How do we solve this triangle using law of sines?












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enter image description here



ABC is a right triangle, $AB perp AC, DE perp BC, |AD| = |BD| = 3$. The area of ABC is $6$ times of the area of BDE. Evaluate $|AC|$.




How do we solve this triangle using law of sines?






geometry trigonometry







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  • $begingroup$
    There's easier ways than the sine rule. is that required?
    $endgroup$
    – TurlocTheRed
    Dec 9 '18 at 4:05
















0












$begingroup$



enter image description here



ABC is a right triangle, $AB perp AC, DE perp BC, |AD| = |BD| = 3$. The area of ABC is $6$ times of the area of BDE. Evaluate $|AC|$.




How do we solve this triangle using law of sines?






geometry trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    There's easier ways than the sine rule. is that required?
    $endgroup$
    – TurlocTheRed
    Dec 9 '18 at 4:05














0












0








0





$begingroup$



enter image description here



ABC is a right triangle, $AB perp AC, DE perp BC, |AD| = |BD| = 3$. The area of ABC is $6$ times of the area of BDE. Evaluate $|AC|$.




How do we solve this triangle using law of sines?






geometry trigonometry







share|cite|improve this question











$endgroup$





enter image description here



ABC is a right triangle, $AB perp AC, DE perp BC, |AD| = |BD| = 3$. The area of ABC is $6$ times of the area of BDE. Evaluate $|AC|$.




How do we solve this triangle using law of sines?






geometry trigonometry




geometry trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 14:32









Dylan

12.9k31027




12.9k31027










asked Dec 8 '18 at 11:21









HamiltonHamilton

1798




1798












  • $begingroup$
    There's easier ways than the sine rule. is that required?
    $endgroup$
    – TurlocTheRed
    Dec 9 '18 at 4:05


















  • $begingroup$
    There's easier ways than the sine rule. is that required?
    $endgroup$
    – TurlocTheRed
    Dec 9 '18 at 4:05
















$begingroup$
There's easier ways than the sine rule. is that required?
$endgroup$
– TurlocTheRed
Dec 9 '18 at 4:05




$begingroup$
There's easier ways than the sine rule. is that required?
$endgroup$
– TurlocTheRed
Dec 9 '18 at 4:05










4 Answers
4






active

oldest

votes


















1












$begingroup$

Let $angle ABC = theta$, then $EB = 3costheta$ and $AB=6$. Since the two triangles are similar, the area scale factor is



$$ frac{A_{BDE}}{A_{ABC}} = left(frac{EB}{AB}right)^2 = left(frac{costheta}{2}right)^2 = frac16 $$



$$ implies cos^2theta = frac23, sin^2theta = frac13, tan^2theta = frac12 $$



Finally



$$ AC = ABtantheta = frac{6}{sqrt{2}} = 3sqrt{2} $$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    We have $$A_{ABC}=6A_{BDE}$$ so we get
    $$6frac{AC}{2}=6frac{DEcdot BE}{2}$$ with
    $$frac{DE}{3}=frac{AC}{BC}$$ and $$frac{BE}{3}=frac{6}{BC}$$ we get
    $$3AC=3cdot 3frac{AC}{BC}cdot frac{18}{BC}$$ so $$BC^2=54$$ and $$6^2+AC^2=54$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Did you use sine law?
      $endgroup$
      – Hamilton
      Dec 8 '18 at 13:46










    • $begingroup$
      No i used the similarity of triangles.
      $endgroup$
      – Dr. Sonnhard Graubner
      Dec 8 '18 at 14:36



















    0












    $begingroup$

    $triangle ABC$ and $triangle DEB$ are similar.$mangle ACB=mangle EDB$. As per sine law,$frac{c}{sinC}=frac{a}{sinA}=frac{b}{sinB}$ in $triangle ABC$. Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{e}{sinE}$in $triangle DEB$.



    For the sake of convenience, $overline{AB}=c, overline{AC}=b,overline{BC}=a$ in $triangle ABC$ and $overline{DB}=e,overline{BE}=d,overline{DE}=b_1$ in $triangle DEB$.



    $mangle E$ and $mangle A$ are $90^circ$ angle in $triangle DEB$ and $triangle ABC$ respectively.So, using sine law,



    $frac{c}{sinC}=frac{b}{sinB}=frac{a}{sinA}=a$ because $sin{90^circ}=1$.



    Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{3}{sinE}=3$ because $sin{E}=sin{90^circ}=1$



    $sin{D}=sin{C}$.



    So, ${d}=3sin{C}$ but $sin{C}=frac6a$.



    So $d=frac{18}{a}$. But area of $triangle ABC$ is $6times$ the area of $triangle DEB$. It can be rewritten as $3b=3db_1$



    So,$b=frac{18}{a}b_1$ $Rightarrow a*sin{B}=frac{54}{a}sin{B}$



    $a^2=54 Rightarrow a=sqrt{54}=7.348469$



    $a^2-c^2=b^2$ So,$54-36=18,Rightarrow b=sqrt{18}$






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      A triangle is solved if there is sufficient information to apply a congruence rule, e.g. SSS, SAS, AAS, and HL.



      We have that the ratio of areas between the larger and smaller triangles is 6 to 1. This implies the ratio if corresponding legs is $sqrt{6}$:1.



      We know the hypotenuse if the smaller triangle us 3. By the length ratio, the hypotenuse of the larger triangle us 3$sqrt{6}$. Now the pythagorean theorem can tell us the length of the remaining unknown side.



      By the sine rule, the sine of either angle divided by the opposite side is the reciprocal if the hypotenuse length. We know each leg length, so we can determine the sine, and by inverting, the angle itself






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        @TuriocTHeRed, You have computed the ratio of corresponding sides of a triangle $sqrt{6}:1$ because the area of larger triangle is 6 times greater than the smaller triangle. I want to know whether this rule is applicable to any triangle or any specific types of triangles
        $endgroup$
        – Dhamnekar Winod
        Dec 9 '18 at 10:27










      • $begingroup$
        More general than that. In Euclidean geometry, the ratio of the areas if two similar figures, whatever the shape, is the square of the length of corresponding parts. For proof, I think you can start by proving it for triangles and squares, then, given that non curved plane figures can be broken up into triangles, generality follows. As for as squares and triangles themselves, I think you can prove algebraically, or straight edge and compass. Euclid has a proof for circles but I don't know it.
        $endgroup$
        – TurlocTheRed
        Dec 9 '18 at 16:09











      Your Answer





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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Let $angle ABC = theta$, then $EB = 3costheta$ and $AB=6$. Since the two triangles are similar, the area scale factor is



      $$ frac{A_{BDE}}{A_{ABC}} = left(frac{EB}{AB}right)^2 = left(frac{costheta}{2}right)^2 = frac16 $$



      $$ implies cos^2theta = frac23, sin^2theta = frac13, tan^2theta = frac12 $$



      Finally



      $$ AC = ABtantheta = frac{6}{sqrt{2}} = 3sqrt{2} $$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Let $angle ABC = theta$, then $EB = 3costheta$ and $AB=6$. Since the two triangles are similar, the area scale factor is



        $$ frac{A_{BDE}}{A_{ABC}} = left(frac{EB}{AB}right)^2 = left(frac{costheta}{2}right)^2 = frac16 $$



        $$ implies cos^2theta = frac23, sin^2theta = frac13, tan^2theta = frac12 $$



        Finally



        $$ AC = ABtantheta = frac{6}{sqrt{2}} = 3sqrt{2} $$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Let $angle ABC = theta$, then $EB = 3costheta$ and $AB=6$. Since the two triangles are similar, the area scale factor is



          $$ frac{A_{BDE}}{A_{ABC}} = left(frac{EB}{AB}right)^2 = left(frac{costheta}{2}right)^2 = frac16 $$



          $$ implies cos^2theta = frac23, sin^2theta = frac13, tan^2theta = frac12 $$



          Finally



          $$ AC = ABtantheta = frac{6}{sqrt{2}} = 3sqrt{2} $$






          share|cite|improve this answer









          $endgroup$



          Let $angle ABC = theta$, then $EB = 3costheta$ and $AB=6$. Since the two triangles are similar, the area scale factor is



          $$ frac{A_{BDE}}{A_{ABC}} = left(frac{EB}{AB}right)^2 = left(frac{costheta}{2}right)^2 = frac16 $$



          $$ implies cos^2theta = frac23, sin^2theta = frac13, tan^2theta = frac12 $$



          Finally



          $$ AC = ABtantheta = frac{6}{sqrt{2}} = 3sqrt{2} $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 11:33









          DylanDylan

          12.9k31027




          12.9k31027























              0












              $begingroup$

              We have $$A_{ABC}=6A_{BDE}$$ so we get
              $$6frac{AC}{2}=6frac{DEcdot BE}{2}$$ with
              $$frac{DE}{3}=frac{AC}{BC}$$ and $$frac{BE}{3}=frac{6}{BC}$$ we get
              $$3AC=3cdot 3frac{AC}{BC}cdot frac{18}{BC}$$ so $$BC^2=54$$ and $$6^2+AC^2=54$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Did you use sine law?
                $endgroup$
                – Hamilton
                Dec 8 '18 at 13:46










              • $begingroup$
                No i used the similarity of triangles.
                $endgroup$
                – Dr. Sonnhard Graubner
                Dec 8 '18 at 14:36
















              0












              $begingroup$

              We have $$A_{ABC}=6A_{BDE}$$ so we get
              $$6frac{AC}{2}=6frac{DEcdot BE}{2}$$ with
              $$frac{DE}{3}=frac{AC}{BC}$$ and $$frac{BE}{3}=frac{6}{BC}$$ we get
              $$3AC=3cdot 3frac{AC}{BC}cdot frac{18}{BC}$$ so $$BC^2=54$$ and $$6^2+AC^2=54$$






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Did you use sine law?
                $endgroup$
                – Hamilton
                Dec 8 '18 at 13:46










              • $begingroup$
                No i used the similarity of triangles.
                $endgroup$
                – Dr. Sonnhard Graubner
                Dec 8 '18 at 14:36














              0












              0








              0





              $begingroup$

              We have $$A_{ABC}=6A_{BDE}$$ so we get
              $$6frac{AC}{2}=6frac{DEcdot BE}{2}$$ with
              $$frac{DE}{3}=frac{AC}{BC}$$ and $$frac{BE}{3}=frac{6}{BC}$$ we get
              $$3AC=3cdot 3frac{AC}{BC}cdot frac{18}{BC}$$ so $$BC^2=54$$ and $$6^2+AC^2=54$$






              share|cite|improve this answer









              $endgroup$



              We have $$A_{ABC}=6A_{BDE}$$ so we get
              $$6frac{AC}{2}=6frac{DEcdot BE}{2}$$ with
              $$frac{DE}{3}=frac{AC}{BC}$$ and $$frac{BE}{3}=frac{6}{BC}$$ we get
              $$3AC=3cdot 3frac{AC}{BC}cdot frac{18}{BC}$$ so $$BC^2=54$$ and $$6^2+AC^2=54$$







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Dec 8 '18 at 11:34









              Dr. Sonnhard GraubnerDr. Sonnhard Graubner

              75.3k42865




              75.3k42865












              • $begingroup$
                Did you use sine law?
                $endgroup$
                – Hamilton
                Dec 8 '18 at 13:46










              • $begingroup$
                No i used the similarity of triangles.
                $endgroup$
                – Dr. Sonnhard Graubner
                Dec 8 '18 at 14:36


















              • $begingroup$
                Did you use sine law?
                $endgroup$
                – Hamilton
                Dec 8 '18 at 13:46










              • $begingroup$
                No i used the similarity of triangles.
                $endgroup$
                – Dr. Sonnhard Graubner
                Dec 8 '18 at 14:36
















              $begingroup$
              Did you use sine law?
              $endgroup$
              – Hamilton
              Dec 8 '18 at 13:46




              $begingroup$
              Did you use sine law?
              $endgroup$
              – Hamilton
              Dec 8 '18 at 13:46












              $begingroup$
              No i used the similarity of triangles.
              $endgroup$
              – Dr. Sonnhard Graubner
              Dec 8 '18 at 14:36




              $begingroup$
              No i used the similarity of triangles.
              $endgroup$
              – Dr. Sonnhard Graubner
              Dec 8 '18 at 14:36











              0












              $begingroup$

              $triangle ABC$ and $triangle DEB$ are similar.$mangle ACB=mangle EDB$. As per sine law,$frac{c}{sinC}=frac{a}{sinA}=frac{b}{sinB}$ in $triangle ABC$. Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{e}{sinE}$in $triangle DEB$.



              For the sake of convenience, $overline{AB}=c, overline{AC}=b,overline{BC}=a$ in $triangle ABC$ and $overline{DB}=e,overline{BE}=d,overline{DE}=b_1$ in $triangle DEB$.



              $mangle E$ and $mangle A$ are $90^circ$ angle in $triangle DEB$ and $triangle ABC$ respectively.So, using sine law,



              $frac{c}{sinC}=frac{b}{sinB}=frac{a}{sinA}=a$ because $sin{90^circ}=1$.



              Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{3}{sinE}=3$ because $sin{E}=sin{90^circ}=1$



              $sin{D}=sin{C}$.



              So, ${d}=3sin{C}$ but $sin{C}=frac6a$.



              So $d=frac{18}{a}$. But area of $triangle ABC$ is $6times$ the area of $triangle DEB$. It can be rewritten as $3b=3db_1$



              So,$b=frac{18}{a}b_1$ $Rightarrow a*sin{B}=frac{54}{a}sin{B}$



              $a^2=54 Rightarrow a=sqrt{54}=7.348469$



              $a^2-c^2=b^2$ So,$54-36=18,Rightarrow b=sqrt{18}$






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                $triangle ABC$ and $triangle DEB$ are similar.$mangle ACB=mangle EDB$. As per sine law,$frac{c}{sinC}=frac{a}{sinA}=frac{b}{sinB}$ in $triangle ABC$. Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{e}{sinE}$in $triangle DEB$.



                For the sake of convenience, $overline{AB}=c, overline{AC}=b,overline{BC}=a$ in $triangle ABC$ and $overline{DB}=e,overline{BE}=d,overline{DE}=b_1$ in $triangle DEB$.



                $mangle E$ and $mangle A$ are $90^circ$ angle in $triangle DEB$ and $triangle ABC$ respectively.So, using sine law,



                $frac{c}{sinC}=frac{b}{sinB}=frac{a}{sinA}=a$ because $sin{90^circ}=1$.



                Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{3}{sinE}=3$ because $sin{E}=sin{90^circ}=1$



                $sin{D}=sin{C}$.



                So, ${d}=3sin{C}$ but $sin{C}=frac6a$.



                So $d=frac{18}{a}$. But area of $triangle ABC$ is $6times$ the area of $triangle DEB$. It can be rewritten as $3b=3db_1$



                So,$b=frac{18}{a}b_1$ $Rightarrow a*sin{B}=frac{54}{a}sin{B}$



                $a^2=54 Rightarrow a=sqrt{54}=7.348469$



                $a^2-c^2=b^2$ So,$54-36=18,Rightarrow b=sqrt{18}$






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $triangle ABC$ and $triangle DEB$ are similar.$mangle ACB=mangle EDB$. As per sine law,$frac{c}{sinC}=frac{a}{sinA}=frac{b}{sinB}$ in $triangle ABC$. Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{e}{sinE}$in $triangle DEB$.



                  For the sake of convenience, $overline{AB}=c, overline{AC}=b,overline{BC}=a$ in $triangle ABC$ and $overline{DB}=e,overline{BE}=d,overline{DE}=b_1$ in $triangle DEB$.



                  $mangle E$ and $mangle A$ are $90^circ$ angle in $triangle DEB$ and $triangle ABC$ respectively.So, using sine law,



                  $frac{c}{sinC}=frac{b}{sinB}=frac{a}{sinA}=a$ because $sin{90^circ}=1$.



                  Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{3}{sinE}=3$ because $sin{E}=sin{90^circ}=1$



                  $sin{D}=sin{C}$.



                  So, ${d}=3sin{C}$ but $sin{C}=frac6a$.



                  So $d=frac{18}{a}$. But area of $triangle ABC$ is $6times$ the area of $triangle DEB$. It can be rewritten as $3b=3db_1$



                  So,$b=frac{18}{a}b_1$ $Rightarrow a*sin{B}=frac{54}{a}sin{B}$



                  $a^2=54 Rightarrow a=sqrt{54}=7.348469$



                  $a^2-c^2=b^2$ So,$54-36=18,Rightarrow b=sqrt{18}$






                  share|cite|improve this answer











                  $endgroup$



                  $triangle ABC$ and $triangle DEB$ are similar.$mangle ACB=mangle EDB$. As per sine law,$frac{c}{sinC}=frac{a}{sinA}=frac{b}{sinB}$ in $triangle ABC$. Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{e}{sinE}$in $triangle DEB$.



                  For the sake of convenience, $overline{AB}=c, overline{AC}=b,overline{BC}=a$ in $triangle ABC$ and $overline{DB}=e,overline{BE}=d,overline{DE}=b_1$ in $triangle DEB$.



                  $mangle E$ and $mangle A$ are $90^circ$ angle in $triangle DEB$ and $triangle ABC$ respectively.So, using sine law,



                  $frac{c}{sinC}=frac{b}{sinB}=frac{a}{sinA}=a$ because $sin{90^circ}=1$.



                  Likewise, $frac{d}{sinD}=frac{b_1}{sinB}=frac{3}{sinE}=3$ because $sin{E}=sin{90^circ}=1$



                  $sin{D}=sin{C}$.



                  So, ${d}=3sin{C}$ but $sin{C}=frac6a$.



                  So $d=frac{18}{a}$. But area of $triangle ABC$ is $6times$ the area of $triangle DEB$. It can be rewritten as $3b=3db_1$



                  So,$b=frac{18}{a}b_1$ $Rightarrow a*sin{B}=frac{54}{a}sin{B}$



                  $a^2=54 Rightarrow a=sqrt{54}=7.348469$



                  $a^2-c^2=b^2$ So,$54-36=18,Rightarrow b=sqrt{18}$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 9 '18 at 3:53

























                  answered Dec 9 '18 at 3:43









                  Dhamnekar WinodDhamnekar Winod

                  386414




                  386414























                      0












                      $begingroup$

                      A triangle is solved if there is sufficient information to apply a congruence rule, e.g. SSS, SAS, AAS, and HL.



                      We have that the ratio of areas between the larger and smaller triangles is 6 to 1. This implies the ratio if corresponding legs is $sqrt{6}$:1.



                      We know the hypotenuse if the smaller triangle us 3. By the length ratio, the hypotenuse of the larger triangle us 3$sqrt{6}$. Now the pythagorean theorem can tell us the length of the remaining unknown side.



                      By the sine rule, the sine of either angle divided by the opposite side is the reciprocal if the hypotenuse length. We know each leg length, so we can determine the sine, and by inverting, the angle itself






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        @TuriocTHeRed, You have computed the ratio of corresponding sides of a triangle $sqrt{6}:1$ because the area of larger triangle is 6 times greater than the smaller triangle. I want to know whether this rule is applicable to any triangle or any specific types of triangles
                        $endgroup$
                        – Dhamnekar Winod
                        Dec 9 '18 at 10:27










                      • $begingroup$
                        More general than that. In Euclidean geometry, the ratio of the areas if two similar figures, whatever the shape, is the square of the length of corresponding parts. For proof, I think you can start by proving it for triangles and squares, then, given that non curved plane figures can be broken up into triangles, generality follows. As for as squares and triangles themselves, I think you can prove algebraically, or straight edge and compass. Euclid has a proof for circles but I don't know it.
                        $endgroup$
                        – TurlocTheRed
                        Dec 9 '18 at 16:09
















                      0












                      $begingroup$

                      A triangle is solved if there is sufficient information to apply a congruence rule, e.g. SSS, SAS, AAS, and HL.



                      We have that the ratio of areas between the larger and smaller triangles is 6 to 1. This implies the ratio if corresponding legs is $sqrt{6}$:1.



                      We know the hypotenuse if the smaller triangle us 3. By the length ratio, the hypotenuse of the larger triangle us 3$sqrt{6}$. Now the pythagorean theorem can tell us the length of the remaining unknown side.



                      By the sine rule, the sine of either angle divided by the opposite side is the reciprocal if the hypotenuse length. We know each leg length, so we can determine the sine, and by inverting, the angle itself






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        @TuriocTHeRed, You have computed the ratio of corresponding sides of a triangle $sqrt{6}:1$ because the area of larger triangle is 6 times greater than the smaller triangle. I want to know whether this rule is applicable to any triangle or any specific types of triangles
                        $endgroup$
                        – Dhamnekar Winod
                        Dec 9 '18 at 10:27










                      • $begingroup$
                        More general than that. In Euclidean geometry, the ratio of the areas if two similar figures, whatever the shape, is the square of the length of corresponding parts. For proof, I think you can start by proving it for triangles and squares, then, given that non curved plane figures can be broken up into triangles, generality follows. As for as squares and triangles themselves, I think you can prove algebraically, or straight edge and compass. Euclid has a proof for circles but I don't know it.
                        $endgroup$
                        – TurlocTheRed
                        Dec 9 '18 at 16:09














                      0












                      0








                      0





                      $begingroup$

                      A triangle is solved if there is sufficient information to apply a congruence rule, e.g. SSS, SAS, AAS, and HL.



                      We have that the ratio of areas between the larger and smaller triangles is 6 to 1. This implies the ratio if corresponding legs is $sqrt{6}$:1.



                      We know the hypotenuse if the smaller triangle us 3. By the length ratio, the hypotenuse of the larger triangle us 3$sqrt{6}$. Now the pythagorean theorem can tell us the length of the remaining unknown side.



                      By the sine rule, the sine of either angle divided by the opposite side is the reciprocal if the hypotenuse length. We know each leg length, so we can determine the sine, and by inverting, the angle itself






                      share|cite|improve this answer









                      $endgroup$



                      A triangle is solved if there is sufficient information to apply a congruence rule, e.g. SSS, SAS, AAS, and HL.



                      We have that the ratio of areas between the larger and smaller triangles is 6 to 1. This implies the ratio if corresponding legs is $sqrt{6}$:1.



                      We know the hypotenuse if the smaller triangle us 3. By the length ratio, the hypotenuse of the larger triangle us 3$sqrt{6}$. Now the pythagorean theorem can tell us the length of the remaining unknown side.



                      By the sine rule, the sine of either angle divided by the opposite side is the reciprocal if the hypotenuse length. We know each leg length, so we can determine the sine, and by inverting, the angle itself







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                      answered Dec 9 '18 at 4:19









                      TurlocTheRedTurlocTheRed

                      856311




                      856311












                      • $begingroup$
                        @TuriocTHeRed, You have computed the ratio of corresponding sides of a triangle $sqrt{6}:1$ because the area of larger triangle is 6 times greater than the smaller triangle. I want to know whether this rule is applicable to any triangle or any specific types of triangles
                        $endgroup$
                        – Dhamnekar Winod
                        Dec 9 '18 at 10:27










                      • $begingroup$
                        More general than that. In Euclidean geometry, the ratio of the areas if two similar figures, whatever the shape, is the square of the length of corresponding parts. For proof, I think you can start by proving it for triangles and squares, then, given that non curved plane figures can be broken up into triangles, generality follows. As for as squares and triangles themselves, I think you can prove algebraically, or straight edge and compass. Euclid has a proof for circles but I don't know it.
                        $endgroup$
                        – TurlocTheRed
                        Dec 9 '18 at 16:09


















                      • $begingroup$
                        @TuriocTHeRed, You have computed the ratio of corresponding sides of a triangle $sqrt{6}:1$ because the area of larger triangle is 6 times greater than the smaller triangle. I want to know whether this rule is applicable to any triangle or any specific types of triangles
                        $endgroup$
                        – Dhamnekar Winod
                        Dec 9 '18 at 10:27










                      • $begingroup$
                        More general than that. In Euclidean geometry, the ratio of the areas if two similar figures, whatever the shape, is the square of the length of corresponding parts. For proof, I think you can start by proving it for triangles and squares, then, given that non curved plane figures can be broken up into triangles, generality follows. As for as squares and triangles themselves, I think you can prove algebraically, or straight edge and compass. Euclid has a proof for circles but I don't know it.
                        $endgroup$
                        – TurlocTheRed
                        Dec 9 '18 at 16:09
















                      $begingroup$
                      @TuriocTHeRed, You have computed the ratio of corresponding sides of a triangle $sqrt{6}:1$ because the area of larger triangle is 6 times greater than the smaller triangle. I want to know whether this rule is applicable to any triangle or any specific types of triangles
                      $endgroup$
                      – Dhamnekar Winod
                      Dec 9 '18 at 10:27




                      $begingroup$
                      @TuriocTHeRed, You have computed the ratio of corresponding sides of a triangle $sqrt{6}:1$ because the area of larger triangle is 6 times greater than the smaller triangle. I want to know whether this rule is applicable to any triangle or any specific types of triangles
                      $endgroup$
                      – Dhamnekar Winod
                      Dec 9 '18 at 10:27












                      $begingroup$
                      More general than that. In Euclidean geometry, the ratio of the areas if two similar figures, whatever the shape, is the square of the length of corresponding parts. For proof, I think you can start by proving it for triangles and squares, then, given that non curved plane figures can be broken up into triangles, generality follows. As for as squares and triangles themselves, I think you can prove algebraically, or straight edge and compass. Euclid has a proof for circles but I don't know it.
                      $endgroup$
                      – TurlocTheRed
                      Dec 9 '18 at 16:09




                      $begingroup$
                      More general than that. In Euclidean geometry, the ratio of the areas if two similar figures, whatever the shape, is the square of the length of corresponding parts. For proof, I think you can start by proving it for triangles and squares, then, given that non curved plane figures can be broken up into triangles, generality follows. As for as squares and triangles themselves, I think you can prove algebraically, or straight edge and compass. Euclid has a proof for circles but I don't know it.
                      $endgroup$
                      – TurlocTheRed
                      Dec 9 '18 at 16:09


















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