$PQRS$ is a trapezium in which $\angle PQR=\angle QRS=90{}^\circ $. $PQ=\left( x+3 \right)$ cm

Question

$PQRS$ is a trapezium in which $\angle PQR=\angle QRS=90{}^\circ $. $PQ=\left( x+3 \right)$ cm, $QR=\left( 3x-1 \right)$ cm and $RS=\left( 7x-9 \right)$ cm. Given that the area of the trapezium is $72$ cm$^{2}$, form an equation in $x$ and show that it reduces to $12{{x}^{2}}-13x-69=0$. Solve the equation $12{{x}^{2}}-13x-69=0$. Given that the length of $PS$ is $10$ cm. Write down, in terms of $x$, the length of the perimeter of the trapezium. Hence, find the perimeter.

Accepted Answer

(a) $12x^2-13x-69=0$ (b) $x=3$ or $x=-\frac{23}{12}$ (c) $11x+3\text{ cm}$, hence $36\text{ cm}$

Geometry Application: Solving a Quadratic in a Trapezium

Geometry Application: Solving a Quadratic in a Trapezium

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