Question

Let $ABC$ be a right angled triangle with AC as its hypotenuse. Then,

• A. $A{C}^3>A{B}^3+A{C}^3$
• B. $A{C}^2>A{B}^2+B{C}^2$
• C. $AC>AB+BC$
• D. None of these

Answer 1

The correct option is A

$A{C}^3>A{B}^3+A{C}^3$

Explanation for the correct option.

In any right-angled triangle, the longest side is the hypotenuse. So, in the right angled triangle $ABC$,

$AC>AB$ and $AC>BC$ as $AC$ is the hypotenuse.

Now, $A{C}^3$ can be written as $A{C}^2\times AC$.

According to the Pythagoras Theorem, in a right-angled triangle

$Hypotenus{e}^2=Perpendicula{r}^2+Bas{e}^2$

$$\begin{gathered} \Rightarrow A{C}^2=A{B}^2+B{C}^2 \\ \Rightarrow A{C}^3=\left(A{B}^2+B{C}^2\right)AC \end{gathered}$$

$\Rightarrow A{C}^3=\left(A{B}^2\times AC\right)+\left(B{C}^2\times AC\right)$ …………… $\left(1\right)$

Now, as $AC>AB$ and $AC>BC$, so

$$\begin{gathered} \left(A{B}^2\times AC\right)+\left(B{C}^2\times AC\right)>\left(A{B}^2\times AB\right)+\left(B{C}^2\times BC\right) \\ \Rightarrow \left(A{B}^2\times AC\right)+\left(B{C}^2\times AC\right)>A{B}^3+B{C}^3 \end{gathered}$$

$\Rightarrow A{C}^3>A{B}^3+B{C}^3$ (from $\left(1\right)$)

Explanation for the incorrect options.

Option B: $A{C}^2>A{B}^2+B{C}^2$, is incorrect because according to the Pythagoras Theorem $A{C}^2=A{B}^2+B{C}^2$.

Option C: $AC>AB+BC$, is incorrect because the sum of any two sides of the triangle is greater than the third side, so $AC<AB+BC$.

Option D: None of these, is incorrect because option A is true.

Hence, the correct option is A.