In the number below, AB=BC=CD. If the area the triangle CDE is 42, what is the area the triangle ADG?

I think triangles space similar. Are there any type of properties of comparable triangles concerning their area. Assist me gain the answer v explanation please

ans :378


As $CE||BF||AG,$

So, $fracDCDE=fracDBDF=fracDADG$

But $DA=DC+CB+BA=3DCimplies DG=3DE$ and $AG=3CE$

Now $ riangle CDE=frac12cdot EC cdot DE=42$(given),

$ riangle ADG =frac12cdot AGcdot DG=frac12cdot3CEcdot 3DE=9cdot frac12cdot CE cdot DE=9cdot 42=378$


The currently $CE$, $BF$, $AG$ space parallel since they have actually a usual perpendicular line.By the intercept theorem, the triangles are indead comparable and the scaling element is 3, hence both height and also base heat of the bigger triangle room 3 time the corresponding length of the smaller, thus lastly the area is 9 ties bigger.

You are watching: In the figure below ab=bc=cd


Yes, the triangles room similar. How have the right to you prove this? (one means would it is in to discover 2 sets of congruent angles)

Now the we recognize the triangles space similar, let"s shot and discover the proportion of similarity. Us could shot and to compare the bases of the triangles, the altitudes (the "right" sides) or the hypotenuses (the "left" sides). Which perform we understand the many about? (Hint: if abdominal muscle = BC = CD, what does the say about BD and also AD?)

Finally, yes, over there is a property regarding comparable triangles and also their areas. Check out if you can deduce what the is - in this instance you understand how huge the base and height of each triangle are in relationship to every other, for this reason it should be straightforward to figure this out.


Because of similar triangles we have that$$ fracDEDG=fracCEAG=frac13longrightarrow fracDEDGcdotfracCEAG=frac19longrightarrowfracdisplaystylefrac DE cdot CE2displaystylefracDGcdot AG2=frac42 riangle ADG=frac19longrightarrow riangle ADG=378.$$



The ideal triangles CDE and also ADG are similar since there is the usual angle in ~ D, the right angle and 3rd angle (determined because triangles have actually 180 degrees). So all sides are in the very same proportion. Due to the fact that $AD=3CD$, then the various other sides, basic $AG=3CE$, and height $GD=3ED$. Utilizing the formula for the area together $frac12$base$cdot$ height, the area that the larger triangle is 9 times the smaller area or $9cdot 42=378$.

If 2 triangles space similar, climate the area ration would the square that the next ratio. Here, side proportion is 3. So, the area ratio would it is in 3^2=9.Therefore, the area that the bigger triangle is 9*42=378

Here is the easiest Way :

since theorem says- In two similar triangles, the proportion of their areas is the square that the ratio of their sides.

area the CDE/area that ADG=(CD/DA)^2

Here in number suppose CD=x,then AD=3x(Already mentioned in inquiry AB=BC=CD)

so area that CDE/area of ADG=(x/3x)^242/area of ADG=1/9area of ADG=378 ANSWER:378

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