Answer:
[tex]\textsf{(a)}\quad r=\sqrt{10}[/tex]
[tex]\textsf{(b)}\quad \textsf{Perimeter}=4\sqrt{10}\; \sf cm[/tex]
[tex]\textsf{(c)}\quad R=5.454 \; \sf cm^2[/tex]
Step-by-step explanation:
Part (a)
The formula for the area of a sector is:
[tex]\boxed{\begin{array}{l}\underline{\sf Area \;of\; a\; sector}\\\\A=\dfrac12 r^2 \theta\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$A$ is the area.}\\ \phantom{ww}\bullet\;\textsf{$r$ is the radius.}\\ \phantom{ww}\bullet\;\textsf{$\theta$ is the angle measured in radians.}\end{array}}[/tex]
Given that the area of the sector is 10 cm² and the central angle, ∠AOB, is 2 radians, then:
- [tex]A = 10[/tex]
- [tex]\theta = 2[/tex]
Substitute these values into the formula and solve for r:
[tex]\begin{aligned}10&=\dfrac{1}{2} r^2 \cdot 2\\\\10&=r^2\\\\\sqrt{r^2}&=\sqrt{10}\\\\r&=\sqrt{10}\end{aligned}[/tex]
Hence, we have proved that [tex]r=\sqrt{10}[/tex].
[tex]\hrulefill[/tex]
Part (b)
To determine the perimeter of the sector OAB, we can use the perimeter of a sector formula:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Perimeter of a sector}}\\\\P=r(2+\theta)\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$P$ is the perimeter.}\\ \phantom{ww}\bullet\;\textsf{$r$ is the radius.}\\ \phantom{ww}\bullet\;\textsf{$\theta$ is the angle measured in radians.}\end{array}}[/tex]
In this case:
Substitute the values into the formula and solve for P:
[tex]\begin{aligned}P&=\sqrt{10}(2+2)\\\\P&=4\sqrt{10}\; \sf cm\end{aligned}[/tex]
Therefore, the perimeter of the sector OAB is:
[tex]\Large\boxed{\boxed{4\sqrt{10}\; \sf cm}}[/tex]
[tex]\hrulefill[/tex]
Part (c)
The shaded segment R is enclosed by the arc AB and the straight line AB. To calculate the area of R, we can subtract the area of isosceles triangle AOB from the area of sector OAB.
The formula for the area of an isosceles triangle, given its congruent side lengths and the measure of the angle (in radians) between the congruent sides, is:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Area of an isosceles triangle}}\\\\A=\dfrac{1}{2}r^2\sin \theta\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$A$ is the area.}\\\phantom{ww}\bullet\;\textsf{$r$ is the congruent side length.}\\\phantom{ww}\bullet\;\textsf{$\theta$ is the angle between the two congruent sides.}\end{array}}[/tex]
In this case:
Substitute the values into the formula and solve for A:
[tex]\begin{aligned}A&=\dfrac{1}{2}\cdot (\sqrt{10})^2\cdot \sin (2)\\\\A&=\dfrac{1}{2}\cdot 10\cdot \sin (2)\\\\A&=5\sin(2)\end{aligned}[/tex]
Now, subtract the area of the isosceles triangle from the area of the sector:
[tex]\begin{aligned}R&=10-5\sin(2)\\\\R&=10-4.546487134...\\\\R&=5.4535128658...\\\\R&=5.454\; \sf cm^2\end{aligned}[/tex]
Therefore, the area of R rounded to 3 decimal places is:
[tex]\Large\boxed{\boxed{R=5.454 \; \sf cm^2}}[/tex]
