The figure below is developed by joining six equilateral triangles ABC, ACD, CDE, DEF, EFG, and also FGH, all of whose edges space 1 unit long. That is offered that HIJKLMB is straight.

You are watching: There are six equilateral triangles in a regular

(a) over there are 5 triangles in the figure that are comparable to CMB. Perform them, making sure that you match corresponding vertices.

(b) uncover the lengths of CM and also EK.

(c) perform the five triangles that are similar to AMB.

(d) find the lengths of CL, HI, IJ, and also JK. I believe that we have to use a concept based on similar triangles, but I am unsure regarding what I should be doing. Any help will be significantly appreciated.

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edited Jun 12 "20 at 10:38 CommunityBot
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request Oct 1 "17 at 14:58 geo_freakgeo_freak
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Since $\Delta CMB\sim\Delta AMH$, us obtain:$$\fracCMAM=\fracBCAH$$ or

$$\fracCMAC-CM=\fracBCAH,$$which claims $$CM=\frac14.$$

$EK=\frac12$ by symmetry.

Now, $\Delta BLC\sim\Delta HLD.$

Thus, $$\fracCLLD=\fracBCHD$$ or$$\fracCL1-CL=\frac12,$$which offers $$CL=\frac13.$$Now, due to the fact that $FI=CM=\frac14$, by the law of cosines we obtain:$$HI=\sqrt1^2+\left(\frac14\right)^2-2\cdot1\cdot\frac14\cdot\frac12=\frac\sqrt134.$$Also, due to the fact that $FI=\frac12$ and also $AJ=\frac13$, us obtain:$$IJ=\sqrt\left(\frac14\right)^2+\left(\frac13\right)^2-2\cdot\frac14\cdot\frac13\cdot\frac12=\frac\sqrt1312$$ and since $EJ=1-\frac13=\frac23,$ us obtain:$$JK=\sqrt\left(\frac23\right)^2+\left(\frac12\right)^2-2\cdot\frac23\cdot\frac12\cdot\frac12=\frac\sqrt136.$$ Done!

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edited Oct 1 "17 in ~ 16:00
answer Oct 1 "17 in ~ 15:05 Michael RozenbergMichael Rozenberg
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Here"s the interesting component (to me): $$|\overlineAP|:|\overlinePQ|:|\overlineQR|:|\overlineAB| = 3:1:2:12$$

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reply Oct 2 "17 at 1:48 BlueBlue
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As $BMC\sim BKE\sim BIG$ we have actually $$CM:KE:IJ=1:2:3.$$ yet $KE=KD=1/2$ (by symmetry), then:$$CM=IF=1\over4,\quad IG=AM=3\over4.$$In addition, $LMC\sim BMA$ and $CL:AB=CM:AM=1:3$, whence: $CL=1/3$.

Finally: $GHI\sim FJI\sim EJK$ and also we know that $FJ=CL=1/3$ and $JE=2/3$, so we have:$$HI:IJ:JK=3:1:2,\quad\hboxwhich entails\quadIJ:HK=1:6\quad\hboxand\quadIJ:HB=1:12.$$so that:$$IJ=HB\over12,\quadJK=HB\over6,\quadIJ=HB\over4,$$and we deserve to compute $HB=\sqrt13$ through Pythagoras" theorem to gain the required lengths.

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answered Oct 1 "17 at 21:24 Intelligenti paucaIntelligenti pauca
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