| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof |
| Type | Geometric proof using coordinates |
| Difficulty | Challenging +1.2 This is a geometric proof requiring the angle-in-a-semicircle theorem (∠ACB = 90°), Pythagoras' theorem, and the area formula for equilateral triangles. While it involves multiple steps and some insight to connect the circle property to the triangle areas, the techniques are all standard AS-level content and the 5-mark allocation suggests a structured but not exceptionally demanding proof. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.05c Area of triangle: using 1/2 ab sin(C) |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | Uses ‘angle in a semicircle’ to | |
| justify ACB = 90° | 2.4 | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Deduces that AB2 = AC2 + BC2 | 2.2a | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| triangle | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| triangles | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| reasoned proof. | 2.1 | R1 |
| Total | 5 | |
| Q | Marking Instructions | AO |
Question 9:
9 | Uses ‘angle in a semicircle’ to
justify ACB = 90° | 2.4 | E1 | Angle ACB = 90° (Angle in a
semicircle)
AB2 = BC2 + AC2 (Pythagoras)
Area of ABK = AB2 sin 60°
1
Area of BCL = BC2 sin 60°
2
1
Area of CAM = AC2 sin 60°
2
1
2
BCL + CAM = sin 60° (BC2 + AC2)
1
= sin 60° (AB2)
2
1
= ABK
2
Deduces that AB2 = AC2 + BC2 | 2.2a | B1
Applies area formula to one
triangle | 1.1a | M1
Applies area formula to all three
triangles | 1.1b | A1
Forms a correct equation involving
BCL + CAM and completes
reasoned proof. | 2.1 | R1
Total | 5
Q | Marking Instructions | AO | Marks | Typical SolutionThe diagram below shows a circle and four triangles.
\begin{tikzpicture}[line join=round, line cap=round, thick]
% Define the radius of the circle
\def\R{2.5}
% Center of the circle
\coordinate (O) at (0,0);
% A and B form the horizontal diameter
\coordinate (A) at (-\R, 0);
\coordinate (B) at (\R, 0);
% C is exactly on the circle.
% The angle -65 is chosen to match the slight right-slant of the original image
\def\angC{-65}
\coordinate (C) at (\angC:\R);
% K forms an outward equilateral triangle KAB on AB
% (Rotates B counter-clockwise around A by 60 degrees)
\coordinate (K) at ($ (A) ! 1 ! 60:(B) $);
% M forms an outward equilateral triangle MAC on AC
% (Rotates C clockwise around A by 60 degrees)
\coordinate (M) at ($ (A) ! 1 ! -60:(C) $);
% L forms an outward equilateral triangle LBC on BC
% (Rotates C counter-clockwise around B by 60 degrees)
\coordinate (L) at ($ (B) ! 1 ! 60:(C) $);
% Draw the circle
\draw (O) circle (\R);
% Draw the inner right-angled triangle ABC
\draw (A) -- (B) -- (C) -- cycle;
% Draw the outer perimeter created by the three equilateral triangles
\draw (M) -- (A) -- (K) -- (B) -- (L) -- (C) -- cycle;
% Add the labels
\node[left=2pt] at (A) {$A$};
\node[right=2pt] at (B) {$B$};
\node[below=3pt] at (C) {$C$};
\node[above=2pt] at (K) {$K$};
\node[below left=2pt] at (M) {$M$};
\node[below right=2pt] at (L) {$L$};
\end{tikzpicture}
$AB$ is a diameter of the circle. $C$ is a point on the circumference of the circle.
Triangles $ABK$, $BCL$ and $CAM$ are equilateral.
Prove that the area of triangle $ABK$ is equal to the sum of the areas of triangle $BCL$ and triangle $CAM$. [5 marks]
\hfill \mbox{\textit{AQA AS Paper 1 2020 Q9 [5]}}