Geometric area leads to quadratic

A question is this type if and only if a geometric context (triangle, trapezium, rectangle, L-shape, playground) is used to set up a quadratic equation or expression for area, which is then solved or optimised.

3 questions · Moderate -0.4

1.02f Solve quadratic equations: including in a function of unknown
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OCR MEI C1 2007 June Q10
5 marks Moderate -0.8
10 The triangle shown in Fig. 10 has height \(( x + 1 ) \mathrm { cm }\) and base \(( 2 x - 3 ) \mathrm { cm }\). Its area is \(9 \mathrm {~cm} ^ { 2 }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{4d8caf0f-7594-42cb-bd40-e6c11e2b6832-3_444_1088_351_715} \captionsetup{labelformat=empty} \caption{Fig. 10}
\end{figure}
  1. Show that \(2 x ^ { 2 } - x - 21 = 0\).
  2. By factorising, solve the equation \(2 x ^ { 2 } - x - 21 = 0\). Hence find the height and base of the triangle. Section B (36 marks)
OCR MEI C1 Q5
5 marks Standard +0.3
5 Fig. 9 shows a trapezium ABCD , with the lengths in centimetres of three of its sides. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{d22f53f5-ba80-4065-a94b-2a9c92c20dfb-1_462_877_1796_684} \captionsetup{labelformat=empty} \caption{Fig. 9}
\end{figure} This trapezium has area \(140 \mathrm {~cm} ^ { 2 }\).
  1. Show that \(x ^ { 2 } + 2 x - 35 = 0\).
  2. Hence find the length of side AB of the trapezium.
OCR MEI C1 2011 January Q9
5 marks Moderate -0.8
9 Fig. 9 shows a trapezium ABCD , with the lengths in centimetres of three of its sides. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{754d34e4-2f47-48b7-9fbb-6caa7ac21eb7-3_464_878_347_632} \captionsetup{labelformat=empty} \caption{Fig. 9}
\end{figure} This trapezium has area \(140 \mathrm {~cm} ^ { 2 }\).
  1. Show that \(x ^ { 2 } + 2 x - 35 = 0\).
  2. Hence find the length of side AB of the trapezium.