This MCQ module is based on: Pythagorean Theorem – Statement & Proof
TOPIC 6 OF 23
Pythagorean Theorem – Statement & Proof
🎓 Class 8
Mathematics
CBSE
Theory
Ch 2 — The Pythagoras Theorem
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Pythagorean Theorem – Statement & Proof
Targeting Class 8 level in General Mathematics, with Basic difficulty.
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2.4 Combining Two Different Squares
In the previous sections, we combined two copies of the same square to make a larger square whose area is the sum of the two. The sidelength of the larger square is the length of the diagonal of either smaller square.
Combining two identical squares: the diagonal of one square becomes the side of the combined square.
But what if we wish to combine two squares of different sizes to make a large square whose area is the sum of the two smaller squares?
Can we combine two different-sized squares into one larger square?
In his Sulba-Sutra (Verse 1.12), Baudhayana provides a truly remarkable solution to this more general problem. He writes:
Baudhayana's Rule (Verse 1.12)
The area of the square produced by the diagonal is the sum of the areas of the squares produced by the two sides.
That is, to combine two different squares: make a right-angled triangle? whose perpendicular sides are the sidelengths of the two squares. The square whose sidelength is the hypotenuse? of this right-angled triangle has an area that is the sum of the areas of the original two squares.
A right triangle with perpendicular sides \(a\) and \(b\): the square on the hypotenuse \(c\) has area equal to the sum of the squares on \(a\) and \(b\).
Baudhayana's Construction Method
Subsequently in his Sulba-Sutra (Verse 2.1), Baudhayana explains how the method works in general:
Baudhayana's Construction (Verse 2.1)
To combine different squares, mark a rectangular portion of the larger square using a side of the smaller square. The diagonal of this rectangle is the side of a square that has area equal to the sum of the areas of the smaller squares.
Let us follow Baudhayana's instructions step by step.
Step 1: Join the Two Squares
Place the two squares side by side, sharing an edge. The larger square has side \(a\) and the smaller has side \(b\).
Step 1: Join the two squares side by side along a common edge.
Step 2: Mark a Rectangle and Draw the Diagonal
Mark a rectangular portion of the larger square using a side of the smaller one. Draw its diagonal. This gives us a right triangle with perpendicular sides \(a\) and \(b\).
Step 2: The diagonal of the rectangle gives the hypotenuse of a right triangle with sides \(a\) and \(b\).
Step 3: Build Four Congruent Right Triangles
Make a 4-sided figure over the hypotenuse by drawing three more copies of this right triangle. We label the triangles T, U, V, W and the regions X.
Steps 3-4: Four congruent right triangles (T, U, V, W) arranged to form a square on the hypotenuse.
Step 4: Verify the Inner Figure is a Square
Since T, U, W, and X are all congruent, the sides of the 4-sided figure all have the same length. To verify it is a square, we check the angles:
Each angle of the inner figure is \(x + (90 - x) = 90^\circ\). So the inner figure is indeed a square.
At each vertex of the inner figure, two acute angles from adjacent right triangles meet. One angle is \(x\) and the other is \(90^\circ - x\). Their sum is \(x + (90^\circ - x) = 90^\circ\). So all four angles of the inner figure are right angles, confirming it is a square.
The Area Proof
Now Baudhayana's assertion becomes clear:
- The area of the square on the hypotenuse = sum of areas of T, U, V, and W
- = sum of the areas of the two given squares
The Fundamental Result
The area of the square on the hypotenuse = the sum of the areas of the squares on \(a\) and \(b\).\[\boxed{a^2 + b^2 = c^2}\]
Combining Two Squares Using Paper
Activity: Combining Two Different Squares
Materials: Two different-sized squares of coloured paper (one red, one yellow), scissors, pencil, ruler.
Predict: Can you cut these two squares into just 3 pieces total and rearrange them to form one larger square?
- Cut out two different-sized squares and join them side by side along one edge.
- Make two cuts to create three pieces: one red square, one yellow triangle, and one combined trapezium.
- Rearrange the three pieces into a larger square.
- Now make a right triangle using the two smaller squares. Draw a square on the hypotenuse.
- Verify that the pieces from both smaller squares cover the hypotenuse square exactly.
Observe: The three pieces rearrange perfectly because the cutting line is precisely the diagonal of the rectangle formed by combining the two squares. This diagonal becomes the side of the new (larger) square.
Explain: This physically demonstrates that \(a^2 + b^2 = c^2\), where \(a\) and \(b\) are the sides of the two original squares and \(c\) is the hypotenuse of the right triangle formed by sides \(a\) and \(b\).
Paper demonstration: two coloured squares cut and rearranged into one larger square on the hypotenuse.
The areas of the squares on the two shorter sides (red + yellow) equal the area of the square on the hypotenuse (green).
2.5 Baudhayana's Theorem on Right-Angled Triangles
Baudhayana-Pythagoras Theorem
If a right-angled triangle has sidelengths \(a\), \(b\), and \(c\), where \(c\) is the length of the hypotenuse, then
\[a^2 + b^2 = c^2\]
Baudhayana was the first person in history to state this theorem in this generality and essentially modern form. The theorem is also known as the Pythagorean Theorem?, after the Greek philosopher-mathematician Pythagoras (c. 500 BCE) who also admired and studied this result, and lived a couple of hundred years after Baudhayana. It is also often called the 'Baudhayana-Pythagoras Theorem' as a transitional name.
Using Baudhayana's Theorem
Make a right-angled triangle in your notebook whose shorter sidelengths are 3 cm and 4 cm. Now measure the length of the hypotenuse. It should read about 5 cm.
A right triangle with sides 3 cm and 4 cm. By Baudhayana's theorem: \(3^2 + 4^2 = 9 + 16 = 25 = 5^2\). So hypotenuse = 5 cm.
Verification
Let \(a = 3\) and \(b = 4\). By Baudhayana's Theorem:\(a^2 + b^2 = c^2\)
\(3^2 + 4^2 = c^2\)
\(9 + 16 = c^2\)
\(25 = c^2\)
So, \(c = 5\) cm.
Interactive: Baudhayana-Pythagoras Verifier
Enter any two sides of a right triangle and find the third. Toggle which side is unknown.
Enter values and click Calculate.
Figure it Out (Section 2.4 – 2.5)
Q1. If a right-angled triangle has shorter sides of lengths 5 cm and 12 cm, then what is the length of its hypotenuse? First draw the right-angled triangle with these sidelengths and measure the hypotenuse, then check your answer using Baudhayana's Theorem.
Solution:
\(a = 5, \; b = 12\)
\(c^2 = a^2 + b^2 = 25 + 144 = 169\)
\(c = \sqrt{169} = 13\) cm.
The hypotenuse is 13 cm.
\(a = 5, \; b = 12\)
\(c^2 = a^2 + b^2 = 25 + 144 = 169\)
\(c = \sqrt{169} = 13\) cm.
The hypotenuse is 13 cm.
Q2. If a right-angled triangle has a short side of length 8 cm and hypotenuse of length 17 cm, what is the length of the third side? Again, try drawing the triangle and measuring, and then check your answer using Baudhayana's Theorem.
Solution:
\(a = 8, \; c = 17\)
\(b^2 = c^2 - a^2 = 289 - 64 = 225\)
\(b = \sqrt{225} = 15\) cm.
The third side is 15 cm.
\(a = 8, \; c = 17\)
\(b^2 = c^2 - a^2 = 289 - 64 = 225\)
\(b = \sqrt{225} = 15\) cm.
The third side is 15 cm.
Q3. Using the constructions you have now seen, how would you construct a square whose area is triple the area of a given square? Five times the area of a given square? (Baudhayana's Sulba-Sutra, Verse 1.10)
Solution (Triple): Given a square of side \(s\), first double it using the diagonal to get side \(s\sqrt{2}\). Then combine this new square (area \(2s^2\)) with the original (area \(s^2\)) using a right triangle with legs \(s\sqrt{2}\) and \(s\). The hypotenuse gives a square of area \(2s^2 + s^2 = 3s^2\).
Solution (Five times): Make a right triangle with legs \(s\) and \(2s\). By Baudhayana's Theorem: \(s^2 + (2s)^2 = s^2 + 4s^2 = 5s^2\). The hypotenuse gives a square of area \(5s^2\).
Solution (Five times): Make a right triangle with legs \(s\) and \(2s\). By Baudhayana's Theorem: \(s^2 + (2s)^2 = s^2 + 4s^2 = 5s^2\). The hypotenuse gives a square of area \(5s^2\).
Q4. Let \(a\), \(b\) and \(c\) denote the length of the sides of a right triangle, with \(c\) being the length of the hypotenuse. Find the missing sidelength in each of the following cases:
(i) \(a = 5, b = 7\) (ii) \(a = 8, b = 12\) (iii) \(a = 9, c = 15\) (iv) \(a = 7, b = 12\) (v) \(a = 1.5, b = 3.5\)
(i) \(a = 5, b = 7\) (ii) \(a = 8, b = 12\) (iii) \(a = 9, c = 15\) (iv) \(a = 7, b = 12\) (v) \(a = 1.5, b = 3.5\)
(i) \(c^2 = 5^2 + 7^2 = 25 + 49 = 74\). \(c = \sqrt{74}\). Since \(8^2 = 64\) and \(9^2 = 81\), \(c\) is between 8 and 9. More precisely \(8.6^2 = 73.96\), so \(c \approx 8.6\).
(ii) \(c^2 = 8^2 + 12^2 = 64 + 144 = 208\). \(c = \sqrt{208}\). Since \(14^2 = 196\) and \(15^2 = 225\), \(c\) is between 14 and 15. More precisely \(c \approx 14.4\).
(iii) \(b^2 = 15^2 - 9^2 = 225 - 81 = 144\). \(b = 12\).
(iv) \(c^2 = 7^2 + 12^2 = 49 + 144 = 193\). \(c = \sqrt{193}\). Since \(13^2 = 169\) and \(14^2 = 196\), \(c\) is between 13 and 14. More precisely \(c \approx 13.9\).
(v) \(c^2 = 1.5^2 + 3.5^2 = 2.25 + 12.25 = 14.5\). \(c = \sqrt{14.5}\). Since \(3.8^2 = 14.44\), \(c \approx 3.8\).
(ii) \(c^2 = 8^2 + 12^2 = 64 + 144 = 208\). \(c = \sqrt{208}\). Since \(14^2 = 196\) and \(15^2 = 225\), \(c\) is between 14 and 15. More precisely \(c \approx 14.4\).
(iii) \(b^2 = 15^2 - 9^2 = 225 - 81 = 144\). \(b = 12\).
(iv) \(c^2 = 7^2 + 12^2 = 49 + 144 = 193\). \(c = \sqrt{193}\). Since \(13^2 = 169\) and \(14^2 = 196\), \(c\) is between 13 and 14. More precisely \(c \approx 13.9\).
(v) \(c^2 = 1.5^2 + 3.5^2 = 2.25 + 12.25 = 14.5\). \(c = \sqrt{14.5}\). Since \(3.8^2 = 14.44\), \(c \approx 3.8\).
Competency-Based Questions
An architect is designing a rectangular room that is 6 m long and 8 m wide. A diagonal support beam needs to be installed from one corner to the opposite corner of the floor. The architect uses Baudhayana's Theorem to calculate measurements.
Q1. What is the length of the diagonal support beam?
L3 ApplyAnswer: (b) 10 m
The diagonal of the rectangle forms the hypotenuse of a right triangle with sides 6 m and 8 m.
\(d^2 = 6^2 + 8^2 = 36 + 64 = 100\)
\(d = \sqrt{100} = 10\) m.
The diagonal of the rectangle forms the hypotenuse of a right triangle with sides 6 m and 8 m.
\(d^2 = 6^2 + 8^2 = 36 + 64 = 100\)
\(d = \sqrt{100} = 10\) m.
Q2. The architect discovers that one wall is not perfectly perpendicular. The diagonal measures 10.2 m instead of 10 m. Analyse whether the corner is truly a right angle.
L4 AnalyseAnswer: If the corner were a perfect right angle, by Baudhayana's Theorem: \(6^2 + 8^2 = 100\), so the diagonal should be exactly 10 m. Since the measured diagonal is 10.2 m, we have \(10.2^2 = 104.04 > 100 = 6^2 + 8^2\). The actual sum of squares of the sides is less than the square of the diagonal, meaning the angle between the walls is greater than 90 degrees. The corner is NOT a right angle -- it is an obtuse angle.
Q3. A student says: "In any triangle, the square of the longest side equals the sum of the squares of the other two sides." Evaluate whether this statement is always true.
L5 EvaluateAnswer: This statement is false in general. Baudhayana's Theorem states that \(a^2 + b^2 = c^2\) only for right-angled triangles. For acute triangles, \(a^2 + b^2 > c^2\), and for obtuse triangles, \(a^2 + b^2 < c^2\). Example: an equilateral triangle with all sides 5 has \(5^2 + 5^2 = 50 \neq 25 = 5^2\).
Q4. Create a method using Baudhayana's Theorem to verify whether a given triangle with sides 7, 24, and 25 is right-angled. Then devise a general test that works for any three given side lengths.
L6 CreateAnswer: Test for (7, 24, 25): The longest side is 25 (potential hypotenuse). Check: \(7^2 + 24^2 = 49 + 576 = 625 = 25^2\). Since the equation holds, the triangle IS right-angled.
General test: Given sides \(p\), \(q\), \(r\): (1) Find the longest side, say \(r\). (2) Compute \(p^2 + q^2\) and \(r^2\). (3) If \(p^2 + q^2 = r^2\), it is right-angled. If \(p^2 + q^2 > r^2\), it is acute. If \(p^2 + q^2 < r^2\), it is obtuse.
General test: Given sides \(p\), \(q\), \(r\): (1) Find the longest side, say \(r\). (2) Compute \(p^2 + q^2\) and \(r^2\). (3) If \(p^2 + q^2 = r^2\), it is right-angled. If \(p^2 + q^2 > r^2\), it is acute. If \(p^2 + q^2 < r^2\), it is obtuse.
Assertion–Reason Questions
Assertion (A): In a right triangle with sides 5, 12, and 13, the hypotenuse is 13.
Reason (R): \(5^2 + 12^2 = 25 + 144 = 169 = 13^2\).
Reason (R): \(5^2 + 12^2 = 25 + 144 = 169 = 13^2\).
Answer: (a) — Both true. \(5^2 + 12^2 = 169 = 13^2\), confirming 13 is the hypotenuse by Baudhayana's Theorem. R directly proves A.
Assertion (A): Baudhayana's Theorem can be used to find the diagonal of a rectangle.
Reason (R): A rectangle has four right angles, and a diagonal divides it into two right triangles.
Reason (R): A rectangle has four right angles, and a diagonal divides it into two right triangles.
Answer: (a) — Both true. Since a rectangle's diagonal creates a right triangle with the sides as legs, Baudhayana's Theorem directly gives: diagonal = \(\sqrt{l^2 + w^2}\). R explains why the theorem is applicable.
Assertion (A): The area of a square on the hypotenuse equals the sum of areas of squares on the other two sides only if the triangle is equilateral.
Reason (R): Baudhayana's Theorem applies only to right-angled triangles.
Reason (R): Baudhayana's Theorem applies only to right-angled triangles.
Answer: (d) — A is false: the property holds for right-angled triangles, not equilateral ones. R is true: Baudhayana's Theorem indeed applies only to right-angled triangles.
Frequently Asked Questions — Chapter 2
What is Pythagorean Theorem - Statement & Proof in NCERT Class 8 Mathematics?
Pythagorean Theorem - Statement & Proof 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 Theorem - Statement & Proof step by step?
To solve problems on Pythagorean Theorem - Statement & Proof, 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 Theorem - Statement & Proof important for the Class 8 board exam?
Pythagorean Theorem - Statement & Proof 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 Theorem - Statement & Proof?
Common mistakes in Pythagorean Theorem - Statement & Proof 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 Theorem - Statement & Proof?
End-of-chapter NCERT exercises for Pythagorean Theorem - Statement & Proof 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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