We have triangle $ABC$, and we construct a circle on side AC to become its diameter.
This circle contains the middle point of side $BC$ and intersects side $AB$ in point D, in ratio of $AD:DB=1:2$. If $AB$ is $3cm$ what is the area of triangle $ABC$?

My attempt: $CD$ seems to be the height of triangle(because it's at $90$ degrees), perpendicular to $AC$. I used that $CD=\sqrt{AD*DB}$. So I have height and base of triangle which turned out that area is $3\sqrt{2}/2$, but solution seems to be $3\sqrt{2}$. Why?

  • $\begingroup$ Why do you think that $CD \perp AB$? $\endgroup$ – Dbchatto67 Apr 21 at 7:54
  • $\begingroup$ Is it not that triangle $ADC$ is a right triangle? $\endgroup$ – Darko Dekan Apr 21 at 8:03
  • $\begingroup$ Why? Can you explain? I can't see any such thing. $\endgroup$ – Dbchatto67 Apr 21 at 8:05
  • $\begingroup$ Angle in a semicircle is always 90 degrees when one side is a diameter of a triangle. $\endgroup$ – Darko Dekan Apr 21 at 8:07
  • $\begingroup$ I used Thales's theorem. $\endgroup$ – Darko Dekan Apr 21 at 8:07

Refer to the figure:

$\hspace{5cm}$![enter image description here

The angles $ADC$ and $AEC$ are right, because both subtend the diameter $AC$.

From $AD:BD=1:2$ and $AB=3$ we can find $AD=1$ and $BD=2$.

The line $AE$ is both height and median, it implies the triangle $ABC$ is isosceles. Hence, $AC=3$.

From the right triangle $ACD$ we can find $CD=\sqrt{AC^2-AD^2}=2\sqrt{2}$.

Finally, the area of the triangle $ABC$ is $\frac12 \cdot AB\cdot CD=\frac12 \cdot 3\cdot 2\sqrt{2}=3\sqrt{2}.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.