 Other leg² = 676 − 100
🔹 Other leg² = 576
🔹 Other leg = √576 = 24 cm
🔒 ❓ Q14. Find the area of a right triangle with legs 12 cm and 16 cm.
📌 ✅ Answer:
🔹 Area = (1/2) × 12 × 16
🔹 Area = (1/2) × 192
🔹 Area = 96 cm²
🔒 ❓ Q15. A triangle has sides 8, 15, 17. Prove it is right-angled.
📌 ✅ Answer:
🔹 8² + 15² = 64 + 225 = 289
🔹 17² = 289
🔹 Since 8² + 15² = 17², it is a right triangle.
🔒 ❓ Q16. Find the side of a rhombus whose diagonals are 24 cm and 10 cm.
📌 ✅ Answer:
🔹 In a rhombus, diagonals bisect at right angle.
🔹 Half diagonals = 24/2 = 12, and 10/2 = 5
🔹 Side² = 12² + 5²
🔹 Side² = 144 + 25
🔹 Side² = 169
🔹 Side = √169 = 13 cm
🔒 ❓ Q17. A ladder is 13 m long and its foot is 5 m from the wall. How high does it reach?
📌 ✅ Answer:
🔹 Height² = 13² − 5²
🔹 Height² = 169 − 25
🔹 Height² = 144
🔹 Height = √144 = 12 m
🔒 ❓ Q18. A rectangular field is 48 m long and 14 m wide. Find its diagonal.
📌 ✅ Answer:
🔹 Diagonal² = 48² + 14²
🔹 Diagonal² = 2304 + 196
🔹 Diagonal² = 2500
🔹 Diagonal = √2500 = 50 m
🔒 ❓ Q19. True/False with reason: “Every Baudhāyana triple is primitive.”
📌 ✅ Answer:
🔹 False.
🔹 Example: (6, 8, 10) is a triple but not primitive.
🔹 Because gcd(6, 8, 10) = 2 > 1.
🔒 ❓ Q20. Give 3 Baudhāyana triples different from (3, 4, 5).
📌 ✅ Answer:
🔹 (5, 12, 13) because 5² + 12² = 13²
🔹 (8, 15, 17) because 8² + 15² = 17²
🔹 (7, 24, 25) because 7² + 24² = 25²
Section C — Detailed / Reasoning (Q21–Q30)
🔒 ❓ Q21. Find the missing side in a right triangle where the hypotenuse is 29 cm and one leg is 20 cm.
📌 ✅ Answer:
🔹 Let the missing leg be x cm.
🔹 By Baudhāyana–Pythagoras theorem: x² + 20² = 29²
🔹 x² + 400 = 841
🔹 x² = 841 − 400
🔹 x² = 441
🔹 x = √441
🔹 x = 21 cm
🔹 Final: The missing side is 21 cm.
🔒 ❓ Q22. A triangle has sides 11 cm, 60 cm, and 61 cm. Decide whether it is right-angled, and identify the hypotenuse.
📌 ✅ Answer:
🔹 Largest side is 61 cm, so if right-angled, hypotenuse must be 61.
🔹 Check: 11² + 60²
🔹 11² = 121
🔹 60² = 3600
🔹 11² + 60² = 121 + 3600 = 3721
🔹 61² = 3721
🔹 Since 11² + 60² = 61², the triangle is right-angled.
🔹 Final: Right-angled triangle, hypotenuse = 61 cm.
🔒 ❓ Q23. A rectangular park is 75 m by 20 m. A boy walks from one corner to the opposite corner. How much does he walk?
📌 ✅ Answer:
🔹 The path is the diagonal of the rectangle.
🔹 Diagonal² = 75² + 20²
🔹 75² = 5625
🔹 20² = 400
🔹 Diagonal² = 5625 + 400
🔹 Diagonal² = 6025
🔹 Diagonal = √6025
🔹 √6025 = √(25×241)
🔹 √6025 = 5√241 m
🔹 Final: Distance walked = 5√241 m (≈ 77.6 m).
🔒 ❓ Q24. Find the diagonal of a square of side 12 cm and then find the area of the square formed on that diagonal.
📌 ✅ Answer:
🔹 Side of given square = 12 cm.
🔹 Diagonal² = 12² + 12²
🔹 Diagonal² = 144 + 144
🔹 Diagonal² = 288
🔹 Diagonal = √288
🔹 √288 = √(144×2)
🔹 Diagonal = 12√2 cm
🔹 Now the new square has side = 12√2 cm.
🔹 Area of new square = (12√2)²
🔹 Area = 12² × (√2)²
🔹 Area = 144 × 2
🔹 Area = 288 cm²
🔹 Final: Diagonal = 12√2 cm and area of square on diagonal = 288 cm².
🔒 ❓ Q25. In a right triangle, legs are (x + 3) cm and (x − 1) cm, and hypotenuse is (x + 5) cm. Find x and all side lengths.
📌 ✅ Answer:
🔹 Use: (x + 3)² + (x − 1)² = (x + 5)²
🔹 Expand (x + 3)²:
🔹 (x + 3)² = x² + 6x + 9
🔹 Expand (x − 1)²:
🔹 (x − 1)² = x² − 2x + 1
🔹 Add them:
🔹 (x² + 6x + 9) + (x² − 2x + 1) = 2x² + 4x + 10
🔹 Expand RHS:
🔹 (x + 5)² = x² + 10x + 25
🔹 Equation:
🔹 2x² + 4x + 10 = x² + 10x + 25
🔹 2x² − x² + 4x − 10x + 10 − 25 = 0
🔹 x² − 6x − 15 = 0
🔹 Solve by quadratic formula:
🔹 x = [6 ± √(36 + 60)] / 2
🔹 x = [6 ± √96] / 2
🔹 x = [6 ± 4√6] / 2
🔹 x = 3 ± 2√6
🔹 Only positive and meaningful side lengths require x − 1 > 0.
🔹 x = 3 + 2√6 works, x = 3 − 2√6 is negative.
🔹 Final: x = 3 + 2√6
🔹 Sides: x + 3 = 6 + 2√6, x − 1 = 2 + 2√6, hypotenuse x + 5 = 8 + 2√6
🔒 ❓ Q26. A rhombus has diagonals 48 cm and 14 cm. Find its side length.
📌 ✅ Answer:
🔹 Diagonals of a rhombus bisect at right angles.
🔹 Half of 48 = 24 cm
🔹 Half of 14 = 7 cm
🔹 Side² = 24² + 7²
🔹 Side² = 576 + 49
🔹 Side² = 625
🔹 Side = √625 = 25 cm
🔹 Final: Side length = 25 cm.
🔒 ❓ Q27. Construct a square whose area equals the difference of areas of squares of side 9 cm and 5 cm. Find the side of the required square.
📌 ✅ Answer:
🔹 Area of square with side 9 = 9² = 81 cm²
🔹 Area of square with side 5 = 5² = 25 cm²
🔹 Difference of areas = 81 − 25
🔹 Difference of areas = 56 cm²
🔹 Let required square side be s cm.
🔹 Then s² = 56
🔹 s = √56
🔹 √56 = √(4×14)
🔹 s = 2√14 cm
🔹 Final: Side of required square = 2√14 cm.
🔒 ❓ Q28. Using a dot grid, make squares of areas 2, 5, and 10 square units (possible on lattice points). Explain why these areas are possible.
📌 ✅ Answer:
🔹 On a dot grid, a tilted square can have side length √k where k is an integer formed as (horizontal step)² + (vertical step)².
🔹 Then area of that square = (side)² = k.
🔹 Area 2: choose steps (1, 1)
🔹 k = 1² + 1² = 2
🔹 So side = √2 and area = 2
🔹 Area 5: choose steps (1, 2)
🔹 k = 1² + 2² = 5
🔹 So side = √5 and area = 5
🔹 Area 10: choose steps (1, 3)
🔹 k = 1² + 3² = 10
🔹 So side = √10 and area = 10
🔹 Final: These areas are possible because they are sums of two squares of integers.
🔒 ❓ Q29. Find the area of an equilateral triangle of side 10 units by using Baudhāyana–Pythagoras theorem to find its height.
📌 ✅ Answer:
🔹 In an equilateral triangle, the altitude bisects the base.
🔹 Base = 10, so half-base = 5.
🔹 Let height be h.
🔹 Right triangle formed has hypotenuse 10 and one leg 5.
🔹 h² + 5² = 10²
🔹 h² + 25 = 100
🔹 h² = 75
🔹 h = √75 = √(25×3) = 5√3
🔹 Area = (1/2) × base × height
🔹 Area = (1/2) × 10 × 5√3
🔹 Area = 5 × 5√3
🔹 Area = 25√3 square units
🔹 Final: Area = 25√3 square units.
🔒 ❓ Q30. A wire forms the perimeter of a right triangle with legs 9 cm and 12 cm. The wire is cut and reshaped into a square. Find the side of the square.
📌 ✅ Answer:
🔹 First find the hypotenuse.
🔹 c² = 9² + 12²
🔹 c² = 81 + 144
🔹 c² = 225
🔹 c = √225 = 15 cm
🔹 Perimeter of triangle = 9 + 12 + 15
🔹 Perimeter = 36 cm
🔹 Same wire makes a square, so perimeter of square = 36 cm.
🔹 Side of square = 36 ÷ 4
🔹 Side of square = 9 cm
🔹 Final: Side of the square = 9 cm.
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