Volume of a pyramid with equilateral triangle as base












1












$begingroup$



The base of a right pyramid is an equilateral triangle of side 4cm
each. Each slant edge is 5cm long. The volume of pyramid is




For this question,




answer provided by sscadda.com is $dfrac{20}{3}sqrt{3}$ (see comments section at bottom)




(a) But this may be wrong because volume is $dfrac{1}{3}×text{base}×text{height}$ and in the solution, slant edge is used in place of height. Am I correct here?



(b) So, for correct solution, we may need to first find height and then use the formula volume is $dfrac{1}{3}×text{base}×text{height}$ , am I correct?



(c) Is the the volume is $dfrac{4sqrt{59}}{3}$ as provided in careerbless.com? If this approach is right, is there any shortcut formula for the same?



(d) If none of my comments is right, please guide how to handle this kind of a question.






geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think you're correct in (a). That is a bad mistake. I get that the pyramid's height is $;frac{sqrt{209}}3;$
    $endgroup$
    – DonAntonio
    Mar 15 '16 at 20:33










  • $begingroup$
    @Joanpemo , please share your steps.
    $endgroup$
    – Kiran
    Mar 15 '16 at 20:34










  • $begingroup$
    By "right pyramid", do you mean that one of the edges is perpendicular to the base?
    $endgroup$
    – robjohn♦
    Jun 5 '17 at 17:59










  • $begingroup$
    The answer should be $frac 43 sqrt{59}$. The elementary approach is best, the height of the pyramid intersects the base at the centroid of the base, so use the $2:1$ ratio and Pythagoras to figure out one of the catheti of the right triangle with the slant edge as its hypotenuse. This allows you to determine the height of the pyramid by another application of Pythagoras. It's all smooth sailing from there. I assume you know how to work this approach but if you want a full answer, let me know.
    $endgroup$
    – Deepak
    Jun 5 '18 at 4:59












  • $begingroup$
    @robjohn A right pyramid is one where the apex lies directly (vertically) over the centroid of the base. mathworld.wolfram.com/RightPyramid.html
    $endgroup$
    – Deepak
    Jun 5 '18 at 5:05


















1












$begingroup$



The base of a right pyramid is an equilateral triangle of side 4cm
each. Each slant edge is 5cm long. The volume of pyramid is




For this question,




answer provided by sscadda.com is $dfrac{20}{3}sqrt{3}$ (see comments section at bottom)




(a) But this may be wrong because volume is $dfrac{1}{3}×text{base}×text{height}$ and in the solution, slant edge is used in place of height. Am I correct here?



(b) So, for correct solution, we may need to first find height and then use the formula volume is $dfrac{1}{3}×text{base}×text{height}$ , am I correct?



(c) Is the the volume is $dfrac{4sqrt{59}}{3}$ as provided in careerbless.com? If this approach is right, is there any shortcut formula for the same?



(d) If none of my comments is right, please guide how to handle this kind of a question.






geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think you're correct in (a). That is a bad mistake. I get that the pyramid's height is $;frac{sqrt{209}}3;$
    $endgroup$
    – DonAntonio
    Mar 15 '16 at 20:33










  • $begingroup$
    @Joanpemo , please share your steps.
    $endgroup$
    – Kiran
    Mar 15 '16 at 20:34










  • $begingroup$
    By "right pyramid", do you mean that one of the edges is perpendicular to the base?
    $endgroup$
    – robjohn♦
    Jun 5 '17 at 17:59










  • $begingroup$
    The answer should be $frac 43 sqrt{59}$. The elementary approach is best, the height of the pyramid intersects the base at the centroid of the base, so use the $2:1$ ratio and Pythagoras to figure out one of the catheti of the right triangle with the slant edge as its hypotenuse. This allows you to determine the height of the pyramid by another application of Pythagoras. It's all smooth sailing from there. I assume you know how to work this approach but if you want a full answer, let me know.
    $endgroup$
    – Deepak
    Jun 5 '18 at 4:59












  • $begingroup$
    @robjohn A right pyramid is one where the apex lies directly (vertically) over the centroid of the base. mathworld.wolfram.com/RightPyramid.html
    $endgroup$
    – Deepak
    Jun 5 '18 at 5:05
















1












1








1


2



$begingroup$



The base of a right pyramid is an equilateral triangle of side 4cm
each. Each slant edge is 5cm long. The volume of pyramid is




For this question,




answer provided by sscadda.com is $dfrac{20}{3}sqrt{3}$ (see comments section at bottom)




(a) But this may be wrong because volume is $dfrac{1}{3}×text{base}×text{height}$ and in the solution, slant edge is used in place of height. Am I correct here?



(b) So, for correct solution, we may need to first find height and then use the formula volume is $dfrac{1}{3}×text{base}×text{height}$ , am I correct?



(c) Is the the volume is $dfrac{4sqrt{59}}{3}$ as provided in careerbless.com? If this approach is right, is there any shortcut formula for the same?



(d) If none of my comments is right, please guide how to handle this kind of a question.






geometry







share|cite|improve this question











$endgroup$





The base of a right pyramid is an equilateral triangle of side 4cm
each. Each slant edge is 5cm long. The volume of pyramid is




For this question,




answer provided by sscadda.com is $dfrac{20}{3}sqrt{3}$ (see comments section at bottom)




(a) But this may be wrong because volume is $dfrac{1}{3}×text{base}×text{height}$ and in the solution, slant edge is used in place of height. Am I correct here?



(b) So, for correct solution, we may need to first find height and then use the formula volume is $dfrac{1}{3}×text{base}×text{height}$ , am I correct?



(c) Is the the volume is $dfrac{4sqrt{59}}{3}$ as provided in careerbless.com? If this approach is right, is there any shortcut formula for the same?



(d) If none of my comments is right, please guide how to handle this kind of a question.






geometry




geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 15 '16 at 20:55









Michael Hardy

1




1










asked Mar 15 '16 at 20:24









KiranKiran

3,2141634




3,2141634












  • $begingroup$
    I think you're correct in (a). That is a bad mistake. I get that the pyramid's height is $;frac{sqrt{209}}3;$
    $endgroup$
    – DonAntonio
    Mar 15 '16 at 20:33










  • $begingroup$
    @Joanpemo , please share your steps.
    $endgroup$
    – Kiran
    Mar 15 '16 at 20:34










  • $begingroup$
    By "right pyramid", do you mean that one of the edges is perpendicular to the base?
    $endgroup$
    – robjohn♦
    Jun 5 '17 at 17:59










  • $begingroup$
    The answer should be $frac 43 sqrt{59}$. The elementary approach is best, the height of the pyramid intersects the base at the centroid of the base, so use the $2:1$ ratio and Pythagoras to figure out one of the catheti of the right triangle with the slant edge as its hypotenuse. This allows you to determine the height of the pyramid by another application of Pythagoras. It's all smooth sailing from there. I assume you know how to work this approach but if you want a full answer, let me know.
    $endgroup$
    – Deepak
    Jun 5 '18 at 4:59












  • $begingroup$
    @robjohn A right pyramid is one where the apex lies directly (vertically) over the centroid of the base. mathworld.wolfram.com/RightPyramid.html
    $endgroup$
    – Deepak
    Jun 5 '18 at 5:05




















  • $begingroup$
    I think you're correct in (a). That is a bad mistake. I get that the pyramid's height is $;frac{sqrt{209}}3;$
    $endgroup$
    – DonAntonio
    Mar 15 '16 at 20:33










  • $begingroup$
    @Joanpemo , please share your steps.
    $endgroup$
    – Kiran
    Mar 15 '16 at 20:34










  • $begingroup$
    By "right pyramid", do you mean that one of the edges is perpendicular to the base?
    $endgroup$
    – robjohn♦
    Jun 5 '17 at 17:59










  • $begingroup$
    The answer should be $frac 43 sqrt{59}$. The elementary approach is best, the height of the pyramid intersects the base at the centroid of the base, so use the $2:1$ ratio and Pythagoras to figure out one of the catheti of the right triangle with the slant edge as its hypotenuse. This allows you to determine the height of the pyramid by another application of Pythagoras. It's all smooth sailing from there. I assume you know how to work this approach but if you want a full answer, let me know.
    $endgroup$
    – Deepak
    Jun 5 '18 at 4:59












  • $begingroup$
    @robjohn A right pyramid is one where the apex lies directly (vertically) over the centroid of the base. mathworld.wolfram.com/RightPyramid.html
    $endgroup$
    – Deepak
    Jun 5 '18 at 5:05


















$begingroup$
I think you're correct in (a). That is a bad mistake. I get that the pyramid's height is $;frac{sqrt{209}}3;$
$endgroup$
– DonAntonio
Mar 15 '16 at 20:33




$begingroup$
I think you're correct in (a). That is a bad mistake. I get that the pyramid's height is $;frac{sqrt{209}}3;$
$endgroup$
– DonAntonio
Mar 15 '16 at 20:33












$begingroup$
@Joanpemo , please share your steps.
$endgroup$
– Kiran
Mar 15 '16 at 20:34




$begingroup$
@Joanpemo , please share your steps.
$endgroup$
– Kiran
Mar 15 '16 at 20:34












$begingroup$
By "right pyramid", do you mean that one of the edges is perpendicular to the base?
$endgroup$
– robjohn♦
Jun 5 '17 at 17:59




$begingroup$
By "right pyramid", do you mean that one of the edges is perpendicular to the base?
$endgroup$
– robjohn♦
Jun 5 '17 at 17:59












$begingroup$
The answer should be $frac 43 sqrt{59}$. The elementary approach is best, the height of the pyramid intersects the base at the centroid of the base, so use the $2:1$ ratio and Pythagoras to figure out one of the catheti of the right triangle with the slant edge as its hypotenuse. This allows you to determine the height of the pyramid by another application of Pythagoras. It's all smooth sailing from there. I assume you know how to work this approach but if you want a full answer, let me know.
$endgroup$
– Deepak
Jun 5 '18 at 4:59






$begingroup$
The answer should be $frac 43 sqrt{59}$. The elementary approach is best, the height of the pyramid intersects the base at the centroid of the base, so use the $2:1$ ratio and Pythagoras to figure out one of the catheti of the right triangle with the slant edge as its hypotenuse. This allows you to determine the height of the pyramid by another application of Pythagoras. It's all smooth sailing from there. I assume you know how to work this approach but if you want a full answer, let me know.
$endgroup$
– Deepak
Jun 5 '18 at 4:59














$begingroup$
@robjohn A right pyramid is one where the apex lies directly (vertically) over the centroid of the base. mathworld.wolfram.com/RightPyramid.html
$endgroup$
– Deepak
Jun 5 '18 at 5:05






$begingroup$
@robjohn A right pyramid is one where the apex lies directly (vertically) over the centroid of the base. mathworld.wolfram.com/RightPyramid.html
$endgroup$
– Deepak
Jun 5 '18 at 5:05












2 Answers
2






active

oldest

votes


















0












$begingroup$

@Kiran An equilateral triangle's area is $;frac{sqrt3}4s^2;$ , with $;s=$ the triangle's side, and then he center of the triangle is of length $;frac s{sqrt3};$ , so taking the straight triangle formed by the pyramid's apex, the center of the base triangle and one of the vertices, the height is, by Pythagoras:



$$sqrt{T^2-frac{s^2}3};;,;;;;T=text{ the slanted's pyramid's side's length}$$



Now input $;s=4;,;;T=5;$$ in the above






share|cite|improve this answer









$endgroup$













  • $begingroup$
    with T=5 AND s=4, we get height as $dfrac{sqrt{59}}{sqrt{3}}$, right?
    $endgroup$
    – Kiran
    Mar 15 '16 at 20:46





















0












$begingroup$

By Heron's Formula, the area of the base is
$$
Delta=sqrt{s(s-a)(s-b)(s-c)}=sqrt{6(6-4)(6-4)(6-4)}=4sqrt3
$$
so the radius of the circumcircle is
$$
R=frac{abc}{4Delta}=frac{4cdot4cdot4}{4cdot4sqrt3}=frac4{sqrt3}
$$
The Pythagorean Theorem says that the altitude of the pyramid is
$$
h=sqrt{5^2-R^2}=sqrt{25-frac{16}3}=sqrt{frac{59}3}
$$
The volume is therefore
$$
V=frac13Delta h=frac43sqrt{59}
$$






share|cite|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    @Kiran An equilateral triangle's area is $;frac{sqrt3}4s^2;$ , with $;s=$ the triangle's side, and then he center of the triangle is of length $;frac s{sqrt3};$ , so taking the straight triangle formed by the pyramid's apex, the center of the base triangle and one of the vertices, the height is, by Pythagoras:



    $$sqrt{T^2-frac{s^2}3};;,;;;;T=text{ the slanted's pyramid's side's length}$$



    Now input $;s=4;,;;T=5;$$ in the above






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      with T=5 AND s=4, we get height as $dfrac{sqrt{59}}{sqrt{3}}$, right?
      $endgroup$
      – Kiran
      Mar 15 '16 at 20:46


















    0












    $begingroup$

    @Kiran An equilateral triangle's area is $;frac{sqrt3}4s^2;$ , with $;s=$ the triangle's side, and then he center of the triangle is of length $;frac s{sqrt3};$ , so taking the straight triangle formed by the pyramid's apex, the center of the base triangle and one of the vertices, the height is, by Pythagoras:



    $$sqrt{T^2-frac{s^2}3};;,;;;;T=text{ the slanted's pyramid's side's length}$$



    Now input $;s=4;,;;T=5;$$ in the above






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      with T=5 AND s=4, we get height as $dfrac{sqrt{59}}{sqrt{3}}$, right?
      $endgroup$
      – Kiran
      Mar 15 '16 at 20:46
















    0












    0








    0





    $begingroup$

    @Kiran An equilateral triangle's area is $;frac{sqrt3}4s^2;$ , with $;s=$ the triangle's side, and then he center of the triangle is of length $;frac s{sqrt3};$ , so taking the straight triangle formed by the pyramid's apex, the center of the base triangle and one of the vertices, the height is, by Pythagoras:



    $$sqrt{T^2-frac{s^2}3};;,;;;;T=text{ the slanted's pyramid's side's length}$$



    Now input $;s=4;,;;T=5;$$ in the above






    share|cite|improve this answer









    $endgroup$



    @Kiran An equilateral triangle's area is $;frac{sqrt3}4s^2;$ , with $;s=$ the triangle's side, and then he center of the triangle is of length $;frac s{sqrt3};$ , so taking the straight triangle formed by the pyramid's apex, the center of the base triangle and one of the vertices, the height is, by Pythagoras:



    $$sqrt{T^2-frac{s^2}3};;,;;;;T=text{ the slanted's pyramid's side's length}$$



    Now input $;s=4;,;;T=5;$$ in the above







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Mar 15 '16 at 20:40









    DonAntonioDonAntonio

    177k1492227




    177k1492227












    • $begingroup$
      with T=5 AND s=4, we get height as $dfrac{sqrt{59}}{sqrt{3}}$, right?
      $endgroup$
      – Kiran
      Mar 15 '16 at 20:46




















    • $begingroup$
      with T=5 AND s=4, we get height as $dfrac{sqrt{59}}{sqrt{3}}$, right?
      $endgroup$
      – Kiran
      Mar 15 '16 at 20:46


















    $begingroup$
    with T=5 AND s=4, we get height as $dfrac{sqrt{59}}{sqrt{3}}$, right?
    $endgroup$
    – Kiran
    Mar 15 '16 at 20:46






    $begingroup$
    with T=5 AND s=4, we get height as $dfrac{sqrt{59}}{sqrt{3}}$, right?
    $endgroup$
    – Kiran
    Mar 15 '16 at 20:46













    0












    $begingroup$

    By Heron's Formula, the area of the base is
    $$
    Delta=sqrt{s(s-a)(s-b)(s-c)}=sqrt{6(6-4)(6-4)(6-4)}=4sqrt3
    $$
    so the radius of the circumcircle is
    $$
    R=frac{abc}{4Delta}=frac{4cdot4cdot4}{4cdot4sqrt3}=frac4{sqrt3}
    $$
    The Pythagorean Theorem says that the altitude of the pyramid is
    $$
    h=sqrt{5^2-R^2}=sqrt{25-frac{16}3}=sqrt{frac{59}3}
    $$
    The volume is therefore
    $$
    V=frac13Delta h=frac43sqrt{59}
    $$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      By Heron's Formula, the area of the base is
      $$
      Delta=sqrt{s(s-a)(s-b)(s-c)}=sqrt{6(6-4)(6-4)(6-4)}=4sqrt3
      $$
      so the radius of the circumcircle is
      $$
      R=frac{abc}{4Delta}=frac{4cdot4cdot4}{4cdot4sqrt3}=frac4{sqrt3}
      $$
      The Pythagorean Theorem says that the altitude of the pyramid is
      $$
      h=sqrt{5^2-R^2}=sqrt{25-frac{16}3}=sqrt{frac{59}3}
      $$
      The volume is therefore
      $$
      V=frac13Delta h=frac43sqrt{59}
      $$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        By Heron's Formula, the area of the base is
        $$
        Delta=sqrt{s(s-a)(s-b)(s-c)}=sqrt{6(6-4)(6-4)(6-4)}=4sqrt3
        $$
        so the radius of the circumcircle is
        $$
        R=frac{abc}{4Delta}=frac{4cdot4cdot4}{4cdot4sqrt3}=frac4{sqrt3}
        $$
        The Pythagorean Theorem says that the altitude of the pyramid is
        $$
        h=sqrt{5^2-R^2}=sqrt{25-frac{16}3}=sqrt{frac{59}3}
        $$
        The volume is therefore
        $$
        V=frac13Delta h=frac43sqrt{59}
        $$






        share|cite|improve this answer











        $endgroup$



        By Heron's Formula, the area of the base is
        $$
        Delta=sqrt{s(s-a)(s-b)(s-c)}=sqrt{6(6-4)(6-4)(6-4)}=4sqrt3
        $$
        so the radius of the circumcircle is
        $$
        R=frac{abc}{4Delta}=frac{4cdot4cdot4}{4cdot4sqrt3}=frac4{sqrt3}
        $$
        The Pythagorean Theorem says that the altitude of the pyramid is
        $$
        h=sqrt{5^2-R^2}=sqrt{25-frac{16}3}=sqrt{frac{59}3}
        $$
        The volume is therefore
        $$
        V=frac13Delta h=frac43sqrt{59}
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jun 6 '18 at 7:35

























        answered Jun 6 '18 at 6:11









        robjohn♦robjohn

        266k27304626




        266k27304626






























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