Forming Equations: WJEC GCSE Higher November 2019 Unit 2 Q11

Higher November 2019 Unit 2 Q11

WJECHigherCurrent spec9 marksForming Equations Similar Shapes Solving Quadratics

11.

(a) The area of the trapezium \(ABCD\) is 25 cm².

Trapezium ABCD with AB = (x + 1) cm, DC = (x + 2) cm and perpendicular height x cm

Diagram not drawn to scale

(i) Show that \(2x^2 + 3x - 50 = 0\). [3]

(ii) Solve the equation \(2x^2 + 3x - 50 = 0\) to calculate the lengths \(AB\) and \(DC\).
Give your answers correct to 1 decimal place. [4]

Length of \(AB\) = …………………… cm Length of \(DC\) = …………………… cm

(b) A rhombus has an area of 36·8 cm².
The rhombus is enlarged by a scale factor of 7.
Calculate the area of the enlarged rhombus. [2]

From the formula list: Area of trapezium \(= \dfrac{1}{2}(a+b)h\); The solutions of \(ax^2 + bx + c = 0\) where \(a \neq 0\) are given by \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Mark scheme (a)

Scheme	Marks	Comments
(i) \(\dfrac{x+1+x+2}{2} \times x\) \((= 25)\)	M1	Missing brackets in the expression \(\dfrac{x(x+1+x+2)}{2}\) may be implied later from correct working.
\(x^2 + x + x^2 + 2x = 50\) OR \(x(2x + 3) = 50\) OR \(\dfrac{2x^2+3x}{2} = 25\) OR \(x^2 + 1.5x = 25\)	m1
\(2x^2 + 3x - 50 = 0\)	A1	Must be convincing. If m1 awarded for \(\dfrac{2x^2+3x}{2} = 25\), a further rearrangement, e.g. 2x² + 3x = 50, must be seen before A1 is awarded.
(ii) \(x = \dfrac{-(3) \pm \sqrt{(3)^2 - 4 \times 2 \times (-50)}}{2 \times 2}\)	M1	May be seen in a(i). Allow one slip in substitution for M1 only, but must be correct formula.
\(= \dfrac{-3 \pm \sqrt{409}}{4}\)	A1
\(x = 4.3(059\ldots)\), \((x = -5.8(059\ldots))\)	A1	CAO.
(AB=) 5.3(cm) AND (DC=) 6.3(cm)	B1	Answers must be to 1 d.p. FT ‘their positive \(x\)’ provided M1 awarded.

Corrected by Westie's Workshop

The board's mark scheme printed “Maybe seen in a(i).”. Changed to “May be seen in a(i).” because it means “may be seen”, written as three words.

Original: see the board's mark scheme PDF (link above).

Mark scheme (b)

Scheme	Marks	Comments
7² × 36·8 OR (7 × √36.8)²	M1
= 1803·2 (cm²)	A1	Allow 1803 (cm²)