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A rectangle DEBC has triangle ABC.AB and AC intersect side DE at points F and G respectively. FG = 4, The perimeter of triangle ABC is double of the perimeter of Triangle AFG. The area of Triangle ABC = 16 sq units. What is the area of DEBC?

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    $\begingroup$ Is A in the interior or exterior of the rectangle. What does a rectangle "having" a triangle mean? $\endgroup$ – fleablood Jan 28 at 19:34
  • $\begingroup$ If you can prove the triangles are symmetric then twice the perimeter means four times the area. $\endgroup$ – fleablood Jan 28 at 19:42
  • $\begingroup$ Sir/Madam, fleablood, The A is actually gets outside of the rectangle if you draw AB and AC lines from points B and C which has to intersects side DE at point F and G. $\endgroup$ – Ghost Jan 28 at 19:45
  • $\begingroup$ The two triangles has three similar angles which makes it similar triangle $\endgroup$ – Ghost Jan 28 at 19:50
  • $\begingroup$ "The A is actually gets outside of the rectangle if you draw AB and AC lines from points B and C which has to intersects side DE at point F and G" Not if you extend $AB$ past $A$ and $AC$ past $A$ and they intersect $DE$ on the other side. $\endgroup$ – fleablood Jan 28 at 19:54
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Since $\Delta ABC\sim\Delta AFG$, we obtain: $$\frac{S_{\Delta AFG}}{S_{\Delta ABC}}=\left(\frac{1}{2}\right)^2,$$ which gives $$S_{\Delta AFG}=4$$ and since $$\frac{S_{\Delta AFC}}{S_{\Delta AFG}}=\frac{AC}{AG}=2,$$ we obtain $$S_{\Delta AFC}=8.$$

Thus, $$4+8=S_{\Delta CFG}=\frac{4\cdot DC}{2},$$ which gives $$DC=6$$ and $$S_{DEBC}=6\cdot8=48.$$

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  • $\begingroup$ I am sorry Sir but I can't understand what exactly did you mean by triangle "AFC" and Triangle "CFG" also DC. It isn't actually matching with my depiction. Also can you please describe?The first line with clarification of above problems? $\endgroup$ – Ghost Jan 28 at 20:28
  • $\begingroup$ Sir An image of your depiction will be very helpful to understand. $\endgroup$ – Ghost Jan 28 at 20:32
  • $\begingroup$ @Ghost $CD$ is an altitude to side $GF$ from $C$ of the $\Delta CFG$. I don't know to draw in the net. I drew it by the given. $\endgroup$ – Michael Rozenberg Jan 28 at 20:42
  • $\begingroup$ $S_{\Delta CFG}=S_{\Delta AFG}+S_{\Delta AFC}.$ $\endgroup$ – Michael Rozenberg Jan 28 at 20:48
  • $\begingroup$ Sir please can you send me a photo of the papers that you've used to draw.It will really helpful. $\endgroup$ – Ghost Jan 29 at 6:23

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