Questions - Edexcel P1 - A-Level Maths

$B C = 8 \mathrm {~m}$
$O A = O B = 3 \mathrm {~m}$
angle $A O B$ is $\frac { 2 \pi } { 3 }$ radians
angle $B C A$ is $\frac { \pi } { 6 }$ radians
1. Calculate (i) the exact area of the sector $A O B X$,
  (ii) the exact perimeter of the sector $A O B X$.
2. Calculate the exact area of the triangle $A O B$.
3. Show that the length $A B$ is $3 \sqrt { 3 } \mathrm {~m}$.
4. Find the total surface area of the pond. Give your answer in $\mathrm { m } ^ { 2 }$ correct to 2 significant figures.

Edexcel P1 2022 October Q9

9. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{db979349-3415-420f-a39f-8cc8c24a69d0-24_889_666_258_703} \captionsetup{labelformat=empty} \caption{Figure 3}

\end{figure} Figure 3 shows a sketch of the curve $C$ with equation $$y = \frac { 1 } { 2 } x ^ { 2 } - 10 x + 22$$

Write $\frac { 1 } { 2 } x ^ { 2 } - 10 x + 22$ in the form $$a ( x + b ) ^ { 2 } + c$$ where $a , b$ and $c$ are constants to be found. The point $M$ is the minimum turning point of $C$, as shown in Figure 3.
Deduce the coordinates of $M$ The line $l$ is the normal to $C$ at the point $P$, as shown in Figure 3.
Given that $l$ has equation $y = k - \frac { 1 } { 8 } x$, where $k$ is a constant,
1. find the coordinates of $P$
2. find the value of $k$ Question 9 continues on the next page \begin{figure}[h]
  \includegraphics[alt={},max width=\textwidth]{db979349-3415-420f-a39f-8cc8c24a69d0-25_903_682_299_605} \captionsetup{labelformat=empty} \caption{Figure 4}
  \end{figure} Figure 4 is a copy of Figure 3. The finite region $R$, shown shaded in Figure 4, is bounded by $l , C$ and the line through $M$ parallel to the $y$-axis.
Identify the inequalities that define $R$.

Edexcel P1 2023 October Q1

Given that

$$y = 5 x ^ { 3 } + \frac { 3 } { x ^ { 2 } } - 7 x \quad x > 0$$ find, in simplest form,

$\frac { \mathrm { d } y } { \mathrm {~d} x }$
$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$

Edexcel P1 2023 October Q2

Given that

$$a = \frac { 1 } { 64 } x ^ { 2 } \quad b = \frac { 16 } { \sqrt { x } }$$ express each of the following in the form $k x ^ { n }$ where $k$ and $n$ are simplified constants.

$a ^ { \frac { 1 } { 2 } }$
$\frac { 16 } { b ^ { 3 } }$
$\left( \frac { a b } { 2 } \right) ^ { - \frac { 4 } { 3 } }$

Edexcel P1 2023 October Q3

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

Write $\frac { 8 - \sqrt { 15 } } { 2 \sqrt { 3 } + \sqrt { 5 } }$ in the form $a \sqrt { 3 } + b \sqrt { 5 }$ where $a$ and $b$ are integers to be found.
Hence, or otherwise, solve $$( x + 5 \sqrt { 3 } ) \sqrt { 5 } = 40 - 2 x \sqrt { 3 }$$ giving your answer in simplest form.

Edexcel P1 2023 October Q4

4. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{c0b4165d-b8bb-419c-b75a-d6c0c2431510-08_687_775_248_646} \captionsetup{labelformat=empty} \caption{Figure 1}

\end{figure} Figure 1 shows a sketch of part of the curve $C$ with equation $y = \frac { 1 } { x + 2 }$

State the equation of the asymptote of $C$ that is parallel to the $y$-axis.
Factorise fully $x ^ { 3 } + 4 x ^ { 2 } + 4 x$ A copy of Figure 1, labelled Diagram 1, is shown on the next page.
On Diagram 1, add a sketch of the curve with equation $$y = x ^ { 3 } + 4 x ^ { 2 } + 4 x$$ On your sketch, state clearly the coordinates of each point where this curve cuts or meets the coordinate axes.
Hence state the number of real solutions of the equation $$( x + 2 ) \left( x ^ { 3 } + 4 x ^ { 2 } + 4 x \right) = 1$$ giving a reason for your answer.
\includegraphics[max width=\textwidth, alt={}]{c0b4165d-b8bb-419c-b75a-d6c0c2431510-09_800_1700_1053_185}
Only use the copy of Diagram 1 if you need to redraw your answer to part (c).

Edexcel P1 2023 October Q5

5. Figure 2 Diagram NOT accurately drawn Figure 2 shows the plan view of a frame for a flat roof.
The shape of the frame consists of triangle $A B D$ joined to triangle $B C D$.
Given that

$B D = x \mathrm {~m}$
$C D = ( 1 + x ) \mathrm { m }$
$B C = 5 \mathrm {~m}$
angle $B C D = \theta ^ { \circ }$
1. show that $\cos \theta ^ { \circ } = \frac { 13 + x } { 5 + 5 x }$

Given also that

$x = 2 \sqrt { 3 }$
angle $B A C = 30 ^ { \circ }$
$A D C$ is a straight line
find the area of triangle $A B C$, giving your answer, in $\mathrm { m } ^ { 2 }$, to one decimal place.

Edexcel P1 2023 October Q6

In this question you must show all stages of your working.

\section*{Solutions relying on calculator technology are not acceptable.} The equation $$4 ( p - 2 x ) = \frac { 12 + 15 p } { x + p } \quad x \neq - p$$ where $p$ is a constant, has two distinct real roots.

Show that $$3 p ^ { 2 } - 10 p - 8 > 0$$
Hence, using algebra, find the range of possible values of $p$

Edexcel P1 2023 October Q7

The curve $C$ has equation $y = \mathrm { f } ( x )$ where $x > 0$

Given that

$f ^ { \prime } ( x ) = \frac { 4 x ^ { 2 } + 10 - 7 x ^ { \frac { 1 } { 2 } } } { 4 x ^ { \frac { 1 } { 2 } } }$
the point $P ( 4 , - 1 )$ lies on $C$
1. (i) find the value of the gradient of $C$ at $P$
  (ii) Hence find the equation of the normal to $C$ at $P$, giving your answer in the form $a x + b y + c = 0$ where $a , b$ and $c$ are integers to be found.
2. Find $\mathrm { f } ( x )$.

Edexcel P1 2023 October Q8

In this question you must show all stages of your working.

\section*{Solutions relying on calculator technology are not acceptable.} The curve $C _ { 1 }$ has equation $$x y = \frac { 15 } { 2 } - 5 x \quad x \neq 0$$ The curve $C _ { 2 }$ has equation $$y = x ^ { 3 } - \frac { 7 } { 2 } x - 5$$

Show that $C _ { 1 }$ and $C _ { 2 }$ meet when $$2 x ^ { 4 } - 7 x ^ { 2 } - 15 = 0$$ Given that $C _ { 1 }$ and $C _ { 2 }$ meet at points $P$ and $Q$
find, using algebra, the exact distance $P Q$

Edexcel P1 2023 October Q9

9. Diagram NOT accurately drawn \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{c0b4165d-b8bb-419c-b75a-d6c0c2431510-24_581_1491_340_296} \captionsetup{labelformat=empty} \caption{Figure 3}

\end{figure} Figure 3 shows the plan view of the area being used for a ball-throwing competition.
Competitors must stand within the circle $C$ and throw a ball as far as possible into the target area, $P Q R S$, shown shaded in Figure 3. Given that

circle $C$ has centre $O$
$P$ and $S$ are points on $C$
$O P Q R S O$ is a sector of a circle with centre $O$
the length of arc $P S$ is 0.72 m
the size of angle $P O S$ is 0.6 radians
1. show that $O P = 1.2 \mathrm {~m}$

Given also that

the target area, $P Q R S$, is $90 \mathrm {~m} ^ { 2 }$
length $P Q = x$ metres
show that

$$5 x ^ { 2 } + 12 x - 1500 = 0$$

Hence calculate the total perimeter of the target area, $P Q R S$, giving your answer to the nearest metre.

Edexcel P1 2023 October Q10

10. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{c0b4165d-b8bb-419c-b75a-d6c0c2431510-28_538_652_255_708} \captionsetup{labelformat=empty} \caption{Figure 4}

\end{figure} Figure 4 shows a sketch of part of the curve $C _ { 1 }$ with equation $$y = 3 \cos \left( \frac { x } { n } \right) ^ { \circ } \quad x \geqslant 0$$ where $n$ is a constant.
The curve $C _ { 1 }$ cuts the positive $x$-axis for the first time at point $P ( 270,0 )$, as shown in Figure 4.

1. State the value of $n$
2. State the period of $C _ { 1 }$ The point $Q$, shown in Figure 4, is a minimum point of $C _ { 1 }$
State the coordinates of $Q$. The curve $C _ { 2 }$ has equation $y = 2 \sin x ^ { \circ } + k$, where $k$ is a constant.
The point $R \left( a , \frac { 12 } { 5 } \right)$ and the point $S \left( - a , - \frac { 3 } { 5 } \right)$, both lie on $C _ { 2 }$
Given that $a$ is a constant less than 90
find the value of $k$.

Edexcel P1 2023 October Q11

11. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{c0b4165d-b8bb-419c-b75a-d6c0c2431510-30_595_869_255_568} \captionsetup{labelformat=empty} \caption{Figure 5}

\end{figure} Figure 5 shows part of the curve $C$ with equation $y = \mathrm { f } ( x )$ where $$f ( x ) = 2 x ^ { 2 } - 12 x + 14$$

Write $2 x ^ { 2 } - 12 x + 14$ in the form $$a ( x + b ) ^ { 2 } + c$$ where $a$, $b$ and $c$ are constants to be found. Given that $C$ has a minimum at the point $P$
state the coordinates of $P$ The line $l$ intersects $C$ at $( - 1,28 )$ and at $P$ as shown in Figure 5.
Find the equation of $l$ giving your answer in the form $y = m x + c$ where $m$ and $c$ are constants to be found. The finite region $R$, shown shaded in Figure 5, is bounded by the $x$-axis, $l$, the $y$-axis, and $C$.
Use inequalities to define the region $R$.

Edexcel P1 2018 Specimen Q1

Given that $y = 4 x ^ { 3 } - \frac { 5 } { x ^ { 2 } } , x \neq 0$, find in their simplest form
1. $\frac { \mathrm { d } y } { \mathrm {~d} x }$,
2. $\int y \mathrm {~d} x$
  a) $y = 4 x ^ { 3 } - 5 x ^ { - 2 }$
  $\frac { d y } { d x } = 12 x ^ { 2 } + 10 x ^ { - 3 }$
  b) $\int 4 x ^ { 3 } - 5 x ^ { - 2 } d x$
  $= \frac { 4 x ^ { 4 } } { 4 } - \frac { 5 x ^ { - 1 } } { - 1 } + c = x ^ { 4 } + 5 x ^ { - 1 } + c$

Edexcel P1 2018 Specimen Q2

2. (a) Given that $3 ^ { - 1.5 } = a \sqrt { 3 }$ find the exact value of $a$
(b) Simplify fully $\frac { \left( 2 x ^ { \frac { 1 } { 2 } } \right) ^ { 3 } } { 4 x ^ { 2 } }$
a) $a = \frac { 3 ^ { - 1.5 } } { \sqrt { 3 } } = \frac { 1 } { 9 }$
b) $\frac { \left( 2 x ^ { 1 / 2 } \right) ^ { 3 } } { 4 x ^ { 2 } } = \frac { 2 ^ { 3 } x ^ { 1 / 2 \times 3 } } { 4 x ^ { 2 } } = 2 x ^ { - 1 / 2 }$ Solve the simultaneous equations $$\begin{aligned} & y + 4 x + 1 = 0
& y ^ { 2 } + 5 x ^ { 2 } + 2 x = 0 \end{aligned}$$ $$\begin{aligned} & \text { (1) } y = - 4 x - 1
& \therefore ( - 4 x - 1 ) ^ { 2 } + 5 x ^ { 2 } + 2 x = 0
& \therefore 16 x ^ { 2 } + 8 x + 1 + 5 x ^ { 2 } + 2 x = 0
& \therefore 21 x ^ { 2 } + 10 x + 1 = 0
& x = \frac { - 10 \pm \sqrt { ( - 10 ) ^ { 2 } - 4 ( 21 ) ( 1 ) } } { 2 \times 21 }
&
& x = - 1 / 7 \quad x = - 1 / 3
& \begin{array} { r l } y = - 4 ( - 1 / 7 ) - 1 & y = - 4 ( - 1 / 3 ) - 1
= & - 1 / 7
( - 1 / 7 , - 3 / 7 ) & = 1 / 3 \end{array}
& ( - 1 / 3,1 / 3 ) \end{aligned}$$

Edexcel P1 2018 Specimen Q4

4. The straight line with equation $y = 4 x + c$, where $c$ is a constant, is a tangent to the curve with equation $y = 2 x ^ { 2 } + 8 x + 3$ Calculate the value of $c$ $$\begin{aligned} & y = m x + c \rightarrow y = 4 x + c \quad ( m = 4 )
& \therefore \frac { d y } { d x } = 4 x + 8 = 4 \quad \text { (Gradient equation) }
& 4 x + 8 = 4
& x = - 1 \rightarrow y = - 3
& \text { At } ( - 1 , - 3 ) - 3 = 4 ( - 1 ) + c
& \quad c = 1
& y = 4 x + 1 \end{aligned}$$ \section*{PMT PhysicsAndMathsTutor.com}

\includegraphics[max width=\textwidth, alt={}]{2217be5e-8edd-413f-9c97-212e585ff58d-09_2258_54_312_34}

Edexcel P1 2018 Specimen Q5

5. (a) On the same axes, sketch the graphs of $y = x + 2$ and $y = x ^ { 2 } - x - 6$ showing the coordinates of all points at which each graph crosses the coordinate axes.
(b) On your sketch, show, by shading, the region $R$ defined by the inequalities $$y < x + 2 \text { and } y > x ^ { 2 } - x - 6$$ (c) Hence, or otherwise, find the set of values of $x$ for which $x ^ { 2 } - 2 x - 8 < 0$
\includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-10_921_1287_699_260} \includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-11_2260_48_313_37} Quadratic: $y = x ^ { 2 } - x - 6 = ( x - 3 ) ( x + 2 )$ $$x = 3 , \quad x = - 2 @ y = 0$$ Linear: $\quad y = x + 2$ $$\begin{array} { l l } x = 0 : & y = 2 \quad ( 0,2 )
y = 0 : & x = - 2 \quad ( - 2,0 ) \end{array}$$ c) $\quad x ^ { 2 } - 2 x - 8 < 0$ $$\begin{aligned} \therefore ( x - 4 ) ( x + 2 ) & < 0
x = 4 \quad x = - 2 &
\therefore - 2 < x < 4 & < 4 \end{aligned}$$

Edexcel P1 2018 Specimen Q6

6. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{2217be5e-8edd-413f-9c97-212e585ff58d-12_440_679_269_630} \captionsetup{labelformat=empty} \caption{Figure 1}

\end{figure} Figure 1 shows a sketch of the curve $C$ with equation $y = \mathrm { f } ( x )$ The curve $C$ passes through the origin and through $( 6,0 )$ The curve $C$ has a minimum at the point $( 3 , - 1 )$ On separate diagrams, sketch the curve with equation

$y = \mathrm { f } ( 2 x )$
$y = \mathrm { f } ( x + p )$, where $p$ is a constant and $0 < p < 3$ On each diagram show the coordinates of any points where the curve intersects the $x$-axis and of any minimum or maximum points.
a) $( 6,0 ) \rightarrow ( 3,0 )$ $$( 3 , - 1 ) - 1 > ( 1.5 , - 1 )$$
\includegraphics[max width=\textwidth, alt={}]{2217be5e-8edd-413f-9c97-212e585ff58d-12_616_772_1624_781}
$$( 1.5 , - 1 )$$ \includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-13_2261_50_312_39}
\includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-13_2637_1835_118_116}

Edexcel P1 2018 Specimen Q7

7. A curve with equation $y = \mathrm { f } ( x )$ passes through the point $( 4,25 )$ Given that $$\mathrm { f } ^ { \prime } ( x ) = \frac { 3 } { 8 } x ^ { 2 } - 10 x ^ { - \frac { 1 } { 2 } } + 1 , \quad x > 0$$ find $\mathrm { f } ( x )$, simplifying each term. $$\begin{aligned} & \therefore F ( x ) = \int F ^ { \prime } ( x ) = \int 3 / 8 x ^ { 2 } - 10 x ^ { - 1 / 2 } + 1 d x
& F ( x ) = \frac { 3 x ^ { 3 } } { 8 ( 3 ) } - \frac { 10 x ^ { 1 / 2 } } { 1 / 2 } + x + c
& F ( x ) = 1 / 8 x ^ { 3 } - 20 x ^ { 1 / 2 } + x + c
& 25 = 1 / 8 ( 4 ) ^ { 3 } - 20 ( 4 ) ^ { 1 / 2 } + 4 + c
& 25 = 8 - 40 + 4 + c
& C = 53
& F ( x ) = 1 / 8 x ^ { 3 } - 20 x ^ { 1 / 2 } + x + 53 \end{aligned}$$ \section*{PMT PhysicsAndMathsTutor.com}

Edexcel P1 2018 Specimen Q8

8. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{2217be5e-8edd-413f-9c97-212e585ff58d-16_769_979_269_479} \captionsetup{labelformat=empty} \caption{Figure 2}

\end{figure} The line $l _ { 1 }$, shown in Figure 2 has equation $2 x + 3 y = 26$
The line $l _ { 2 }$ passes through the origin $O$ and is perpendicular to $l _ { 1 }$

Find an equation for the line $l _ { 2 }$ The line $l _ { 2 }$ intersects the line $l _ { 1 }$ at the point $C$. Line $l _ { 1 }$ crosses the $y$-axis at the point $B$ as shown in Figure 2.
Find the area of triangle $O B C$. Give your answer in the form $\frac { a } { b }$, where $a$ and $b$ are integers to be found.
a) $L _ { 1 } : 2 x + 3 y = 26$
\includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-16_188_820_2039_159}
\includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-16_129_631_2268_468} $$\begin{gathered} y = 3 / 2 x + 0
y = 3 / 2 x \end{gathered}$$ \includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-17_2257_51_315_34}
b) $A = \frac { 6 x h } { 2 } \quad L _ { 1 } : 2 x + 3 y = 26$ $$L _ { 2 } : y = 3 / 2 x$$ At B: $x = 0 : 0 + 3 y = 26 y = 26 / 3$ At C: $2 x + 3 \left( \frac { 3 x } { 2 } \right) = 26$
$\therefore x = 4$
$A = \frac { 4 \times 26 } { 3 } = 52 / 3$
VIAV SIHI NI IIIIM IONOO VIIV SIHI NI JIIIM ION OC VEYV SIHI NI JIIYM ION OO
\section*{PMT PhysicsAndMathsTutor.com}
\includegraphics[max width=\textwidth, alt={}]{2217be5e-8edd-413f-9c97-212e585ff58d-19_2255_54_312_34}
\includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-19_113_61_2604_1884}

Edexcel P1 2018 Specimen Q9

9. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{2217be5e-8edd-413f-9c97-212e585ff58d-20_693_1038_267_450} \captionsetup{labelformat=empty} \caption{Figure 3}

\end{figure} A sketch of part of the curve $C$ with equation $$y = 20 - 4 x - \frac { 18 } { x } , \quad x > 0$$ is shown in Figure 3. Point $A$ lies on $C$ and has $x$ coordinate equal to 2

Show that the equation of the normal to $C$ at $A$ is $y = - 2 x + 7$. The normal to $C$ at $A$ meets $C$ again at the point $B$, as shown in Figure 3 .
Use algebra to find the coordinates of $B$.
a) $y = 20 - 4 ( 2 ) - \frac { 18 } { 2 } \quad \therefore \quad y = 3$ $$\begin{aligned} & y = 20 - 4 x - 18 x ^ { - 1 }
& \therefore \frac { d y } { d x } = - 4 + 18 x ^ { - 2 } \end{aligned}$$
$x = 2 \quad \frac { d y } { d x } = - 4 + \frac { 18 } { 2 ^ { 2 } } = 1 / 2$ $$m = - 2$$
\includegraphics[max width=\textwidth, alt={}]{2217be5e-8edd-413f-9c97-212e585ff58d-20_2258_50_313_1980}
9 continued
\includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-21_2253_51_315_35} $$\begin{aligned} \therefore \quad y & = - 2 x + c
3 & = - 2 ( 2 ) + c
c & = 7
y & = - 2 x + 7 \end{aligned}$$ b) $20 - 4 x - \frac { 18 } { x } = - 2 x + 7$ $$\begin{aligned} \therefore & 13 - 2 x - \frac { 18 } { x } = 0
& 13 x - 2 x ^ { 2 } - 18 = 0
& 0 = 2 x ^ { 2 } - 13 x + 18
\therefore & x = \frac { - b \pm \sqrt { b ^ { 2 } - 4 a c } } { 2 a }
& a = 2 \quad b = - 13 \quad c = 18
& x = 2 \quad x = a / 2
\therefore & y = 3 \quad \therefore y = - 2
B = & ( a / 2 , - 2 ) \end{aligned}$$ "

VIIV SIHI NI IIIIM I I N O C VI4V SIHI NI IIIHM ION OO V34V SIHI NI JIIYM ION OC

Edexcel P1 2018 Specimen Q10

10.

\includegraphics[max width=\textwidth, alt={}]{2217be5e-8edd-413f-9c97-212e585ff58d-23_2255_50_315_37}

\begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{2217be5e-8edd-413f-9c97-212e585ff58d-23_411_1065_252_277} \captionsetup{labelformat=empty} \caption{Figure 4}

\end{figure} The triangle $X Y Z$ in Figure 4 has $X Y = 6 \mathrm {~cm} , Y Z = 9 \mathrm {~cm} , Z X = 4 \mathrm {~cm}$ and angle $Z X Y = \alpha$. The point $W$ lies on the line $X Y$. The circular arc $Z W$, in Figure 4, is a major arc of the circle with centre $X$ and radius 4 cm .

Show that, to 3 significant figures, $\alpha = 2.22$ radians.
Find the area, in $\mathrm { cm } ^ { 2 }$, of the major sector $X Z W X$. The region, shown shaded in Figure 4, is to be used as a design for a logo. \section*{Calculate}
the area of the logo
the perimeter of the logo.
\includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-23_383_705_1813_182} Cosine Rule: $$\begin{aligned} & \cos A = \frac { b ^ { 2 } + c ^ { 2 } - a ^ { 2 } } { 2 b c }
& \cos A = \frac { 4 ^ { 2 } + 6 ^ { 2 } - 9 ^ { 2 } } { 2 ( 4 ) ( 6 ) }
& A = 2.22 \text { radians } \end{aligned}$$ b)
\includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-24_323_429_302_353} $$\begin{aligned} \text { Area } & = 1 / 2 r ^ { 2 } \theta
& = 1 / 2 ( 4 ) ^ { 2 } ( 2 \pi - 2.22 )
& = 32.5 \mathrm {~cm} ^ { 2 } \end{aligned}$$ c) Area (Logo) $=$ Area ( $\Delta$ ) + Area (Sector)
\includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-24_193_268_840_620}
\includegraphics[max width=\textwidth, alt={}, center]{2217be5e-8edd-413f-9c97-212e585ff58d-24_186_390_934_1028} $$\begin{aligned} & = 1 / 2 ( 4 ) ( 6 ) \sin 2.22
\text { Area } ( \log 0 ) & = 1 / 2 ( 4 ) ( 6 ) \sin 2.22 + 1 / 2 ( 4 ) ^ { 2 } ( 2 \pi - 2.22 )
& = 42.1 \mathrm {~cm} ^ { 2 } \end{aligned}$$ d) $P = 9 + 2 +$ Arc length $$\begin{aligned} P & = 9 + 2 + 4 ( 2 \pi - 2.22 )
& = 27.3 \mathrm {~cm} \end{aligned}$$ \section*{PMT PhysicsAndMathsTutor.com}
VIUV SIHI NI IIIUM ION OC VIUV SIHI NI JIIIM ION OC V34V SIHI NI EIIYM ION OC

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