Challenging +1.2 This question requires setting up equations using circle geometry (two triangles sharing base BD), applying the area formula for triangles, and solving a resulting equation. It involves multiple steps including using Pythagoras/perpendicular from centers, but follows a fairly standard approach for intersection problems. The algebra is moderately involved but not exceptionally challenging, placing it above average difficulty.
8 A circle with centre \(A\) and radius 8 cm and a circle with centre \(C\) and radius 12 cm intersect at points B and D .
Quadrilateral \(A B C D\) has area \(60 \mathrm {~cm} ^ { 2 }\).
Determine the two possible values for the length AC.
## Question 8:
Sketch diagram consistent with information in the question | B1 | AO 2.5; Triangle ADC or ABC or quadrilateral ABCD with 8 and 12 indicated e.g. side lengths or radii
$60 = 2 \times \dfrac{1}{2} \times 8 \times 12 \times \sin B$ | M1 | AO 3.1a; M1 implies previous B1; next 3 marks can be for angles $B$ or $D$
$\sin B = \dfrac{5}{8}$ so $B = 38.7°$ (0.675 rads) | A1 | AO 1.1a; One value of $B$ or $\cos B$; $\cos B = \dfrac{\sqrt{39}}{8}$
OR $B = 141.3°$ (2.47 rads) | A1 | AO 3.2a; Other value of $B$ or $\cos B$; $\cos B = -\dfrac{\sqrt{39}}{8}$
$AC^2 = 8^2 + 12^2 - 2\times8\times12\cos 38.7 = 58.1$; $AC = 7.62$ cm | M1, A1 | AO 3.1a; Use of cosine rule; Accept 7.6 www
$AC^2 = 8^2 + 12^2 - 2\times8\times12\cos 141.3 = 357.9$; $AC = 18.9$ cm | A1 [7] | AO 1.1; Accept 19 www; A0 if more than 2 answers
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8 A circle with centre $A$ and radius 8 cm and a circle with centre $C$ and radius 12 cm intersect at points B and D .
Quadrilateral $A B C D$ has area $60 \mathrm {~cm} ^ { 2 }$.\\
Determine the two possible values for the length AC.
\hfill \mbox{\textit{OCR MEI Paper 3 2023 Q8 [7]}}