Possible lengths of an altitude of a triangle












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I've been preparing for a competition and there is this problem that I cannot solve. Can you please help me and also tell me how to do similar problems if they appear in the future?
Problem:

Two of the altitudes of a triangle are 11 units and 10 units. Which of the following can't be the length of an altitude:

(A) 5 units (B) 6 units (c) 7 units (D) 10 units (E) 100 units






triangle







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    0












    $begingroup$


    I've been preparing for a competition and there is this problem that I cannot solve. Can you please help me and also tell me how to do similar problems if they appear in the future?
    Problem:

    Two of the altitudes of a triangle are 11 units and 10 units. Which of the following can't be the length of an altitude:

    (A) 5 units (B) 6 units (c) 7 units (D) 10 units (E) 100 units






    triangle







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I've been preparing for a competition and there is this problem that I cannot solve. Can you please help me and also tell me how to do similar problems if they appear in the future?
      Problem:

      Two of the altitudes of a triangle are 11 units and 10 units. Which of the following can't be the length of an altitude:

      (A) 5 units (B) 6 units (c) 7 units (D) 10 units (E) 100 units






      triangle







      share|cite|improve this question









      $endgroup$




      I've been preparing for a competition and there is this problem that I cannot solve. Can you please help me and also tell me how to do similar problems if they appear in the future?
      Problem:

      Two of the altitudes of a triangle are 11 units and 10 units. Which of the following can't be the length of an altitude:

      (A) 5 units (B) 6 units (c) 7 units (D) 10 units (E) 100 units






      triangle




      triangle






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      asked Jan 3 at 1:40









      The ThunderkingThe Thunderking

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          $begingroup$

          Suppose the three sides of the triangle are $a$, $b$ and $c$, and the corresponding altitudes of this triangle are $h_a=11$, $h_b=10$ and $h_c$. Then $ah_a=bh_b=ch_c = 2A$, where $A$ is the area of the triangle. So, $11a=10b$. Write $a = 10k$. From the triangular inequality we have $(11-10)k<c<(11+10)k$, i.e. $k<c<21k$. Plus, we know that $2A=ah_a=110k$. Thus, $h_c$ must satisfy $frac {110k} {21k} < h_c < frac {110k} k$, which is $5frac 5 {21} < c < 110$.






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            Sorry for my previous misinterpretation.



            The answer is $a)5$.



            Let,the area be $k$ and the unknown altitude is "p" and it's corresponding base is $c$,the other two sides are $a$ and $b$.
            Now,we can say,
            $$frac{1}{2}times a times 11 =frac{1}{2}times b times 10 =frac{1}{2}times c times p =k$$
            so,$$frac{1}{2}times a times 11=kimplies a=frac{2k}{11}$$
            similarly,$$b=frac{2k}{10}~~and~~c=frac{2k}{p}$$
            now,according to inequality of sides law,
            $$a+bgt c$$
            $$implies frac{2k}{11} + frac{2k}{10} gt frac{2k}{p}$$
            $$pgt frac{110}{21}=5.5$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Actually, the $10$ and $11$ are the altitudes, not the side lengths, with the $100$ in the answer options also being an altitude. However, you are correct about using the triangle inequality, with the solution using the triangle area formula of $frac{1}{2}$ the product of base length & altitude length, that the area must obviously be the same for all cases, and then doing the appropriate calculations.
              $endgroup$
              – John Omielan
              Jan 3 at 1:52












            • $begingroup$
              thanks @JohnOmielan for pointing that out.it is fixed now.
              $endgroup$
              – Rakibul Islam Prince
              Jan 3 at 2:34











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            2 Answers
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            2 Answers
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            active

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            2












            $begingroup$

            Suppose the three sides of the triangle are $a$, $b$ and $c$, and the corresponding altitudes of this triangle are $h_a=11$, $h_b=10$ and $h_c$. Then $ah_a=bh_b=ch_c = 2A$, where $A$ is the area of the triangle. So, $11a=10b$. Write $a = 10k$. From the triangular inequality we have $(11-10)k<c<(11+10)k$, i.e. $k<c<21k$. Plus, we know that $2A=ah_a=110k$. Thus, $h_c$ must satisfy $frac {110k} {21k} < h_c < frac {110k} k$, which is $5frac 5 {21} < c < 110$.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Suppose the three sides of the triangle are $a$, $b$ and $c$, and the corresponding altitudes of this triangle are $h_a=11$, $h_b=10$ and $h_c$. Then $ah_a=bh_b=ch_c = 2A$, where $A$ is the area of the triangle. So, $11a=10b$. Write $a = 10k$. From the triangular inequality we have $(11-10)k<c<(11+10)k$, i.e. $k<c<21k$. Plus, we know that $2A=ah_a=110k$. Thus, $h_c$ must satisfy $frac {110k} {21k} < h_c < frac {110k} k$, which is $5frac 5 {21} < c < 110$.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Suppose the three sides of the triangle are $a$, $b$ and $c$, and the corresponding altitudes of this triangle are $h_a=11$, $h_b=10$ and $h_c$. Then $ah_a=bh_b=ch_c = 2A$, where $A$ is the area of the triangle. So, $11a=10b$. Write $a = 10k$. From the triangular inequality we have $(11-10)k<c<(11+10)k$, i.e. $k<c<21k$. Plus, we know that $2A=ah_a=110k$. Thus, $h_c$ must satisfy $frac {110k} {21k} < h_c < frac {110k} k$, which is $5frac 5 {21} < c < 110$.






                share|cite|improve this answer











                $endgroup$



                Suppose the three sides of the triangle are $a$, $b$ and $c$, and the corresponding altitudes of this triangle are $h_a=11$, $h_b=10$ and $h_c$. Then $ah_a=bh_b=ch_c = 2A$, where $A$ is the area of the triangle. So, $11a=10b$. Write $a = 10k$. From the triangular inequality we have $(11-10)k<c<(11+10)k$, i.e. $k<c<21k$. Plus, we know that $2A=ah_a=110k$. Thus, $h_c$ must satisfy $frac {110k} {21k} < h_c < frac {110k} k$, which is $5frac 5 {21} < c < 110$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 3 at 2:10

























                answered Jan 3 at 2:04









                fantasiefantasie

                36718




                36718























                    1












                    $begingroup$

                    Sorry for my previous misinterpretation.



                    The answer is $a)5$.



                    Let,the area be $k$ and the unknown altitude is "p" and it's corresponding base is $c$,the other two sides are $a$ and $b$.
                    Now,we can say,
                    $$frac{1}{2}times a times 11 =frac{1}{2}times b times 10 =frac{1}{2}times c times p =k$$
                    so,$$frac{1}{2}times a times 11=kimplies a=frac{2k}{11}$$
                    similarly,$$b=frac{2k}{10}~~and~~c=frac{2k}{p}$$
                    now,according to inequality of sides law,
                    $$a+bgt c$$
                    $$implies frac{2k}{11} + frac{2k}{10} gt frac{2k}{p}$$
                    $$pgt frac{110}{21}=5.5$$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Actually, the $10$ and $11$ are the altitudes, not the side lengths, with the $100$ in the answer options also being an altitude. However, you are correct about using the triangle inequality, with the solution using the triangle area formula of $frac{1}{2}$ the product of base length & altitude length, that the area must obviously be the same for all cases, and then doing the appropriate calculations.
                      $endgroup$
                      – John Omielan
                      Jan 3 at 1:52












                    • $begingroup$
                      thanks @JohnOmielan for pointing that out.it is fixed now.
                      $endgroup$
                      – Rakibul Islam Prince
                      Jan 3 at 2:34
















                    1












                    $begingroup$

                    Sorry for my previous misinterpretation.



                    The answer is $a)5$.



                    Let,the area be $k$ and the unknown altitude is "p" and it's corresponding base is $c$,the other two sides are $a$ and $b$.
                    Now,we can say,
                    $$frac{1}{2}times a times 11 =frac{1}{2}times b times 10 =frac{1}{2}times c times p =k$$
                    so,$$frac{1}{2}times a times 11=kimplies a=frac{2k}{11}$$
                    similarly,$$b=frac{2k}{10}~~and~~c=frac{2k}{p}$$
                    now,according to inequality of sides law,
                    $$a+bgt c$$
                    $$implies frac{2k}{11} + frac{2k}{10} gt frac{2k}{p}$$
                    $$pgt frac{110}{21}=5.5$$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Actually, the $10$ and $11$ are the altitudes, not the side lengths, with the $100$ in the answer options also being an altitude. However, you are correct about using the triangle inequality, with the solution using the triangle area formula of $frac{1}{2}$ the product of base length & altitude length, that the area must obviously be the same for all cases, and then doing the appropriate calculations.
                      $endgroup$
                      – John Omielan
                      Jan 3 at 1:52












                    • $begingroup$
                      thanks @JohnOmielan for pointing that out.it is fixed now.
                      $endgroup$
                      – Rakibul Islam Prince
                      Jan 3 at 2:34














                    1












                    1








                    1





                    $begingroup$

                    Sorry for my previous misinterpretation.



                    The answer is $a)5$.



                    Let,the area be $k$ and the unknown altitude is "p" and it's corresponding base is $c$,the other two sides are $a$ and $b$.
                    Now,we can say,
                    $$frac{1}{2}times a times 11 =frac{1}{2}times b times 10 =frac{1}{2}times c times p =k$$
                    so,$$frac{1}{2}times a times 11=kimplies a=frac{2k}{11}$$
                    similarly,$$b=frac{2k}{10}~~and~~c=frac{2k}{p}$$
                    now,according to inequality of sides law,
                    $$a+bgt c$$
                    $$implies frac{2k}{11} + frac{2k}{10} gt frac{2k}{p}$$
                    $$pgt frac{110}{21}=5.5$$






                    share|cite|improve this answer











                    $endgroup$



                    Sorry for my previous misinterpretation.



                    The answer is $a)5$.



                    Let,the area be $k$ and the unknown altitude is "p" and it's corresponding base is $c$,the other two sides are $a$ and $b$.
                    Now,we can say,
                    $$frac{1}{2}times a times 11 =frac{1}{2}times b times 10 =frac{1}{2}times c times p =k$$
                    so,$$frac{1}{2}times a times 11=kimplies a=frac{2k}{11}$$
                    similarly,$$b=frac{2k}{10}~~and~~c=frac{2k}{p}$$
                    now,according to inequality of sides law,
                    $$a+bgt c$$
                    $$implies frac{2k}{11} + frac{2k}{10} gt frac{2k}{p}$$
                    $$pgt frac{110}{21}=5.5$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 3 at 2:22

























                    answered Jan 3 at 1:43









                    Rakibul Islam PrinceRakibul Islam Prince

                    1,010211




                    1,010211












                    • $begingroup$
                      Actually, the $10$ and $11$ are the altitudes, not the side lengths, with the $100$ in the answer options also being an altitude. However, you are correct about using the triangle inequality, with the solution using the triangle area formula of $frac{1}{2}$ the product of base length & altitude length, that the area must obviously be the same for all cases, and then doing the appropriate calculations.
                      $endgroup$
                      – John Omielan
                      Jan 3 at 1:52












                    • $begingroup$
                      thanks @JohnOmielan for pointing that out.it is fixed now.
                      $endgroup$
                      – Rakibul Islam Prince
                      Jan 3 at 2:34


















                    • $begingroup$
                      Actually, the $10$ and $11$ are the altitudes, not the side lengths, with the $100$ in the answer options also being an altitude. However, you are correct about using the triangle inequality, with the solution using the triangle area formula of $frac{1}{2}$ the product of base length & altitude length, that the area must obviously be the same for all cases, and then doing the appropriate calculations.
                      $endgroup$
                      – John Omielan
                      Jan 3 at 1:52












                    • $begingroup$
                      thanks @JohnOmielan for pointing that out.it is fixed now.
                      $endgroup$
                      – Rakibul Islam Prince
                      Jan 3 at 2:34
















                    $begingroup$
                    Actually, the $10$ and $11$ are the altitudes, not the side lengths, with the $100$ in the answer options also being an altitude. However, you are correct about using the triangle inequality, with the solution using the triangle area formula of $frac{1}{2}$ the product of base length & altitude length, that the area must obviously be the same for all cases, and then doing the appropriate calculations.
                    $endgroup$
                    – John Omielan
                    Jan 3 at 1:52






                    $begingroup$
                    Actually, the $10$ and $11$ are the altitudes, not the side lengths, with the $100$ in the answer options also being an altitude. However, you are correct about using the triangle inequality, with the solution using the triangle area formula of $frac{1}{2}$ the product of base length & altitude length, that the area must obviously be the same for all cases, and then doing the appropriate calculations.
                    $endgroup$
                    – John Omielan
                    Jan 3 at 1:52














                    $begingroup$
                    thanks @JohnOmielan for pointing that out.it is fixed now.
                    $endgroup$
                    – Rakibul Islam Prince
                    Jan 3 at 2:34




                    $begingroup$
                    thanks @JohnOmielan for pointing that out.it is fixed now.
                    $endgroup$
                    – Rakibul Islam Prince
                    Jan 3 at 2:34


















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