CBSE Class 8 Mathematics The Baudhayana Pythagoras Theorem Very Short Answer Questions with Answers

Very Short Answer Questions

Question 1. In many of his constructions Baudhāyana uses ‘east-west’ and ‘north-south’ lines. What are these lines, and how are they related to each other?
Answer:
‘East-west’ lines are horizontal lines and ‘north-south’ lines are vertical lines.
They are drawn perpendicular to each other.

Teacher's Note:
1. These lines cut the figures into congruent triangles.
2. Perpendicular means meeting at 90°.

Question 2. A square is made up of 2 small congruent triangles. How many of these small triangles make up the square drawn on its diagonal?
Answer:
Four. The square on the diagonal is made of 4 triangles congruent to the 2 in the original square, so its area is double.

Teacher's Note:
1. 4 triangles is twice 2 triangles.
2. This is the reason behind Verse 1.9.

Question 3. In Baudhāyana’s doubling construction, a square of area 5 sq. units is followed by two more squares, each drawn on the diagonal of the one before it. Write the areas of these two squares.
Answer:
Each square on a diagonal has double the area of the square before it.
So the areas are 2 × 5 = 10 sq. units and 2 × 10 = 20 sq. units.

Teacher's Note:
1. Double at every step.
2. After n steps the area is multiplied by 2 n times.

Question 4. Describe in one line how a square paper can be folded to obtain a square of half its area.
Answer:
Fold the paper inwards so that each crease passes through the midpoints of two adjacent sides; the inner square PQRS formed by the creases has half the area of the paper.

Teacher's Note:
1. The four folded corners exactly cover the inner square.
2. This reverses the doubling construction.

Question 5. A square has sidelength 10 cm. What fraction of its area is the area of a square of sidelength 5 cm?
Answer:
Area of the small square = 5 × 5 = 25 cm², area of the large square = 10 × 10 = 100 cm².
25 ÷ 100 = \( \frac{1}{4} \), so it is one-fourth of the area.

Teacher's Note:
1. Half the side gives a quarter of the area.
2. Four such small squares fill the big one.

Question 6. Write the relation between the hypotenuse c and the equal sides a of an isosceles right triangle.
Answer:
c² = 2a², that is, c = a√2.

Teacher's Note:
1. The square on the hypotenuse is double the square on a side.
2. This is the theorem with a = b.

Question 7. Between which two whole numbers does √2 lie? Give the reason in one line.
Answer:
√2 lies between 1 and 2, because 1² = 1 is less than 2 and 2² = 4 is greater than 2.

Teacher's Note:
1. Square the candidates and compare with 2.
2. 1 is a lower bound and 2 is an upper bound.

Question 8. What is meant by a lower bound and an upper bound for √2? Write one such pair with one digit after the decimal point.
Answer:
A lower bound is a number smaller than √2, and an upper bound is a number larger than √2.
One pair is 1.4 and 1.5, since 1.4² = 1.96 < 2 and 1.5² = 2.25 > 2.

Teacher's Note:
1. Bounds trap a number between them.
2. Closer bounds give a better estimate.

Question 9. In Śulba-Sūtra Verse 1.12, Baudhāyana says that the area of the square produced by the diagonal is equal to what?
Answer:
It is equal to the sum of the areas of the squares produced by the two sides, that is, by the two perpendicular sides of the right triangle.

Teacher's Note:
1. This is the general Baudhāyana–Pythagoras theorem.
2. Verse 1.9 is the special case of two equal sides.

Question 10. A rectangular courtyard measures 15 m by 20 m. Find the length of its diagonal.
Answer:
The diagonal is the hypotenuse of a right triangle with sides 15 m and 20 m.
15² + 20² = 225 + 400 = 625 = 25², so the diagonal is 25 m.

Teacher's Note:
1. (15, 20, 25) = 5 × (3, 4, 5).
2. The sides of a rectangle meet at right angles.

Question 11. Is (6, 8, 10) a primitive Baudhāyana triple? Give a reason.
Answer:
No. 6, 8 and 10 have the common factor 2, so (6, 8, 10) is a scaled version of (3, 4, 5) and is not primitive.

Teacher's Note:
1. Primitive means no common factor greater than 1.
2. (6, 8, 10) = 2 × (3, 4, 5).

Question 12. Which book about Fermat’s Last Theorem did the ten-year-old Andrew Wiles read in 1963, and who wrote it?
Answer:
He read The Last Problem, written by Eric Bell.

Teacher's Note:
1. The book inspired Wiles to try the problem himself.
2. He proved the theorem in 1994.

Question 13. A square has an area of 18 sq. units. Find the area and the exact sidelength of the square drawn on its diagonal.
Answer:
The square on the diagonal has double the area: 2 × 18 = 36 sq. units.
Its sidelength is √36 = 6 units.

Teacher's Note:
1. Double the area first, then take the square root.
2. The original side √18 is not a whole number, but the new side is.

Question 14. The square drawn on the diagonal of a square has an area of 50 sq. units. Find the area and the sidelength of the original square.
Answer:
The original square has half that area: \( \frac{50}{2} \) = 25 sq. units.
Its sidelength is √25 = 5 units.

Teacher's Note:
1. Work backwards: halve the area.
2. Check: diagonal² = 5² + 5² = 50.

Question 15. A square paper has an area of 64 cm². The midpoints of its sides are joined to form a square, and the midpoints of that square are joined to form one more square. Write the areas of these two squares.
Answer:
Each midpoint square has half the area of the square before it.
So the areas are 32 cm² and 16 cm².

Teacher's Note:
1. Halve the area at each step.
2. Two steps give one-fourth of the area.

Question 16. Find the exact length of the hypotenuse of an isosceles right triangle whose equal sides are each 11 units long.
Answer:
c² = 2a² = 2 × 11² = 242, so the hypotenuse is √242 = 11√2 units.

Teacher's Note:
1. Exact form means leave the square root.
2. c = a√2 gives 11√2 at once.

Question 17. The hypotenuse of an isosceles right triangle is √50 units. Find the length of each of its equal sides.
Answer:
c² = 2a² gives 50 = 2a², so a² = 25 and a = 5 units.

Teacher's Note:
1. Square the hypotenuse first: (√50)² = 50.
2. Then halve and take the square root.

Question 18. Given that 1.41² = 1.9881 and 1.42² = 2.0164, write bounds for √2 with two digits after the decimal point.
Answer:
1.9881 < 2 < 2.0164, so 1.41 < √2 < 1.42.

Teacher's Note:
1. The squares trap 2, so the numbers trap √2.
2. The gap is now only 0.01.

Question 19. A right-angled triangle has one shorter side of length 16 cm and a hypotenuse of length 34 cm. Find the third side.
Answer:
Third side = √(34² − 16²) = √(1156 − 256) = √900 = 30 cm.

Teacher's Note:
1. Subtract the squares to find a shorter side.
2. (16, 30, 34) = 2 × (8, 15, 17).

Question 20. Two squares of areas 36 sq. units and 64 sq. units are combined into one square by Baudhāyana’s method. Write the area and the sidelength of the square obtained.
Answer:
Area = 36 + 64 = 100 sq. units, so the sidelength is √100 = 10 units.

Teacher's Note:
1. The sides are 6 and 8, giving the triple (6, 8, 10).
2. Combine areas by adding them.

Question 21. Write any two scaled versions of the primitive Baudhāyana triple (8, 15, 17).
Answer:
Taking k = 2 and k = 3 gives (16, 30, 34) and (24, 45, 51); both satisfy a² + b² = c².

Teacher's Note:
1. Multiply all three numbers by the same k.
2. Check: 16² + 30² = 1156 = 34².

Question 22. The triple (21, 28, 35) is not primitive. Find the primitive Baudhāyana triple from which it is obtained.
Answer:
The common factor is 7. Dividing each number by 7 gives the primitive triple (3, 4, 5).

Teacher's Note:
1. Divide by the HCF of the three numbers.
2. (21, 28, 35) = 7 × (3, 4, 5).

Question 23. The number 169 is an odd square. Find the value of n for which 169 is the nth odd number, and write the Baudhāyana triple it produces.
Answer:
2n − 1 = 169 gives n = 85.
(n − 1)² + (2n − 1) = n² gives 84² + 13² = 85², so the triple is (13, 84, 85).

Teacher's Note:
1. 169 = 13², so 13 is one of the sides.
2. The other shorter side is one less than the hypotenuse.

Question 24. Why was nobody ever able to check Fermat’s own proof of his Last Theorem?
Answer:
Fermat never wrote his proof down. He only noted in the margin of a book that the margin was too small to contain it, and no proof of his was ever found.

Teacher's Note:
1. A claim needs a written proof to be checked.
2. The first complete proof came from Wiles in 1994.

Question 25. The equal sides of an isosceles right triangle are each 15 units long. Between which two consecutive whole numbers does its hypotenuse lie?
Answer:
c² = 2 × 225 = 450. Since 21² = 441 and 22² = 484, the hypotenuse lies between 21 and 22 units.

Teacher's Note:
1. Trap 450 between consecutive squares.
2. √450 is about 21.2.

Question 26. Find the exact sidelength of a square whose area is equal to the difference of the areas of two squares of sidelengths 10 units and 6 units.
Answer:
Required area = 10² − 6² = 100 − 36 = 64 sq. units, so the sidelength is √64 = 8 units.

Teacher's Note:
1. Subtract the areas, not the sides.
2. (6, 8, 10) is a Baudhāyana triple.

Question 27. The sides of a triangle measure 9, 12 and 16 units. Is it a right-angled triangle? Justify your answer by calculation.
Answer:
The two shorter sides give 9² + 12² = 81 + 144 = 225, but the longest side gives 16² = 256.
As 225 ≠ 256, the triangle is not right-angled.

Teacher's Note:
1. Compare the sum of the two smaller squares with the largest square.
2. (9, 12, 15) would be right-angled, but not (9, 12, 16).

Question 28. A right triangle has shorter sides a and b and hypotenuse c. Explain in one line why c is always the longest of the three sides.
Answer:
Since c² = a² + b², c² is greater than both a² and b², so c is greater than both a and b.

Teacher's Note:
1. A bigger square has a bigger side.
2. This helps to spot the hypotenuse quickly.

Question 29. The diagonals of a rhombus are 18 units and 24 units long. Find the sidelength of the rhombus.
Answer:
The diagonals bisect each other at right angles, so a side is the hypotenuse of a right triangle with legs 9 and 12.
9² + 12² = 81 + 144 = 225 = 15², so the sidelength is 15 units.

Teacher's Note:
1. Use half of each diagonal.
2. (9, 12, 15) = 3 × (3, 4, 5).

Question 30. Find the exact length of the altitude of an equilateral triangle whose sidelength is 8 units.
Answer:
The altitude bisects the opposite side, giving a right triangle with hypotenuse 8 and one side 4.
Altitude = √(8² − 4²) = √48 = 4√3 units.

Teacher's Note:
1. The altitude splits the triangle into two congruent right triangles.
2. √48 = √(16 × 3) = 4√3.