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# ​ Math Question

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Thanks in advance for the help!

$$The\;area\;of\;\bigtriangleup BDE\;is\;48\;cm^2.\;\\The\;point\;D\;divides\;\overset-{AB}\;so\;that\;\frac{\overset-{AD}}{\overset-{DB}}\;=\;\frac15,\;\\and\;point\;E\;divides\;\overset-{BC}\;so\;that\;\frac{\overset-{BE}}{\overset-{EC}}=\frac23.\;\\~\\The\;area\;of\;\bigtriangleup ABC,\;in\;cm^2,\;is$$

I have edited this to improve the presentation.   Melody

Jan 6, 2019
edited by Melody  Jan 12, 2019

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since ad/db is 1/5, then the ratio of the heights of the two triangles if you turned it so that line BEC is horizontal would be 5/(1+5) or 5/6, and the ratio of the bases is 2/(2+3) or 2/5. then, the ratio of the areas DEB/ABC = 5/6*2/5 or 1/3, so the area of triangle ABC is just 48*3 or 144.

Jan 12, 2019
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Thanks!

Guest Jan 12, 2019