O-LevelMathematicsAlgebra and graphs4024/22 · Oct/Nov 2015 · Q712 Marks
ABD and ACE are straight lines. BD = 12 cm and CE = 4 cm. AB = x cm and AC = (2x – 5) cm. Angle BAC = 0°.
- (a)Show that area of triangle ABC / area of triangle ADE = (AB * AC) / (AD * AE).[2]
- (b)It is given that area of triangle ABC / area of triangle ADE = 1/3. Using the result from part (a), form an equation in x and show that it simplifies to 2x² – 19x + 6 = 0.[3]
- (c)(i)Solve the equation 2x² - 19x + 6 = 0, giving your answers correct to 2 decimal places.[3]
- (c)(ii)State, with a reason, which of these solutions does not apply to triangle ABC.[1]
- (d)Given that 0 = 25, calculate BC.[3]
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Mark scheme
Marking points from the official mark scheme (Oct/Nov 2015). The mark or mark type is shown in brackets.
- [2] M1 for the area sine formula
- Fully shown
- [3] B1 for both x + 12 and 4+2x-5 and M1 for x(2x-5) / ((x+12)(4+2x-5)) = 1/3
- 2x² – 19x + 6 (= 0) correctly obtained
- [3] B1 for √((-19)²-4×2×6) soi and B1 for (-(-19)±√their313) / (2×2) soi and M1 for both real values of (p ± √q) / r
- 9.17 0.33
- [1] 0.33 with reason
- [3ft] M2 for (BC²=) c(i)² + (2c(i)-5)²-2×c(i)×(2c(i)-5)×cos25 or M1 for correct formula with one error and A1 ft for correct evaluation from their M1 SC1 for x² + (2x-5)² – 2x(2x-5)cos25 oe
- 6.35
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