**Q****uestion**

A piece of wire of length 24cm is bent to form the perimeter of a sector of a circle of radius r cm.

**(i)Show that the area of the sector ,\(Acm^{2}\) is given by \(A=12r-r^{2}\)**

(ii)Express A in the form \(a-\left ( r-b \right )^{2}\),where a and b are constants.

**(iii)Given that r can vary,state the greatest value of A and find the corresponding angle of the sector.**

**▶️Answer/Explanation**

(i) Show that the area of the sector \(A\) is given by \(A = 12r – r^2\)

The perimeter of the sector is the length of the wire, which is equal to the sum of the arc length and the two radii.

\( \text{Perimeter} = \text{Arc Length} + 2r \)

Since the arc length is a fraction of the total circumference (which is \(2\pi r\)), we can express the arc length as \(\frac{\theta}{360} \times 2\pi r\), where \(\theta\) is the central angle of the sector.

So, we have the equation:

\( 24 = \frac{\theta}{360} \times 2\pi r + 2r \)

\( \theta = \frac{360}{\pi} (24 – 2r) \)

\( A = \frac{\theta}{360} \times \pi r^2 \)

\( A = \frac{360}{\pi} (24 – 2r) \times \frac{\pi}{360} \times \pi r^2 \)

\( A = (24 – 2r)r = 12r – r^2 \)

(ii) Express \(A\) in the form \(a – (r – b)^2\), where \(a\) and \(b\) are constants:

\( A = 12r – r^2 \)

\( A = -(r^2 – 12r + 36) + 36 \)

\( A = -(r – 6)^2 + 36 \)

So, \(A\) can be expressed in the form \(a – (r – b)^2\) with \(a = 36\) and \(b = 6\).

(iii) Given that \(r\) can vary, state the greatest value of \(A\) and find the corresponding angle of the sector:

The expression \(-(r – 6)^2 + 36\) indicates that \(A\) is maximized when \((r – 6)^2\) is minimized, which occurs when \(r = 6\). So, the greatest value of \(A\) is \(36\), and it occurs when \(r = 6\).

\( \theta = \frac{360}{\pi} (24 – 2r) \)

Substitute \(r = 6\):

\( \theta = \frac{360}{\pi} (24 – 2(6)) = \frac{360}{\pi} \times 12 \)

So, the corresponding angle \(\theta\) is \(120^\circ\).

Therefore, the greatest value of \(A\) is \(36\) square centimeters, and the corresponding angle of the sector is \(120^\circ\).

**Question**

The line with gradient -2 passing through the point P(3t,2t)intersects the x-axis at A and the y-axis at B.

**(i)Find the area of triangle AOB in terms of t.**

(ii)The line through P perpendicular to AB intersects the x-axis at C.

Show that the mid-point of PC lies on the line y=x.

**▶️Answer/Explanation**

(i) Find the area of triangle \(AOB\) in terms of \(t\):

Given the equation \(y – 2t = -2(x – 3t)\), we can simplify it to \(y + 2x = 8t\).

The x-intercept (\(A\)):

– Set \(y = 0\): \(0 + 2x = 8t\)

– Solve for \(x\): \(x = 4t\)

– So, the coordinates of \(A\) are \((4t, 0)\).

The y-intercept (\(B\)):

– Set \(x = 0\): \(y + 2(0) = 8t\)

– Solve for \(y\): \(y = 8t\)

– So, the coordinates of \(B\) are \((0, 8t)\).

Now, the area of triangle \(AOB\) is given by the formula:

\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)

In this case, the base is the distance between points \(A\) and \(B\), which is \(4t\), and the height is the y-coordinate of \(B\), which is \(8t\).

\( \text{Area} = \frac{1}{2} \times 4t \times 8t = 16t^2 \)

So, the area of triangle \(AOB\) is \(16t^2\).

(ii) The line through \(P\) perpendicular to \(AB\) intersects the \(x\)-axis at \(C\).

The slope of the line AB is the negative reciprocal of the slope of the line passing through P:

\(m_{AB} \cdot m_{PC} = -1\)

\((-2) \cdot m_{PC} = -1\)

\(m_{PC} = \frac{1}{2}\)

Now, the line through P with slope \(\frac{1}{2}\) has the equation in point-slope form:

\(y – 2t = \frac{1}{2}(x – 3t)\)

\(y = \frac{1}{2}x + \frac{5}{2}t\)

When this line intersects the \(x\)-axis (\(y = 0\)), we can find point C:

\(0 = \frac{1}{2}x + \frac{5}{2}t\)

\(x = -5t\)

This gives us the coordinates of point C as \((-5t, 0)\).

Now, the mid-point of PC is the average of the x-coordinates and y-coordinates:

\(M\left(\frac{-5t + 3t}{2}, \frac{0 + 2t}{2}\right)\)

\(M(-t, t)\)

To show that this point lies on the line \(y = x\), we can check if \(y = x\) for these coordinates:

\(t = -t\)

This is true, so the mid-point of PC, \(M(-t, t)\), lies on the line \(y = x\).

**Question **

The diagram shows triangle ABC in which AB is [perpendicular to BC .The length of AB is 4cm and angle CAB is \(\alpha\) radians.The arc DE with cent6re A and radius 2cm meets AC at D and AB at E .Find in terms of \(\alpha \)

(i)the area of the shaded region.

(ii)the perimeter of the shaded region.

**Answer/Explanation**

(i)Area \(\Delta =\frac{1}{2}\times 4\times 4\tan \alpha \)

Area Sector\(=\frac{1}{2}\times 2^{2}\alpha \)

Shaded area \(=8\tan \alpha -2\alpha \)

(ii)\(DC=\frac{4}{\cos \alpha }-2\)

arc DE=\(2\alpha\)

Perimeter =\(\frac{4}{\cos \alpha }+4\tan \alpha +2\alpha\)