Question 6

We can use Baudhayana's Theorem (also known as the Pythagorean Theorem) to find new side lengths for our squares.

Step 1 — Finding the side for triple area

Let us start with a square named ABCD. Let its side length be $a$. The area of this square is $a^2$. We want a new square with an area of $3a^2$. This means the side of the new square must be $\sqrt{3a^2}$, which is $a\sqrt{3}$.

Let us draw the diagonal AC of the square ABCD. In triangle ABC, angle B is a right angle. Using Baudhayana's Theorem (hypotenuse squared equals sum of squares of other two sides): $AC^2 = AB^2 + BC^2$ $AC^2 = a^2 + a^2$ $AC^2 = 2a^2$ $AC = \sqrt{2a^2}$

$\boxed{AC = a\sqrt{2}}$ Now, let us construct a rectangle ACEF. One side of this rectangle is AC, which is $a\sqrt{2}$. The other side, AF, is equal to $a$. Consider the right-angled triangle AFE (right-angled at F). The hypotenuse AE will be the side of our new square. Using Baudhayana's Theorem in triangle AFE: $AE^2 = AF^2 + EF^2$ Since EF is equal to AC, we have: $AE^2 = a^2 + (a\sqrt{2})^2$ $AE^2 = a^2 + 2a^2$ $AE^2 = 3a^2$ $AE = \sqrt{3a^2}$ $\boxed{AE = a\sqrt{3}}$

Step 2 — Calculating the area for triple area square

The side of our new square is AE, which is $a\sqrt{3}$. Let us name this new square AEGH. The area of square AEGH is the square of its side length. $Area(AEGH) = (AE)^2$ $Area(AEGH) = (a\sqrt{3})^2$ $Area(AEGH) = 3a^2$ The area of the original square ABCD is $a^2$. So, the area of square AEGH is 3 times the area of square ABCD.

Step 3 — Finding the side for five times area

Let us again start with the square ABCD with side length $a$. Its area is $a^2$. We want a new square with an area of $5a^2$. This means the side of this new square must be $\sqrt{5a^2}$, which is $a\sqrt{5}$.

Let us place another square, CFED, next to square ABCD. They share the side CD. Both squares have side length $a$. This creates a larger rectangle named ABFE. The side EF of this rectangle is equal to $a$. The side BF of this rectangle is the sum of BC and CF. $BF = BC + CF$ $BF = a + a$

$\boxed{BF = 2a}$ Now, consider the right-angled triangle BFE (right-angled at F). The hypotenuse BE will be the side of our new square. Using Baudhayana's Theorem in triangle BFE: $BE^2 = EF^2 + BF^2$ $BE^2 = a^2 + (2a)^2$ $BE^2 = a^2 + 4a^2$ $BE^2 = 5a^2$ $BE = \sqrt{5a^2}$ $\boxed{BE = a\sqrt{5}}$

Step 4 — Calculating the area for five times area square

The side of our new square is BE, which is $a\sqrt{5}$. Let us name this new square BEFG. The area of square BEFG is the square of its side length. $Area(BEFG) = (BE)^2$ $Area(BEFG) = (a\sqrt{5})^2$ $Area(BEFG) = 5a^2$ The area of the original square ABCD is $a^2$. So, the area of square BEFG is 5 times the area of square ABCD.

Answer

(a) To construct a square whose area is triple the area of a given square: Start with a square of side $a$. Find its diagonal, which is $a\sqrt{2}$. Then, form a right-angled triangle using this diagonal ($a\sqrt{2}$) as one leg and the original side ($a$) as the other leg. The hypotenuse of this triangle will be $a\sqrt{3}$, which is the side of the new square. (b) To construct a square whose area is five times the area of a given square: Start with a square of side $a$. Place another identical square next to it, sharing a side, to form a rectangle with sides $a$ and $2a$. The diagonal of this rectangle will be $a\sqrt{5}$, which is the side of the new square.