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27/12/2017 at 17:34

What fraction of the area of rectangle ABCD is the area of inscribed triangle DEF?
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Alone 27/12/2017 at 17:58

S_ABCD=\(\left(\dfrac{3}{2}.w+\dfrac{w}{2}\right).w=\dfrac{4}{2}.w^2=2w^2\)

Draw FK perpendicular with CD.We have:AD=FK

So S_DEF=\(\dfrac{w.\dfrac{3}{2}w}{2}=\dfrac{3}{4}.w^2\)

So the answer is:\(\dfrac{S_{ABCD}}{S_{DEF}}=\dfrac{2.w^2}{\dfrac{3}{4}.w^2}=\dfrac{2}{\dfrac{3}{4}}=\dfrac{8}{3}\)
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FA Liên Quân Garena 01/01/2018 at 10:27

S_ABCD= $(\frac{3}{2} . w + \frac{w}{2}) . w = \frac{4}{2} . w^{2} = 2 w^{2}$

Draw FK perpendicular with CD.We have:AD=FK

So S_DEF= $\frac{w . \frac{3}{2} w}{2} = \frac{3}{4} . w^{2}$

So the answer is: $\frac{S_{A B C D}}{S_{D E F}} = \frac{2. w^{2}}{\frac{3}{4} . w^{2}} = \frac{2}{\frac{3}{4}} = \frac{8}{3}$

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