Finding the area of a sector within a triangle. 
For part (a) of the question I am getting an answer of $33.6^{\circ}$ of $0.586 radians$ which I am pretty sure is right.
Part (b) is where I am having difficulties because my answer is not matching the mark scheme.
For b (i) I have an answer of $4.516 cm$
for (ii) I have an answer of $$\frac{9*2.49\cdot sin(0.586)}{2}$$
Then I am stuck.
Please could someone give a model answer.
Thanks!
 A: Your answers for part (a) and b(i) are correct.
For b(ii) you can use the formula for the area of a sector with radius r and angle θ as $0.5r^2θ$. Here, you have a radius of 3, and an angle found in part (a).
For part b(iii), consider M to be the midpoint of AB and then consider one half of the shaded region, say CMXC. Now this area would just be the area of the sector ACX - area of triangle ACM. This is just answer from $b(ii)-0.5\cdot AC\cdot AM\cdot \sin \theta$.
So your final answer for R would just be twice this calculated value.
Hope that helped :)
A: $|AB|=5, |AC|=|BC|=3=|AX|=|YB|$ and since $|ABC|$ is isosceles, sum of angles in $|ABC|$ is $180°$, then $2\theta_1+\theta_2= 180°$, from cosine rule
$$ \cos(\theta_2) = \frac{ 3^2+3^2-5^2}{2×3^2}$$
$$\theta_2 = \cos^{-1}(\frac{-7}{18})$$
$$\theta_1 = 90° - \frac{1}{2}\theta_2$$
$$\theta_1 = 90° -\frac{1}{2} \cos^{-1}(\frac{-7}{18})°$$
Area of triangle $|ABC|$, the semi-perimeter here is $s = \frac{5+3+3}{2}$, from heron's formula
$$k_1 = \sqrt{\frac{11}{2}(\frac{11}{2}-5)(\frac{11}{2}-3)^2}$$
Perimeter of shape $|CYX| = |YX| +|CY|+|CX|$, $|CY| = |CX|$ now a sector with radius $3$ and angle $[90 -\frac{1}{2} \cos^{-1}(\frac{-7}{18})]°$ in degree, has curved length as
$$|CY| = |CX| = 3×\frac{\pi}{180}[90 -\frac{1}{2} \cos^{-1}(\frac{-7}{18})]°$$
The length $|YX| = |AB|-|AY|-|XB|$, $|AY|=|XB|$ meaning that $|YX| = 5-2|AY|$, and also $|AB|-|AX|= |AY|$ meaning that $5-3 = |AY|$, then the length $|YX|$ is
$$|YX| = 5-2(5-3) = 1$$
$$|CYX| = |YX| +2|CY|$$
$$ k_2 = 1+6\frac{\pi}{180}[90 -\frac{1}{2} \cos^{-1}(\frac{-7}{18})]°$$
The area of sector $|ACX| = \theta_1 × \frac{|AX|^2}{2}$
$$k_3 = \frac{9}{2}\cdot \frac{\pi}{180}[90 -\frac{1}{2} \cos^{-1}(\frac{-7}{18})]°$$
The area of sector $|CYX| = |ABC| -|ACY| -|BCX|$, $|ACY|=|BCX|$ meaning that $|CYX| = k_1 -2|ACY|$, and also $|ABC| -|ACX| = |BCX|$ meaning that $k_1 - k_3 = |BCX|$, then the sector $|CYX|$ is
$$k_4 = k_1 - 2(k_1 - k_3 )$$
$$k_4 =  \sqrt{\frac{11}{2}(\frac{11}{2}-5)(\frac{11}{2}-3)^2}-2\sqrt{\frac{11}{2}(\frac{11}{2}-5)(\frac{11}{2}-3)^2}+2\frac{9}{2}\cdot \frac{\pi}{180}[90 -\frac{1}{2} \cos^{-1}(\frac{-7}{18})]°$$
$$k_4 = -\sqrt{\frac{11}{2}(\frac{11}{2}-5)(\frac{11}{2}-3)^2}+2\frac{9}{2}\cdot \frac{\pi}{180}[90 -\frac{1}{2} \cos^{-1}(\frac{-7}{18})]°$$
A: $CAB=0.586 rad.$, $r=3$:
$\rightarrow CX=CY=3\times 0.586=1.752$ cm
I-$XY=2(3-2.5)=1\rightarrow P_{R}=2\times 1.752+1=4.504$
II-$A_{ACX}=(0.586\times3^2)/2=2.623$ $cm^2$
$h_{ABC}=3 \sin (33.6)=1.66$
$A_{ABC}/2=(1.66\times 2.5)/2=2.075$
III-$A_R=(2.623-2.075)\times 2=1.097$ $cm^2$
