Formula: Baudhāyana's Theorem: a^2 + b^2 = c^2. (i) a = 5, b = 7 We need to find c. c^2 = 5^2 + 7^2 = 25 + 49 = 74. c = √(74). (ii) a = 8, b = 12 We need to find c. c^2 = 8^2 + 12^2 = 64 + 144 = 208. c = √(208) = √(16 × 13) = 4√(13). (iii) a = 9, c = 15 We need to find b. b^2 = c^2 - a^2 = 15^2 -…

The Baudhayana-Pythagoras Theorem Class 8 Mathematics NCERT Solutions

Q1Figure it Out 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

Given: Two identical square papers. Let the side length of each square be 's'. Each square is cut into two identical rectangles. The images suggest the cut is made through the middle, so each of the four resulting pieces is a rectangle of dimensions

s \times \frac{s}{2}

. The total area of the four pieces is

s^2 + s^2 = 2s^2

.

To Do: Arrange these four rectangular pieces to form a single square. The new square must have an area of

2s^2

.

Solution: Yes, the pieces can be arranged to form a larger square. The arrangement is a 'pinwheel' pattern that forms a larger square with a square-shaped hole in the center.

Steps for Arrangement:

Take one rectangle and place it with its longer side 's' oriented vertically.
Take a second rectangle and place it to the right of the first one, but rotated 90 degrees, so its shorter side 's/2' is vertical. Align the bottom edges of the two rectangles.
Take a third rectangle and place it below the second one, rotated so its longer side 's' is horizontal. Align the right edges of the second and third rectangles.
Take the fourth rectangle and place it to the left of the third one, rotated so its shorter side 's/2' is horizontal. Its top edge will align with the bottom edge of the first rectangle, and its left edge will align with the left edge of the first rectangle.

Result: This arrangement forms a large square with an empty square hole in the middle.

The outer side length of this new square is $s + \frac{s}{2} = \frac{3s}{2}$ .
The side length of the inner empty square is $s - \frac{s}{2} = \frac{s}{2}$ .
The area of the arrangement is the area of the large square minus the area of the hole: Area = $(\frac{3s}{2})^2 - (\frac{s}{2})^2 = \frac{9s^2}{4} - \frac{s^2}{4} = \frac{8s^2}{4} = 2s^2$ .

This area is double the area of one of the original squares. Thus, the pieces have been arranged to form a square shape with double the area.

Q2Figure it Out 1

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

Given: An isosceles right triangle with two equal sides of length 'a'. To Find: The length of the hypotenuse 'c', and bounds for 'c' with one decimal place.

Formula: For an isosceles right triangle, the relationship between the equal sides 'a' and the hypotenuse 'c' is given by the Baudhāyana-Pythagoras theorem as

c^2 = a^2 + a^2 = 2a^2

. Therefore,

c = a\sqrt{2}

. We use the approximation

\sqrt{2} \approx 1.414

.

Solutions:

(i) a = 3

Length of hypotenuse: $c = 3\sqrt{2}$ .
Value: $c \approx 3 \times 1.414 = 4.242$ .
Bounds: We check squares of numbers around 4.2. $4.2^2 = 17.64$ and $4.3^2 = 18.49$ . Since $c^2 = 3^2 \times 2 = 18$ , we have $17.64 < 18 < 18.49$ . Therefore, $4.2 < c < 4.3$ .

(ii) a = 4

Length of hypotenuse: $c = 4\sqrt{2}$ .
Value: $c \approx 4 \times 1.414 = 5.656$ .
Bounds: We check squares of numbers around 5.6. $5.6^2 = 31.36$ and $5.7^2 = 32.49$ . Since $c^2 = 4^2 \times 2 = 32$ , we have $31.36 < 32 < 32.49$ . Therefore, $5.6 < c < 5.7$ .

(iii) a = 6

Length of hypotenuse: $c = 6\sqrt{2}$ .
Value: $c \approx 6 \times 1.414 = 8.484$ .
Bounds: We check squares of numbers around 8.4. $8.4^2 = 70.56$ and $8.5^2 = 72.25$ . Since $c^2 = 6^2 \times 2 = 72$ , we have $70.56 < 72 < 72.25$ . Therefore, $8.4 < c < 8.5$ .

(iv) a = 8

Length of hypotenuse: $c = 8\sqrt{2}$ .
Value: $c \approx 8 \times 1.414 = 11.312$ .
Bounds: We check squares of numbers around 11.3. $11.3^2 = 127.69$ and $11.4^2 = 129.96$ . Since $c^2 = 8^2 \times 2 = 128$ , we have $127.69 < 128 < 129.96$ . Therefore, $11.3 < c < 11.4$ .

(v) a = 9

Length of hypotenuse: $c = 9\sqrt{2}$ .
Value: $c \approx 9 \times 1.414 = 12.726$ .
Bounds: We check squares of numbers around 12.7. $12.7^2 = 161.29$ and $12.8^2 = 163.84$ . Since $c^2 = 9^2 \times 2 = 162$ , we have $161.29 < 162 < 163.84$ . Therefore, $12.7 < c < 12.8$ .

Q3Figure it Out 1

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

Given: An isosceles right triangle with hypotenuse

c = 10

.

To Find: The length of the other two equal sidelengths, 'a'.

Formula: For an isosceles right triangle,

c^2 = 2a^2

.

Solution: We substitute the given value of the hypotenuse into the formula:

10^2 = 2a^2

100 = 2a^2

Now, we solve for

a^2

:

a^2 = \frac{100}{2}

a^2 = 50

Finally, we find 'a' by taking the square root:

a = \sqrt{50}

To simplify the square root, we can factor 50:

a = \sqrt{25 \times 2}

a = 5\sqrt{2}

Using the hint: The hint suggests finding the area of the square whose diagonal is the hypotenuse. The area of the square built on the hypotenuse is

c^2 = 10^2 = 100

. This area is double the area of the square built on the side 'a'. So, the area of the square with side 'a' is

a^2 = \frac{100}{2} = 50

. This gives

a = \sqrt{50} = 5\sqrt{2}

.

Final Answer: The length of each of the other two sidelengths is

5\sqrt{2}

units.

Q1Figure it Out 2

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

Given: A right-angled triangle with shorter sides

a = 5 \text{ cm}

and

b = 12 \text{ cm}

.

To Find: The length of the hypotenuse, 'c'.

Formula: Baudhāyana's Theorem:

a^2 + b^2 = c^2

.

Solution: We substitute the given side lengths into the formula:

c^2 = 5^2 + 12^2

c^2 = 25 + 144

c^2 = 169

To find 'c', we take the square root of 169:

c = \sqrt{169}

c = 13

The student is also asked to draw the triangle and measure the hypotenuse, which should give a measurement of approximately 13 cm.

Final Answer: The length of the hypotenuse is 13 cm.

Q2Figure it Out 2

If a right-angled triangle has a short side of length 8 cm and 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

Given: A right-angled triangle with one short side

a = 8 \text{ cm}

and hypotenuse

c = 17 \text{ cm}

.

To Find: The length of the other short side, 'b'.

Formula: Baudhāyana's Theorem:

a^2 + b^2 = c^2

, which can be rearranged to

b^2 = c^2 - a^2

.

Solution: We substitute the given lengths into the rearranged formula:

b^2 = 17^2 - 8^2

b^2 = 289 - 64

b^2 = 225

To find 'b', we take the square root of 225:

b = \sqrt{225}

b = 15

The student is also asked to draw the triangle and measure the side, which should give a measurement of approximately 15 cm.

Final Answer: The length of the third side is 15 cm.

Q3Figure it Out 2

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

Given: A square with side length 'a' and area

A = a^2

.

1. To Construct a Square with Triple the Area ( $3a^2$ ):

The side of the new square must be

s_3 = \sqrt{3a^2} = a\sqrt{3}

. We can construct this length using the Baudhāyana-Pythagoras theorem twice.

Step 1: Construct a length of $a\sqrt{2}$ . Draw the diagonal of the original square of side 'a'. The length of this diagonal is $d = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2}$ .
Step 2: Construct a length of $a\sqrt{3}$ . Construct a new right-angled triangle where the two shorter sides are the side of the original square ('a') and the diagonal we just found (' $a\sqrt{2}$ '). The hypotenuse 'c' of this new triangle will have a length given by: $c^2 = a^2 + (a\sqrt{2})^2 = a^2 + 2a^2 = 3a^2$ $c = \sqrt{3a^2} = a\sqrt{3}$
Step 3: Construct the final square. Construct a square with side length 'c' (the hypotenuse from Step 2). The area of this square will be $c^2 = 3a^2$ , which is triple the area of the original square.

2. To Construct a Square with Five Times the Area ( $5a^2$ ):

The side of the new square must be

s_5 = \sqrt{5a^2} = a\sqrt{5}

.

Step 1: Construct a length of '2a'. Double the side 'a' of the original square to get a length of '2a'.
Step 2: Construct a length of $a\sqrt{5}$ . Construct a right-angled triangle with the shorter sides having lengths 'a' and '2a'. The hypotenuse 'c' of this triangle will have a length given by: $c^2 = a^2 + (2a)^2 = a^2 + 4a^2 = 5a^2$ $c = \sqrt{5a^2} = a\sqrt{5}$
Step 3: Construct the final square. Construct a square on the hypotenuse 'c' from Step 2. The area of this square will be $c^2 = 5a^2$ , which is five times the area of the original square.

Q4Figure it Out 2

Let a, b and c denote the length 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

Formula: Baudhāyana's Theorem:

a^2 + b^2 = c^2

.

(i) a = 5, b = 7

We need to find c.
$c^2 = 5^2 + 7^2 = 25 + 49 = 74$ .
$c = \sqrt{74}$ .

(ii) a = 8, b = 12

We need to find c.
$c^2 = 8^2 + 12^2 = 64 + 144 = 208$ .
$c = \sqrt{208} = \sqrt{16 \times 13} = 4\sqrt{13}$ .

(iii) a = 9, c = 15

We need to find b.
$b^2 = c^2 - a^2 = 15^2 - 9^2 = 225 - 81 = 144$ .
$b = \sqrt{144} = 12$ .

(iv) a = 7, b = 12

We need to find c.
$c^2 = 7^2 + 12^2 = 49 + 144 = 193$ .
$c = \sqrt{193}$ . (193 is a prime number).

(v) a = 1.5, b = 3.5

We need to find c.
$c^2 = (1.5)^2 + (3.5)^2 = 2.25 + 12.25 = 14.5$ .
$c = \sqrt{14.5} = \sqrt{\frac{29}{2}}$ .

Final Answers:

(i)

c = \sqrt{74}

(ii)

c = 4\sqrt{13}

(iii)

b = 12

(iv)

c = \sqrt{193}

(v)

c = \sqrt{14.5}

Q1Figure it Out 3

Find 5 more Baudhāyana triples using this idea.

Solution

Given: The method described in the text generates Baudhāyana triples using the identity

(n-1)^2 + (2n-1) = n^2

. A triple is found when the term

(2n-1)

is a perfect square. Let

2n-1 = k^2

, where 'k' is an odd integer. This means

n = \frac{k^2+1}{2}

. The resulting triple is

(n-1, \sqrt{2n-1}, n)

, which simplifies to

(\frac{k^2-1}{2}, k, \frac{k^2+1}{2})

. Conventionally, we list the smaller sides first.

The text has already used

k=3

to get

(4, 3, 5)

or

(3, 4, 5)

, and

k=5

to get

(12, 5, 13)

or

(5, 12, 13)

. We will find 5 more by choosing the next 5 odd integers for 'k'.

Solution:

Let k = 7: $a = k = 7$ $b = \frac{7^2-1}{2} = \frac{49-1}{2} = \frac{48}{2} = 24$ $c = \frac{7^2+1}{2} = \frac{49+1}{2} = \frac{50}{2} = 25$ Triple: (7, 24, 25)
Let k = 9: $a = k = 9$ $b = \frac{9^2-1}{2} = \frac{81-1}{2} = \frac{80}{2} = 40$ $c = \frac{9^2+1}{2} = \frac{81+1}{2} = \frac{82}{2} = 41$ Triple: (9, 40, 41)
Let k = 11: $a = k = 11$ $b = \frac{11^2-1}{2} = \frac{121-1}{2} = \frac{120}{2} = 60$ $c = \frac{11^2+1}{2} = \frac{121+1}{2} = \frac{122}{2} = 61$ Triple: (11, 60, 61)
Let k = 13: $a = k = 13$ $b = \frac{13^2-1}{2} = \frac{169-1}{2} = \frac{168}{2} = 84$ $c = \frac{13^2+1}{2} = \frac{169+1}{2} = \frac{170}{2} = 85$ Triple: (13, 84, 85)
Let k = 15: $a = k = 15$ $b = \frac{15^2-1}{2} = \frac{225-1}{2} = \frac{224}{2} = 112$ $c = \frac{15^2+1}{2} = \frac{225+1}{2} = \frac{226}{2} = 113$ Triple: (15, 112, 113)

Q2Figure it Out 3

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

Answer: No, this method does not yield non-primitive Baudhāyana triples. It only generates primitive triples.

Reasoning: A Baudhāyana triple is primitive if the greatest common divisor (GCD) of its three numbers is 1. The triples generated by this method are of the form

(k, \frac{k^2-1}{2}, \frac{k^2+1}{2})

, where k is an odd integer. Let's denote the two larger numbers as

b = \frac{k^2-1}{2}

and

c = \frac{k^2+1}{2}

.

As the hint suggests, let's find the difference between the hypotenuse 'c' and the longer leg 'b':

c - b = \frac{k^2+1}{2} - \frac{k^2-1}{2} = \frac{(k^2+1) - (k^2-1)}{2} = \frac{2}{2} = 1

Since the difference between two of the numbers in the triple (b and c) is 1, they cannot have any common divisor greater than 1. In other words,

\text{gcd}(b, c) = 1

.

If a triple

(a, b, c)

had a common divisor

d > 1

, then

d

would have to divide all three numbers. This means

d

would have to divide

b

and

c

. But we have just shown that the only common divisor of

b

and

c

is 1. Therefore, the only common divisor for the entire triple is 1.

Thus, all triples generated by this specific method are primitive.

Q3Figure it Out 3

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

Solution

Answer: Yes, there are primitive Baudhāyana triples that cannot be obtained through this method.

Reasoning: The method described generates triples of the form

(k, \frac{k^2-1}{2}, \frac{k^2+1}{2})

. A key characteristic of these triples is that the hypotenuse and the longer of the two shorter sides are consecutive integers (their difference is 1).

However, there exist primitive Baudhāyana triples where this is not the case. We can find primitive triples where the difference between the hypotenuse and the next longest side is greater than 1.

Examples:

The triple (8, 15, 17):
- This is a Baudhāyana triple because $8^2 + 15^2 = 64 + 225 = 289 = 17^2$ .
- It is primitive because the greatest common divisor of 8, 15, and 17 is 1.
- The difference between the hypotenuse and the next longest side is $17 - 15 = 2$ . Since the difference is not 1, this triple cannot be generated by the given method.
The triple (20, 21, 29):
- This is a Baudhāyana triple because $20^2 + 21^2 = 400 + 441 = 841 = 29^2$ .
- It is primitive because $\text{gcd}(20, 21, 29) = 1$ .
- The difference between the hypotenuse and the next longest side is $29 - 21 = 8$ . This triple also cannot be generated by the given method.

Therefore, the method is a way to generate one specific family of primitive triples, but not all of them.

Q1Figure it Out 4

Find the diagonal of a square with sidelength 5 cm.

Solution

Given: A square with side length

s = 5 \text{ cm}

.

To Find: The length of the diagonal, 'd'.

Solution: The diagonal of a square divides it into two congruent isosceles right-angled triangles. The sides of the square are the two shorter sides of the triangle, and the diagonal is the hypotenuse.

Using the Baudhāyana-Pythagoras theorem, where

a=s

and

b=s

:

d^2 = s^2 + s^2 = 2s^2

Substitute the given side length

s=5

cm:

d^2 = 2 \times 5^2

d^2 = 2 \times 25

d^2 = 50

Now, take the square root to find 'd':

d = \sqrt{50}

d = \sqrt{25 \times 2}

d = 5\sqrt{2}

Final Answer: The length of the diagonal is

5\sqrt{2}

cm.

Q2Figure it Out 4

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

Solution

Given: A rhombus with diagonals

d_1 = 24

units and

d_2 = 70

units.

To Find: The sidelength of the rhombus, 's'.

Properties of a Rhombus: The diagonals of a rhombus are perpendicular bisectors of each other. This means they intersect at a right angle (

90^\circ

) and cut each other into two equal halves.

Solution: The diagonals divide the rhombus into four congruent right-angled triangles. The hypotenuse of each triangle is a side of the rhombus. The other two sides of each triangle are half the lengths of the diagonals.

Half of the first diagonal: $\frac{d_1}{2} = \frac{24}{2} = 12$ units.
Half of the second diagonal: $\frac{d_2}{2} = \frac{70}{2} = 35$ units.

Let these be the shorter sides, 'a' and 'b', of one of the right-angled triangles. The side of the rhombus 's' is the hypotenuse 'c'.

Using the Baudhāyana-Pythagoras theorem,

s^2 = a^2 + b^2

:

s^2 = 12^2 + 35^2

s^2 = 144 + 1225

s^2 = 1369

To find 's', we take the square root:

s = \sqrt{1369}

We can test values.

30^2 = 900

and

40^2 = 1600

. The number ends in 9, so the root must end in 3 or 7. Let's try 37:

37^2 = 1369

. So,

s = 37

.

Final Answer: The sidelength of the rhombus is 37 units.

Q3Figure it Out 4

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

Solution

Answer: Yes, the hypotenuse is always the longest side of a right-angled triangle.

Justification: Let a right-angled triangle have shorter sides of length 'a' and 'b', and a hypotenuse of length 'c'. The lengths a, b, and c must all be positive numbers.

According to the Baudhāyana-Pythagoras theorem:

c^2 = a^2 + b^2

Compare c with a: Since 'b' is a length, $b > 0$ , which means $b^2 > 0$ . Therefore, $c^2 = a^2 + b^2 > a^2$ . Since both 'c' and 'a' are positive, we can take the square root of both sides of the inequality: $\sqrt{c^2} > \sqrt{a^2}$ $c > a$
Compare c with b: Similarly, since 'a' is a length, $a > 0$ , which means $a^2 > 0$ . Therefore, $c^2 = a^2 + b^2 > b^2$ . Taking the square root of both sides: $\sqrt{c^2} > \sqrt{b^2}$ $c > b$

Since the length of the hypotenuse 'c' is greater than the length of side 'a' and also greater than the length of side 'b', it is the longest side of the right-angled triangle.

Q4Figure it Out 4

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

Solution

Answer: True.

Reasoning: Let

(a, b, c)

be any Baudhāyana triple. By definition, a, b, and c are positive integers such that

a^2 + b^2 = c^2

.

Let

d = \text{gcd}(a, b, c)

, which is the greatest common divisor of the three integers.

Case 1: The triple is primitive. If $d = 1$ , the numbers have no common factor greater than 1. By definition, the triple $(a, b, c)$ is a primitive Baudhāyana triple.
Case 2: The triple is not primitive. If $d > 1$ , the triple is not primitive. Since $d$ is a common divisor, we can write: $a = d \cdot a'$ , $b = d \cdot b'$ , and $c = d \cdot c'$ , where $a'$ , $b'$ , and $c'$ are integers.

Now, let's check if the new triple $(a', b', c')$ is a Baudhāyana triple. We substitute the expressions for a, b, and c into the theorem: $(d a')^2 + (d b')^2 = (d c')^2$ $d^2 (a')^2 + d^2 (b')^2 = d^2 (c')^2$ Dividing the entire equation by $d^2$ (which is not zero): $(a')^2 + (b')^2 = (c')^2$ This shows that $(a', b', c')$ is also a Baudhāyana triple.

Furthermore, the greatest common divisor of $a'$ , $b'$ , and $c'$ is $\text{gcd}(\frac{a}{d}, \frac{b}{d}, \frac{c}{d}) = \frac{\text{gcd}(a,b,c)}{d} = \frac{d}{d} = 1$ . This means the triple $(a', b', c')$ is primitive.

Therefore, the original non-primitive triple $(a, b, c)$ is a scaled version of the primitive triple $(a', b', c')$ , with a scaling factor of $d$ .

Conclusion: Every Baudhāyana triple is either primitive itself or is a multiple (a scaled version) of a primitive triple.

Q5Figure it Out 4

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

Solution

Concept: For a rectangle to have integer sidelengths and an integer diagonal, the two sidelengths (a, b) and the diagonal (c) must form a right-angled triangle. Therefore, we need to find sets of three integers (a, b, c) that satisfy the Baudhāyana-Pythagoras theorem,

a^2 + b^2 = c^2

. These sets are known as Baudhāyana triples.

Examples: Here are 5 examples based on common Baudhāyana triples:

Triple (3, 4, 5): A rectangle with sidelengths 3 units and 4 units. Its diagonal will be 5 units, since $3^2 + 4^2 = 9 + 16 = 25 = 5^2$ .
Triple (5, 12, 13): A rectangle with sidelengths 5 units and 12 units. Its diagonal will be 13 units, since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$ .
Triple (8, 15, 17): A rectangle with sidelengths 8 units and 15 units. Its diagonal will be 17 units, since $8^2 + 15^2 = 64 + 225 = 289 = 17^2$ .
Triple (7, 24, 25): A rectangle with sidelengths 7 units and 24 units. Its diagonal will be 25 units, since $7^2 + 24^2 = 49 + 576 = 625 = 25^2$ .
Triple (6, 8, 10): (This is a scaled version of the 3, 4, 5 triple) A rectangle with sidelengths 6 units and 8 units. Its diagonal will be 10 units, since $6^2 + 8^2 = 36 + 64 = 100 = 10^2$ .

Q6Figure it Out 4

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

Solution

Problem Analysis:

Area of the larger square (side 7): $7^2 = 49$ square units.
Area of the smaller square (side 5): $5^2 = 25$ square units.
Difference in areas: $49 - 25 = 24$ square units.

We need to construct a square with an area of 24 square units. Let the side of this required square be 'a'. Then

a^2 = 24

. This can be written as

a^2 = 7^2 - 5^2

, or

a^2 + 5^2 = 7^2

. This is the Baudhāyana-Pythagoras theorem for a right-angled triangle with shorter side 'a', other shorter side 5, and hypotenuse 7.

So, the problem is to construct the side 'a' of such a triangle.

Steps of Construction:

Draw a straight line, let's call it L.
Mark a point B on the line L. At point B, construct a line segment BA perpendicular to L, with length 5 units.
Set your compass to a radius of 7 units. Place the compass point at A and draw an arc that intersects the line L.
Let the point of intersection on line L be C. (Note: The arc will intersect the line at two points; either can be chosen as C).
Join A and C. The triangle ABC is a right-angled triangle with $\angle B = 90^\circ$ , side AB = 5, and hypotenuse AC = 7. The side BC is the required length 'a'.
By construction and the theorem, the length of BC satisfies $BC^2 + 5^2 = 7^2$ , so $BC^2 = 24$ .
To construct the final square, measure the length of the segment BC. Construct a square with this side length. This square will have an area of 24 square units.

Q7Figure it Out 4

(i) Using the dots of a grid as the vertices, can you create a square that has an area of (a) 2 sq. units,…

(i)

Using the dots of a grid as the vertices, can you create a square that has an area of (a) 2 sq. units, (b) 3 sq. units, (c) 4 sq.units, and (d) 5 sq. unit?

(ii)

Suppose the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner?

Solution

Concept: Let the grid points be integer coordinates

(x, y)

. The side of a square on the grid connects two points, say

(x_1, y_1)

and

(x_2, y_2)

. The length of the side squared, which is the area of the square, is given by the distance formula squared: Area =

(x_2 - x_1)^2 + (y_2 - y_1)^2

. Let

\Delta x = x_2 - x_1

and

\Delta y = y_2 - y_1

. These are integers representing the horizontal and vertical separation of the vertices of a side. The area of the square is

\Delta x^2 + \Delta y^2

. So, an integer area is possible if and only if it can be written as the sum of two perfect squares.

(i) Checking Specific Areas:

(a) Area = 2 sq. units: Can we find integers $\Delta x, \Delta y$ such that $\Delta x^2 + \Delta y^2 = 2$ ? Yes, if $\Delta x = 1$ and $\Delta y = 1$ , then $1^2 + 1^2 = 2$ . This corresponds to a 'tilted' square whose side connects a point to its diagonal neighbor. Yes, it is possible.
(b) Area = 3 sq. units: Can we find integers $\Delta x, \Delta y$ such that $\Delta x^2 + \Delta y^2 = 3$ ? The only perfect squares less than 3 are 0 and 1. No sum of two of these (0+0, 0+1, 1+1) equals 3. No, it is not possible.
(c) Area = 4 sq. units: Can we find integers $\Delta x, \Delta y$ such that $\Delta x^2 + \Delta y^2 = 4$ ? Yes, if $\Delta x = 2$ and $\Delta y = 0$ , then $2^2 + 0^2 = 4$ . This corresponds to an 'axis-aligned' square with side length 2. Yes, it is possible.
(d) Area = 5 sq. units: Can we find integers $\Delta x, \Delta y$ such that $\Delta x^2 + \Delta y^2 = 5$ ? Yes, if $\Delta x = 2$ and $\Delta y = 1$ (or vice versa), then $2^2 + 1^2 = 4+1=5$ . Yes, it is possible.

(ii) Possible Integer-Valued Areas:

The possible integer-valued areas of squares that can be created on a grid are all the integers that can be expressed as the sum of two perfect squares (including 0).

Examples of possible areas: $1 = 1^2 + 0^2$ $2 = 1^2 + 1^2$ $4 = 2^2 + 0^2$ $5 = 2^2 + 1^2$ $8 = 2^2 + 2^2$ $9 = 3^2 + 0^2$ $10 = 3^2 + 1^2$ $13 = 3^2 + 2^2$
Examples of impossible areas: 3, 6, 7, 11, 12, 15, 19, 21, 22, 23, 24.

(Note: A number can be written as the sum of two squares if and only if its prime factorization contains no prime of the form

4k+3

raised to an odd power.)

Q8Figure it Out 4

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

Given: An equilateral triangle with side length

s = 6

units.

To Find: The area of the triangle.

Solution:

Step 1: Find the height of the triangle. Let the equilateral triangle be ABC, with

AB = BC = CA = 6

. Draw an altitude (height) from vertex A to the opposite side BC. Let the point where it meets BC be D.

Justification from the hint: In $\triangle ADB$ and $\triangle ADC$ , we have:
- $AB = AC$ (sides of an equilateral triangle)
- $AD = AD$ (common side)
- $\angle ADB = \angle ADC = 90^\circ$ (AD is an altitude) By the Right-angle-Hypotenuse-Side (RHS) congruence criterion, $\triangle ADB \cong \triangle ADC$ . Therefore, $BD = DC$ . This shows the altitude bisects the base BC.
Since $BC = 6$ , we have $BD = \frac{6}{2} = 3$ units.
Now, consider the right-angled triangle ADB. The sides are $AD$ (the height, h), $BD=3$ , and the hypotenuse $AB=6$ . Using the Baudhāyana-Pythagoras theorem: $AD^2 + BD^2 = AB^2$ $h^2 + 3^2 = 6^2$ $h^2 + 9 = 36$ $h^2 = 36 - 9 = 27$ $h = \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3}$ units.

Step 2: Calculate the area of the triangle. The formula for the area of a triangle is

\frac{1}{2} \times \text{base} \times \text{height}

.

Base = $BC = 6$ units
Height = $h = 3\sqrt{3}$ units

Area = \frac{1}{2} \times 6 \times 3\sqrt{3}

Area = 3 \times 3\sqrt{3}

Area = 9\sqrt{3}

Final Answer: The area of the equilateral triangle is

9\sqrt{3}

square units.

The Baudhayana-Pythagoras TheoremClass 8 Mathematics NCERT Solutions

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

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.

3

4

6

8

9

Solution

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

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

If a right-angled triangle has a short side of length 8 cm and 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

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

Let a, b and c denote the length 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:

a=5, b=7

a=8, b=12

a=9, c=15

a=7, b=12

a=1.5, b=3.5

Solution

Find 5 more Baudhāyana triples using this idea.

Solution

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

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

Solution

Find the diagonal of a square with sidelength 5 cm.

Solution

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

Solution

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

Solution

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

Solution

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

Solution

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

Solution

(i) Using the dots of a grid as the vertices, can you create a square that has an area of (a) 2 sq. units,…

Using the dots of a grid as the vertices, can you create a square that has an area of (a) 2 sq. units, (b) 3 sq. units, (c) 4 sq.units, and (d) 5 sq. unit?

Suppose the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner?

Solution

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