IT-4

Chapter: SQUARES AND SQUARE ROOTS
Class: 8 (Class 8)
Category: in_text

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Question

Context: So, the new square has double the area of the original square, because the original square is made up of two small triangles, while the new square is made up of four small triangles.

Q. Moreover, all these small triangles are congruent to each other. Can you explain why?

Question diagram(s):

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All the "small triangles" are congruent because they are all right-angled isosceles triangles with the same side lengths.

Step 1 — Understanding the triangles in the original square

Let us consider the original square. Let its side length be $s$. The area of this square is $s \times s = s^2$. The diagram shows this square is divided into two triangles by one of its diagonals. A square has four equal sides and all its angles are $90^\circ$. When we draw a diagonal in a square, it divides the square into two triangles. Each of these two triangles has two sides equal to the side of the square, which is $s$. The angle between these two sides is $90^\circ$. Since two sides and the included angle are equal (SAS congruence rule), these two triangles are congruent to each other. Also, because two sides are equal, these are isosceles triangles. Since one angle is $90^\circ$, they are right-angled isosceles triangles. Using the Pythagorean theorem, the third side (hypotenuse) of each triangle is:

$\text{Hypotenuse}^2 = s^2 + s^2$

$\text{Hypotenuse}^2 = 2s^2$

$\boxed{\text{Hypotenuse} = s\sqrt{2}}$

So, each of these two "small triangles" has sides of length $s, s, s\sqrt{2}$. Their angles are $90^\circ, 45^\circ, 45^\circ$.

Step 2 — Understanding the triangles in the new square

The problem states that the new square has double the area of the original square. Area of original square = $s^2$. So, the area of the new square is $2 \times s^2 = \mathbf{2s^2}$. Let the side length of the new square be $S$. Then, its area is $S^2$.

$S^2 = 2s^2$

To find the side length $S$, we take the square root of both sides:

$S = \sqrt{2s^2}$

$\boxed{S = s\sqrt{2}}$

So, the side length of the new square is $s\sqrt{2}$. The diagram shows this new square is divided into four triangles by its two diagonals. The diagonals of a square are equal in length and they bisect (cut into two equal halves) each other at right angles. The length of each diagonal of the new square is $S \times \sqrt{2}$.

$\text{Diagonal length} = (s\sqrt{2}) \times \sqrt{2}$

$= s \times (\sqrt{2} \times \sqrt{2})$

$\boxed{\text{Diagonal length} = 2s}$

Each of the four triangles formed by these diagonals has two sides that are half the length of a diagonal. Half the diagonal length = $\frac{2s}{2} = \mathbf{s}$. The angle between these two sides is $90^\circ$ (where the diagonals intersect). So, these four triangles are also right-angled isosceles triangles. Their two equal sides (legs) are $s$, and their third side (hypotenuse) is the side of the new square, which is $S = s\sqrt{2}$. Therefore, each of these four "small triangles" also has sides of length $s, s, s\sqrt{2}$ and angles $90^\circ, 45^\circ, 45^\circ$.

Step 3 — Explaining congruence

From Step 1, we found that the two "small triangles" making up the original square have sides $s, s, s\sqrt{2}$. From Step 2, we found that the four "small triangles" making up the new square also have sides $s, s, s\sqrt{2}$. Since all these triangles have the same three side lengths, they are all congruent to each other by the SSS (Side-Side-Side) congruence rule. Alternatively, since they are all right-angled isosceles triangles with legs of length $s$, they are congruent by the SAS (Side-Angle-Side) rule.

Answer

All the small triangles are congruent to each other because they are all right-angled isosceles triangles, each having two sides of length $s$ and a hypotenuse of length $s\sqrt{2}$, where $s$ is the side length of the original square.