Volume of a pyramid with equilateral triangle as base
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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
edited Mar 15 '16 at 20:55
Michael Hardy
1
asked Mar 15 '16 at 20:24
Kiran
3,2141633
|
show 2 more comments
up vote
1
down vote
favorite
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
edited Mar 15 '16 at 20:55
Michael Hardy
1
asked Mar 15 '16 at 20:24
Kiran
3,2141633
I think you're correct in (a). That is a bad mistake. I get that the pyramid's height is $;frac{sqrt{209}}3;$
– DonAntonio
Mar 15 '16 at 20:33
@Joanpemo , please share your steps.
– Kiran
Mar 15 '16 at 20:34
By "right pyramid", do you mean that one of the edges is perpendicular to the base?
– robjohn♦
Jun 5 '17 at 17:59
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.
– Deepak
Jun 5 at 4:59
@robjohn A right pyramid is one where the apex lies directly (vertically) over the centroid of the base. mathworld.wolfram.com/RightPyramid.html
– Deepak
Jun 5 at 5:05
|
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
edited Mar 15 '16 at 20:55
Michael Hardy
1
asked Mar 15 '16 at 20:24
Kiran
3,2141633
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
edited Mar 15 '16 at 20:55
Michael Hardy
1
asked Mar 15 '16 at 20:24
Kiran
3,2141633
edited Mar 15 '16 at 20:55
Michael Hardy
1
asked Mar 15 '16 at 20:24
Kiran
3,2141633
edited Mar 15 '16 at 20:55
Michael Hardy
1
edited Mar 15 '16 at 20:55
Michael Hardy
1
edited Mar 15 '16 at 20:55
Michael Hardy
1
1
asked Mar 15 '16 at 20:24
Kiran
3,2141633
asked Mar 15 '16 at 20:24
Kiran
3,2141633
asked Mar 15 '16 at 20:24
Kiran
3,2141633
3,2141633
I think you're correct in (a). That is a bad mistake. I get that the pyramid's height is $;frac{sqrt{209}}3;$
– DonAntonio
Mar 15 '16 at 20:33
@Joanpemo , please share your steps.
– Kiran
Mar 15 '16 at 20:34
By "right pyramid", do you mean that one of the edges is perpendicular to the base?
– robjohn♦
Jun 5 '17 at 17:59
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.
– Deepak
Jun 5 at 4:59
@robjohn A right pyramid is one where the apex lies directly (vertically) over the centroid of the base. mathworld.wolfram.com/RightPyramid.html
– Deepak
Jun 5 at 5:05
|
show 2 more comments
I think you're correct in (a). That is a bad mistake. I get that the pyramid's height is $;frac{sqrt{209}}3;$
– DonAntonio
Mar 15 '16 at 20:33
@Joanpemo , please share your steps.
– Kiran
Mar 15 '16 at 20:34
By "right pyramid", do you mean that one of the edges is perpendicular to the base?
– robjohn♦
Jun 5 '17 at 17:59
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.
– Deepak
Jun 5 at 4:59
@robjohn A right pyramid is one where the apex lies directly (vertically) over the centroid of the base. mathworld.wolfram.com/RightPyramid.html
– Deepak
Jun 5 at 5:05
I think you're correct in (a). That is a bad mistake. I get that the pyramid's height is $;frac{sqrt{209}}3;$
– DonAntonio
Mar 15 '16 at 20:33
I think you're correct in (a). That is a bad mistake. I get that the pyramid's height is $;frac{sqrt{209}}3;$
– DonAntonio
Mar 15 '16 at 20:33
@Joanpemo , please share your steps.
– Kiran
Mar 15 '16 at 20:34
@Joanpemo , please share your steps.
– Kiran
Mar 15 '16 at 20:34
By "right pyramid", do you mean that one of the edges is perpendicular to the base?
– robjohn♦
Jun 5 '17 at 17:59
By "right pyramid", do you mean that one of the edges is perpendicular to the base?
– robjohn♦
Jun 5 '17 at 17:59
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.
– Deepak
Jun 5 at 4:59
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.
– Deepak
Jun 5 at 4:59
@robjohn A right pyramid is one where the apex lies directly (vertically) over the centroid of the base. mathworld.wolfram.com/RightPyramid.html
– Deepak
Jun 5 at 5:05
@robjohn A right pyramid is one where the apex lies directly (vertically) over the centroid of the base. mathworld.wolfram.com/RightPyramid.html
– Deepak
Jun 5 at 5:05
|
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
0
down vote
@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;$$ in the above
answered Mar 15 '16 at 20:40
DonAntonio
176k1491224
with T=5 AND s=4, we get height as $dfrac{sqrt{59}}{sqrt{3}}$, right?
– Kiran
Mar 15 '16 at 20:46
add a comment |
up vote
0
down vote
By Heron's Formula, the area of the base is
$$
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}
$$
answered Jun 6 at 6:11
robjohn♦
263k27301623
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
@Kiran An equilateral triangle's area is $;frac{sqrt3}4s^2;$ , with $;s=$ the triangle's side, and then he center of the triangle is of length $;frac s{sqrt3};$ , so taking the straight triangle formed by the pyramid's apex, the center of the base triangle and one of the vertices, the height is, by Pythagoras:
$$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
answered Mar 15 '16 at 20:40
DonAntonio
176k1491224
with T=5 AND s=4, we get height as $dfrac{sqrt{59}}{sqrt{3}}$, right?
– Kiran
Mar 15 '16 at 20:46
add a comment |
up vote
0
down vote
@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;$$ in the above
answered Mar 15 '16 at 20:40
DonAntonio
176k1491224
with T=5 AND s=4, we get height as $dfrac{sqrt{59}}{sqrt{3}}$, right?
– Kiran
Mar 15 '16 at 20:46
add a comment |
up vote
0
down vote
up vote
0
down vote
@Kiran An equilateral triangle's area is $;frac{sqrt3}4s^2;$ , with $;s=$ the triangle's side, and then he center of the triangle is of length $;frac s{sqrt3};$ , so taking the straight triangle formed by the pyramid's apex, the center of the base triangle and one of the vertices, the height is, by Pythagoras:
$$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
answered Mar 15 '16 at 20:40
DonAntonio
176k1491224
@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;$$ in the above
answered Mar 15 '16 at 20:40
DonAntonio
176k1491224
answered Mar 15 '16 at 20:40
DonAntonio
176k1491224
answered Mar 15 '16 at 20:40
DonAntonio
176k1491224
answered Mar 15 '16 at 20:40
DonAntonio
176k1491224
176k1491224
with T=5 AND s=4, we get height as $dfrac{sqrt{59}}{sqrt{3}}$, right?
– Kiran
Mar 15 '16 at 20:46
add a comment |
with T=5 AND s=4, we get height as $dfrac{sqrt{59}}{sqrt{3}}$, right?
– Kiran
Mar 15 '16 at 20:46
with T=5 AND s=4, we get height as $dfrac{sqrt{59}}{sqrt{3}}$, right?
– Kiran
Mar 15 '16 at 20:46
with T=5 AND s=4, we get height as $dfrac{sqrt{59}}{sqrt{3}}$, right?
– Kiran
Mar 15 '16 at 20:46
add a comment |
up vote
0
down vote
By Heron's Formula, the area of the base is
$$
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}
$$
answered Jun 6 at 6:11
robjohn♦
263k27301623
add a comment |
up vote
0
down vote
By Heron's Formula, the area of the base is
$$
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}
$$
answered Jun 6 at 6:11
robjohn♦
263k27301623
add a comment |
up vote
0
down vote
up vote
0
down vote
By Heron's Formula, the area of the base is
$$
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}
$$
answered Jun 6 at 6:11
robjohn♦
263k27301623
By Heron's Formula, the area of the base is
$$
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}
$$
answered Jun 6 at 6:11
robjohn♦
263k27301623
edited Jun 6 at 7:35
answered Jun 6 at 6:11
robjohn♦
263k27301623
answered Jun 6 at 6:11
robjohn♦
263k27301623
answered Jun 6 at 6:11
robjohn♦
263k27301623
263k27301623
add a comment |
add a comment |
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I think you're correct in (a). That is a bad mistake. I get that the pyramid's height is $;frac{sqrt{209}}3;$
– DonAntonio
Mar 15 '16 at 20:33
@Joanpemo , please share your steps.
– Kiran
Mar 15 '16 at 20:34
By "right pyramid", do you mean that one of the edges is perpendicular to the base?
– robjohn♦
Jun 5 '17 at 17:59
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.
– Deepak
Jun 5 at 4:59
@robjohn A right pyramid is one where the apex lies directly (vertically) over the centroid of the base. mathworld.wolfram.com/RightPyramid.html
– Deepak
Jun 5 at 5:05