Higher November 2020 Unit 1 Q8
8. A circle, centre \(O\), has a radius of 4 cm.
\(A\) and \(B\) are points on the circumference of the circle.
Lines \(PA\) and \(PB\) are both tangents to the circle.
\(PB\) = 12 cm.
Diagram not drawn to scale
(a) What is the length of \(PA\)?
State the circle theorem you have used to find your answer. [1]
State the circle theorem you have used to find your answer. [1]
\(PA\) = ……………………
Circle theorem: ……………………
(b) What is the size of \(P\hat{A}O\)?
State the circle theorem you have used to find your answer. [1]
State the circle theorem you have used to find your answer. [1]
\(P\hat{A}O\) = ……………………
Circle theorem: ……………………
(c) Calculate the area of the quadrilateral \(PAOB\). [2]
| Scheme | Marks | Comments |
|---|---|---|
| PA = 12(cm) AND correct theorem given, e.g. ‘tangents from an external point are equal in length’. | E1 | Must use the words ‘tangents’ AND ‘equal (identical/same)’. Do not accept e.g. ‘PA = PB’. (E0) Accept alternative correct answers. |
| Scheme | Marks | Comments |
|---|---|---|
| PÂO = 90(°) AND correct theorem given, e.g. ‘the tangent at any point on a circle is perpendicular to the radius at that point’. | E1 | Must use the words ‘tangent’ AND ‘radius (diameter)’ Allow e.g. ‘radius and tangent meet at 90’. (E1) Do not accept e.g. ‘PA and OA meet at 90’. (E0) |
| Scheme | Marks | Comments |
|---|---|---|
| (Area PAOB =) \(2 \times \dfrac{12 \times 4}{2}\) or equivalent. | M1 | OR FT \(\dfrac{\text{‘their PA’} \times 4}{2} + \dfrac{12 \times 4}{2}\) M0 for 48 × 2 or 12 × 4 × 2 (= 96) |
| = 48 (cm2) | A1 | An unsupported final answer of 48 gains both marks. If no marks gained allow SC1 for sight of 24(cm2) OR a correct evaluation of (‘their PA’ × 4) / 2. |