Homesweet Learning helps students learn!
AMC10 2020a Test Paper
Complete the quiz in 75 minutes. Do the quiz as if you are taking the real test. You score will be compared with other students taking the same test to give you a ranking among your peers.
To get a human or AI tutor to help you, click Register
Sample Question 12:

Triangle \(AMC\) is isosceles with \(AM = AC\). Medians \(\overline{MV}\) and \(\overline{CU}\) are perpendicular to each other, and \(MV=CU=12\). What is the area of \(\triangle AMC?\)

\(\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192\)




Answer Keys

Question 12: C


Solutions

Question 12
Step 1: Note that quadrilateral \(UVCM\) has perpendicular diagonals. Therefore, the area of this quadrilateral could be found as half of the product of the length of the diagonals.

Step 2: Realize that triangle \(AUV\) has \(\frac{1}{4}\) the area of triangle \(AMC\) based on similarity rules. So, \(UVCM\) has \(\frac{3}{4}\) of the area of \(AMC\).

Step 3: Setting up the equation based on step 2, we obtain: \(\frac{1}{2} \times 12 \times 12 = \frac{3}{4} \times [AMC]\). Solving this equation gives us: \(72 = \frac{3}{4} \times [AMC]\).

Step 4: Solving this equation further, we find that the area of \(AMC\) equals to 96. The area of triangle AMC is 96 square units or \(\boxed{\textbf{(C)}\ 96}\).