This MCQ module is based on: Pythagorean Triples & Chapter Exercises
TOPIC 7 OF 23
Pythagorean Triples & Chapter Exercises
🎓 Class 8
Mathematics
CBSE
Theory
Ch 2 — The Pythagoras Theorem
⏱ ~35 min
🌐 Language: [gtranslate]
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2.5 Right-Triangles Having Integer Sidelengths
In his Sulba-Sutra (Verse 1.13), Baudhayana lists a number of integer triples \((a, b, c)\) that form the sidelengths of a right triangle and therefore satisfy \(a^2 + b^2 = c^2\). These include:
| a | b | c | Check: \(a^2+b^2\) | \(c^2\) | Match? |
|---|---|---|---|---|---|
| 3 | 4 | 5 | 9+16=25 | 25 | Yes |
| 5 | 12 | 13 | 25+144=169 | 169 | Yes |
| 8 | 15 | 17 | 64+225=289 | 289 | Yes |
| 7 | 24 | 25 | 49+576=625 | 625 | Yes |
| 12 | 35 | 37 | 144+1225=1369 | 1369 | Yes |
| 15 | 36 | 39 | 225+1296=1521 | 1521 | Yes |
Definition
Such triples \((a, b, c)\) that form the sidelengths of a right triangle (equivalently, satisfy \(a^2 + b^2 = c^2\)) are called Baudhayana triples?. They are also called Baudhayana-Pythagoras triples, Pythagorean triples, and right-angled triangle triples.
Think About It: List down all the Baudhayana triples with numbers less than or equal to 20.
Generating Infinitely Many Triples
Is there an unending sequence of Baudhayana triples? Mathematicians have found a method to generate all such triples. Let us take a few steps in this direction.
We have seen that \((3, 4, 5)\) is a Baudhayana triple.
Think About It: Is \((30, 40, 50)\) a Baudhayana triple? Check: \(30^2 + 40^2 = 900 + 1600 = 2500 = 50^2\). Yes!
Is \((300, 400, 500)\) a Baudhayana triple? Yes, since \(300^2 + 400^2 = 250000 = 500^2\).
The list of Baudhayana triples with numbers less than or equal to 20 contains the following:
(3, 4, 5), (6, 8, 10), (9, 12, 15), (12, 16, 20), (5, 12, 13), (8, 15, 17)
Pattern Observation: Do you see any pattern among them? All these triples can be obtained by multiplying each term of \((3, 4, 5)\) or \((5, 12, 13)\) or \((8, 15, 17)\) by a certain positive integer.
Scaled Triples
Conjecture
If \((a, b, c)\) is a Baudhayana triple, then \((ka, kb, kc)\) is also a Baudhayana triple, where \(k\) is any positive integer.Proof: Since \((a, b, c)\) is a triple, we have \(a^2 + b^2 = c^2\). Now check:
\((ka)^2 + (kb)^2 = k^2a^2 + k^2b^2 = k^2(a^2 + b^2) = k^2 c^2 = (kc)^2\).
So \((ka, kb, kc)\) is indeed a Baudhayana triple.
We call \((ka, kb, kc)\) a scaled version of \((a, b, c)\).
Primitive vs Scaled Triples
A Baudhayana triple that does not have any common factor greater than 1 is called a primitive Baudhayana triple?.For example, \((3, 4, 5)\) is primitive, whereas \((9, 12, 15)\) is not (common factor 3).
Similarly, \((5, 12, 13)\) is primitive, but \((10, 24, 26)\) is not.
Generating Triples Using Odd Number Sums
We know the relation between the sum of consecutive odd numbers and square numbers:
Key Identity
\(1 + 3 + 5 + \cdots + (2n-1) = n^2\)The sum of the first \(n\) odd numbers is \(n^2\).
This means: \((n-1)^2 + (2n-1) = n^2\)
If the \(n\)th odd number, \(2n - 1\), is also a perfect square, then we have a sum of two square numbers equal to another square number -- giving us a Baudhayana triple!
Example: \(n = 5\)
The 5th odd number is \(2 \times 5 - 1 = 9\). Is 9 a perfect square? Yes, \(9 = 3^2\). So we have:
\(1 + 3 + 5 + 7 + 9 = 5^2\)
\(4^2 + 3^2 = 5^2\)
This gives the triple \((3, 4, 5)\).
We could also verify using the equation \((n-1)^2 + (2n-1) = n^2\). Since we took the 5th odd number, \(n = 5\):
\((5-1)^2 + 9 = 5^2\)
\(16 + 9 = 25\) ... \(4^2 + 3^2 = 5^2\)
Example: \(n = 13\)
The 13th odd number is \(2 \times 13 - 1 = 25\). Is 25 a perfect square? Yes, \(25 = 5^2\). So:
\(1 + 3 + 5 + \cdots + 23 + 25 = 13^2\)
\(12^2 + 5^2 = 13^2\)
This gives the triple \((5, 12, 13)\).
Substituting \(n = 13\) in the equation: \((13-1)^2 + 25 = 13^2\), i.e., \(144 + 25 = 169\). Confirmed!
Figure it Out (Section 2.5)
Q1. Find 5 more Baudhayana triples using the odd-number-sum method.
We look for values of \(n\) where \(2n - 1\) is a perfect square:
\(n = 5\): \(2n-1 = 9 = 3^2\). Triple: \((3, 4, 5)\).
\(n = 13\): \(2n-1 = 25 = 5^2\). Triple: \((5, 12, 13)\).
\(n = 25\): \(2n-1 = 49 = 7^2\). \((24)^2 + 7^2 = 576 + 49 = 625 = 25^2\). Triple: \((7, 24, 25)\).
\(n = 41\): \(2n-1 = 81 = 9^2\). \(40^2 + 9^2 = 1600 + 81 = 1681 = 41^2\). Triple: \((9, 40, 41)\).
\(n = 61\): \(2n-1 = 121 = 11^2\). \(60^2 + 11^2 = 3600 + 121 = 3721 = 61^2\). Triple: \((11, 60, 61)\).
\(n = 85\): \(2n-1 = 169 = 13^2\). \(84^2 + 13^2 = 7056 + 169 = 7225 = 85^2\). Triple: \((13, 84, 85)\).
\(n = 113\): \(2n-1 = 225 = 15^2\). \(112^2 + 15^2 = 12544 + 225 = 12769 = 113^2\). Triple: \((15, 112, 113)\).
\(n = 5\): \(2n-1 = 9 = 3^2\). Triple: \((3, 4, 5)\).
\(n = 13\): \(2n-1 = 25 = 5^2\). Triple: \((5, 12, 13)\).
\(n = 25\): \(2n-1 = 49 = 7^2\). \((24)^2 + 7^2 = 576 + 49 = 625 = 25^2\). Triple: \((7, 24, 25)\).
\(n = 41\): \(2n-1 = 81 = 9^2\). \(40^2 + 9^2 = 1600 + 81 = 1681 = 41^2\). Triple: \((9, 40, 41)\).
\(n = 61\): \(2n-1 = 121 = 11^2\). \(60^2 + 11^2 = 3600 + 121 = 3721 = 61^2\). Triple: \((11, 60, 61)\).
\(n = 85\): \(2n-1 = 169 = 13^2\). \(84^2 + 13^2 = 7056 + 169 = 7225 = 85^2\). Triple: \((13, 84, 85)\).
\(n = 113\): \(2n-1 = 225 = 15^2\). \(112^2 + 15^2 = 12544 + 225 = 12769 = 113^2\). Triple: \((15, 112, 113)\).
Q2. Does this method yield non-primitive Baudhayana triples? [Hint: Observe that among the triples generated, one of the smaller sidelengths is one less than the hypotenuse.]
Solution: In every triple generated by this method, the larger leg is exactly \(n - 1\) and the hypotenuse is \(n\), so the larger leg is always one less than the hypotenuse. All triples from this method have the form \((m, \frac{m^2-1}{2}, \frac{m^2+1}{2})\) where \(m\) is odd. These are always primitive when \(m\) is an odd prime. For composite odd \(m\), some may be non-primitive. For instance, if \(m = 9\): triple is \((9, 40, 41)\), which IS primitive. But for \(m = 15\): triple is \((15, 112, 113)\), also primitive. This particular method always produces primitive triples.
Q3. Are there primitive triples that cannot be obtained through this method? If yes, give examples.
Solution: Yes. The triple \((8, 15, 17)\) is primitive (no common factor), but in this triple the difference between the hypotenuse and the larger leg is \(17 - 15 = 2\), not 1. Similarly, \((20, 21, 29)\) is primitive with difference \(29 - 21 = 8\). The odd-number-sum method only generates triples where the hypotenuse exceeds the larger leg by exactly 1. So triples like \((8, 15, 17)\) cannot be obtained by this method.
2.6 A Long-Standing Open Problem
The study of Baudhayana triples inspired the great French mathematician Fermat? -- who lived during the 17th century -- to make a general statement about the sum of powers of positive integers.
We have seen that there are an infinite number of square numbers that can be written as a sum of two square numbers. This made Fermat wonder if there is a perfect cube that can be written as a sum of two perfect cubes, a fourth power that can be written as a sum of two fourth powers, and so on.
In other words, he wondered if there is a solution to the equation
\(x^n + y^n = z^n\)
where \(x\), \(y\), and \(z\) are natural numbers, and \(n > 2\).
Historical Note
In the margin of a book that dealt with properties and patterns of positive integers (like Baudhayana triples), Fermat wrote that one cannot find a single perfect cube that is a sum of two perfect cubes, a fourth power that is a sum of two fourth powers, and so on. He claimed the equation has no solution for powers greater than 2.
Fermat's Last Theorem
The equation \(x^n + y^n = z^n\) has no solution in positive integers when \(n > 2\).No one could ever find Fermat's proof of this statement, which became known as Fermat's Last Theorem.
After Fermat's death, many great mathematicians tried their hand at proving this theorem. There followed more than 300 years of failed attempts.
In 1963, a 10-year-old boy named Andrew Wiles read a book about Fermat's Last Theorem and its history. Despite reading about the failures of so many great mathematicians, this young boy resolved to prove this theorem. He eventually did prove it in 1994!
2.7 Further Applications of the Baudhayana-Pythagoras Theorem
A Problem from Bhaskaracharya's Lilavati
Here is a translation of a classic problem from Bhaskaracharya's (Bhaskara II) Lilavati:
Lilavati Problem
"In a lake surrounded by chakra and krauncha birds, there is a lotus flower peeping out of the water, with the tip of its stem 1 unit above the water. On being swayed by a gentle breeze, the tip touches the water 3 units away from its original position. Quickly tell the depth of the lake."
The lotus stem problem: stem length = depth + 1 unit. When swayed, it forms a right triangle.
Solution
Let \(x\) be the depth of the lake. The total length of the stem is \(x + 1\).When the lotus is swayed, the stem makes a right triangle with:
- Vertical side = \(x\) (depth)
- Horizontal side = 3 (distance from original position)
- Hypotenuse = \(x + 1\) (total stem length)
By Baudhayana's Theorem:
\(3^2 + x^2 = (x+1)^2\)
\(9 + x^2 = x^2 + 2x + 1\)
\(9 = 2x + 1\)
\(x = 4\)
The depth of the lake is 4 units.
Figure it Out (Section 2.7 -- Final Exercises)
Q1. Find the diagonal of a square with sidelength 5 cm.
The diagonal of a square with side \(s\) is \(s\sqrt{2}\).
Diagonal = \(5\sqrt{2} = 5 \times 1.414... \approx 7.07\) cm.
Diagonal = \(5\sqrt{2} = 5 \times 1.414... \approx 7.07\) cm.
Q2. Find the missing sidelengths in the following right triangles:
Triangle (7, 9, ?): \(c^2 = 7^2 + 9^2 = 49 + 81 = 130\). \(c = \sqrt{130} \approx 11.4\).
Triangle (4, 10, ?): \(c^2 = 4^2 + 10^2 = 16 + 100 = 116\). \(c = \sqrt{116} \approx 10.8\).
Triangle (40, ?, 41): \(b^2 = 41^2 - 40^2 = 1681 - 1600 = 81\). \(b = 9\).
Triangle (?, ?, \(\sqrt{200}\)) [isosceles right triangle]: \(2a^2 = 200\), so \(a^2 = 100\), \(a = 10\). Both legs are 10.
Triangle (10, \(\sqrt{150}\), ?): \(c^2 = 100 + 150 = 250\). \(c = \sqrt{250} = 5\sqrt{10} \approx 15.8\).
Triangle (27, 45, ?): \(c^2 = 729 + 2025 = 2754\). \(c = \sqrt{2754} \approx 52.5\).
Alternatively: \(27 = 9 \times 3\), \(45 = 9 \times 5\). This is a scaled \((3, 5, ?)\) triangle with \(k=9\). Hypotenuse = \(9\sqrt{34} \approx 52.5\).
Triangle (4, 10, ?): \(c^2 = 4^2 + 10^2 = 16 + 100 = 116\). \(c = \sqrt{116} \approx 10.8\).
Triangle (40, ?, 41): \(b^2 = 41^2 - 40^2 = 1681 - 1600 = 81\). \(b = 9\).
Triangle (?, ?, \(\sqrt{200}\)) [isosceles right triangle]: \(2a^2 = 200\), so \(a^2 = 100\), \(a = 10\). Both legs are 10.
Triangle (10, \(\sqrt{150}\), ?): \(c^2 = 100 + 150 = 250\). \(c = \sqrt{250} = 5\sqrt{10} \approx 15.8\).
Triangle (27, 45, ?): \(c^2 = 729 + 2025 = 2754\). \(c = \sqrt{2754} \approx 52.5\).
Alternatively: \(27 = 9 \times 3\), \(45 = 9 \times 5\). This is a scaled \((3, 5, ?)\) triangle with \(k=9\). Hypotenuse = \(9\sqrt{34} \approx 52.5\).
Q3. Find the sidelength of a rhombus whose diagonals are of length 24 units and 70 units.
Solution: The diagonals of a rhombus bisect each other at right angles. Half-diagonals: \(\frac{24}{2} = 12\) and \(\frac{70}{2} = 35\).
Each side forms a right triangle with these half-diagonals as legs:
\(s^2 = 12^2 + 35^2 = 144 + 1225 = 1369\)
\(s = \sqrt{1369} = 37\) units.
The sidelength of the rhombus is 37 units.
Each side forms a right triangle with these half-diagonals as legs:
\(s^2 = 12^2 + 35^2 = 144 + 1225 = 1369\)
\(s = \sqrt{1369} = 37\) units.
The sidelength of the rhombus is 37 units.
Q4. Is the hypotenuse the longest side of a right triangle? Justify your answer.
Solution: Yes! Since \(c^2 = a^2 + b^2\), and \(a^2 > 0\) and \(b^2 > 0\), we have \(c^2 > a^2\) and \(c^2 > b^2\). Since all lengths are positive, \(c > a\) and \(c > b\). The hypotenuse is always the longest side.
Q5. True or False -- Every Baudhayana triple is either a primitive triple or a scaled version of a primitive triple.
True. Every Baudhayana triple \((a, b, c)\) can be divided by the GCD of its three terms. If \(d = \gcd(a, b, c)\), then \((\frac{a}{d}, \frac{b}{d}, \frac{c}{d})\) is a primitive triple. So \((a, b, c) = d \times (\frac{a}{d}, \frac{b}{d}, \frac{c}{d})\), which is a scaled version of a primitive triple (or the primitive triple itself if \(d = 1\)).
Q6. Give 5 examples of rectangles whose sidelengths and diagonals are all integers.
We need Baudhayana triples \((a, b, c)\) where \(a\) and \(b\) are the sides and \(c\) is the diagonal:
1. Rectangle \(3 \times 4\), diagonal = 5
2. Rectangle \(5 \times 12\), diagonal = 13
3. Rectangle \(8 \times 15\), diagonal = 17
4. Rectangle \(7 \times 24\), diagonal = 25
5. Rectangle \(20 \times 21\), diagonal = 29
1. Rectangle \(3 \times 4\), diagonal = 5
2. Rectangle \(5 \times 12\), diagonal = 13
3. Rectangle \(8 \times 15\), diagonal = 17
4. Rectangle \(7 \times 24\), diagonal = 25
5. Rectangle \(20 \times 21\), diagonal = 29
Q7. Construct a square whose area is equal to the difference of the areas of squares of sidelengths 5 units and 7 units.
Solution: We need a square with area \(= 7^2 - 5^2 = 49 - 25 = 24\).
Side of this square = \(\sqrt{24} = 2\sqrt{6} \approx 4.9\) units.
Construction method: Since \(7^2 = 5^2 + (\sqrt{24})^2\), draw a right triangle with hypotenuse 7 and one leg 5. The other leg will be \(\sqrt{24}\). Build a square on this leg.
Side of this square = \(\sqrt{24} = 2\sqrt{6} \approx 4.9\) units.
Construction method: Since \(7^2 = 5^2 + (\sqrt{24})^2\), draw a right triangle with hypotenuse 7 and one leg 5. The other leg will be \(\sqrt{24}\). Build a square on this leg.
Q8. (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?
(ii) Suppose the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner?
(ii) Suppose the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner?
(i)
(a) Area 2: Tilt a unit square by 45 degrees. Vertices at grid points (0,1), (1,0), (0,-1), (-1,0). Side = \(\sqrt{2}\), area = 2. YES.
(b) Area 3: We need vertices at grid points forming a square with area 3. No -- on a square grid, areas of grid-vertex squares are always of the form \(a^2 + b^2\) where \(a, b\) are integers. Since 3 cannot be written as a sum of two perfect squares (check: \(0+3\) fails, \(1+2\) fails), area 3 is NOT possible.
(c) Area 4: A \(2 \times 2\) aligned square. YES.
(d) Area 5: Side vector (1, 2). Vertices at (0,0), (1,2), (3,1), (2,-1). Area = \(1^2 + 2^2 = 5\). YES.
(ii) The possible integer-valued areas are all numbers that can be expressed as a sum of two squares of integers: 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, ... These are numbers of the form \(a^2 + b^2\).
(a) Area 2: Tilt a unit square by 45 degrees. Vertices at grid points (0,1), (1,0), (0,-1), (-1,0). Side = \(\sqrt{2}\), area = 2. YES.
(b) Area 3: We need vertices at grid points forming a square with area 3. No -- on a square grid, areas of grid-vertex squares are always of the form \(a^2 + b^2\) where \(a, b\) are integers. Since 3 cannot be written as a sum of two perfect squares (check: \(0+3\) fails, \(1+2\) fails), area 3 is NOT possible.
(c) Area 4: A \(2 \times 2\) aligned square. YES.
(d) Area 5: Side vector (1, 2). Vertices at (0,0), (1,2), (3,1), (2,-1). Area = \(1^2 + 2^2 = 5\). YES.
(ii) The possible integer-valued areas are all numbers that can be expressed as a sum of two squares of integers: 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, ... These are numbers of the form \(a^2 + b^2\).
Q9. 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.]
Solution: In an equilateral triangle with side 6, the altitude bisects the base.
Half-base = 3, hypotenuse = 6.
By Baudhayana's Theorem: \(h^2 + 3^2 = 6^2\)
\(h^2 = 36 - 9 = 27\)
\(h = \sqrt{27} = 3\sqrt{3}\)
Area = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 3\sqrt{3} = 9\sqrt{3}\)
Since \(\sqrt{3} \approx 1.732\): Area \(\approx 15.59\) sq. units.
Half-base = 3, hypotenuse = 6.
By Baudhayana's Theorem: \(h^2 + 3^2 = 6^2\)
\(h^2 = 36 - 9 = 27\)
\(h = \sqrt{27} = 3\sqrt{3}\)
Area = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 3\sqrt{3} = 9\sqrt{3}\)
Since \(\sqrt{3} \approx 1.732\): Area \(\approx 15.59\) sq. units.
Interactive: Baudhayana Triple Finder
Enter two integers and check whether they, along with a third, form a Baudhayana triple.
Enter two positive integers and click Check.
Activity: Discovering Triples with the Odd-Number Method
Materials: Notebook, calculator (optional).
Predict: Using the formula \((n-1)^2 + (2n-1) = n^2\), for which values of \(n\) (between 1 and 100) does \(2n - 1\) turn out to be a perfect square?
- List all perfect squares that are odd and less than 200: 1, 9, 25, 49, 81, 121, 169.
- For each, find \(n\): if \(2n - 1 = m^2\), then \(n = \frac{m^2 + 1}{2}\).
- For \(m = 3\): \(n = 5\), triple = \((3, 4, 5)\). For \(m = 5\): \(n = 13\), triple = \((5, 12, 13)\). Continue...
- Verify each triple: check \(a^2 + b^2 = c^2\).
- Try to find a primitive triple NOT generated by this method.
Observe: Every triple generated has the form where the hypotenuse is exactly 1 more than the larger leg. The triple \((8, 15, 17)\) has hypotenuse 2 more than the larger leg, so it cannot come from this method.
Explain: The formula \((n-1)^2 + (2n-1) = n^2\) always gives triples where \(c - b = 1\) (since \(c = n\) and the larger leg \(= n - 1\)). Triples with \(c - b > 1\) require different generation methods.
Summary
- The Baudhayana-Pythagoras Theorem is one of the most fundamental theorems in geometry. It expresses the relationship among the three sides of a right-angled triangle.
- If \(a\), \(b\), \(c\) are the sidelengths of a right-angled triangle, where \(c\) is the length of hypotenuse, then \(a^2 + b^2 = c^2\).
- In an isosceles right triangle with sidelengths \(a\), \(a\), \(c\), we have \(a^2 + a^2 = 2a^2 = c^2\), i.e., \(c = a\sqrt{2}\).
- The number \(\sqrt{2}\) lies between 1.414 and 1.415. However, it cannot be expressed as a terminating decimal. It also cannot be expressed as a fraction \(\frac{m}{n}\) with \(m\), \(n\) positive integers.
- A triple \((a, b, c)\) of positive integers satisfying \(a^2 + b^2 = c^2\) is called a Baudhayana triple. Examples include \((3, 4, 5)\), \((5, 12, 13)\), and \((8, 15, 17)\). Infinitely many such triples can be constructed.
- The equation \(a^n + b^n = c^n\) has no solution in positive integers when \(n > 2\). This is known as Fermat's Last Theorem. It was proven by Andrew Wiles in 1994.
Puzzle Time: Find the Colours!
A classic logic puzzle to test your reasoning skills.
There are 3 closed boxes -- one containing only red balls, the second containing only blue balls, and the third containing only green balls. The boxes are labelled RED, BLUE, and GREEN such that no box has the correct label. We need to find which label goes with which box. How can this be done if we are allowed to open only one box?
Solution: Since ALL labels are wrong:
Step 1: Open the box labelled "GREEN". It does NOT contain green balls. Suppose it contains RED balls.
Step 2: Now the box labelled "GREEN" actually has red balls. The box labelled "RED" cannot have red balls (wrong label) and cannot have green balls (because the box labelled "BLUE" must have the remaining colour). So the box labelled "RED" has BLUE balls.
Step 3: That leaves green balls for the box labelled "BLUE".
The same logic works regardless of what we find in the first box we open. Opening just one box is sufficient! The key insight is that since ALL labels are wrong, finding the contents of one box determines the contents of the other two.
Step 1: Open the box labelled "GREEN". It does NOT contain green balls. Suppose it contains RED balls.
Step 2: Now the box labelled "GREEN" actually has red balls. The box labelled "RED" cannot have red balls (wrong label) and cannot have green balls (because the box labelled "BLUE" must have the remaining colour). So the box labelled "RED" has BLUE balls.
Step 3: That leaves green balls for the box labelled "BLUE".
The same logic works regardless of what we find in the first box we open. Opening just one box is sufficient! The key insight is that since ALL labels are wrong, finding the contents of one box determines the contents of the other two.
Competency-Based Questions
A group of students is studying Baudhayana triples for a mathematics project. They are trying to classify triples, generate new ones, and apply the theorem to real-world problems.
Q1. Which of the following is a Baudhayana triple?
L3 ApplyAnswer: (b) (6, 8, 10)
Check: \(6^2 + 8^2 = 36 + 64 = 100 = 10^2\). This is a scaled version of (3, 4, 5) with \(k=2\). The others fail: \(16+25 \neq 36\); \(25+36 \neq 49\); \(4+9 \neq 16\).
Check: \(6^2 + 8^2 = 36 + 64 = 100 = 10^2\). This is a scaled version of (3, 4, 5) with \(k=2\). The others fail: \(16+25 \neq 36\); \(25+36 \neq 49\); \(4+9 \neq 16\).
Q2. A ladder 13 m long leans against a wall with its foot 5 m from the base. How high up the wall does it reach? Analyse what happens to this height if the foot is moved 1 m further from the wall.
L4 AnalyseAnswer: Original position: \(h^2 + 5^2 = 13^2\), so \(h^2 = 169 - 25 = 144\), \(h = 12\) m.
New position (foot at 6 m): \(h'^2 + 6^2 = 13^2\), so \(h'^2 = 169 - 36 = 133\), \(h' = \sqrt{133} \approx 11.53\) m.
The ladder drops by about \(12 - 11.53 = 0.47\) m. Moving the foot 1 m outward does NOT cause the height to drop by 1 m -- the relationship is non-linear because of the squares.
New position (foot at 6 m): \(h'^2 + 6^2 = 13^2\), so \(h'^2 = 169 - 36 = 133\), \(h' = \sqrt{133} \approx 11.53\) m.
The ladder drops by about \(12 - 11.53 = 0.47\) m. Moving the foot 1 m outward does NOT cause the height to drop by 1 m -- the relationship is non-linear because of the squares.
Q3. A student claims that (9, 12, 15) is a primitive Baudhayana triple. Evaluate this claim.
L5 EvaluateAnswer: The claim is incorrect. While \(9^2 + 12^2 = 81 + 144 = 225 = 15^2\) confirms it IS a Baudhayana triple, it is NOT primitive. The GCD of 9, 12, and 15 is 3. Dividing by 3 gives \((3, 4, 5)\), which is the primitive triple. So \((9, 12, 15) = 3 \times (3, 4, 5)\) is a scaled version.
Q4. Create a general formula for finding the area of an equilateral triangle with side \(s\), using Baudhayana's Theorem to determine the height. Show all steps.
L6 CreateAnswer: An altitude of an equilateral triangle bisects the base (by symmetry/congruent triangles).
Half-base = \(\frac{s}{2}\), hypotenuse = \(s\).
By Baudhayana's Theorem: \(h^2 + \left(\frac{s}{2}\right)^2 = s^2\)
\(h^2 = s^2 - \frac{s^2}{4} = \frac{3s^2}{4}\)
\(h = \frac{s\sqrt{3}}{2}\)
Area = \(\frac{1}{2} \times s \times \frac{s\sqrt{3}}{2} = \frac{s^2\sqrt{3}}{4}\)
General formula: Area of equilateral triangle with side \(s\) = \(\frac{s^2\sqrt{3}}{4}\).
Half-base = \(\frac{s}{2}\), hypotenuse = \(s\).
By Baudhayana's Theorem: \(h^2 + \left(\frac{s}{2}\right)^2 = s^2\)
\(h^2 = s^2 - \frac{s^2}{4} = \frac{3s^2}{4}\)
\(h = \frac{s\sqrt{3}}{2}\)
Area = \(\frac{1}{2} \times s \times \frac{s\sqrt{3}}{2} = \frac{s^2\sqrt{3}}{4}\)
General formula: Area of equilateral triangle with side \(s\) = \(\frac{s^2\sqrt{3}}{4}\).
Assertion–Reason Questions
Assertion (A): \((6, 8, 10)\) is a Baudhayana triple but not a primitive one.
Reason (R): A primitive Baudhayana triple has no common factor greater than 1 among all three numbers.
Reason (R): A primitive Baudhayana triple has no common factor greater than 1 among all three numbers.
Answer: (a) — Both true. \(6^2+8^2=100=10^2\) so it IS a triple. GCD(6,8,10)=2, so it is not primitive. R defines what primitive means and directly explains why A is true.
Assertion (A): Fermat's Last Theorem states that \(x^3 + y^3 = z^3\) has no solution in positive integers.
Reason (R): Fermat's Last Theorem applies to all integer exponents \(n > 2\), not just \(n = 3\).
Reason (R): Fermat's Last Theorem applies to all integer exponents \(n > 2\), not just \(n = 3\).
Answer: (a) — Both true. A is a specific case (\(n=3\)) of the general theorem (\(n>2\)). R provides the full statement which includes \(n=3\) as a special case, thus explaining A.
Frequently Asked Questions — Chapter 2
What is Pythagorean Triples & Chapter Exercises in NCERT Class 8 Mathematics?
Pythagorean Triples & Chapter Exercises is a key concept covered in NCERT Class 8 Mathematics, Chapter 2: Chapter 2. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Pythagorean Triples & Chapter Exercises step by step?
To solve problems on Pythagorean Triples & Chapter Exercises, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 8 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 2: Chapter 2?
The essential formulas of Chapter 2 (Chapter 2) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Pythagorean Triples & Chapter Exercises important for the Class 8 board exam?
Pythagorean Triples & Chapter Exercises is part of the NCERT Class 8 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Pythagorean Triples & Chapter Exercises?
Common mistakes in Pythagorean Triples & Chapter Exercises include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Pythagorean Triples & Chapter Exercises?
End-of-chapter NCERT exercises for Pythagorean Triples & Chapter Exercises cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 2, and solve at least one previous-year board paper to consolidate your understanding.
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