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09 Oct 2019, 02:24
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In the figure, CD = 2AB and AB is parallel to CD. If the area of ABE is 3, what is the area of CED?
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09 Oct 2019, 03:14
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As AB||CD, the triangles ABE and CDE are similar and their areas will be in a ratio that will be square of the ratio of sides..
Now, CD is similar to AB, so ratio of sides = \(AB:CD = AB:2AB=1:2\)..
SO the area of ABE and CDE will b ein the ratio \(1^2:2^2=1:4\)
If the area of ABE is 3, the area of CED will be 4*3=12 square units
A
11 Oct 2019, 11:55
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\(\angle E = \angle E\) common in both the \(\Delta's\)
\(\angle A = \angle D\) since AB // DC and \(\angle A\) and \(\angle D\) are corresponding \(\angle's\)
\(\angle B = \angle C\)since AB // DC and \(\angle B\) and \(\angle C\) are corresponding \(\angle's\)
Now in two similar \(\Delta's\):
Ratio of square of the sides = ratio of area of the triangles
Since we know \(\frac{CD}{AB} =\frac{2}{1}\) we can write \(\frac{CD^2}{AB^2}\) =\(\frac{4}{1}\)=\(\frac{area \hspace{1mm} of \hspace{1mm}\Delta CED }{area \hspace{1mm} of \hspace{1mm} \Delta ABE}\) ,given area of \(\Delta\) ABE =3
\(\frac{4}{1}\) = \(\frac{area \hspace{1mm} of \hspace{1mm}\Delta CED }{3}\)
area of \(\Delta\)CED =12
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