Chapter 2 The Baudhāyana-Pythagoras Theorem (Class 8 - Latest Maths NCERT (Ganita Prakash II) Solutions)

To Find:

A method to arrange the four triangular pieces (1, 2, 3, and 4) obtained from two identical squares to form a single larger square with double the area.

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Solution:

Each original square is cut along its diagonal into two identical isosceles right-angled triangles. Since we have two such squares, we have a total of four identical triangles labeled 1, 2, 3, and 4.

To create a square with double the area, we can arrange them as follows:

1. Place the four triangles such that their right-angled vertices (the $90^\circ$ corners) meet at a single central point.

2. The hypotenuse (the longest side) of each triangle will now form the outer boundary or the "sides" of the new larger square.

3. Because we have used the entire area of two squares to form this new shape without any overlaps or gaps, the area of the resulting square is exactly double the area of one original square.

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Indian Perspective:

This geometric transformation is a practical application of the Baudhayana-Pythagoras Theorem found in the ancient Indian Sulba Sutras. Indian mathematicians used such "cut and move" methods (known as Yuktibhasha logic) to demonstrate that the area of a square built on the diagonal of another square is twice the area of the original square.

Given:

The length of two equal sides (legs) of an isosceles right triangle is $a$.

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To Find:

The length of the hypotenuse $h$ and its decimal bounds.

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Solution:

By the Baudhayana-Pythagoras Theorem, for a right-angled triangle with sides $a, a$ and hypotenuse $h$:

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(i) For side $a = 3$:

$h = 3\sqrt{2} = \sqrt{18}$

Since $4.2^2 = 17.64$ and $4.3^2 = 18.49$, we have:

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(ii) For side $a = 4$:

$h = 4\sqrt{2} = \sqrt{32}$

Since $5.6^2 = 31.36$ and $5.7^2 = 32.49$, we have:

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(iii) For side $a = 6$:

$h = 6\sqrt{2} = \sqrt{72}$

Since $8.4^2 = 70.56$ and $8.5^2 = 72.25$, we have:

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(iv) For side $a = 8$:

$h = 8\sqrt{2} = \sqrt{128}$

Since $11.3^2 = 127.69$ and $11.4^2 = 129.96$, we have:

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(v) For side $a = 9$:

$h = 9\sqrt{2} = \sqrt{162}$

Since $12.7^2 = 161.29$ and $12.8^2 = 163.84$, we have:

Given:

Hypotenuse $h = 10$ units.

The triangle is an isosceles right-angled triangle.

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To Find:

The length of the other two equal sides (legs), let's call them $x$.

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Solution:

Let the length of each of the two equal sides be $x$.

By the Baudhayana-Pythagoras Theorem:

To find the value of $\sqrt{50}$:

We know $7^2 = 49$ and $8^2 = 64$. So $\sqrt{50}$ is slightly more than $7$.

Approximating to two decimal places: $x \approx 7.07$ units.

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Alternate Solution (Using the Hint):

If we take two such isosceles right triangles and join them along their hypotenuses, they form a square with side length $x$ and diagonal $10$.

Area of this square = $x^2$.

Also, the area of a square with diagonal $d$ is given by $\frac{1}{2}d^2$.

Area = $\frac{1}{2} \times 10^2 = \frac{100}{2} = 50$.

Since Area = $x^2 = 50$, then $x = \sqrt{50} \approx 7.07$.

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Indian Perspective:

This problem demonstrates the relationship between the side of a Chaturasra (square) and its Akshnaya-rajju (diagonal) as described in the Baudhayana Sulba Sutras. Ancient Indian architects used these ratios to construct sacrificial altars (Vedi) with precise dimensions.

Given:

Length of the first shorter side (base) = $5$ cm

Length of the second shorter side (perpendicular) = $12$ cm

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To Find:

The length of the hypotenuse.

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Construction and Measurement:

1. Draw a horizontal line segment $AB = 5$ cm.

2. At point $A$, construct a $90^\circ$ angle and draw a vertical segment $AC = 12$ cm.

3. Join points $B$ and $C$ to form the hypotenuse.

4. Upon measuring the segment $BC$ with a ruler, we find that the length is exactly $13$ cm.

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Solution (Verification using Baudhāyana’s Theorem):

According to Baudhāyana’s Theorem (popularly known as the Pythagoras Theorem in modern Indian schools), in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Substituting the given values:

$\text{Hypotenuse}^2 = 5^2 + 12^2$

$\text{Hypotenuse}^2 = 25 + 144$

$\text{Hypotenuse}^2 = 169$

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Indian Perspective:

This set of numbers $(5, 12, 13)$ is known as a Baudhāyana-Pythagorean Triple. The ancient Indian text Śulba Sūtras lists several such triples used for constructing precise sacrificial altars (Vedis). The theorem states that the akṣṇayā-rajju (diagonal rope) of a rectangle produces an area equal to the sum of the areas produced by its sides.

Given:

Length of the hypotenuse = $17$ cm

Length of one shorter side = $8$ cm

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To Find:

The length of the third side (the other shorter side).

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Construction and Measurement:

1. Draw a line segment of $8$ cm.

2. Construct a perpendicular line at one of the endpoints.

3. Use a compass set to $17$ cm, place the needle on the other endpoint of the $8$ cm segment, and draw an arc to intersect the perpendicular line.

4. Upon measuring the vertical segment formed, the length is found to be $15$ cm.

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Solution (Verification using Baudhāyana’s Theorem):

Let the third side be $x$.

$64 + x^2 = 289$

$x^2 = 225$

Thus, the third side is $15$ cm.

Indian Context:

The Baudhāyana Śulba-Sūtra (Verse 1.10) provides a direct method for constructing a square that is a multiple of another square's area. This is achieved by repeatedly applying the property of the diagonal of a rectangle.

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1. Constructing a square with Triple the Area:

Let the side of the given square be $a$. Its area is $a^2$. We want to construct a square with area $3a^2$.

Step 1: Construct a square with Double the Area. This is done by taking the diagonal of the original square (side $a$). The diagonal is $\sqrt{a^2 + a^2} = \sqrt{2}a$. A square built on this diagonal has area $2a^2$.

Step 2: Construct a Rectangle with one side equal to $a$ and the other side equal to $\sqrt{2}a$ (the diagonal found in Step 1).

Step 3: The diagonal of this new rectangle will have a length $d$ where:

Conclusion: A square constructed with side length $d$ (the diagonal of the $a \times \sqrt{2}a$ rectangle) will have an area of exactly $3a^2$.

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2. Constructing a square with Five times the Area:

To get an area of $5a^2$, we need a side length of $\sqrt{5}a$.

Step 1: Take the given square side $a$ and double it to get a segment of length $2a$.

Step 2: Construct a Right-angled Triangle with one side $a$ and the other side $2a$.

Step 3: The hypotenuse $h$ of this triangle will be:

Conclusion: A square constructed with side length $h$ will have an area of $5a^2$. Alternatively, following Baudhāyana's logic of samāsa (combining areas), one can keep adding the area of the original square to the previously formed square's diagonal.

To Find:

The missing side length in various right-angled triangles using the given side lengths.

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Methodology (Indian Perspective):

In India, the relationship between the sides of a right-angled triangle was first described in the Baudhāyana Śulba Sūtras. The theorem states that in a right-angled triangle, the square of the hypotenuse ($c$) is equal to the sum of the squares of the other two sides ($a$ and $b$).

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(i) Given: $a = 5, b = 7$

To find the missing hypotenuse $c$:

$c^2 = 5^2 + 7^2$

$c^2 = 25 + 49$

$c = \sqrt{74}$

Since $74$ is not a perfect square, we can approximate it: $\mathbf{c \approx 8.60}$ units.

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(ii) Given: $a = 8, b = 12$

To find the hypotenuse $c$:

$c^2 = 8^2 + 12^2$

$c^2 = 64 + 144$

$c = \sqrt{208} = \sqrt{16 \times 13}$

$\mathbf{c = 4\sqrt{13} \approx 14.42}$ units.

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(iii) Given: $a = 9, c = 15$

To find the missing side $b$:

$9^2 + b^2 = 15^2$

$81 + b^2 = 225$

$b^2 = 144$

$b = \sqrt{144}$

$\mathbf{b = 12}$ units.

Note: $(9, 12, 15)$ is a Baudhāyana Triple, as it is a multiple of the basic $(3, 4, 5)$ triple.

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(iv) Given: $a = 7, b = 12$

To find the hypotenuse $c$:

$c^2 = 7^2 + 12^2$

$c^2 = 49 + 144$

$c = \sqrt{193}$

Approximating the value: $\mathbf{c \approx 13.89}$ units.

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(v) Given: $a = 1.5, b = 3.5$

To find the hypotenuse $c$:

$c^2 = (1.5)^2 + (3.5)^2$

$c^2 = 2.25 + 12.25$

$c = \sqrt{14.5}$

$\mathbf{c \approx 3.81}$ units.

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Summary Table:

Case	Missing Side	Value
(i)	$c$	$\sqrt{74} \approx 8.60$
(ii)	$c$	$4\sqrt{13} \approx 14.42$
(iii)	$b$	$12$
(iv)	$c$	$\sqrt{193} \approx 13.89$
(v)	$c$	$\sqrt{14.5} \approx 3.81$

Given:

The method to generate Baudhāyana triples (Pythagorean triples) based on an odd number $a$, where the other two sides are $b = \frac{a^2 - 1}{2}$ and $c = \frac{a^2 + 1}{2}$.

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Solution:

To find five more triples, we will choose odd numbers for $a$ starting from $7$ (assuming $3$ and $5$ were already discussed).

1. For $a = 7$:

$b = \frac{7^2 - 1}{2} = \frac{48}{2} = 24$

$c = \frac{7^2 + 1}{2} = \frac{50}{2} = 25$

Triple: $(7, 24, 25)$

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2. For $a = 9$:

$b = \frac{9^2 - 1}{2} = \frac{80}{2} = 40$

$c = \frac{9^2 + 1}{2} = \frac{82}{2} = 41$

Triple: $(9, 40, 41)$

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3. For $a = 11$:

$b = \frac{11^2 - 1}{2} = \frac{120}{2} = 60$

$c = \frac{11^2 + 1}{2} = \frac{122}{2} = 61$

Triple: $(11, 60, 61)$

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4. For $a = 13$:

$b = \frac{13^2 - 1}{2} = \frac{168}{2} = 84$

$c = \frac{13^2 + 1}{2} = \frac{170}{2} = 85$

Triple: $(13, 84, 85)$

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5. For $a = 15$:

$b = \frac{15^2 - 1}{2} = \frac{224}{2} = 112$

$c = \frac{15^2 + 1}{2} = \frac{226}{2} = 113$

Triple: $(15, 112, 113)$

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Summary Table:

Value of $a$ (Odd)	Value of $b$ ($\frac{a^2-1}{2}$)	Value of $c$ ($\frac{a^2+1}{2}$)	Baudhāyana Triple
7	24	25	(7, 24, 25)
9	40	41	(9, 40, 41)
11	60	61	(11, 60, 61)
13	84	85	(13, 84, 85)
15	112	113	(15, 112, 113)

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Indian Perspective:

This systematic method of generating triples was well-known to ancient Indian mathematicians. The Śulba Sūtras describe these relationships to ensure the geometric accuracy of large structures. It is worth noting that for all these triples, the difference between the hypotenuse and the larger side is exactly $1$.

Solution:

A primitive triple is a set of three positive integers $(a, b, c)$ such that they have no common divisor other than $1$.

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In the method provided, the relationship between the side $b$ and the hypotenuse $c$ is:

If $b$ and $c$ had a common factor $d$ ($d > 1$), then $d$ would also have to divide their difference.

Since the only positive integer that divides $1$ is $1$ itself, the greatest common divisor of $b$ and $c$ must be $1$. Because $gcd(b, c) = 1$, it follows that $gcd(a, b, c) = 1$.

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Conclusion:

No, this method does not yield non-primitive Baudhāyana triples. It always generates primitive triples because the hypotenuse and the largest leg are always consecutive integers.

Solution:

Yes, there are many primitive Baudhāyana triples that cannot be obtained through this specific method.

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The method discussed only generates triples where the difference between the hypotenuse ($c$) and the largest leg ($b$) is exactly $1$. Any primitive triple where $c - b > 1$ cannot be obtained using this formula.

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Examples:

1. Triple $(8, 15, 17)$:

This is a primitive triple as $gcd(8, 15, 17) = 1$. However, the difference between the hypotenuse and the largest side is:

Since the difference is not $1$, this triple cannot be generated by the "odd number" method described earlier.

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2. Triple $(20, 21, 29)$:

This is another primitive triple. The difference is:

Clearly, this also does not fit the pattern where the side is one less than the hypotenuse.

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Indian Perspective:

While the Śulba Sūtras highlight the $c = b + 1$ pattern for its ease of construction, Indian mathematicians of the later classical period, such as Brahmagupta, provided more general algebraic solutions that could generate all possible primitive and non-primitive triples, including those where the difference is greater than $1$.

Given:

Side length of the square ($s$) = $5$ cm.

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To Find:

The length of the diagonal ($d$).

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Solution:

In a square, the diagonal divides it into two identical isosceles right-angled triangles. By the Baudhāyana-Pythagoras Theorem, the square of the diagonal is equal to the sum of the squares of the two sides.

Substituting the given side length:

We can simplify $\sqrt{50}$ as $\sqrt{25 \times 2}$, which gives $5\sqrt{2}$. Using the approximation $\sqrt{2} \approx 1.414$:

$d \approx 5 \times 1.414 = 7.07$ cm.

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Indian Perspective:

This calculation is a direct application of the rule mentioned in the Sulba Sutras: "The diagonal of a square produces an area twice as much as the side of the square." This means $d^2 = 2s^2$, which we used to find the length in Indian geometry centuries before the same concept was popularized in the west.

Solution:

We will apply the theorem $a^2 + b^2 = c^2$ to find the missing side in each triangle shown in the image.

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Triangle 1 (Top-Left): Sides $7$ and $9$ are given.

We need to find the hypotenuse $c$.

$c^2 = 7^2 + 9^2 = 49 + 81 = 130$

$\mathbf{c = \sqrt{130} \approx 11.40}$

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Triangle 2 (Top-Right): Side $4$ and Hypotenuse $10$ are given.

We need to find the base $b$.

$4^2 + b^2 = 10^2 \Rightarrow 16 + b^2 = 100$

$b^2 = 100 - 16 = 84$

$\mathbf{b = \sqrt{84} = 2\sqrt{21} \approx 9.17}$

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Triangle 3 (Middle-Left): Side $40$ and Hypotenuse $41$ are given.

We need to find the perpendicular $a$.

$a^2 + 40^2 = 41^2 \Rightarrow a^2 + 1600 = 1681$

$\mathbf{a = \sqrt{81} = 9}$

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Triangle 4 (Middle-Right): Side $10$ and Hypotenuse $\sqrt{200}$ are given.

We need to find the base $b$.

$10^2 + b^2 = (\sqrt{200})^2 \Rightarrow 100 + b^2 = 200$

$b^2 = 100$

$\mathbf{b = 10}$

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Triangle 5 (Bottom-Left): Side $10$ and Side $\sqrt{150}$ are given.

We need to find the hypotenuse $c$.

$c^2 = 10^2 + (\sqrt{150})^2 = 100 + 150$

$c^2 = 250$

$\mathbf{c = \sqrt{250} = 5\sqrt{10} \approx 15.81}$

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Triangle 6 (Bottom-Right): Side $27$ and Hypotenuse $45$ are given.

We need to find the base $b$.

$27^2 + b^2 = 45^2 \Rightarrow 729 + b^2 = 2025$

$b^2 = 2025 - 729 = 1296$

$\mathbf{b = \sqrt{1296} = 36}$

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Alternate Solution for Triangle 6:

Notice that $27 = 3 \times 9$ and $45 = 5 \times 9$. This is a scaled version of the basic Baudhāyana Triple $(3, 4, 5)$. The missing side must be $4 \times 9 = 36$. Using such triples is a common technique in Indian school mathematics to solve problems quickly.

Given:

A rhombus with diagonal $d_1 = 24$ units and diagonal $d_2 = 70$ units.

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To Find:

The length of the side of the rhombus.

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Solution:

We know that the diagonals of a rhombus bisect each other at right angles ($90^\circ$). This property allows us to use the Baudhāyana-Pythagoras Theorem.

Let the side of the rhombus be $s$. The diagonals are divided into two equal halves at the intersection point.

In the right-angled triangle formed by the half-diagonals and the side $s$:

$s^2 = 144 + 1225$

The sidelength of the rhombus is $37$ units.

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Indian Perspective:

In Indian geometry, a rhombus is often referred to as a Samachatur bhuja (a quadrilateral with equal sides). The relationship between diagonals and sides was crucial for ancient land surveyors using ropes (rajju) to measure fields accurately.

Solution:

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

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Justification:

1. Geometric Reason: In any triangle, the side opposite the largest angle is always the longest side. In a right-angled triangle, one angle is $90^\circ$, and the sum of the other two angles is $90^\circ$ (since the total sum is $180^\circ$). Therefore, the $90^\circ$ angle is the largest angle, and the side opposite to it, the hypotenuse, is the longest.

2. Algebraic Reason (Baudhāyana's Theorem): Let the sides be $a$ and $b$, and the hypotenuse be $c$.

Since $a^2$ and $b^2$ are positive quantities, $c^2$ must be greater than $a^2$ and also greater than $b^2$.

This mathematically proves that $c$ is longer than both $a$ and $b$.

Solution:

The statement is True.

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Justification:

A primitive triple is a Baudhāyana triple where the three integers have no common factor other than $1$ (e.g., $3, 4, 5$). Any other triple is formed by multiplying a primitive triple by a constant $k$. For example, if we take $(3, 4, 5)$ and multiply by $2$, we get $(6, 8, 10)$, which is a scaled version. This principle is used extensively in Indian mathematics to derive various sets of triples from a single base set.

Solution:

A rectangle with sidelengths $a, b$ and diagonal $d$ forms a right-angled triangle. Thus, we need to find sets of Baudhāyana triples ($a, b, d$).

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Five examples of such rectangles (Sides $\times$ Diagonal) are:

1. Side $3$ and $4$, Diagonal $5$

2. Side $5$ and $12$, Diagonal $13$

3. Side $8$ and $15$, Diagonal $17$

4. Side $7$ and $24$, Diagonal $25$

5. Side $20$ and $21$, Diagonal $29$

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Indian Perspective:

In the Indian school curriculum, these triples are often memorized as they help in solving geometry and trigonometry problems much faster during competitive examinations like the NTSE or Olympiads.

To Find:

A square with area equal to $7^2 - 5^2$.

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Solution:

Area of first square = $7 \times 7 = 49$ square units.

Area of second square = $5 \times 5 = 25$ square units.

Difference in areas = $49 - 25 = 24$ square units.

We need to construct a square with area $24$. The side of this square would be $\sqrt{24}$ units.

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Construction (Based on Baudhāyana's Śulba Sūtras):

Ancient Indian mathematicians used geometric constructions for differences. To construct a square with area equal to the difference of two given squares:

1. Draw a line segment of length $7$ units (representing the hypotenuse of a right triangle).

2. Draw another segment of length $5$ units starting from one end of the first segment.

3. Arrange them such that the segment of $5$ units is one side (leg) and the segment of $7$ units is the hypotenuse.

4. The third side (the other leg) will be of length $x$. By Baudhāyana's Theorem:

5. The square built on this third side $x$ will have an area of exactly $24$ square units.

To Find:

Possibility of creating squares with specific integer areas on a dot grid and identifying the general rule for possible areas.

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Solution (i):

In a dot grid, if we connect two dots that are $a$ units apart horizontally and $b$ units apart vertically, the length of the segment (side of the square) $s$ is given by Baudhāyana’s Theorem:

For an area to be possible, it must be expressible as the sum of two squares of integers ($a^2 + b^2$).

Target Area	Sum of Squares ($a^2 + b^2$)	Possible?	Construction
(a) 2 sq. units	$1^2 + 1^2 = 1 + 1 = 2$	Yes	Connect dots across a $1 \times 1$ diagonal.
(b) 3 sq. units	No integer sum exists	No	Cannot be formed on an integer grid.
(c) 4 sq. units	$2^2 + 0^2 = 4 + 0 = 4$	Yes	Connect dots 2 units apart horizontally.
(d) 5 sq. units	$2^2 + 1^2 = 4 + 1 = 5$	Yes	Connect dots across a $2 \times 1$ diagonal.

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Solution (ii):

For an indefinitely extending grid, the possible integer-valued areas of squares are those integers that can be written as the sum of two squares of non-negative integers ($a^2 + b^2$).

Examples of such areas include: $1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, \dots$ and so on.

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Indian Perspective:

This type of grid-based geometry is deeply rooted in Indian culture through the art of Kolam or Rangoli. In South India, women draw intricate patterns using a grid of dots (Pulli). Understanding the distances between these dots using Baudhāyana’s logic helps in creating perfectly symmetrical geometric shapes that have been part of Indian households for generations.

Given:

An equilateral triangle with side length $s = 6$ units.

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To Find:

The area of the triangle.

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Construction and Proof:

Let the triangle be $\triangle ABC$ with $AB = BC = CA = 6$. Draw an altitude $AD$ from vertex $A$ to the base $BC$.

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Solution:

Step 1: Find the height (Altitude $AD$)

In right-angled $\triangle ABD$, by Baudhāyana’s Theorem:

Step 2: Calculate the Area

Using $\sqrt{3} \approx 1.732$, $\text{Area} \approx 9 \times 1.732 = 15.588$ sq. units.

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Alternate Solution:

Using the direct formula for the area of an equilateral triangle:

$\text{Area} = \frac{\sqrt{3}}{4}s^2$

$\text{Area} = \frac{\sqrt{3}}{4} \times 6^2 = \frac{\sqrt{3}}{4} \times 36 = 9\sqrt{3} \text{ sq. units}$.

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Indian Perspective:

Calculations involving the altitude (Lamba) and base (Adhara) of triangles (Tribhuja) were documented extensively by Indian mathematicians like Bhāskara II in his work Lilavati. These geometric principles were essential for Vastu Shastra to ensure the balanced proportions of temples and homes.