Squares and Square Roots | FIO

Question 6

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

Understand the Question

To construct a square whose area is a multiple of the area of a given square of side $a$ $a$ (original area $= a^2$ $= a^{2}$ ):
- For triple the area ( $3a^2$ ), the required side length is $\sqrt{3a^2} = a\sqrt{3}$ .
- For five times the area ( $5a^2$ ), the required side length is $\sqrt{5a^2} = a\sqrt{5}$ .
We use Baudhāyana's Theorem (Pythagoras theorem: $\text{Hypotenuse}^2 = \text{Base}^2 + \text{Perpendicular}^2$ ) to construct segments of lengths $a\sqrt{3}$ and $a\sqrt{5}$ as hypotenuses of suitable right-angled triangles.

(a) How would you construct a square whose area is triple the area of a given square?

Step 1 · Construct Side Length $a\sqrt{3}$

Let the given square be $\text{ABCD}$ with side length $a$ and area $a^2$ .

Diagram 1 In right $\Delta \text{ABC}$ (with $\angle \text{B} = 90^\circ$ ):

\begin{aligned} \text{AC}^2 &= \text{AB}^2 + \text{BC}^2 \\ &= a^2 + a^2 \\ &= 2a^2 \\ \text{AC} &= \sqrt{2a^2} = a\sqrt{2} \end{aligned}

Now, construct rectangle $\text{ACEF}$ with side $\text{EF} = \text{AC} = a\sqrt{2}$ and side $\text{AF} = a$ .

In right $\Delta \text{AFE}$ (with $\angle \text{F} = 90^\circ$ ), by Baudhāyana's Theorem:

\begin{aligned} \text{AE}^2 &= \text{AF}^2 + \text{EF}^2 \\ &= a^2 + (a\sqrt{2})^2 \\ &= a^2 + 2a^2 \\ &= 3a^2 \\ \text{AE} &= \sqrt{3a^2} = a\sqrt{3} \end{aligned}

Step 2 · Calculate the Area of the New Square

Constructing square $\text{AEGH}$ with side length $\text{AE} = a\sqrt{3}$ :

\begin{aligned} \text{Area}(\text{AEGH}) &= (\text{AE})^2 \\ &= (a\sqrt{3})^2 \\ &= 3a^2 \end{aligned}

Thus, the area of square $\text{AEGH}$ is $3$ times the area of square $\text{ABCD}$ .

Answer

(a) Construct a right-angled triangle with legs $a$ and $a\sqrt{2}$ (the diagonal of the given square). Its hypotenuse $a\sqrt{3}$ forms the side of a square whose area is $3a^2$ .

(b) How would you construct a square whose area is five times the area of a given square?

Step 1 · Construct Side Length $a\sqrt{5}$

Let the given square be $\text{ABCD}$ with side length $a$ and area $a^2$ .

Diagram 2 Place an identical square $\text{CFED}$ of side $a$ next to $\text{ABCD}$ , sharing side $\text{CD}$ to form rectangle $\text{ABFE}$ with side $\text{EF} = a$ and base $\text{BF}$ :

\begin{aligned} \text{BF} &= \text{BC} + \text{CF} \\ &= a + a \\ &= 2a \end{aligned}

In right $\Delta \text{BFE}$ (with $\angle \text{F} = 90^\circ$ ), by Baudhāyana's Theorem:

\begin{aligned} \text{BE}^2 &= \text{EF}^2 + \text{BF}^2 \\ &= a^2 + (2a)^2 \\ &= a^2 + 4a^2 \\ &= 5a^2 \\ \text{BE} &= \sqrt{5a^2} = a\sqrt{5} \end{aligned}

Step 2 · Calculate the Area of the New Square

Constructing square $\text{BEFG}$ with side length $\text{BE} = a\sqrt{5}$ :

\begin{aligned} \text{Area}(\text{BEFG}) &= (\text{BE})^2 \\ &= (a\sqrt{5})^2 \\ &= 5a^2 \end{aligned}

Thus, the area of square $\text{BEFG}$ is $5$ times the area of square $\text{ABCD}$ .

Answer

(b) Construct a rectangle of dimensions $a$ and $2a$ (by placing two identical squares side by side). Its diagonal $a\sqrt{5}$ forms the side of a square whose area is $5a^2$ .

Common Mistakes

Side Scaling vs. Area Scaling: Scaling side length by $n$ multiplies area by $n^2$ . To triple the area, the side must be $a\sqrt{3}$ , not $3a$ (which would give an area of $9a^2$ ).
Incorrect Right-Triangle Legs: For $\sqrt{5}a$ , the legs must be $a$ and $2a$ (since $a^2 + (2a)^2 = 5a^2$ ), not $a$ and $4a$ .

Question 6

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