Hey guys,
The last question of the chapter has stumped me once again! I need to find the area of part of this triangle. I’m not sure how to attack the question considering that I’m not given the radius of the circle and I’m not sure how the measurement of 4cm will help me?
Thanks, Scratch Cat
One way is to find the difference in areas of a right-angled triangle and a circular sector.
Finding both areas requires finding the radius of the circle, which may be found through the triangle
$$\begin{align*} \frac{r}{r+4} &= \sin 35^\circ\\ r &= (r+4)\sin 35^\circ\\ (1-\sin 35^\circ)r &= 4\sin 35^\circ\\ r &= \frac{4\sin 35^\circ}{1-\sin 35^\circ} \end{align*}$$
Step 1: Using our knowledge that $∠BAC = ∠DAE = 35^o$ and that $∠ADE = 90^o$ (tangentiality), we can calculate section $AD = AE/cos(35) \approx 4.88$
Step 2: Using Thales' Theorem, we know that $∠ADF=90^o$.
Thus, we can simply calculate $AF=AD/cos(35) \approx 5.96$
Then, using simple trigonometry again we obtain $AB = AF/cos(35)\approx7.28 $
Step 3: Invoking Thales' theorem again, we obtain that $∠ABC = 90^o$.
We then calculate $AC=AB/cos(35)\approx8.88$
Similarly, $FB=AF*tan(35)\approx4.17$
Step 4: The area of ABC is then = $0.5*FB*AC = 18.54$
The radius of the large circle = $8.88-4 \approx 4.88$
The area of the large circle = $\pi*r^2 \approx 74.91$
Step 5: However, we only need to know the part within the triangle. As $∠ABC = 90^o$ and $∠BAC=35^o$, $∠BCA = 55^o$.
Thus, the relevant area $ \approx 55/360*74.91 = 11.44$
Answer: The answer is then $18.54-11.44 = 7.10cm^2$
