O-LevelMathematicsAlgebra and graphs4024/21 · Oct/Nov 2021 · Q612 Marks

6 (a) [Figure diagram of a right-angled triangle with sides 8, 2x, 3x+5] The diagram shows a right-angled triangle, with dimensions given in centimetres. (i) Show that 5x²+30x-39 = 0. (ii) Solve the equation 5x²+30x-39 = 0. Show your working and give your answers correct to 2 decimal places. x = or x = (iii) Calculate the area of the triangle. cm² (b) [Figure diagram of a different right-angled triangle with angle 35° and hypotenuse 12cm] The diagram shows a different right-angled triangle. The length of the hypotenuse is 12 cm. Calculate the shortest distance from A to BC. cm

(a)(i)Show that 5x²+30x-39 = 0.[3]
(a)(ii)Solve the equation 5x²+30x-39 = 0. Show your working and give your answers correct to 2 decimal places.[3]
(a)(iii)Calculate the area of the triangle.[2]
(b)The diagram shows a different right-angled triangle. The length of the hypotenuse is 12 cm. Calculate the shortest distance from A to BC.[4]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.

[3] (a)(i) 8²+(2x)² = (3x+5)²
[3] M1
[3] 9x² +15x+15x+25
[3] B1 Expansion of brackets
[3] 64+4x² = 9x² +15x+15x+25 leading to 5x² +30x-39=0
[3] A1
[3] (a)(ii) 1.10 and -7.10
[3] B2
[3] B1 for √30²-4×5×-39 oe
[3] B1 for (-30±√their 1680) / (2×5) oe
[3] B1 for (x+30/10)² - 39/5 oe
[2] (a)(iii) 8.78 to 8.8[0]
[2] M1 for 1/2 × 8 × 2 × their1.10
[4] (b) 5.63 to 5.64 nfww
[4] M1 for correct method to find AB or AC
[4] M1 for cos 35 = [AC]/12 oe or cos(90-35) = [AB]/12 oe
[4] M2 for [ ]= their AC×sin 35 oe or [ ]= their AB×sin(90-35) oe
[4] M2 for [ ]= 1/2 × 12 × their AC × sin 35 or [ ]= 1/2 × 12 × their AB × sin(90-35) oe
[4] M1 for sin 35 = [ ] / their AC oe or sin(90-35)= [ ] / their AB oe
[4] M1 for 1/2 × d × 12 = 1/2 × 12 × their AC × sin 35 oe or 1/2 × d × 12 = 1/2 × 12 × their AB × sin(90-35) oe

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