CBSE Class 8 Mathematics The Baudhayana Pythagoras Theorem MCQs Set 11

Answer: (c) a diagonal of the given square

Explanation:
1. Verse 1.9 says that the diagonal of a square produces a square of double the area of the original square.
2. So the new square is built on a diagonal of the given square.
3. A square built on a side would only give back the same square.

Teacher's Note:
1. Remember: diagonal doubles the area.
2. This is the starting idea of the whole chapter.

Answer: (b) four times the original area

Explanation:
1. Let the side be s, so the area is s².
2. New side = 2s, so new area = 2s × 2s = 4s².
3. The area becomes four times the original area.

Teacher's Note:
1. Area grows by the square of the scale factor.
2. That is why doubling the side cannot double the area.

Answer: (d) one-half of the area of the sheet

Explanation:
1. The creases cut off four congruent corner triangles.
2. When these are folded in, they exactly cover the inner square PQRS, which is also made of four such triangles.
3. So PQRS has one-half of the area of the sheet.

Teacher's Note:
1. Joining midpoints of a square halves its area.
2. This is the doubling construction done in reverse.

Answer: (a) opposite the right angle

Explanation:
1. By definition, the hypotenuse is the side opposite the right angle.
2. It is also the longest side, but it may be drawn in any direction and need not have a whole-number length.

Teacher's Note:
1. Find the right angle first; the hypotenuse is across from it.
2. The side opposite the smallest angle is the shortest side.

Answer: (b) √2 units

Explanation:
1. Two such triangles make a unit square of area 1.
2. The square on the hypotenuse (the diagonal) has double this area, so c² = 2.
3. Hence c = √2 units.

Teacher's Note:
1. The hypotenuse here is the diagonal of a unit square.
2. √2 is about 1.414, longer than a side but shorter than 2.

Answer: (c) a² + b² = c²

Explanation:
1. The squares on the two shorter sides together have the same area as the square on the hypotenuse.
2. So a² + b² = c².

Teacher's Note:
1. The theorem is about areas of squares, so the sides are squared.
2. a + b = c is false: two sides of a triangle add to more than the third.

Answer: (d) have no common factor greater than 1

Explanation:
1. A triple is primitive when its three numbers have no common factor greater than 1.
2. For example, (3, 4, 5) is primitive, while (9, 12, 15) has the common factor 3 and is not.

Teacher's Note:
1. Primitive triples cannot be scaled down any further.
2. (3, 4, 5) already shows that the numbers need not all be odd or prime.

Answer: (d) A is false, but R is true.

Explanation:
1. A is false: doubling the side makes the area 2 × 2 = 4 times as large.
2. R is true: it is Baudhāyana’s rule from Verse 1.9, the correct way to double an area.

Teacher's Note:
1. Do not confuse doubling the side with doubling the area.
2. Use the diagonal, not a doubled side, to double a square.

Answer: (c) Both Statement I and Statement II are true.

Explanation:
1. Statement I is exactly what Fermat claimed for powers greater than 2. True.
2. Statement II is true: after more than 300 years, Andrew Wiles proved it in 1994.
3. So both statements are true.

Teacher's Note:
1. For n = 2 there are many solutions: the Baudhāyana triples.
2. The claim is only for n greater than 2.

Answer: (b) 20 m

Explanation:
1. The bar is the hypotenuse of a right triangle with shorter sides 12 m and 16 m.
2. 12² + 16² = 144 + 256 = 400 = 20².
3. So the bar is 20 m long.

Teacher's Note:
1. The width and height of a rectangle meet at a right angle.
2. (12, 16, 20) is (3, 4, 5) scaled by 4.

Answer: (a) (i)-(q), (ii)-(s), (iii)-(p), (iv)-(r)

Explanation:
1. Baudhāyana wrote the Śulba-Sūtra, c. 800 BCE: (i)-(q).
2. Pythagoras was a Greek philosopher-mathematician, c. 500 BCE: (ii)-(s).
3. Andrew Wiles proved Fermat’s Last Theorem in 1994: (iii)-(p).
4. Bhāskarāchārya wrote the Līlāvatī: (iv)-(r). So the answer is (a).

Teacher's Note:
1. Link each name with one clear fact.
2. Baudhāyana came about 300 years before Pythagoras.

Answer: (d) 13

Explanation:
1. The shorter sides are AB = 5 and BC = 12, with the right angle at B.
2. AC² = 5² + 12² = 25 + 144 = 169 = 13².
3. So the hypotenuse AC = 13.

Teacher's Note:
1. (5, 12, 13) is a triple from Baudhāyana’s own list.
2. The hypotenuse must be longer than 12, which rules out 7 and 11.

Answer: (b) 1.414 and 1.415

Explanation:
1. All four pairs trap √2, because 1.414² = 1.999396 < 2 < 2.002225 = 1.415².
2. The gap between the bounds is 1, 0.1, 0.01 and 0.001 for the pairs 1–2, 1.4–1.5, 1.41–1.42 and 1.414–1.415.
3. The narrowest pair is 1.414 and 1.415.

Teacher's Note:
1. Closer bounds means a smaller gap.
2. Each new decimal place makes the gap ten times smaller.

Answer: (c) (9, 16, 25)

Explanation:
1. 9² + 16² = 81 + 256 = 337, but 25² = 625. They are not equal.
2. 8² + 15² = 289 = 17², 7² + 24² = 625 = 25² and 12² + 35² = 1369 = 37².
3. So (9, 16, 25) is not a Baudhāyana triple.

Teacher's Note:
1. Always square and add the two smaller numbers.
2. 9, 16, 25 are themselves squares, which is a trap.

Answer: (a) Only Statement I is true.

Explanation:
1. Statement I: (3k)² + (4k)² = 9k² + 16k² = 25k² = (5k)². True.
2. Statement II: a scaled triple such as (6, 8, 10) has the common factor 2, so it is not primitive. False.
3. So only Statement I is true.

Teacher's Note:
1. Scaling keeps the triple but adds a common factor.
2. A scaled triple (k > 1) is never primitive.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. If √2 = \( \frac{m}{n} \), then 2n² = m².
2. By R, the prime 2 occurs an even number of times in m² and in n², so it occurs an odd number of times in 2n².
3. The same number cannot have both, so A is true, and R is exactly what proves it.

Teacher's Note:
1. This proof is due to Euclid (c. 300 BCE).
2. Counting the factor 2 on both sides gives the contradiction.

Answer: (c) A is true, but R is false.

Explanation:
1. 15² + 36² = 225 + 1296 = 1521 = 39², so A is true.
2. 15, 36 and 39 have the common factor 3, so the triple is not primitive. R is false.

Teacher's Note:
1. (15, 36, 39) = 3 × (5, 12, 13).
2. Check for a common factor before calling a triple primitive.

Answer: (b) Only Statement II is true.

Explanation:
1. Statement I: halving the side makes the area \( \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \) of the original. False.
2. Statement II: four quarter-area squares fill the given square exactly. True.
3. So only Statement II is true.

Teacher's Note:
1. Halving the side quarters the area.
2. To halve the area, join the midpoints of the sides instead.

Answer: (b) 11 and 12 units

Explanation:
1. c² = 2a² = 2 × 64 = 128.
2. 11² = 121 < 128 < 144 = 12².
3. So the hypotenuse √128 lies between 11 and 12 units.

Teacher's Note:
1. Trap c² between two consecutive square numbers.
2. √128 is about 11.3.

Answer: (d) 51 m

Explanation:
1. The diagonal is the hypotenuse of a right triangle with shorter sides 24 m and 45 m.
2. 24² + 45² = 576 + 2025 = 2601 = 51².
3. So the path is 51 m long.

Teacher's Note:
1. (24, 45, 51) is (8, 15, 17) scaled by 3.
2. Check the answer by squaring: 51 × 51 = 2601.

Answer: (a) (i)-(r), (ii)-(p), (iii)-(q), (iv)-(s)

Explanation:
1. 3² + 4² = 25 = 5²: (i)-(r).
2. 8² + 15² = 64 + 225 = 289 = 17²: (ii)-(p).
3. 12² + 35² = 144 + 1225 = 1369 = 37²: (iii)-(q).
4. 15² + 36² = 225 + 1296 = 1521 = 39²: (iv)-(s). The answer is (a).

Teacher's Note:
1. These are the triples listed in Verse 1.13.
2. The hypotenuse is always larger than both shorter sides.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. A is Fermat’s Last Theorem for n = 3, which Wiles proved. True.
2. R is true, because every triple can be scaled to give new ones.
3. R is about squares and cannot explain a fact about cubes, so R is not the explanation of A.

Teacher's Note:
1. The contrast between squares and cubes is what made Fermat curious.
2. Two true statements need not be cause and effect.

Answer: (d) Neither Statement I nor Statement II is true.

Explanation:
1. Statement I is false: Baudhāyana (c. 800 BCE) stated it first, a couple of hundred years before Pythagoras (c. 500 BCE).
2. Statement II is false: the relation holds only for right-angled triangles (for example, 9² + 12² ≠ 16²).
3. So neither statement is true.

Teacher's Note:
1. The order of the names is reversed in Statement I.
2. The theorem needs a right angle.

Answer: (b) 20

Explanation:
1. The right angle is at P, so RQ = 25 is the hypotenuse and PR = 15 is a shorter side.
2. PQ² = 25² − 15² = 625 − 225 = 400.
3. So PQ = 20.

Teacher's Note:
1. To find a shorter side, subtract the squares.
2. (15, 20, 25) is (3, 4, 5) scaled by 5.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. R is true: from a² + b² = c², dividing by f² gives (a/f)² + (b/f)² = (c/f)².
2. Repeating this, the numbers keep getting smaller, so we must stop at a triple with no common factor, a primitive one.
3. So A is true and R explains it.

Teacher's Note:
1. Every triple shrinks down to a primitive one.
2. Example: (9, 12, 15) ÷ 3 = (3, 4, 5).

Answer: (a) Only Statement I is true.

Explanation:
1. The triple produced has shorter side n − 1 and hypotenuse n, so Statement I is true.
2. (8, 15, 17) is primitive, but neither 8 nor 15 is one less than 17, so it cannot come from this method. Statement II is false.
3. So only Statement I is true.

Teacher's Note:
1. One counter-example is enough to show a statement is false.
2. Triples like (3, 4, 5), (5, 12, 13), (7, 24, 25) come from this method.

Answer: (c) (i)-(s), (ii)-(q), (iii)-(p), (iv)-(r)

Explanation:
1. (i) Half-diagonals are 5 and 12; 5² + 12² = 169 = 13², so the side is 13: (i)-(s).
2. (ii) 33² + 56² = 1089 + 3136 = 4225 = 65²: (ii)-(q).
3. (iii) 41² − 40² = 1681 − 1600 = 81 = 9²: (iii)-(p).
4. (iv) 28² + 45² = 784 + 2025 = 2809 = 53²: (iv)-(r). The answer is (c).

Teacher's Note:
1. In a rhombus, use half of each diagonal.
2. Subtract squares when the hypotenuse is given.

Answer: (d) A is false, but R is true.

Explanation:
1. A is false: 1.414² = 1.999396, which is not 2, so 1.414 is only a lower bound.
2. R is true, and it shows that √2 lies strictly between 1.414 and 1.415.
3. In fact √2 = 1.41421356… never ends.

Teacher's Note:
1. A bound is not the exact value.
2. No terminating decimal squares to exactly 2.

Answer: (a) 3 units

Explanation:
1. The two squares are the squares on the perpendicular sides of a right triangle.
2. The square on the hypotenuse has area 2 + 7 = 9 sq. units.
3. Its sidelength is √9 = 3 units.

Teacher's Note:
1. Add the areas, not the sides.
2. 9 is the area; the side is its square root.

Answer: (c) 12 sq. units

Explanation:
1. A square drawn on the diagonal of a square has double its area.
2. Square II = 2 × 3 = 6 sq. units.
3. Square III = 2 × 6 = 12 sq. units.

Teacher's Note:
1. Each step doubles the area, so two steps multiply it by 4.
2. The figure shows square III made of twice as many small triangles as II.