# CIE A level -Pure Mathematics 1 : Topic :  1.4 Circular measure: solving problems concerning the arc length and sector area of a circle : Exam Questions Paper 1

Question

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.

(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.

(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.

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

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