Sample Question 9:

The two legs of a right triangle, which are altitudes, have lengths \(2\sqrt3\) and \(6\). How long is the third altitude of the triangle?

\(\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5\)

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Answer Keys

Question 9: C

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Solutions

Question 9
Step 1: Calculate the area of the triangle using the given lengths of the two legs, which are also altitudes. The formula for the area of a triangle is 1/2 base * height. Here, it's \(\frac{6\times 2\sqrt{3}}{2} =6\sqrt{3}\).

Step 2: To find the length of the hypotenuse, apply the Pythagorean Theorem. The formula is \( \sqrt{{base}^2 + {height}^2}\), so here it gives \( \sqrt{(2\sqrt{3})^2 + 6^2} = 4\sqrt{3}\).

Step 3: Now, drop an altitude from the right angle to the hypotenuse. This forms two right-angled triangles.

Step 4: Let \(h\) be the third altitude of the triangle. Now the triangle can be described in another way using h and the hypotenuse.

Step 5: The area of the triangle from this perspective is \(\frac{4\sqrt{3}h}{2}\). Set this equal to the previous area calculation, \(6\sqrt{3}\), to solve for \(h\). From this, you get \(h = \frac{6 \cdot 2}{4}\), finally \(h =3\).

So, the length of the third altitude is 3. Your answer is \(\textbf{(C)}\ 3\).