(a) Show that \(\dfrac{\sqrt{63}}{\sqrt7} + \sqrt{147} + \sqrt{48}\) can be written in the form \(a + b\sqrt3\), where \(a\) and \(b\) are integers. [3]
(b)
Diagram not drawn to scale
The area of this trapezium is \((6\sqrt2 - 1)\) cm2.
Find the height of the trapezium. Give your answer in its simplest form. [5]
Mark scheme (a)
Scheme
Marks
Comments
\(3 + 11\sqrt3\)
B3
B2 for \(k + 11\sqrt3\) or for any two correct terms in \(3 + 7\sqrt3 + 4\sqrt3\) or B1 for \(\dfrac{\sqrt7\sqrt9}{\sqrt7}\) oe si or \(7\sqrt3\) or \(4\sqrt3\) seen
Mark scheme (b)
Scheme
Marks
Comments
\(\dfrac12\times(7 - 2\sqrt2 + 5 + \sqrt2)\times h = 6\sqrt2 - 1\) or \(\dfrac12\times h\times(3\sqrt2 - 2) + (7 - 2\sqrt2)\times h = 6\sqrt2 - 1\)
FT \(\text{'their}\left(\dfrac{12 - \sqrt2}{2}\right)\text{'}\,h = 6\sqrt2 - 1\); \(\text{'their}\left(\dfrac{12 - \sqrt2}{2}\right)\text{'}\) must be of the form \(a + b\sqrt2\) where \(a\) and \(b\) are non-zero;
FT \(\dfrac{2(6\sqrt2 - 1)}{\text{'their}(12 - \sqrt2)\text{'}}\times\dfrac{\text{'their}(12 + \sqrt2)\text{'}}{\text{'their}(12 + \sqrt2)\text{'}}\) multiplies out their expression for \(h\) or ½\(h\)
\((h =)\ \sqrt2\)
A1
CAO
Alternative method
Scheme
Marks
Comments
\(\dfrac12\times(7 - 2\sqrt2 + 5 + \sqrt2)\times h = 6\sqrt2 - 1\) or \(\dfrac12\times h\times(3\sqrt2 - 2) + (7 - 2\sqrt2)\times h = 6\sqrt2 - 1\)
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