Question

Explain the concept of the Pythagorean Theorem.

Answer 1

Step 1: State the Pythagorean Theorem

In a right-angled triangle, the square of the length of the hypotenuse is the sum of the square of the base and the square of the perpendicular.

Thus, ${\left(\mathrm{Hypotenuse}\right)}^2={\left(\mathrm{Base}\right)}^2+{\left(\mathrm{Perpendicular}\right)}^2$.

Step 2: Prove the Pythagorean Theorem

Consider the right-angled triangle, $∆ABC$, such that $\angle B={90}^{\circ }$.

Now, a perpendicular is drawn from the point $B$ to the side $CA$.

It is known from the theorem that, The perpendicular drawn from the vertex of the right angle of a right-angled triangle to the hypotenuse then both sides of the perpendicular are similar to the whole triangle and to each other.

Therefore, $∆ABD~∆ABC$.

Thus, according to the condition of similarity, $\frac{AD}{AB}=\frac{AB}{AC}$.

$\Rightarrow A{B}^2=AD\times AC.....................\left(1\right)$.

Therefore, $∆DBC~∆ABC$.

Thus, according to the condition of similarity, $\frac{CD}{BC}=\frac{BC}{AC}$.

$\Rightarrow B{C}^2=CD\times AC.....................\left(2\right)$.

Step 3: Add equation $\left(1\right)$ and $\left(2\right)$.

$$\begin{gathered} A{B}^2+B{C}^2=\left(AD\times AC\right)+\left(CD\times AC\right) \\ \Rightarrow A{B}^2+B{C}^2=AC\left(AD+CD\right) \end{gathered}$$

From the given figure it is clear that, $\left(AD+CD\right)=AC$.

$$\begin{gathered} \therefore A{B}^2+B{C}^2=AC\times AC \\ \Rightarrow A{B}^2+B{C}^2=A{C}^2 \\ \Rightarrow A{C}^2=A{B}^2+B{C}^2 \end{gathered}$$

Hence, it is proven that, ${\left(\mathrm{Hypotenuse}\right)}^{\mathbf{2}}\mathbf{=}{\left(\mathrm{Base}\right)}^{\mathbf{2}}\mathbf{+}{\left(\mathrm{Perpendicular}\right)}^{\mathbf{2}}$.