Class 8 Maths Ganita Prakash Part 2 Chapter 2: The Baudhāyana-Pythagoras Theorem MCQs with Answers

Let the original side be a units.
Original area = a² square units.
New side = 2a units.
New area = (2a)² = 4a² square units.
Therefore, doubling the side makes the area four times the original area.

Let the original side be a and its diagonal be d.
Original area A = a².
The diagonal divides the square into two right triangles, so d² = a² + a² = 2a².
The new square has side d and area d² = 2A.

Let the original side be a. The four corner triangles are right triangles with perpendicular sides a/2 and a/2.
Area of one corner triangle = (1/2) × (a/2) × (a/2) = a²/8.
Area of the inner square = a² − 4 × a²/8 = a²/2.
Therefore, it occupies half the original area.

Let the original side be a units, so the original area is a² square units.
New side = a/2 units.
New area = (a/2)² = a²/4 square units.
Thus, four such smaller squares have the same total area as the original square.

The hypotenuse is the side opposite the right angle.
Here, the right angle is at B, and the side opposite B is AC.
AB and BC meet at B and form the right angle. Therefore, AC is the hypotenuse.

Let the hypotenuse be c units.
The two perpendicular sides are each 1 unit long.
c² = 1² + 1² = 2
c = √2.
The positive square root is taken because c is a length. Therefore, the hypotenuse is √2 units.

1.414² = 1.999396, which is less than 2.
1.415² = 2.002225, which is greater than 2.
Since both bounds are positive, their squares place √2 between them.
Therefore, 1.414 < √2 < 1.415.

Suppose √2 = m/n, where m and n are positive integers.
Then m² = 2n².
Every prime factor occurs an even number of times in a square. The number of factors 2 is even in m² but odd in 2n².
This is impossible, so √2 cannot be expressed as such a fraction.

Let the hypotenuse be c units.
c² = 4² + 4² = 16 + 16 = 32
c = √32 = 4√2.
Therefore, the hypotenuse is 4√2 units. Its square equals the sum of the squares of the two perpendicular sides.

Let each equal side be a units.
For an isosceles right triangle, 2a² = c².
2a² = 10² = 100
a² = 50, so a = √50.
Therefore, each equal side is √50 units long; halving the hypotenuse would not give the correct side length.

Let the hypotenuse be c units.
c² = 12² + 12² = 288.
Since 16² = 256 and 17² = 289, we have 256 < 288 < 289. Therefore, 16 < c < 17, so the hypotenuse lies between 16 and 17 units.

Let each equal side be a units.
2a² = (√72)² = 72
a² = 72 ÷ 2 = 36
a = 6.
Therefore, each equal side is 6 units long. The two squares of area 36 together equal the hypotenuse square’s area of 72.

The square on the hypotenuse has the same area as the two squares on the perpendicular sides together.
Their areas are c², a², and b².
Therefore, a² + b² = c². This relation applies when c is the hypotenuse of a right triangle.

The rearrangement preserves the total area of the two original squares.
Total area = 3² + 4² = 9 + 16 = 25 cm².
Let the larger square’s side be s cm.
s² = 25, so s = 5.
Therefore, the larger square has side length 5 cm.

Let the hypotenuse be c cm.
By the Baudhāyana-Pythagoras theorem, c² = 5² + 12².
c² = 25 + 144 = 169
c = √169 = 13.
Therefore, the hypotenuse is 13 cm, which is longer than either perpendicular side.

Let the unknown perpendicular side be b cm.
8² + b² = 17²
b² = 289 − 64 = 225
b = √225 = 15.
Therefore, the other side is 15 cm. The hypotenuse square’s area is reduced by the known side’s square area.

Let the hypotenuse be c units.
c² = 1.5² + 3.5²
c² = 2.25 + 12.25 = 14.5
c = √14.5.
Therefore, the exact hypotenuse is √14.5 units. The theorem adds the squares of the lengths, rather than adding the lengths themselves.

A primitive triple satisfies a² + b² = c² and has no common factor greater than 1.
5² + 12² = 25 + 144 = 169 = 13².
The HCF of 5, 12, and 13 is 1.
Each of the other triples has a common factor greater than 1, so none of them is primitive.

All three side lengths are multiplied by the same factor.
6 × (3, 4, 5) = (18, 24, 30).
Also, 18² + 24² = 324 + 576 = 900 = 30².
Therefore, the resulting triple is a scaled version of (3, 4, 5).

For the triple (7, 24, 25),
7² + 24² = 49 + 576 = 625 = 25².
Thus, it satisfies the required relation. The other options give unequal values when the squares of the two smaller numbers are added and compared with the largest number’s square.

(3k)² + (4k)² = 9k² + 16k²
= 25k²
= (5k)².
Therefore, each positive integer k gives a right-triangle triple. There is no largest positive integer, so these scaled triples continue without end. For k greater than 1, they are not primitive.

2n − 1 = 49 gives n = 25.
Then (n − 1)² + 49 = n² becomes 24² + 7² = 25².
Since 576 + 49 = 625, the three side lengths are 7, 24, and 25.
Therefore, the resulting triple is (7, 24, 25).

In this method, the identity has the form (n − 1)² + an odd square = n².
Therefore, one side must equal n − 1 when the hypotenuse is n.
For (8, 15, 17), neither 8 nor 15 equals 16.
The other triples each have a side one less than the hypotenuse.

Fermat’s Last Theorem states that xⁿ + yⁿ = zⁿ has no positive-integer solution for integer powers n greater than 2.
The triple (3, 4, 5) works for squares because 3² + 4² = 5².
For cubes, 3³ + 4³ = 91, while 5³ = 125.

Let the lake’s depth be x units. The stem’s length is x + 1 units.
The tilted stem forms the hypotenuse of a right triangle with perpendicular sides x and 3.
x² + 3² = (x + 1)²
x² + 9 = x² + 2x + 1
2x = 8, so x = 4.
Therefore, the lake is 4 units deep.

Let the diagonal be d cm. It is the hypotenuse of a right triangle with perpendicular sides 5 cm each.
d² = 5² + 5² = 50
d = √50 = 5√2.
Therefore, the diagonal is 5√2 cm long.

The diagonals of a rhombus bisect each other at right angles. Their half-lengths are 12 and 35 units.
A side of the rhombus is the hypotenuse of the resulting right triangle.
s² = 12² + 35² = 144 + 1225 = 1369
s = 37 units.

The larger square has area 7² = 49 square units, and the smaller has area 5² = 25 square units.
Required area = 49 − 25 = 24 square units.
Let the new side be s units.
s² = 24, so s = √24 units.

A side joining grid dots has integer horizontal and vertical changes, say p and q.
The square’s area equals its side length squared, so the area is p² + q².
The areas 2, 4, and 5 equal 1² + 1², 2² + 0², and 1² + 2².
No two integer squares add to 3, so 3 square units is impossible.

The altitude of an equilateral triangle bisects its base, giving two base segments of 3 units.
Let the altitude be h units.
h² + 3² = 6², so h² = 27 and h = 3√3.
Area = (1/2) × Base × Height
= (1/2) × 6 × 3√3 = 9√3 square units.