A geometry problem by Swapan Bagchi

Let the sides of the triangle be $x$ and $y$ . By considering the area of the triangle we get:

$\dfrac{xy}{2}=\dfrac{10 \times 6}{2} \Rightarrow xy=60$

Also because the triangle is right-angled we have:

$x^2+y^2=10^2 \Rightarrow x^2+y^2=100$

Now by GM-QM inequality we have:

$\sqrt{xy} \leq \sqrt{\dfrac{x^2+y^2}{2}}$

Putting in our values for $xy$ and $x^2+y^2$ we get:

$\sqrt{60} \leq \sqrt{\dfrac{100}{2}} \Rightarrow \sqrt{60} \leq \sqrt{50} \Rightarrow 60 \leq 50$

This is clearly a contradiction so no such triangle exists.

If we divide the hypothenuse into lengths of $x$ and $10-x$ , then by Geometric Means:

$\frac{6}{x} = \frac{10-x}{6} \Rightarrow 36 = -x^2 + 10x \Rightarrow x^2 -10x + 36 = 0 \Rightarrow x = \frac{10 \pm \sqrt{10^2 - 4(1)(36)}}{2} = \frac{10 \pm \sqrt{-44}}{2} \Rightarrow \boxed{x = 5 \pm \sqrt{11}i}$

which means this right triangle configuration is impossible since $x \notin (0, 10)$ .

$\mathbb{Q.} \mathbb{E.} \mathbb{D.}$

If a right triangle has a hypotenuse of 10, it means the other sides should be 6 and 8. Thus, the area should be 24. with the given dimensions of the altitude and the hypotenuse the area is 30.