(b) Show that the triangle below is not isosceles. You must use an algebraic method and not trial and improvement. [5]
Diagram not drawn to scale
Mark scheme (a)
Scheme
Marks
Comments
\(8a + 22\) OR \(2(4a + 11)\)
B2
B1 for one of the following: \(8a + k\) where \(k \ne 0\) \(ka + 22\) where \(k \ne 0\) \(15a - 6 - 7a + 28\)
Mark final answer for B2 only -1 for incorrect further work
Mark scheme (b)
Scheme
Marks
Comments
\(5x + 7 + 6x - 9 + 8x - 27 = 180\)
M1
May be seen in stages Allow M1 for (180-7+9+27) ÷ (5+6+8)
\(19x = 209\)
A1
Implies M1 Allow 209 ÷ 19
\(x = 11\)
A1
FT from \(ax = b\) OR ‘their 209’ ÷ ‘their 19’ provided M1 awarded For FT, accept answer written as a correctly simplified, improper fraction (unless it gives a whole number), mixed number or decimal. If answer is only given as a decimal, it must be correct to 1 decimal place – rounded or truncated.
Method to calculate all three angles e.g. \(5\times11 + 7\) (= 62) \(6\times11 - 9\) (= 57) \(8\times11 - 27\) (= 61) OR calculates two angles by substitution and the third by subtracting sum from 180
M1
If \(x \ne 11\) FT ‘their \(x\)’ provided \(x \geqslant 4\) and M1 previously awarded.
M1 implied if three angles correctly found (without working shown) from FT of ‘their \(x\)’
All three angles correctly calculated: 62(°), 57(°), 61(°) (so not isosceles)
A1
CAO
If no marks, award: SC2 for \(x = 11\), obtained with no working shown or trials used AND angles of 62(°), 57(°), 61(°) (so not isosceles) OR SC1 for \(x = 11\) obtained with no working shown or trials used.
Alternative method – showing that if two angles are equal, angle sum is not 180(°)
Scheme
Marks
Comments
One pair of angles equated e.g. \(6x - 9 = 5x + 7\)
M1
Correct value of \(x\) obtained e.g. \(x = 16\) AND correct angle sum calculated e.g. 275(°)
A1
Both other pairs of angles equated e.g. \(8x - 27 = 5x + 7\) AND \(8x - 27 = 6x - 9\)
M1
Both equations solved correctly \(x = 11\frac13\) AND \(x = 9\) AND correct angle sums calculated e.g. \(186\frac13\)(°) and 142(°)
A2
A1 for one correct value of \(x\) and the correct angle sum provided M1 awarded.
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