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In the figure below, AB=BC=CD. If the area of triangle CDE is 42, what is the area of triangle ADG? enter image description here

I think triangles are similar. Are there any properties of similar triangles regarding their area. Help me get the answer with explanation please

ans :378

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  • $\begingroup$ Have you learnt similar triangles? $\endgroup$ – JSCB Sep 7 '12 at 12:47
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As $CE||BF||AG,$

So, $\frac{DC}{DE}=\frac{DB}{DF}=\frac{DA}{DG}$

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

Now $\triangle CDE=\frac{1}{2}\cdot EC \cdot DE=42$(given),

$\triangle ADG =\frac{1}{2}\cdot AG\cdot DG=\frac{1}{2}\cdot3CE\cdot 3DE=9\cdot \frac{1}{2}\cdot CE \cdot DE=9\cdot 42=378$

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  • $\begingroup$ You say: 9 x (1/2) x CE x DE = 9 x 42 =378, but what happened to the (1/2)? EDIT: Nevermind, I figured it out. $\endgroup$ – user3025403 Jun 25 '15 at 23:26
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The lines $CE$, $BF$, $AG$ are parallel because they have a common perpendicular line. By the intercept theorem, the triangles are indead similar and the scaling factor is 3, hence both height and base line of the bigger triangle are 3 times the corresponding length of the smaller, thus finally the area is 9 ties bigger.

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Yes, the triangles are similar. How can you prove this? (one way would be to find 2 sets of congruent angles)

Now that we know the triangles are similar, let's try and find the ratio of similarity. We could try and compare the bases of the triangles, the altitudes (the "right" sides) or the hypotenuses (the "left" sides). Which do we know the most about? (Hint: if AB = BC = CD, what does that say about BD and AD?)

Finally, yes, there is a property regarding similar triangles and their areas. See if you can deduce what it is - in this case you know how big the base and height of each triangle are in proportion to each other, so it should be easy to figure this out.

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Because of similar triangles we have that $$ \frac{DE}{DG}=\frac{CE}{AG}=\frac{1}{3}\longrightarrow \frac{DE}{DG}\cdot\frac{CE}{AG}=\frac{1}{9}\longrightarrow\frac{\displaystyle\frac{ DE \cdot CE}{2}}{\displaystyle\frac{DG\cdot AG}{2}}=\frac{42}{\triangle ADG}=\frac{1}9\longrightarrow \triangle ADG=378.$$

Q.E.D.

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The right triangles CDE and ADG are similar since there is the common angle at D, the right angle and third angle (determined since triangles have 180 degrees). So all sides are in the same proportion. Since $AD=3CD$, then the other sides, base $AG=3CE$, and height $GD=3ED$. Using the formula for the area as $\frac{1}{2}$base$\cdot$ height, the area of the larger triangle is 9 times the smaller area or $9\cdot 42=378$.

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If two triangles are similar, then the area ration would the square of the side ratio. Here, side ratio is 3. So, the area ratio would be 3^2=9. Therefore, the area of the larger triangle is 9*42=378

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Here is the Easiest Way :

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

area of CDE/area of ADG=(CD/DA)^2

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

so area of CDE/area of ADG=(x/3x)^2 42/area of ADG=1/9 area of ADG=378 ANSWER:378

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  • $\begingroup$ Its really the easiest way to solve such type of questions $\endgroup$ – Pallavi Oct 29 '13 at 23:53

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