Questions
Points A and B lie on the circumference of a circle of radius rcm with centre at O .
The sector OAB is shown on the following diagram. The angle AÔB is denoted as \(\theta\) and is measured in radians.
The perimeter of the sector is 10 cm and the area of the sector is 6.25 cm2 .
(a) Show that \(4r^{2}-20r+25=0\) .
(b) Hence, or otherwise, find the value of r and the value of \(\theta\) .
▶️Answer/Explanation
Detailed solution
(a) Show that \( 4r^2 – 20r + 25 = 0 \)
The sector OAB includes two radii (OA and OB) and the arc AB. The perimeter of the sector is the sum of these lengths:
– Length of OA = \( r \) cm,
– Length of OB = \( r \) cm,
– Length of arc AB = the arc length, which for a sector with radius \( r \) and angle \( \theta \) radians is \( r \theta \).
So, the perimeter is:
\[
\text{Perimeter} = r + r + r \theta = 2r + r \theta
\]
We’re told the perimeter is 10 cm:
\[
2r + r \theta = 10 \quad (1)
\]
The area of the sector is given by the formula \( \frac{1}{2} r^2 \theta \) for a sector with radius \( r \) and angle \( \theta \) radians. We’re told the area is 6.25 cm²:
\[
\text{Area} = \frac{1}{2} r^2 \theta = 6.25 \quad (2)
\]
Now, solve these equations. From (2), solve for \( \theta \):
\[
\frac{1}{2} r^2 \theta = 6.25
\]
\[
r^2 \theta = 12.5
\]
\[
\theta = \frac{12.5}{r^2} \quad (3)
\]
Substitute \( \theta \) from (3) into (1):
\[
2r + r \cdot \frac{12.5}{r^2} = 10
\]
Simplify the second term:
\[
r \cdot \frac{12.5}{r^2} = \frac{12.5}{r}
\]
So:
\[
2r + \frac{12.5}{r} = 10
\]
To eliminate the fraction, multiply through by \( r \) (assuming \( r \neq 0 \), which is true since \( r \) is a radius):
\[
2r \cdot r + \frac{12.5}{r} \cdot r = 10r
\]
\[
2r^2 + 12.5 = 10r
\]
Move all terms to one side:
\[
2r^2 – 10r + 12.5 = 0
\]
To match the given equation \( 4r^2 – 20r + 25 = 0 \), multiply the entire equation by 2 to clear the decimal:
\[
2 \cdot (2r^2 – 10r + 12.5) = 2 \cdot 0
\]
\[
4r^2 – 20r + 25 = 0
\]
This confirms the equation \( 4r^2 – 20r + 25 = 0 \), as required!
(b) Hence, or Otherwise, Find the Value of \( r \) and the Value of \( \theta \)
Now, solve the quadratic equation \( 4r^2 – 20r + 25 = 0 \). Use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \), where \( a = 4 \), \( b = -20 \), \( c = 25 \):
– Discriminant: \( b^2 – 4ac = (-20)^2 – 4 \cdot 4 \cdot 25 = 400 – 400 = 0 \),
– Since the discriminant is 0, there is one real solution (a repeated root):
\[
r = \frac{-(-20) \pm \sqrt{0}}{2 \cdot 4} = \frac{20}{8} = 2.5
\]
So, \( r = 2.5 \) cm. A radius of 2.5 cm is positive, which makes sense for a circle.
Now find \( \theta \) using equation (2):
\[
\frac{1}{2} r^2 \theta = 6.25
\]
Substitute \( r = 2.5 \):
\[
\frac{1}{2} (2.5)^2 \theta = 6.25
\]
\[
(2.5)^2 = 6.25
\]
\[
\frac{1}{2} \cdot 6.25 \cdot \theta = 6.25
\]
\[
3.125 \theta = 6.25
\]
\[
\theta = \frac{6.25}{3.125}
\]
Simplify:
\[
\frac{6.25}{3.125} = \frac{6.25 \div 1.25}{3.125 \div 1.25} = \frac{5}{2.5} = 2
\]
So, \( \theta = 2 \) radians. Let’s verify with the perimeter:
\[
2r + r \theta = 2 \cdot 2.5 + 2.5 \cdot 2 = 5 + 5 = 10
\]
This matches the given perimeter of 10 cm. The area check:
\[
\frac{1}{2} r^2 \theta = \frac{1}{2} \cdot 6.25 \cdot 2 = 6.25
\]
………………………….Markscheme………………………
Ans:
(a) \(2r+r\theta =10\)
\(\frac{1}{2}r^{2}\theta =6.25\)
attempt to eliminate \(\theta\) to obtain an equation in r
correct intermediate equation in r
\(10-2r=\frac{25}{2r}\) OR \(\frac{10}{r}-2=\frac{25}{2r^{2}}\) OR \(\frac{1}{2}r^{2}(\frac{10}{r}-2)=6.25\) OR \(12.5+2r^{2}=10r\)
\(4r^{2}-20r+25=0\)
(b) attempt to solve quadratic by factorizing or use of formula or completing the square
\((2r-5)^{2}=0\) OR \(r=\frac{20\pm \sqrt{(-20)^{2}-4(4)(25)}}{2(4)}\left ( =\frac{20\pm \sqrt{400-400}}{8} \right )\)
\(r=\frac{5}{2}\)
attempt to substitute their value or r into their perimeter or area equation
\(\theta =\frac{10-2\left ( \frac{5}{2} \right )}{\left ( \frac{5}{2} \right )}\) OR \(\theta =\frac{25}{2\left ( \frac{5}{2} \right )^{2}}\)
\(\theta=2\)