Question
Determine whether the triangle having sides $(a - 1)\ cm$, $2\sqrt{\text{a}}\text{ cm}$ and $(a + 1)\ cm$ is a right angled triangle.
✓
Answer
Sides of a triangle are $(a - 1)cm$, $2\sqrt{\text{a}}\text{ cm}$ and $(a + 1)cm$
Let, $AB = (a - 1)cm, BC = (a + 1)cm$ and $\text{AC}=2\sqrt{\text{a}}$
Now $AB^2 = (a - 1)^2$
$= a^2 - 2a + 1 BC^2$
$= (a + 1)^2 = a^2 + 2a + 1$
$\text{AC}^2=\big(2\sqrt{\text{a}}\big)^2=4\text{a}$
Now $AB^2 + AC^2 = a^2 - 2a + 1 + 4a = a^2 + 2a + 1 = BC^2$
$\therefore\triangle\text{ABC}$ is a right triangle right angle at $\angle\text{A}$ (By converse of pythagoras theorem)
Let, $AB = (a - 1)cm, BC = (a + 1)cm$ and $\text{AC}=2\sqrt{\text{a}}$
Now $AB^2 = (a - 1)^2$
$= a^2 - 2a + 1 BC^2$
$= (a + 1)^2 = a^2 + 2a + 1$
$\text{AC}^2=\big(2\sqrt{\text{a}}\big)^2=4\text{a}$
Now $AB^2 + AC^2 = a^2 - 2a + 1 + 4a = a^2 + 2a + 1 = BC^2$
$\therefore\triangle\text{ABC}$ is a right triangle right angle at $\angle\text{A}$ (By converse of pythagoras theorem)
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