Isosceles triangle height












2














I was doing a problem in which I was told the lengths of all sides of an isosceles triangle and asked to find the area.



I solved the problem by dividing the isosceles triangle into two equal triangles to find the height which I used in the area formula for the original triangle.



Looking at the answer, my method resulted in the correct value but, it seems I could have used the legs of the isosceles triangle (both 8) as the base and height and skipped finding the height. Why does that shortcut work?
The question is below:



enter image description here










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  • 2




    That only works if the isosceles triangle is also a right angled triangle. I presume the third side was given to be $8sqrt 2$?
    – Deepak
    39 mins ago












  • Might be a coincidence of the mathematics but it's hard to say without knowing your work explicitly. Or as Deepak said, could be an isosceles right triangle. Could you edit your original post to include the original problem and how you worked it using your shortcut?
    – Eevee Trainer
    38 mins ago










  • Also, note that with the triangle lengths available, regardless of it being any specific type of triangle, you didn't need to determine the height. Instead, you could have directly solved the problem using these values in Heron's formula. I just saw you added the question itself. In your case, you could have also used the Pythagorean Theorem to determine the other side lengths, then getting the area using those $2$ values, with it being simpler in this case then Heron's formula.
    – John Omielan
    35 mins ago












  • @Deepak you were right. It makes sense once you recognize that the triangle is a right angle and it's drawn out. I can see the height is one of the legs. Thanks
    – Crt
    31 mins ago
















2














I was doing a problem in which I was told the lengths of all sides of an isosceles triangle and asked to find the area.



I solved the problem by dividing the isosceles triangle into two equal triangles to find the height which I used in the area formula for the original triangle.



Looking at the answer, my method resulted in the correct value but, it seems I could have used the legs of the isosceles triangle (both 8) as the base and height and skipped finding the height. Why does that shortcut work?
The question is below:



enter image description here










share|cite|improve this question









New contributor




Crt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 2




    That only works if the isosceles triangle is also a right angled triangle. I presume the third side was given to be $8sqrt 2$?
    – Deepak
    39 mins ago












  • Might be a coincidence of the mathematics but it's hard to say without knowing your work explicitly. Or as Deepak said, could be an isosceles right triangle. Could you edit your original post to include the original problem and how you worked it using your shortcut?
    – Eevee Trainer
    38 mins ago










  • Also, note that with the triangle lengths available, regardless of it being any specific type of triangle, you didn't need to determine the height. Instead, you could have directly solved the problem using these values in Heron's formula. I just saw you added the question itself. In your case, you could have also used the Pythagorean Theorem to determine the other side lengths, then getting the area using those $2$ values, with it being simpler in this case then Heron's formula.
    – John Omielan
    35 mins ago












  • @Deepak you were right. It makes sense once you recognize that the triangle is a right angle and it's drawn out. I can see the height is one of the legs. Thanks
    – Crt
    31 mins ago














2












2








2







I was doing a problem in which I was told the lengths of all sides of an isosceles triangle and asked to find the area.



I solved the problem by dividing the isosceles triangle into two equal triangles to find the height which I used in the area formula for the original triangle.



Looking at the answer, my method resulted in the correct value but, it seems I could have used the legs of the isosceles triangle (both 8) as the base and height and skipped finding the height. Why does that shortcut work?
The question is below:



enter image description here










share|cite|improve this question









New contributor




Crt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I was doing a problem in which I was told the lengths of all sides of an isosceles triangle and asked to find the area.



I solved the problem by dividing the isosceles triangle into two equal triangles to find the height which I used in the area formula for the original triangle.



Looking at the answer, my method resulted in the correct value but, it seems I could have used the legs of the isosceles triangle (both 8) as the base and height and skipped finding the height. Why does that shortcut work?
The question is below:



enter image description here







geometry triangle






share|cite|improve this question









New contributor




Crt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Crt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 35 mins ago





















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asked 46 mins ago









Crt

1155




1155




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Crt is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.








  • 2




    That only works if the isosceles triangle is also a right angled triangle. I presume the third side was given to be $8sqrt 2$?
    – Deepak
    39 mins ago












  • Might be a coincidence of the mathematics but it's hard to say without knowing your work explicitly. Or as Deepak said, could be an isosceles right triangle. Could you edit your original post to include the original problem and how you worked it using your shortcut?
    – Eevee Trainer
    38 mins ago










  • Also, note that with the triangle lengths available, regardless of it being any specific type of triangle, you didn't need to determine the height. Instead, you could have directly solved the problem using these values in Heron's formula. I just saw you added the question itself. In your case, you could have also used the Pythagorean Theorem to determine the other side lengths, then getting the area using those $2$ values, with it being simpler in this case then Heron's formula.
    – John Omielan
    35 mins ago












  • @Deepak you were right. It makes sense once you recognize that the triangle is a right angle and it's drawn out. I can see the height is one of the legs. Thanks
    – Crt
    31 mins ago














  • 2




    That only works if the isosceles triangle is also a right angled triangle. I presume the third side was given to be $8sqrt 2$?
    – Deepak
    39 mins ago












  • Might be a coincidence of the mathematics but it's hard to say without knowing your work explicitly. Or as Deepak said, could be an isosceles right triangle. Could you edit your original post to include the original problem and how you worked it using your shortcut?
    – Eevee Trainer
    38 mins ago










  • Also, note that with the triangle lengths available, regardless of it being any specific type of triangle, you didn't need to determine the height. Instead, you could have directly solved the problem using these values in Heron's formula. I just saw you added the question itself. In your case, you could have also used the Pythagorean Theorem to determine the other side lengths, then getting the area using those $2$ values, with it being simpler in this case then Heron's formula.
    – John Omielan
    35 mins ago












  • @Deepak you were right. It makes sense once you recognize that the triangle is a right angle and it's drawn out. I can see the height is one of the legs. Thanks
    – Crt
    31 mins ago








2




2




That only works if the isosceles triangle is also a right angled triangle. I presume the third side was given to be $8sqrt 2$?
– Deepak
39 mins ago






That only works if the isosceles triangle is also a right angled triangle. I presume the third side was given to be $8sqrt 2$?
– Deepak
39 mins ago














Might be a coincidence of the mathematics but it's hard to say without knowing your work explicitly. Or as Deepak said, could be an isosceles right triangle. Could you edit your original post to include the original problem and how you worked it using your shortcut?
– Eevee Trainer
38 mins ago




Might be a coincidence of the mathematics but it's hard to say without knowing your work explicitly. Or as Deepak said, could be an isosceles right triangle. Could you edit your original post to include the original problem and how you worked it using your shortcut?
– Eevee Trainer
38 mins ago












Also, note that with the triangle lengths available, regardless of it being any specific type of triangle, you didn't need to determine the height. Instead, you could have directly solved the problem using these values in Heron's formula. I just saw you added the question itself. In your case, you could have also used the Pythagorean Theorem to determine the other side lengths, then getting the area using those $2$ values, with it being simpler in this case then Heron's formula.
– John Omielan
35 mins ago






Also, note that with the triangle lengths available, regardless of it being any specific type of triangle, you didn't need to determine the height. Instead, you could have directly solved the problem using these values in Heron's formula. I just saw you added the question itself. In your case, you could have also used the Pythagorean Theorem to determine the other side lengths, then getting the area using those $2$ values, with it being simpler in this case then Heron's formula.
– John Omielan
35 mins ago














@Deepak you were right. It makes sense once you recognize that the triangle is a right angle and it's drawn out. I can see the height is one of the legs. Thanks
– Crt
31 mins ago




@Deepak you were right. It makes sense once you recognize that the triangle is a right angle and it's drawn out. I can see the height is one of the legs. Thanks
– Crt
31 mins ago










3 Answers
3






active

oldest

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3














Pythagoras' theorem ($a^2 + b^2 = c^2$) is generally used to compute one "missing" side when two others are known in a triangle known to be a right triangle.



However, there is a converse of Pythagoras' theorem, whereby you can show a given triangle is right if the side lengths satisfy the relation.



In this case, $8^2 + 8^2 = (8sqrt 2)^2$, so the triangle is right-angled, and you can immediately find the area as $frac 12 (8)(8) = 32$.



(In this case, you don't need the converse. It's already given to be a right triangle. But in the case you're given the side lengths of $(8,8,8sqrt 2)$, you can apply the converse to simplify your work a bit).






share|cite|improve this answer































    1














    If you know the lengths $L_1$, $L_2$ of two legs of a triangle and the angle between them $theta$, then you can calculate the area as
    $$A = frac12L_1L_2sintheta$$
    In your case $theta$ was 90 degrees, making the formula $A = frac12L_1L_2$.






    share|cite|improve this answer








    New contributor




    Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.


























      0














      After the edit to the OP, yeah, as pointed out by Deepak in the comments: it is because the triangle is not just any isosceles triangle, but an isosceles right triangle. That the question specifies this also may be indicative that your "shortcut" was the intended method (though kudos to you for finding an additional method either way!).



      As is probably obvious whenever you draw right triangles, its area can be given by



      $$A = frac12 ab$$



      where $a,b$ are the legs of the triangle. In right triangles, the legs can be used as the height and the base. To actually further this discussion and extend to isosceles right triangles, suppose you have only the hypotenuse $h$. Since the triangle is isosceles and right, the legs are equal ($a=b$) and are given by $h/sqrt 2$. In that light we could make this even shorter by noting:



      $$A = frac12 ab = frac12 frac{h}{sqrt 2} frac{h}{sqrt 2} = frac{h^2}{4}$$



      so in a sense you don't even need to find the legs: in an isosceles right triangle, the hypotenuse uniquely determines the legs, and vice versa.






      share|cite|improve this answer





















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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

        oldest

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        active

        oldest

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        active

        oldest

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        3














        Pythagoras' theorem ($a^2 + b^2 = c^2$) is generally used to compute one "missing" side when two others are known in a triangle known to be a right triangle.



        However, there is a converse of Pythagoras' theorem, whereby you can show a given triangle is right if the side lengths satisfy the relation.



        In this case, $8^2 + 8^2 = (8sqrt 2)^2$, so the triangle is right-angled, and you can immediately find the area as $frac 12 (8)(8) = 32$.



        (In this case, you don't need the converse. It's already given to be a right triangle. But in the case you're given the side lengths of $(8,8,8sqrt 2)$, you can apply the converse to simplify your work a bit).






        share|cite|improve this answer




























          3














          Pythagoras' theorem ($a^2 + b^2 = c^2$) is generally used to compute one "missing" side when two others are known in a triangle known to be a right triangle.



          However, there is a converse of Pythagoras' theorem, whereby you can show a given triangle is right if the side lengths satisfy the relation.



          In this case, $8^2 + 8^2 = (8sqrt 2)^2$, so the triangle is right-angled, and you can immediately find the area as $frac 12 (8)(8) = 32$.



          (In this case, you don't need the converse. It's already given to be a right triangle. But in the case you're given the side lengths of $(8,8,8sqrt 2)$, you can apply the converse to simplify your work a bit).






          share|cite|improve this answer


























            3












            3








            3






            Pythagoras' theorem ($a^2 + b^2 = c^2$) is generally used to compute one "missing" side when two others are known in a triangle known to be a right triangle.



            However, there is a converse of Pythagoras' theorem, whereby you can show a given triangle is right if the side lengths satisfy the relation.



            In this case, $8^2 + 8^2 = (8sqrt 2)^2$, so the triangle is right-angled, and you can immediately find the area as $frac 12 (8)(8) = 32$.



            (In this case, you don't need the converse. It's already given to be a right triangle. But in the case you're given the side lengths of $(8,8,8sqrt 2)$, you can apply the converse to simplify your work a bit).






            share|cite|improve this answer














            Pythagoras' theorem ($a^2 + b^2 = c^2$) is generally used to compute one "missing" side when two others are known in a triangle known to be a right triangle.



            However, there is a converse of Pythagoras' theorem, whereby you can show a given triangle is right if the side lengths satisfy the relation.



            In this case, $8^2 + 8^2 = (8sqrt 2)^2$, so the triangle is right-angled, and you can immediately find the area as $frac 12 (8)(8) = 32$.



            (In this case, you don't need the converse. It's already given to be a right triangle. But in the case you're given the side lengths of $(8,8,8sqrt 2)$, you can apply the converse to simplify your work a bit).







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 25 mins ago

























            answered 30 mins ago









            Deepak

            16.8k11436




            16.8k11436























                1














                If you know the lengths $L_1$, $L_2$ of two legs of a triangle and the angle between them $theta$, then you can calculate the area as
                $$A = frac12L_1L_2sintheta$$
                In your case $theta$ was 90 degrees, making the formula $A = frac12L_1L_2$.






                share|cite|improve this answer








                New contributor




                Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.























                  1














                  If you know the lengths $L_1$, $L_2$ of two legs of a triangle and the angle between them $theta$, then you can calculate the area as
                  $$A = frac12L_1L_2sintheta$$
                  In your case $theta$ was 90 degrees, making the formula $A = frac12L_1L_2$.






                  share|cite|improve this answer








                  New contributor




                  Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.





















                    1












                    1








                    1






                    If you know the lengths $L_1$, $L_2$ of two legs of a triangle and the angle between them $theta$, then you can calculate the area as
                    $$A = frac12L_1L_2sintheta$$
                    In your case $theta$ was 90 degrees, making the formula $A = frac12L_1L_2$.






                    share|cite|improve this answer








                    New contributor




                    Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    If you know the lengths $L_1$, $L_2$ of two legs of a triangle and the angle between them $theta$, then you can calculate the area as
                    $$A = frac12L_1L_2sintheta$$
                    In your case $theta$ was 90 degrees, making the formula $A = frac12L_1L_2$.







                    share|cite|improve this answer








                    New contributor




                    Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    share|cite|improve this answer



                    share|cite|improve this answer






                    New contributor




                    Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    answered 31 mins ago









                    Erik Parkinson

                    1164




                    1164




                    New contributor




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                    New contributor





                    Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    Erik Parkinson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                        0














                        After the edit to the OP, yeah, as pointed out by Deepak in the comments: it is because the triangle is not just any isosceles triangle, but an isosceles right triangle. That the question specifies this also may be indicative that your "shortcut" was the intended method (though kudos to you for finding an additional method either way!).



                        As is probably obvious whenever you draw right triangles, its area can be given by



                        $$A = frac12 ab$$



                        where $a,b$ are the legs of the triangle. In right triangles, the legs can be used as the height and the base. To actually further this discussion and extend to isosceles right triangles, suppose you have only the hypotenuse $h$. Since the triangle is isosceles and right, the legs are equal ($a=b$) and are given by $h/sqrt 2$. In that light we could make this even shorter by noting:



                        $$A = frac12 ab = frac12 frac{h}{sqrt 2} frac{h}{sqrt 2} = frac{h^2}{4}$$



                        so in a sense you don't even need to find the legs: in an isosceles right triangle, the hypotenuse uniquely determines the legs, and vice versa.






                        share|cite|improve this answer


























                          0














                          After the edit to the OP, yeah, as pointed out by Deepak in the comments: it is because the triangle is not just any isosceles triangle, but an isosceles right triangle. That the question specifies this also may be indicative that your "shortcut" was the intended method (though kudos to you for finding an additional method either way!).



                          As is probably obvious whenever you draw right triangles, its area can be given by



                          $$A = frac12 ab$$



                          where $a,b$ are the legs of the triangle. In right triangles, the legs can be used as the height and the base. To actually further this discussion and extend to isosceles right triangles, suppose you have only the hypotenuse $h$. Since the triangle is isosceles and right, the legs are equal ($a=b$) and are given by $h/sqrt 2$. In that light we could make this even shorter by noting:



                          $$A = frac12 ab = frac12 frac{h}{sqrt 2} frac{h}{sqrt 2} = frac{h^2}{4}$$



                          so in a sense you don't even need to find the legs: in an isosceles right triangle, the hypotenuse uniquely determines the legs, and vice versa.






                          share|cite|improve this answer
























                            0












                            0








                            0






                            After the edit to the OP, yeah, as pointed out by Deepak in the comments: it is because the triangle is not just any isosceles triangle, but an isosceles right triangle. That the question specifies this also may be indicative that your "shortcut" was the intended method (though kudos to you for finding an additional method either way!).



                            As is probably obvious whenever you draw right triangles, its area can be given by



                            $$A = frac12 ab$$



                            where $a,b$ are the legs of the triangle. In right triangles, the legs can be used as the height and the base. To actually further this discussion and extend to isosceles right triangles, suppose you have only the hypotenuse $h$. Since the triangle is isosceles and right, the legs are equal ($a=b$) and are given by $h/sqrt 2$. In that light we could make this even shorter by noting:



                            $$A = frac12 ab = frac12 frac{h}{sqrt 2} frac{h}{sqrt 2} = frac{h^2}{4}$$



                            so in a sense you don't even need to find the legs: in an isosceles right triangle, the hypotenuse uniquely determines the legs, and vice versa.






                            share|cite|improve this answer












                            After the edit to the OP, yeah, as pointed out by Deepak in the comments: it is because the triangle is not just any isosceles triangle, but an isosceles right triangle. That the question specifies this also may be indicative that your "shortcut" was the intended method (though kudos to you for finding an additional method either way!).



                            As is probably obvious whenever you draw right triangles, its area can be given by



                            $$A = frac12 ab$$



                            where $a,b$ are the legs of the triangle. In right triangles, the legs can be used as the height and the base. To actually further this discussion and extend to isosceles right triangles, suppose you have only the hypotenuse $h$. Since the triangle is isosceles and right, the legs are equal ($a=b$) and are given by $h/sqrt 2$. In that light we could make this even shorter by noting:



                            $$A = frac12 ab = frac12 frac{h}{sqrt 2} frac{h}{sqrt 2} = frac{h^2}{4}$$



                            so in a sense you don't even need to find the legs: in an isosceles right triangle, the hypotenuse uniquely determines the legs, and vice versa.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 28 mins ago









                            Eevee Trainer

                            4,6371634




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