| Date | May 2011 | Marks available | 2 | Reference code | 11M.1.sl.TZ2.15 |
| Level | SL only | Paper | 1 | Time zone | TZ2 |
| Command term | Solve | Question number | 15 | Adapted from | N/A |
Question
In the diagram, \({\text{B}}\hat {\text{A}}{\text{C}} = {90^ \circ }\) . The length of the three sides are \(x{\text{ cm}}\), \((x + 7){\text{ cm}}\) and \((x + 8){\text{ cm}}\).
Write down and simplify a quadratic equation in \(x\) which links the three sides of the triangle.
Solve the quadratic equation found in part (a).
Write down the value of the perimeter of the triangle.
Markscheme
\({(x + 8)^2} = {(x + 7)^2} + {x^2}\) (A1)
Note: Award (A1) for a correct equation.
\({x^2} + 16x + 64 = {x^2} + 14x + 49 + {x^2}\) (A1)
Note: Award (A1) for correctly removed parentheses.
\({x^2} - 2x - 15 = 0\) (A1) (C3)
Note: Accept any equivalent form.
[3 marks]
\(x = 5\), \(x = - 3\) (A1)(ft)(A1)(ft) (C2)
Notes: Accept (A1)(ft) only from the candidate’s quadratic equation.
[2 marks]
\(30{\text{ cm}}\) (A1)(ft) (C1)
Note: Follow through from a positive answer found in part (b).
[1 mark]
Examiners report
This question proved to be difficult for the majority of candidates. Many simply were unable to see that, to relate the three given lengths, a Pythagorean equation needed to be produced. Indeed, many did not seem to appreciate the concept of a quadratic equation and, as a consequence, either wrote down a linear equation linking one length to the sum of the other two lengths or multiplied all three lengths together. For the minority who stated a correct Pythagorean equation, many could not remove brackets successfully and arrived at \({x^2} = 15\) . Consequently, very few candidates earned more than one mark for part (a). Where the correct quadratic equation was seen in part (a), many were able to solve this quadratic correctly in part (b) and arrive at the required value of \(x = 5\) for the answer for part (c).
This question proved to be difficult for the majority of candidates. Many simply were unable to see that, to relate the three given lengths, a Pythagorean equation needed to be produced. Indeed, many did not seem to appreciate the concept of a quadratic equation and, as a consequence, either wrote down a linear equation linking one length to the sum of the other two lengths or multiplied all three lengths together. For the minority who stated a correct Pythagorean equation, many could not remove brackets successfully and arrived at \({x^2} = 15\) . Consequently, very few candidates earned more than one mark for part (a). Where the correct quadratic equation was seen in part (a), many were able to solve this quadratic correctly in part (b) and arrive at the required value of \(x = 5\) for the answer for part (c).
This question proved to be difficult for the majority of candidates. Many simply were unable to see that, to relate the three given lengths, a Pythagorean equation needed to be produced. Indeed, many did not seem to appreciate the concept of a quadratic equation and, as a consequence, either wrote down a linear equation linking one length to the sum of the other two lengths or multiplied all three lengths together. For the minority who stated a correct Pythagorean equation, many could not remove brackets successfully and arrived at \({x^2} = 15\) . Consequently, very few candidates earned more than one mark for part (a). Where the correct quadratic equation was seen in part (a), many were able to solve this quadratic correctly in part (b) and arrive at the required value of \(x = 5\) for the answer for part (c).
