Question

In triangle ABC, the measure of side AC is $21\text{inches},CB$ is $12\text{inches}$, and AB is $3\sqrt{33}\text{inches}$ as given in the figure below. What kind of triangle is this and what is the measure of $\angle ABC$?

Answer 1

Given:
$AC=21\text{in}$
$AB=3\sqrt{33}\text{in}$
$CB=12\text{in}$
$21>3\sqrt{33}>12$ (Arranging the sides in descending order)
${21}^2=441,3{\sqrt{33}}^2=297,{12}^2=144$ (Finding the squares of the sides)
$441=297+144$ (Analyzing the squares of the sides)
${21}^2=3{\sqrt{33}}^2+{12}^2$
We can observe that the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides.
Hence, $\Delta ABC$ is a right triangle. (Using converse of the Pythagorean theorem)
And AC is the hypotenuse, so $\angle ABC={90}^{\circ }$. (Angle opposite the hypotenuse is a right angle)
Hence, $\Delta ABC$ is a right triangle and $\angle ABC={90}^{\circ }$.
$\to$ Option C is correct.