2
$\begingroup$

Can anyone help me with this problem? I assume that similar triangles are used to solve the problem, but I can't find the solution.

Problem: ABC is an equilateral triangle with sides equal to 2 cm. BC is extended its own length to D, and E is the midpoint of AB. ED meets AC at F. Find the area of the quadrilateral BEFC in square centimeters in simplest radical form.

TriangleareaTrifurcation

$\endgroup$
  • 1
    $\begingroup$ Your title says you want the triangle's area, but the question says you want the quadrilateral's. Please clarify. $\endgroup$ – jpmc26 Apr 15 '16 at 3:50
  • $\begingroup$ @jpmc26, the title apparently refers to that part of the area of EBD that's inside ABC, ie, the intersection's area $\endgroup$ – James Waldby - jwpat7 Apr 15 '16 at 4:20
  • 1
    $\begingroup$ @jwpat7 Probably, but the OP should still clarify. No sense leaving around potentially confusing things. $\endgroup$ – jpmc26 Apr 15 '16 at 4:32
  • $\begingroup$ @teammember introduced area labeling , hope it is OK. $\endgroup$ – Narasimham Apr 16 '16 at 3:31
2
$\begingroup$

Notice that $\triangle ABC$ and $\triangle BDE$ are each half of $\triangle ABD$. Therefore $\triangle AEF$ and $\triangle CDF$ have the same area.

Now $\triangle BCF$ has the same area as $\triangle CDF$ because $BC=CD$, and $\triangle BEF$ has the same area as $\triangle AEF$ because $BE=AE$.

Therefore $\triangle AEF$, $\triangle BEF$, and $\triangle BCF$ all have the same area, and $BCFE$ is two-thirds of $\triangle ABC$.

$\endgroup$
1
$\begingroup$

Construction of side $ FB $ is crucial to calculate the three areas involved.

Areas

$$ p=q ;\, r= s $$

due to same base and altitude.

Apply area formula $ A= \frac12\, a\,b\, \sin C $ to find

$$ ( p+q+r) $$

and

$$(q+r +s )$$

combination triangles. In the triangles included angle is same. First case has (2,2) as sides and other has (1,4). The common part is $(q+r) $ so the uncommon parts $p,s$ areas are of equal area. In fact you can take cardboard to cut flip it, to see that

$$ p=s $$

So

$$ p =q;\, r=s;\, p=s;\, \rightarrow \, p= q = r =s $$

In other words, $ EF,FB $ * trifurcate * the area of the given equilateral triangle $ABC$ so also $ FB,FC$ * trifurcate * $ FBD$

So the required area is $\frac23$ of the given equilateral triangle area

$$ \frac23 \frac{\sqrt3}{4} 2^2 = \frac23 \sqrt3. $$

$\endgroup$
  • $\begingroup$ Thank you for help, but I can't understand what you said. I am slow. $\endgroup$ – user321645 Apr 15 '16 at 2:30
  • $\begingroup$ I shall make an explanatory sketch later. $\endgroup$ – Narasimham Apr 15 '16 at 4:30
1
$\begingroup$

Note that $\angle A = \pi/3$ because $\triangle ABC$ is equilateral. Hence we can compute:

$$ S_{\triangle ABC} = \frac{2\times 2 \sin(\pi/3)}{2} = \sqrt{3} $$

and

$$ S_{\triangle AEF} = \frac{1\times \overline{AF}\sin(\pi/3)}{2}. $$

So we have to calculate $\overline{AF}$, but it is a simple application of Menelao's Theorem:

$$1=\frac{\overline{BE}\cdot \overline{AF}\cdot \overline{CD}}{\overline{AE}\cdot\overline{FC}\cdot\overline{BD}} = \frac{1\cdot x\cdot 2}{1\cdot (2-x)\cdot 4}, $$

where $x=\overline{AF}$. Solvin this equation for $x$ gives $x=4/3$

Finally, the area $S$ we are looking for is the diference $S_{\triangle ABC}-S_{\triangle AEF}$. Then

$$ S = \sqrt{3}-\frac{4\sqrt{3}}{4\cdot 3} = \frac{2\sqrt{3}}{3}. $$

$\endgroup$
0
$\begingroup$

$\triangle EBD = \triangle ABC -- \triangle EBD$ has 1/2 the height and 2x the base.

$\triangle AEF = \triangle CDF$

The distance from AE to F is 1/2 the distance from AC to F.

$\triangle ABF = 2\times\triangle BCF\\ \triangle BFE = \triangle AFE$

$BCFE$ has 2/3 the area as $\triangle ABC.$

$\dfrac{2 \sqrt3}{3}$

$\endgroup$
  • $\begingroup$ Thank you for helping. I don't get this part: "△AEF=△CDF The distance from AE to F is 1/2 the distance from AC to F?" $\endgroup$ – user321645 Apr 14 '16 at 23:34
  • $\begingroup$ The area of triangle AEF = area of triangle CDF. This is true because Area of ABC = area of BED and both triangles share the same BCFE quadrangle. So the areas not covered by that quadrangle are equal. Next if AEF has the same area as CDF and AE is 1/2 the length of CD, Then the height of AEF must be twice the height of CDF. $\endgroup$ – Doug M Apr 14 '16 at 23:57
  • $\begingroup$ Thank you very much for the detailed explanation! $\endgroup$ – user321645 Apr 15 '16 at 2:44
  • $\begingroup$ I still have hard time to understand part of your answer. Why △ABF=2×△BCF△BFE=△AFE? $\endgroup$ – user321645 Apr 15 '16 at 3:10
  • $\begingroup$ ABF = 2 x BCF because both triangles have bases of equal length, And ACF has 2x the height as BCF. E is on the midpoint of AB, AEF and BEF have bases of equal length and have the same height. $\endgroup$ – Doug M Apr 15 '16 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy