This MCQ module is based on: Doubling a Square & Discovering Square Root 2
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Doubling a Square & Discovering Square Root 2
🎓 Class 8
Mathematics
CBSE
Theory
Ch 2 — The Pythagoras Theorem
⏱ ~30 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Doubling a Square & Discovering Square Root 2
Targeting Class 8 level in General Mathematics, with Basic difficulty.
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2.1 Doubling a Square
In Baudhayana's Sulba-Sutra? (c. 800 BCE), Baudhayana considers a fascinating question:
Key Question
How can one construct a square having double the area of a given square?
A first guess might be to simply double the length of each side. If the original square has side \(s\), the doubled-side square has side \(2s\). But its area is \((2s)^2 = 4s^2\), which is four times the area, not double!
Doubling each side produces a square with 4 times the area, not 2 times.
So doubling the side gives 4 times the area. The correct answer comes from Baudhayana himself, in Verse 1.9 of his Sulba-Sutra:
Baudhayana's Rule (Verse 1.9)
The diagonal of a square produces a square of double the area of the original square.
The key idea is to construct a new square on the diagonal? of the original square. This new (tilted) square has exactly twice the area of the original.
The tilted square constructed on the diagonal has double the area of the original.
Why Does This Work?
To understand why the new (dotted) square has double the area, draw horizontal and vertical grid lines through the original square. The extensions of the vertical and horizontal sides of the original square pass through the vertices of the dotted square.
The original square is made of 2 small triangles; the new square is made of 4 such triangles. So the new square has double the area.
The original square is made of 2 small congruent triangles, and the new (tilted) square is made of 4 such congruent triangles. Since all four triangles are congruent (they share the same shape and size), the new square has exactly twice the area of the original.
Think About It: Why should the extension of the vertical and horizontal sides of the original square pass through the vertices of the dotted square? Hint: From the diagonal property of a square, the line that bisects an angle passes through the opposite vertex. Argue why the vertical and horizontal sides of the original square bisect the two angles of the dotted square.
Sequence of Doubled Squares
We can repeat this process to create a sequence of squares, each having double the area of the previous one:
Successive squares with areas A, 2A, and 4A, made of 2, 4, and 8 small congruent triangles.
Doubling a Square Using Paper
Activity: Doubling a Square with Paper Cutting
Materials: Two identical square pieces of paper, scissors, pencil, ruler.
Predict: If you cut both squares along their diagonals, can you rearrange the triangles to form a bigger square? How much bigger will it be?
- Cut out two identical squares of paper.
- Square 1: Draw both diagonals. Label the four triangles as 1, 2, 3, 4.
- Square 2: Draw both diagonals. Label the four triangles as 5, 6, 7, 8.
- Cut Square 2 along its diagonals to get triangles 5, 6, 7, 8.
- Place pieces 5, 6, 7, and 8 around Square 1 to form a larger square with double the area.
Observe: The new square has side length equal to the diagonal of the original square. Since each original square provides 4 triangles, and the new square uses all 8 triangles (from both squares), its area = 2 times the area of one original square.
Explain: The arrangement works because the triangles from Square 2 fit perfectly into the gaps created by tilting Square 1's orientation by 45 degrees.
Pieces 5, 6, 7, 8 from Square 2 placed around Square 1 to form a square with double the area.
2.2 Halving a Square
Now suppose we are given a square and want to construct a square whose area is half that of the original square. How would you do it?
One way is to reverse the construction of the previous section. Draw a tilted smaller square inside the larger one, connecting the midpoints of the sides:
The tilted inner square (connecting midpoints) has half the area of the original. The 4 corner triangles together equal the inner square in area.
Adding east-west and north-south lines reveals that the inner square consists of 4 small triangles, while the outer square consists of 8 such triangles. So the inner square has exactly half the area.
Halving a Square Using Paper
Cut out a square from a piece of paper. Fold the square paper inward, so that the crease lines pass through the midpoints of the sides. The resulting shape PQRS is a square with half the area.
Folding the square paper inward through midpoints creates PQRS, a square with half the area.
Think About It: Why is PQRS a square? Why is its area half that of the original paper? Hint: Connect QS and PR, find the different angles formed, and then use triangle congruence.
2.3 Hypotenuse of an Isosceles Right Triangle
Recall that in a right triangle, the side opposite to the right angle is called the hypotenuse?.
Consider an isosceles right triangle? with each equal side of length 1 unit. What is the length of its hypotenuse?
An isosceles right triangle with equal sides of length 1. What is the hypotenuse?
We know that a square of side 1 unit is made of two such isosceles right triangles. We also know that the square constructed on the diagonal of this square has twice the area of the original.
Square PEAR with side \(a\). The diagonal has length \(c\). The square on the diagonal (REST) has area \(= 2 \times\) Area of PEAR.
Let \(c\) be the length of the hypotenuse (the diagonal ER). Then:
Derivation
Area of the square on the diagonal = \(c \times c = c^2\).This area = \(2 \times\) Area of PEAR = \(2 \times a \times a = 2a^2\).
So, \(c^2 = 2a^2\).
For \(a = 1\): \(c^2 = 2\), therefore \(c = \sqrt{2}\).
The hypotenuse of an isosceles right triangle with equal sides 1 is \(\sqrt{2}\) units.
The hypotenuse of an isosceles right triangle with equal sides of length 1 is \(\sqrt{2}\).
General Formula
For an isosceles right triangle with equal sides of length \(a\), the hypotenuse \(c\) satisfies:
\(c^2 = 2a^2\), so \(c = a\sqrt{2}\)
Example 1
Find the hypotenuse of an isosceles right triangle whose equal sides have length 12.
Solution
Using the formula: \(c = \sqrt{2} \times 12^2 = \sqrt{2 \times 144} = \sqrt{288}\).We check: \(16^2 = 256\) and \(17^2 = 289\).
So \(\sqrt{288}\) lies between 16 and 17.
The hypotenuse is between 16 and 17 units.
Example 2
If the hypotenuse of an isosceles right triangle is \(\sqrt{72}\), find the other two sides.
Solution
We have \(c = \sqrt{72}\). Using \(c^2 = 2a^2\):\((\sqrt{72})^2 = 2a^2\)
\(72 = 2a^2\)
\(a^2 = \frac{72}{2} = 36\)
\(a = \sqrt{36} = 6\)
Each of the other two sides has length 6 units.
2.3.1 Decimal Representation of \(\sqrt{2}\)
What is the value of \(\sqrt{2}\)? Let us find its decimal representation using a step-by-step argument.
Step 1: Is \(\sqrt{2}\) less than or greater than 1?
A square of side 1 has area \(1^2 = 1\) sq. unit. A square of side \(\sqrt{2}\) has area 2 sq. units. Since \(2 > 1\), we have \(\sqrt{2} > 1\).
Step 2: Is \(\sqrt{2}\) less than or greater than 2?
A square of side 2 has area \(2^2 = 4\) sq. units. Since \(4 > 2\), we have \(\sqrt{2} < 2\). Therefore: \(1 < \sqrt{2} < 2\).
Step 3: Narrowing to one decimal place
| Check | Square | Compare with 2 |
|---|---|---|
| \(1.1^2\) | 1.21 | < 2 |
| \(1.2^2\) | 1.44 | < 2 |
| \(1.3^2\) | 1.69 | < 2 |
| \(1.4^2\) | 1.96 | < 2 |
| \(1.5^2\) | 2.25 | > 2 |
So \(1.4 < \sqrt{2} < 1.5\).
Step 4: Narrowing to two decimal places
| Check | Square | Compare with 2 |
|---|---|---|
| \(1.41^2\) | 1.9881 | < 2 |
| \(1.42^2\) | 2.0164 | > 2 |
So \(1.41 < \sqrt{2} < 1.42\).
Step 5: Even more precision
| Check | Square | Compare with 2 |
|---|---|---|
| \(1.411^2\) | 1.990921 | < 2 |
| \(1.412^2\) | 1.993744 | < 2 |
| \(1.413^2\) | 1.996569 | < 2 |
| \(1.414^2\) | 1.999396 | < 2 |
| \(1.415^2\) | 2.002225 | > 2 |
So \(1.414 < \sqrt{2} < 1.415\).
Key Insight
The decimal expansion of \(\sqrt{2}\) never terminates. It goes on forever: \(\sqrt{2} = 1.41421356...\) This means \(\sqrt{2}\) cannot be expressed as a terminating decimal. In fact, it cannot even be expressed as a fraction \(\frac{m}{n}\) where \(m\) and \(n\) are counting numbers.
Historical Note
This beautiful proof that \(\sqrt{2}\) cannot be expressed as a fraction was given by Euclid in his book Elements (c. 300 BCE). The key argument uses the prime factorisation of a square number: in the equation \(2n^2 = m^2\), the prime 2 would occur an odd number of times on the left and an even number of times on the right, which is impossible.
The number \(\sqrt{2}\) cannot be expressed as a terminating decimal or a fraction. But we can think of it as a certain non-terminating decimal: \(\sqrt{2} = 1.41421356...\)
Interactive: Square Doubler
Enter a side length and see how the diagonal square has exactly double the area.
Enter a value and click Calculate.
Figure it Out (Section 2.1 – 2.3)
Q1. Earlier, we saw a method to create a square with double the area of a given square paper. Here is another method: Two identical square papers are cut as shown (into a triangle and a trapezium each). Can you arrange these pieces to create a square with double the area of either square?
Solution: Yes! Each square is cut along a diagonal into a triangle and a trapezium (pieces 1, 2, 3, 4 from one square and 5, 6, 7, 8 from the other). Place the 4 triangles at the corners and the 4 trapeziums along the sides, or alternatively rearrange all 8 pieces to form a large square whose side equals the diagonal of the original square. The new square has area = \(2 \times\) area of one original square.
Q2. The length of the two equal sides of an isosceles right triangle is given. Find the length of the hypotenuse. Find bounds on the hypotenuse such that they have at least one digit after the decimal point.
(i) 3 (ii) 4 (iii) 6 (iv) 8 (v) 9
(i) 3 (ii) 4 (iii) 6 (iv) 8 (v) 9
Using \(c = a\sqrt{2}\) where \(c^2 = 2a^2\):
(i) \(a = 3\): \(c^2 = 2 \times 9 = 18\). Since \(4^2 = 16\) and \(4.3^2 = 18.49\), we get \(4.2 < c < 4.3\).
(ii) \(a = 4\): \(c^2 = 2 \times 16 = 32\). Since \(5.6^2 = 31.36\) and \(5.7^2 = 32.49\), we get \(5.6 < c < 5.7\).
(iii) \(a = 6\): \(c^2 = 72\). Since \(8.4^2 = 70.56\) and \(8.5^2 = 72.25\), we get \(8.4 < c < 8.5\).
(iv) \(a = 8\): \(c^2 = 128\). Since \(11.3^2 = 127.69\) and \(11.4^2 = 129.96\), we get \(11.3 < c < 11.4\).
(v) \(a = 9\): \(c^2 = 162\). Since \(12.7^2 = 161.29\) and \(12.8^2 = 163.84\), we get \(12.7 < c < 12.8\).
(i) \(a = 3\): \(c^2 = 2 \times 9 = 18\). Since \(4^2 = 16\) and \(4.3^2 = 18.49\), we get \(4.2 < c < 4.3\).
(ii) \(a = 4\): \(c^2 = 2 \times 16 = 32\). Since \(5.6^2 = 31.36\) and \(5.7^2 = 32.49\), we get \(5.6 < c < 5.7\).
(iii) \(a = 6\): \(c^2 = 72\). Since \(8.4^2 = 70.56\) and \(8.5^2 = 72.25\), we get \(8.4 < c < 8.5\).
(iv) \(a = 8\): \(c^2 = 128\). Since \(11.3^2 = 127.69\) and \(11.4^2 = 129.96\), we get \(11.3 < c < 11.4\).
(v) \(a = 9\): \(c^2 = 162\). Since \(12.7^2 = 161.29\) and \(12.8^2 = 163.84\), we get \(12.7 < c < 12.8\).
Q3. The hypotenuse of an isosceles right triangle is 10. What are its other two sidelengths? [Hint: Find the area of the square composed of two such right triangles.]
Solution: Using \(c^2 = 2a^2\):
\(10^2 = 2a^2\)
\(100 = 2a^2\)
\(a^2 = 50\)
\(a = \sqrt{50} = 5\sqrt{2}\)
Since \(7.0^2 = 49\) and \(7.1^2 = 50.41\), we get \(a\) is between 7.0 and 7.1.
Each of the other two sides is \(\sqrt{50} \approx 7.07\) units.
\(10^2 = 2a^2\)
\(100 = 2a^2\)
\(a^2 = 50\)
\(a = \sqrt{50} = 5\sqrt{2}\)
Since \(7.0^2 = 49\) and \(7.1^2 = 50.41\), we get \(a\) is between 7.0 and 7.1.
Each of the other two sides is \(\sqrt{50} \approx 7.07\) units.
Competency-Based Questions
A school's math club is designing a large mural on a square board of side 2 m. They want to create a tilted inner square by connecting the midpoints of the sides, and also construct a larger outer square on the diagonal of the original board.
Q1. What is the area of the tilted inner square formed by connecting the midpoints of the 2 m square board?
L3 ApplyAnswer: (b) 2 m²
The inner square formed by connecting midpoints has half the area of the original: \(\frac{1}{2} \times 2^2 = \frac{1}{2} \times 4 = 2\) m².
The inner square formed by connecting midpoints has half the area of the original: \(\frac{1}{2} \times 2^2 = \frac{1}{2} \times 4 = 2\) m².
Q2. The club wants to build a larger square frame on the diagonal of the original 2 m board. What will be the side length of this larger square?
L4 AnalyseAnswer: The diagonal of the 2 m square = \(2\sqrt{2}\) m. The larger square built on the diagonal has side = \(2\sqrt{2} \approx 2.83\) m, and area = \((2\sqrt{2})^2 = 8\) m², which is double the original area of 4 m².
Q3. A student claims that tripling each side of a square triples its area. Evaluate this claim and explain the error.
L5 EvaluateAnswer: The claim is incorrect. If the original side is \(s\), the original area is \(s^2\). Tripling the side gives \((3s)^2 = 9s^2\), which is 9 times the area, not 3 times. In general, if each side is multiplied by \(k\), the area is multiplied by \(k^2\).
Q4. Design a method to construct a square whose area is exactly 3 times that of a given square of side \(s\). Describe the steps clearly.
L6 CreateAnswer: Step 1: Build a square of side \(s\) (area \(s^2\)). Step 2: Double it using the diagonal method to get a square of area \(2s^2\) with side \(s\sqrt{2}\). Step 3: Now combine the original square (area \(s^2\)) and the doubled square (area \(2s^2\)) using Baudhayana's method for combining two different squares. Form a right triangle with perpendicular sides \(s\) and \(s\sqrt{2}\). The hypotenuse has length \(\sqrt{s^2 + 2s^2} = s\sqrt{3}\). Build a square on this hypotenuse: area \(= 3s^2\).
Assertion–Reason Questions
Assertion (A): The diagonal of a square of side 5 cm is \(5\sqrt{2}\) cm.
Reason (R): The diagonal of a square produces a square with double the area of the original.
Reason (R): The diagonal of a square produces a square with double the area of the original.
Answer: (a) — Both are true. The diagonal of a side-5 square forms a right isosceles triangle with legs 5, so hypotenuse = \(5\sqrt{2}\). R explains this because the square on the diagonal has area \(2 \times 25 = 50\), so diagonal = \(\sqrt{50} = 5\sqrt{2}\).
Assertion (A): \(\sqrt{2}\) can be written as the fraction \(\frac{1414}{1000}\).
Reason (R): The decimal expansion of \(\sqrt{2}\) is non-terminating and non-repeating.
Reason (R): The decimal expansion of \(\sqrt{2}\) is non-terminating and non-repeating.
Answer: (d) — A is false: \(\frac{1414}{1000} = 1.414\), but \(1.414^2 = 1.999396 \neq 2\). So \(\frac{1414}{1000}\) is only an approximation, not equal to \(\sqrt{2}\). R is true: \(\sqrt{2}\) cannot be expressed as any fraction \(\frac{m}{n}\).
Assertion (A): An isosceles right triangle with hypotenuse 10 has equal sides of length \(5\sqrt{2}\).
Reason (R): In an isosceles right triangle, \(c^2 = 2a^2\).
Reason (R): In an isosceles right triangle, \(c^2 = 2a^2\).
Answer: (a) — Both true, R explains A. From \(c^2 = 2a^2\): \(100 = 2a^2\), \(a^2 = 50\), \(a = \sqrt{50} = 5\sqrt{2}\).
Frequently Asked Questions — Chapter 2
What is Doubling a Square & Discovering Square Root 2 in NCERT Class 8 Mathematics?
Doubling a Square & Discovering Square Root 2 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 Doubling a Square & Discovering Square Root 2 step by step?
To solve problems on Doubling a Square & Discovering Square Root 2, 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 Doubling a Square & Discovering Square Root 2 important for the Class 8 board exam?
Doubling a Square & Discovering Square Root 2 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 Doubling a Square & Discovering Square Root 2?
Common mistakes in Doubling a Square & Discovering Square Root 2 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 Doubling a Square & Discovering Square Root 2?
End-of-chapter NCERT exercises for Doubling a Square & Discovering Square Root 2 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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