# 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.

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 :)

• For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – mag Mar 31 at 8:49

$$|AB|=5, |AC|=|BC|=3=|AX|=|YB|$$ 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 $$k_4 = k_1 - 2(k_1 - k_3 )$$ $$k_4 = \sqrt{\frac{11}{2}(\frac{11}{2}-5)(\frac{11}{2}-3)^2}-2\sqrt{\frac{11}{2}(\frac{11}{2}-5)(\frac{11}{2}-3)^2}+2\frac{9}{2}\cdot \frac{\pi}{180}[90 -\frac{1}{2} \cos^{-1}(\frac{-7}{18})]°$$

$$k_4 = -\sqrt{\frac{11}{2}(\frac{11}{2}-5)(\frac{11}{2}-3)^2}+2\frac{9}{2}\cdot \frac{\pi}{180}[90 -\frac{1}{2} \cos^{-1}(\frac{-7}{18})]°$$

$$CAB=0.586 rad.$$, $$r=3$$:

$$\rightarrow CX=CY=3\times 0.586=1.752$$ 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-$$A_R=(2.623-2.075)\times 2=1.097$$ $$cm^2$$