The diagram shows a figure $ABCD$, in which $BC=13$ m, $CD=6$ m, $\angle BAD=90{}^\circ$ and

Question

The diagram shows a figure $ABCD$, in which $BC=13$ m, $CD=6$ m, $\angle BAD=90{}^\circ$ and $\angle ADC=\angle ABC=\theta$. Show that the perimeter, $L$ metres, of $ABCD$ is given by[2] $$L = 19 + 19\cos\theta + 19\sin\theta$$ By expressing $L$ in the form of $19 + R\cos(\theta - \alpha)$, where $R > 0$ and $\alpha$ is acute, find the maximum possible perimeter of $ABCD$.[4] Find the values of $\theta$ for which $L = 45$ m.[3]

Accepted Answer

(a) $L=19+19\cos\theta+19\sin\theta\text{ m}$ (b) $R=19\sqrt2,\ \alpha=45^\circ,\ L_{\max}=45.9\text{ m}$ (c) $\theta=30.4^\circ\text{ or }59.6^\circ$

2025 SCGS PRELIMS P1 Q11

2025 SCGS PRELIMS P1 Q11

Solution:

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