6743
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2011-02-01T00:00:01
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<p>The diagram has rotational symmetry of order four about $D$.<br></br>
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<mdo:image alt="" height="361" src="square.jpg" width="358"></mdo:image><br></br>
If angle $ABC$ is $15^{\circ}$ and the area of $ABEF$ is $24$cm$^2$, what, in cm, is the length of $CD$?<br></br>
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If you liked this problem, <a href="http://nrich.maths.org/public/viewer.php?obj_id=542&part=">here is an NRICH task</a> which challenges you to use similar mathematical ideas.</p>
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<mdoxml version="1.0"><p>$ABEF$ is a square, and its area is $4\times$ area$(BDA) = 4\times\frac{1}{2}(BD\times DA) = 2BD^2 = 24$cm$^2$, so $BD=\sqrt{12}=2\sqrt{3}$cm.<br></br>
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The angle $ABD$ is $45^{\circ}$ so the angle $CBD$ is $45^{\circ}-15^{\circ}=30^{\circ}$.<br></br>
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Therefore $\tan{30^{\circ}}=\frac{1}{\sqrt{3}}=\frac{CD}{BD}=\frac{CD}{2\sqrt{3}}$ so $CD=2$cm.<br></br>
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Weekly Problem 43 - 2009
Measures and Mensuration
Area
2D Geometry, Shape and Space
Angles