## Online Course Discussion Forum

### MC II-A final quiz (Sensored to whoever student that hasn't taked the test yet)

Hello!

So basically I finished the test and found out I got problem 2 wrong. I got 2 square root 3 so I simplified it to square root 12 and entered that as my answer. However, the answer said that you just put in the 3 in the 2 square root 3. Did I do something wrong? Thanks!

Your friend,

Yoyo Yang

P.S. Here is the problem: In triangle ABCABC, AB=ACAB=AC, and BDBD is the altitude on ACAC. Given that BD=3–√3BD=3, and ∠DBC=60 degrees, the area of △ABC△ABC is sqrt(K)

for an integer KK. What is KK?

Thank you again!

P.P.S. Ignore the little blue letters

P.P.P.S. Can you reattempt the quiz as many times as you want?

P.P.P.P.S. Happy Early Thanksgiving! Eat a turkey!Hi Yoyo!!

I have the feeling that you forgot to divide by $2$ when you were finding that area. Note in the solution we are finding the area of $\triangle ABC$ using $BC$ as base and $AE$ as altitude.

Here's another way to look at it. Since $\triangle DBC$ is a 30-60-90 triangle, $CD = 3$. You can split it in three congruent triangles by drawing the altitude from $A$ of $\triangle ABC$. Then each of the triangles has area $\dfrac{1}{2}\cdot 3\cdot \sqrt{3}$, so $\triangle ABC$ has area $\sqrt{3}$.

Normally you could only attempt the quiz one time. We opened it again for you if you want to try it again later.

Happy Thanksgiving!!

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