Higher November 2019 Unit 2 Q17
17. Triangle \(DEC\) lies within a square \(ABCD\), as shown below.
\(DE\) = 8 cm and \(EC\) = 9 cm.
\(D\hat{E}C\) = 75°.
Diagram not drawn to scale
Calculate the area of the shaded region. [7]
From the formula list: In any triangle \(ABC\): Cosine rule \(a^2 = b^2 + c^2 - 2bc\cos A\); Area of triangle \(= \dfrac{1}{2}ab\sin C\)
| Scheme | Marks | Comments |
|---|---|---|
| For an attempt to subtract the area of a triangle from the area of square, with use of cosine rule and area of a triangle formula (½absinC). | S1 | |
| (Area of square or CD2 =) 8² + 9² – 2×8×9×cos75° | M1 | Allow (CD =) √[8² + 9² – 2×8×9×cos75°] |
| CD2 = 107.7(30…) OR CD = 10.37(9…cm) OR CD = 10.38(…cm) OR CD = √[107.7(30…)] | A1 | Allow CD = 10.4 (cm) |
| Area of square = 107.7(30…cm2) | A1 | Allow an answer in the range 107.5(cm2) to 108.2(cm2). May be implied in further working. |
| (Area of triangle = ) ½×8×9×sin75° | M1 | |
| = 34.77(…cm2) OR 34.8(cm2) OR 9√6+9√2(cm2) | A1 | Accept an answer in the range 34.6(cm2) to 35(cm2). |
| (Area of the shaded region=) answer in the range of 72.9 (cm²) to 73 (cm²) | B1 | CAO. |