How to find the length of one of the sides of a triangle given the area

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The triangles are drawn to scale. The first triangle has side lengths of 17, 17, 16 while the second triangle has side lengths of 17,17,$k$. The triangles have the same area.



Find the value of $k$ algebraically.



So for the first triangle, I know that the height of the triangle splits the base into two parts of 8 each. sO then using pythagorean theorem, I get the height to be 15 and then the area is $frac12 16 *15 = 120$



For the second triangle, I'm not sure what to do and I don't know what solving for $k$ algebraically means.






triangle area







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    up vote
    4
    down vote

    favorite












    enter image description here



    The triangles are drawn to scale. The first triangle has side lengths of 17, 17, 16 while the second triangle has side lengths of 17,17,$k$. The triangles have the same area.



    Find the value of $k$ algebraically.



    So for the first triangle, I know that the height of the triangle splits the base into two parts of 8 each. sO then using pythagorean theorem, I get the height to be 15 and then the area is $frac12 16 *15 = 120$



    For the second triangle, I'm not sure what to do and I don't know what solving for $k$ algebraically means.






    triangle area







    share|cite|improve this question

























      up vote
      4
      down vote

      favorite









      up vote
      4
      down vote

      favorite











      enter image description here



      The triangles are drawn to scale. The first triangle has side lengths of 17, 17, 16 while the second triangle has side lengths of 17,17,$k$. The triangles have the same area.



      Find the value of $k$ algebraically.



      So for the first triangle, I know that the height of the triangle splits the base into two parts of 8 each. sO then using pythagorean theorem, I get the height to be 15 and then the area is $frac12 16 *15 = 120$



      For the second triangle, I'm not sure what to do and I don't know what solving for $k$ algebraically means.






      triangle area







      share|cite|improve this question















      enter image description here



      The triangles are drawn to scale. The first triangle has side lengths of 17, 17, 16 while the second triangle has side lengths of 17,17,$k$. The triangles have the same area.



      Find the value of $k$ algebraically.



      So for the first triangle, I know that the height of the triangle splits the base into two parts of 8 each. sO then using pythagorean theorem, I get the height to be 15 and then the area is $frac12 16 *15 = 120$



      For the second triangle, I'm not sure what to do and I don't know what solving for $k$ algebraically means.






      triangle area




      triangle area






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 3 hours ago









      gt6989b

      31.9k22351




      31.9k22351










      asked 3 hours ago









      user130306

      33717




      33717




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          4
          down vote



          accepted










          HINT



          Exactly the same idea. Split in two parts using the height, and in the half-triangle you have the hypotenuse of $17$ and one of the legs is $k/2$.



          1. By the Pythagorean theorem, height $h$ satisfies $h^2 + (k/2)^2 = 17^2$, can you find $h(k)$?

          2. Now the area of the big triangle is $k cdot h(k) /2$, but you already know this is $120$, can you solve for $k$?

          Remark It's obvious one of the answers will be $k=16$ because then the triangles are identical. Are there other values?



          Update



          You have $$k = sqrt4left(17^2 - h^2right) = 2sqrt17^2 - h^2,$$ hence the final equation is
          $$
          120 = k(h) cdot h /2
          = frach2 cdot 2sqrt17^2 - h^2
          = h sqrt17^2 - h^2
          $$



          To solve this, square both sides to get
          $$
          120^2 = h^2 left(17^2 - h^2right)
          $$
          and let $z = h^2$ to get a quadratic in $z$.






          share|cite|improve this answer






















          • well I have that $frack^2h2 = 120$ so $k^2h = 240$ but now I don't know what to do
            – user8290579
            3 hours ago










          • I know that $k^2 = frac240h$ and plugged that into the first equation but that doesn't help at all
            – user8290579
            3 hours ago










          • i don't know how to find h*k
            – user8290579
            3 hours ago










          • @user8290579 see update
            – gt6989b
            3 hours ago










          • I'm sorry maybe I'm just not understanding but with your update I get that $k^2 = 1156-4h^2$ I can't take the square root of both sides and get a simple answer because square root doesn't distribute over addition or subtraction right?
            – user130306
            3 hours ago

















          up vote
          0
          down vote













          My first inclination would be to use Heron's formula for the area of $Delta ABC = (ABC)$ whose side lengths $a$, $b$, and $c$ opposite angles $A$, $B$, and $C$ are $a=k$, $b=17$, and $c=17$ and whose semiperimeter $s$ therefore is $s = frac34+k2$.



          That is, $$(ABC) = s(s-a)(s-b)(s-c) = 120$$ by Heron's formula, using the values given above for the parameters, and the area calculated using the other triangle.



          This is an expression in $k$ which is cubic, but it will have only 1 real solution. Solve it!






          share|cite|improve this answer




















          • I got $k^2 (17k+578) = 960$ How do I solve from here?
            – user130306
            2 hours ago










          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          4
          down vote



          accepted










          HINT



          Exactly the same idea. Split in two parts using the height, and in the half-triangle you have the hypotenuse of $17$ and one of the legs is $k/2$.



          1. By the Pythagorean theorem, height $h$ satisfies $h^2 + (k/2)^2 = 17^2$, can you find $h(k)$?

          2. Now the area of the big triangle is $k cdot h(k) /2$, but you already know this is $120$, can you solve for $k$?

          Remark It's obvious one of the answers will be $k=16$ because then the triangles are identical. Are there other values?



          Update



          You have $$k = sqrt4left(17^2 - h^2right) = 2sqrt17^2 - h^2,$$ hence the final equation is
          $$
          120 = k(h) cdot h /2
          = frach2 cdot 2sqrt17^2 - h^2
          = h sqrt17^2 - h^2
          $$



          To solve this, square both sides to get
          $$
          120^2 = h^2 left(17^2 - h^2right)
          $$
          and let $z = h^2$ to get a quadratic in $z$.






          share|cite|improve this answer






















          • well I have that $frack^2h2 = 120$ so $k^2h = 240$ but now I don't know what to do
            – user8290579
            3 hours ago










          • I know that $k^2 = frac240h$ and plugged that into the first equation but that doesn't help at all
            – user8290579
            3 hours ago










          • i don't know how to find h*k
            – user8290579
            3 hours ago










          • @user8290579 see update
            – gt6989b
            3 hours ago










          • I'm sorry maybe I'm just not understanding but with your update I get that $k^2 = 1156-4h^2$ I can't take the square root of both sides and get a simple answer because square root doesn't distribute over addition or subtraction right?
            – user130306
            3 hours ago














          up vote
          4
          down vote



          accepted










          HINT



          Exactly the same idea. Split in two parts using the height, and in the half-triangle you have the hypotenuse of $17$ and one of the legs is $k/2$.



          1. By the Pythagorean theorem, height $h$ satisfies $h^2 + (k/2)^2 = 17^2$, can you find $h(k)$?

          2. Now the area of the big triangle is $k cdot h(k) /2$, but you already know this is $120$, can you solve for $k$?

          Remark It's obvious one of the answers will be $k=16$ because then the triangles are identical. Are there other values?



          Update



          You have $$k = sqrt4left(17^2 - h^2right) = 2sqrt17^2 - h^2,$$ hence the final equation is
          $$
          120 = k(h) cdot h /2
          = frach2 cdot 2sqrt17^2 - h^2
          = h sqrt17^2 - h^2
          $$



          To solve this, square both sides to get
          $$
          120^2 = h^2 left(17^2 - h^2right)
          $$
          and let $z = h^2$ to get a quadratic in $z$.






          share|cite|improve this answer






















          • well I have that $frack^2h2 = 120$ so $k^2h = 240$ but now I don't know what to do
            – user8290579
            3 hours ago










          • I know that $k^2 = frac240h$ and plugged that into the first equation but that doesn't help at all
            – user8290579
            3 hours ago










          • i don't know how to find h*k
            – user8290579
            3 hours ago










          • @user8290579 see update
            – gt6989b
            3 hours ago










          • I'm sorry maybe I'm just not understanding but with your update I get that $k^2 = 1156-4h^2$ I can't take the square root of both sides and get a simple answer because square root doesn't distribute over addition or subtraction right?
            – user130306
            3 hours ago












          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          HINT



          Exactly the same idea. Split in two parts using the height, and in the half-triangle you have the hypotenuse of $17$ and one of the legs is $k/2$.



          1. By the Pythagorean theorem, height $h$ satisfies $h^2 + (k/2)^2 = 17^2$, can you find $h(k)$?

          2. Now the area of the big triangle is $k cdot h(k) /2$, but you already know this is $120$, can you solve for $k$?

          Remark It's obvious one of the answers will be $k=16$ because then the triangles are identical. Are there other values?



          Update



          You have $$k = sqrt4left(17^2 - h^2right) = 2sqrt17^2 - h^2,$$ hence the final equation is
          $$
          120 = k(h) cdot h /2
          = frach2 cdot 2sqrt17^2 - h^2
          = h sqrt17^2 - h^2
          $$



          To solve this, square both sides to get
          $$
          120^2 = h^2 left(17^2 - h^2right)
          $$
          and let $z = h^2$ to get a quadratic in $z$.






          share|cite|improve this answer














          HINT



          Exactly the same idea. Split in two parts using the height, and in the half-triangle you have the hypotenuse of $17$ and one of the legs is $k/2$.



          1. By the Pythagorean theorem, height $h$ satisfies $h^2 + (k/2)^2 = 17^2$, can you find $h(k)$?

          2. Now the area of the big triangle is $k cdot h(k) /2$, but you already know this is $120$, can you solve for $k$?

          Remark It's obvious one of the answers will be $k=16$ because then the triangles are identical. Are there other values?



          Update



          You have $$k = sqrt4left(17^2 - h^2right) = 2sqrt17^2 - h^2,$$ hence the final equation is
          $$
          120 = k(h) cdot h /2
          = frach2 cdot 2sqrt17^2 - h^2
          = h sqrt17^2 - h^2
          $$



          To solve this, square both sides to get
          $$
          120^2 = h^2 left(17^2 - h^2right)
          $$
          and let $z = h^2$ to get a quadratic in $z$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 hours ago

























          answered 3 hours ago









          gt6989b

          31.9k22351




          31.9k22351











          • well I have that $frack^2h2 = 120$ so $k^2h = 240$ but now I don't know what to do
            – user8290579
            3 hours ago










          • I know that $k^2 = frac240h$ and plugged that into the first equation but that doesn't help at all
            – user8290579
            3 hours ago










          • i don't know how to find h*k
            – user8290579
            3 hours ago










          • @user8290579 see update
            – gt6989b
            3 hours ago










          • I'm sorry maybe I'm just not understanding but with your update I get that $k^2 = 1156-4h^2$ I can't take the square root of both sides and get a simple answer because square root doesn't distribute over addition or subtraction right?
            – user130306
            3 hours ago
















          • well I have that $frack^2h2 = 120$ so $k^2h = 240$ but now I don't know what to do
            – user8290579
            3 hours ago










          • I know that $k^2 = frac240h$ and plugged that into the first equation but that doesn't help at all
            – user8290579
            3 hours ago










          • i don't know how to find h*k
            – user8290579
            3 hours ago










          • @user8290579 see update
            – gt6989b
            3 hours ago










          • I'm sorry maybe I'm just not understanding but with your update I get that $k^2 = 1156-4h^2$ I can't take the square root of both sides and get a simple answer because square root doesn't distribute over addition or subtraction right?
            – user130306
            3 hours ago















          well I have that $frack^2h2 = 120$ so $k^2h = 240$ but now I don't know what to do
          – user8290579
          3 hours ago




          well I have that $frack^2h2 = 120$ so $k^2h = 240$ but now I don't know what to do
          – user8290579
          3 hours ago












          I know that $k^2 = frac240h$ and plugged that into the first equation but that doesn't help at all
          – user8290579
          3 hours ago




          I know that $k^2 = frac240h$ and plugged that into the first equation but that doesn't help at all
          – user8290579
          3 hours ago












          i don't know how to find h*k
          – user8290579
          3 hours ago




          i don't know how to find h*k
          – user8290579
          3 hours ago












          @user8290579 see update
          – gt6989b
          3 hours ago




          @user8290579 see update
          – gt6989b
          3 hours ago












          I'm sorry maybe I'm just not understanding but with your update I get that $k^2 = 1156-4h^2$ I can't take the square root of both sides and get a simple answer because square root doesn't distribute over addition or subtraction right?
          – user130306
          3 hours ago




          I'm sorry maybe I'm just not understanding but with your update I get that $k^2 = 1156-4h^2$ I can't take the square root of both sides and get a simple answer because square root doesn't distribute over addition or subtraction right?
          – user130306
          3 hours ago










          up vote
          0
          down vote













          My first inclination would be to use Heron's formula for the area of $Delta ABC = (ABC)$ whose side lengths $a$, $b$, and $c$ opposite angles $A$, $B$, and $C$ are $a=k$, $b=17$, and $c=17$ and whose semiperimeter $s$ therefore is $s = frac34+k2$.



          That is, $$(ABC) = s(s-a)(s-b)(s-c) = 120$$ by Heron's formula, using the values given above for the parameters, and the area calculated using the other triangle.



          This is an expression in $k$ which is cubic, but it will have only 1 real solution. Solve it!






          share|cite|improve this answer




















          • I got $k^2 (17k+578) = 960$ How do I solve from here?
            – user130306
            2 hours ago














          up vote
          0
          down vote













          My first inclination would be to use Heron's formula for the area of $Delta ABC = (ABC)$ whose side lengths $a$, $b$, and $c$ opposite angles $A$, $B$, and $C$ are $a=k$, $b=17$, and $c=17$ and whose semiperimeter $s$ therefore is $s = frac34+k2$.



          That is, $$(ABC) = s(s-a)(s-b)(s-c) = 120$$ by Heron's formula, using the values given above for the parameters, and the area calculated using the other triangle.



          This is an expression in $k$ which is cubic, but it will have only 1 real solution. Solve it!






          share|cite|improve this answer




















          • I got $k^2 (17k+578) = 960$ How do I solve from here?
            – user130306
            2 hours ago












          up vote
          0
          down vote










          up vote
          0
          down vote









          My first inclination would be to use Heron's formula for the area of $Delta ABC = (ABC)$ whose side lengths $a$, $b$, and $c$ opposite angles $A$, $B$, and $C$ are $a=k$, $b=17$, and $c=17$ and whose semiperimeter $s$ therefore is $s = frac34+k2$.



          That is, $$(ABC) = s(s-a)(s-b)(s-c) = 120$$ by Heron's formula, using the values given above for the parameters, and the area calculated using the other triangle.



          This is an expression in $k$ which is cubic, but it will have only 1 real solution. Solve it!






          share|cite|improve this answer












          My first inclination would be to use Heron's formula for the area of $Delta ABC = (ABC)$ whose side lengths $a$, $b$, and $c$ opposite angles $A$, $B$, and $C$ are $a=k$, $b=17$, and $c=17$ and whose semiperimeter $s$ therefore is $s = frac34+k2$.



          That is, $$(ABC) = s(s-a)(s-b)(s-c) = 120$$ by Heron's formula, using the values given above for the parameters, and the area calculated using the other triangle.



          This is an expression in $k$ which is cubic, but it will have only 1 real solution. Solve it!







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          Circulwyrd

          992722




          992722











          • I got $k^2 (17k+578) = 960$ How do I solve from here?
            – user130306
            2 hours ago
















          • I got $k^2 (17k+578) = 960$ How do I solve from here?
            – user130306
            2 hours ago















          I got $k^2 (17k+578) = 960$ How do I solve from here?
          – user130306
          2 hours ago




          I got $k^2 (17k+578) = 960$ How do I solve from here?
          – user130306
          2 hours ago

















           

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