CBSE Class 8 Mathematics Part II Chapter 2 Questions and Answers

Class 8 Maths

The Baudhayana-Pythagoras Theorem Class 8 Maths Ganita Prakash Part 2 Chapter 2 NCERT Solutions

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Figure it Out (Page 39)

1. Earlier, we saw a method to create a square with double the area of a given square paper. There is another method to do this in which two identical square papers are cut in the following way.

Can you arrange these pieces to create a square with double the area of either square?

Solution:
A diagonal divides a square into two equal-area triangles:
△1 = △2 , △3 = △4
Original square = △1 + △2 = △3 + △4
Resultant square = △1 + △2 + △3 + △4
Resultant square = 2 × Original Square
Hence, the given triangles can be arranged to create a square with double the area of either square.

2. The length of the two equal sides of an isosceles right triangle is given. Find the length of the hypotenuse. Find bounds on the length of the hypotenuse such that they have at least one digit after the decimal point.
(i) 3   (ii) 4   (iii) 6   (iv) 8   (v) 9

Solution:

Let a be the length of the equal sides of an isosceles triangle and c the length of the hypotenuse.
Area of SQVU = 2 × Area of PQRS
So, c² = 2a².
c = √(2a²) = a√2

(i) a = 3
Using the formula, we get:
c = a√2 = 3√2

(ii) a = 4
Using the formula, we get:
c = a√2 = 4√2

(iii) a = 6
c = a√2 = 6√2

(iv) a = 8
c = a√2 = 8√2

(v) a = 9
c = a√2 = 9√2

3. 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:

Let a be the length of the equal sides and 10 be the length of the hypotenuse of the isosceles right triangle.
Area of REST = 2 × Area of PEAR
10² = 2 × a²
100 = 2 × a²
50 = a²
a = √50 = √(5 × 5 × 2) = 5√2
Hence, the other two side lengths are 5√2 each.

Figure it Out (Page 47)

1. 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 Baudhāyana’s Theorem.

Solution:
Draw a right-angled triangle with side a = 5 cm and b = 12 cm. On measuring, the hypotenuse (c) is approximately 13 cm.

Using Baudhāyana’s Theorem:
a² + b² = c²
5² + 12² = c²
25 + 144 = c²
169 = c²
c = 13
Therefore, the hypotenuse is 13 cm.

2. If a right-angled triangle has a short side of length 8 cm and a 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 Baudhāyana’s Theorem.

Solution:
Draw a right-angled triangle with one side (a) 8 cm and hypotenuse (c) 17 cm. On measuring, the third side (b) will be close to 15 cm.

Using Baudhāyana’s Theorem:
a² + b² = c²
8² + b² = 17²
64 + b² = 289
b² = 289 – 64 = 225
b = 15
Therefore, the third side is 15 cm.

3. 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? (Baudhāyana’s Śulba-Sūtra, Verse 1.10)

Solution:
(A) Draw a square ABCD with side a.
Area of ABCD = a²
Join AC.
By Baudhāyana’s Theorem:
AC = √(a² + a²) = √(2a²) = a√2
At point C, draw a line perpendicular to AC.
Mark a point E such that CE = a.
Now in triangle ACE, using Baudhāyana’s Theorem:
AC² + CE² = AE²
(a√2)² + a² = AE²
2a² + a² = AE²
3a² = AE²
AE = √(3a²) = a√3
Construct square AEGH on side AE.
Area of AEGH = (a√3)² = 3a²
∴ Area of the new square = 3 × Area of the original square.

(B) Draw two squares ABCD and DCFE, each of sides a and area a².
Join them together to form a rectangle ABFE such that:
EF = a and BF = BC + CF = a + a = 2a
Draw the diagonal BE of the rectangle ABFE.
In △BFE, using Baudhāyana’s Theorem:
BF² + EF² = BE²
(2a)² + a² = BE²
4a² + a² = BE²
5a² = BE²
BE = √(5a²) = a√5
Construct a square BEGH using BE as one side.
Area of square BEGH = BE² = 5a²
∴ Area of BEGH = 5 × Area of ABCD.

4. Let a, b and c denote the lengths 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

Solution:

(i) a = 5, b = 7
Using Baudhāyana’s Theorem:
a² + b² = c²
5² + 7² = c²
25 + 49 = c² → c² = 74 → c = √74.

(ii) a = 8, b = 12
Using Baudhāyana’s Theorem:
a² + b² = c²
8² + 12² = c²
64 + 144 = c² → c² = 208
c = √208 = √(2 × 2 × 2 × 2 × 13) = 4√13.

(iii) a = 9, c = 15
Using Baudhāyana’s Theorem:
a² + b² = c²
9² + b² = 15²
81 + b² = 225 → b² = 225 – 81 = 144 → b = 12.

(iv) a = 7, b = 12
Using Baudhāyana’s Theorem:
a² + b² = c²
7² + 12² = c²
49 + 144 = c² → c² = 193 → c = √193.

(v) a = 1.5, b = 3.5
Using Baudhāyana’s Theorem:
a² + b² = c²
1.5² + 3.5² = c²
2.25 + 12.25 = c² → c² = 14.5 → c = √14.5.

Figure it Out (Page 50)

1. Find 5 more Baudhāyana triples using this idea.

Solution:

• (i) 49 is an odd square. It is the 25th odd number (49 = 2 × 25 – 1).
  So, (1 + 3 + 5 + … + 47) + 49 = 25² → 24² + 7² = 25² → (24, 7, 25)
• (ii) 81 is an odd square. It is the 41st odd number (81 = 2 × 41 – 1).
  So, (1 + 3 + 5 + … + 79) + 81 = 41² → 40² + 9² = 41² → (40, 9, 41)
• (iii) 121 is an odd number. It is the 61st odd number (121 = 2 × 61 – 1).
  So, (1 + 3 + 5 + … + 119) + 121 = 61² → 60² + 11² = 61² → (60, 11, 61)
• (iv) 169 is an odd square. It is the 85th odd number (169 = 2 × 85 – 1).
  So, (1 + 3 + 5 + … + 167) + 169 = 85² → 84² + 13² = 85² → (84, 13, 85)
• (v) 225 is an odd square. It is the 113th odd number (225 = 2 × 113 – 1).
  So, (1 + 3 + 5 + … + 223) + 225 = 113² → 112² + 15² = 113² → (112, 15, 113)

2. Does this method yield non-primitive Baudhāyana triples?
[Hint: Observe that among the triples generated, one of the smaller sidelengths is one less than the hypotenuse.]

Solution:
The method is based on the equation (n − 1)² + (2n − 1) = n².
If (2n – 1) is a perfect square, say k², then:
(n − 1)² + k² = n²
Thus, it generates triples of the form (n − 1, k, n).
Here, n and n − 1 are consecutive integers, so gcd(n, n − 1) = 1.
Also, k² = 2n − 1 is odd, so k is odd.
Thus, the three numbers have no common factor greater than 1.
Hence, all triples obtained by this method are primitive (it does not yield non-primitive triples).

3. Are there primitive triples that cannot be obtained through this method? If yes, give examples.

Solution:
The method generates triples of the form (n − 1, k, n), meaning one side is always one less than the hypotenuse.
However, not all primitive triples satisfy this condition.
For example, (8, 15, 17) and (20, 21, 29) do not satisfy this condition.
Hence, such primitive triples cannot be generated by this method.

Figure it Out (Page 52)

1. Find the diagonal of a square with sidelength 5 cm.

Solution:
Let the sides of the square be 5 cm, and its diagonal be d.
Since the diagonal forms a right triangle with the sides, using Baudhāyana’s Theorem:
5² + 5² = d²
25 + 25 = d² → d² = 50
d = √50 = 5√2 ≈ 5 × 1.414 = 7.07 cm.
Therefore, the diagonal of the square is 7.07 cm.

2. Find the missing sidelengths in the following right triangles:

Solution:

(i)

Using Baudhāyana’s Theorem:
7² + 9² = a²
49 + 81 = a² → a² = 130 → a = √130.

(ii)

Using Baudhāyana’s Theorem:
b² + 40² = 41²
b² + 1600 = 1681
b² = 1681 – 1600 = 81 → b = 9.

(iii)

Using Baudhāyana’s Theorem:
10² + (√150)² = c²
100 + 150 = c² → c² = 250
c = √250 = √(5 × 5 × 5 × 2) = 5√10.

(iv)

Using Baudhāyana’s Theorem:
4² + 10² = d²
16 + 100 = d² → d² = 116 → d = √116 (or 2√29).

(v)

Using Baudhāyana’s Theorem:
10² + e² = (√200)²
100 + e² = 200
e² = 200 – 100 = 100 → e = 10.

(vi)

Using Baudhāyana’s Theorem:
f² + 27² = 45²
f² + 729 = 2025
f² = 2025 – 729 = 1296 → f = 36.

3. Find the sidelength of a rhombus whose diagonals are of length 24 units and 70 units.

Solution:
Diagonal 1 = 24 units; Diagonal 2 = 70 units.
In a rhombus, diagonals bisect each other at right angles.
Semi-diagonals = 24/2 = 12 units and 70/2 = 35 units.
Using Baudhāyana’s Theorem:
12² + 35² = Side²
144 + 1225 = Side²
1369 = Side² → Side = √1369 = 37 units.
Therefore, the side length of the rhombus is 37 units.

4. Is the hypotenuse the longest side of a right triangle? Justify your answer.

Solution:
If the legs are a and b and hypotenuse is c, then:
c² = a² + b²
Since a² > 0 and b² > 0, c² > a² and c² > b².
Taking positive square roots: c > a and c > b.
Therefore, the hypotenuse is always the longest side.

5. True or False — Every Baudhāyana triple is either a primitive triple or a scaled version of a primitive triple.

Solution:
True. Every Baudhāyana triple can be expressed as a primitive triple multiplied by a common integer factor k: (a, b, c) = (kx, ky, kz).

6. Give 5 examples of rectangles whose sidelengths and diagonals are all integers.

Solution:
Using Baudhāyana triples (length, breadth, diagonal):

1. 3 × 4, diagonal = 5
2. 5 × 12, diagonal = 13
3. 8 × 15, diagonal = 17
4. 7 × 24, diagonal = 25
5. 20 × 21, diagonal = 29

7. Construct a square whose area is equal to the difference of the areas of squares of sidelengths 5 units and 7 units.

Solution:
Area of the required square = 7² – 5² = 49 – 25 = 24 sq. units.
(i) Construct square PQRS with side 2 units → Diagonal PR = √(2² + 2²) = 2√2 units.
(ii) At R, draw RT ⊥ PR such that RT = 4 units.
(iii) Join PT. By Baudhāyana’s Theorem:
PR² + RT² = PT² → (2√2)² + 4² = PT² → 8 + 16 = 24 = PT² → PT = √24.
(iv) Construct square PTUV on side PT.
Area of PTUV = (√24)² = 24 sq. units.

8. Find the area of an equilateral triangle with sidelength 6 units.
[Hint: Show that an altitude bisects the opposite side. Use this to find the height.]

Solution:
Let △ABC be an equilateral triangle with sides = 6 units.
Let AD be the altitude on BC.
By RHS congruence (△ADB ≅ △ADC), BD = CD = 6 / 2 = 3 units.
In △ADC, using Baudhāyana’s Theorem:
CD² + AD² = AC²
3² + AD² = 6²
9 + AD² = 36 → AD² = 27 → AD = √27 = 3√3 units.

Area of △ABC = (1/2) × Base × Height = (1/2) × 6 × 3√3 = 9√3 sq. units.