Finding the length of the side of the equilateral triangle Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Find side of an equilateral triangle inscribed in a rhombusTo calculate side of the Equilateral triangleFind the side of an equilateral triangle inscribed in a circle.Finding the length of a side of an equilateral triangleProblem on Equilateral Triangle and points$ABCD$ is a square with side-length $1$ and equilateral triangles $Delta AYB$ and $Delta CXD$ are inside the square. What is the length of $XY$?Area of the largest square inscribed in an equilateral triangle that is itself inscribed in a circle of radius $r$Largest rectangle that can fit isnide of equilateral triangleArea of enclosed overlapping circles within an equilateral triangleChanging an equilateral triangle with side length 100 to side length 1
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Finding the length of the side of the equilateral triangle
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Find side of an equilateral triangle inscribed in a rhombusTo calculate side of the Equilateral triangleFind the side of an equilateral triangle inscribed in a circle.Finding the length of a side of an equilateral triangleProblem on Equilateral Triangle and points$ABCD$ is a square with side-length $1$ and equilateral triangles $Delta AYB$ and $Delta CXD$ are inside the square. What is the length of $XY$?Area of the largest square inscribed in an equilateral triangle that is itself inscribed in a circle of radius $r$Largest rectangle that can fit isnide of equilateral triangleArea of enclosed overlapping circles within an equilateral triangleChanging an equilateral triangle with side length 100 to side length 1
$begingroup$
Here, ABCD is a rectangle, and BC = 3 cm. An Equilateral triangle XYZ is inscribed inside the rectangle as shown in the figure where YE = 2 cm. YE is perpendicular to DC. Calculate the length of the side of the equilateral triangle XYZ.
geometry triangles rectangles
edited Nov 21 '14 at 17:06
ProgramFOX
1055
asked Nov 21 '14 at 17:03
BIKASH JENABIKASH JENA
371
$endgroup$
add a comment |
$begingroup$
Here, ABCD is a rectangle, and BC = 3 cm. An Equilateral triangle XYZ is inscribed inside the rectangle as shown in the figure where YE = 2 cm. YE is perpendicular to DC. Calculate the length of the side of the equilateral triangle XYZ.
geometry triangles rectangles
edited Nov 21 '14 at 17:06
ProgramFOX
1055
asked Nov 21 '14 at 17:03
BIKASH JENABIKASH JENA
371
$endgroup$
$begingroup$
@ProgramFOX: Good edit.
$endgroup$
– André Nicolas
Nov 21 '14 at 17:07
add a comment |
$begingroup$
Here, ABCD is a rectangle, and BC = 3 cm. An Equilateral triangle XYZ is inscribed inside the rectangle as shown in the figure where YE = 2 cm. YE is perpendicular to DC. Calculate the length of the side of the equilateral triangle XYZ.
geometry triangles rectangles
edited Nov 21 '14 at 17:06
ProgramFOX
1055
asked Nov 21 '14 at 17:03
BIKASH JENABIKASH JENA
371
$endgroup$
Here, ABCD is a rectangle, and BC = 3 cm. An Equilateral triangle XYZ is inscribed inside the rectangle as shown in the figure where YE = 2 cm. YE is perpendicular to DC. Calculate the length of the side of the equilateral triangle XYZ.
geometry triangles rectangles
geometry triangles rectangles
edited Nov 21 '14 at 17:06
ProgramFOX
1055
asked Nov 21 '14 at 17:03
BIKASH JENABIKASH JENA
371
edited Nov 21 '14 at 17:06
ProgramFOX
1055
asked Nov 21 '14 at 17:03
BIKASH JENABIKASH JENA
371
edited Nov 21 '14 at 17:06
ProgramFOX
1055
edited Nov 21 '14 at 17:06
ProgramFOX
1055
edited Nov 21 '14 at 17:06
ProgramFOX
1055
1055
asked Nov 21 '14 at 17:03
BIKASH JENABIKASH JENA
371
asked Nov 21 '14 at 17:03
BIKASH JENABIKASH JENA
371
asked Nov 21 '14 at 17:03
BIKASH JENABIKASH JENA
371
371
$begingroup$
@ProgramFOX: Good edit.
$endgroup$
– André Nicolas
Nov 21 '14 at 17:07
add a comment |
$begingroup$
@ProgramFOX: Good edit.
$endgroup$
– André Nicolas
Nov 21 '14 at 17:07
$begingroup$
@ProgramFOX: Good edit.
$endgroup$
– André Nicolas
Nov 21 '14 at 17:07
$begingroup$
@ProgramFOX: Good edit.
$endgroup$
– André Nicolas
Nov 21 '14 at 17:07
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Consider the reference system with the origin in $Z$ in which $DC$ is the real axis, and let $EZ=a$. Then we have $Y=a+2i$ and:
$$e^pi i/6(a+2i) = X $$
so:
$$Im left[left(frac12+fracsqrt32iright)cdotleft(a+2iright)right]=3, $$
or:
$$ 1+fracsqrt32a = 3 $$
so $a = frac4sqrt3$, and by the Pythagorean theorem:
$$ ZY^2 = a^2 + 4 = frac163+4 = frac283 $$
so the side length is $2sqrtfrac73$.
answered Nov 21 '14 at 17:14
Jack D'AurizioJack D'Aurizio
292k33284673
$endgroup$
add a comment |
$begingroup$
Hint: Extend $EY$ to meet $AB$ at $F$. Drop a vertical line from $X$, hitting $CD$ at $G$. You now have three right triangles with the hypotenuse being the side of the equilateral triangle.
answered Nov 21 '14 at 17:14
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
So you obtain three equations by pythagora formula and two other equations FY+YE=3 and GZ+ZE=XF. The system can be solved uniquely.
$endgroup$
– Idris
Nov 21 '14 at 17:32
add a comment |
$begingroup$
Let $|XY|=|YZ|=|ZX|=a$,
$angle EYZ=theta$, $|FY|=1$.
Then $angle XYF=120^circ-theta$,
beginalign
triangle EYZ:quad
acostheta&=2
tag1label1
,\
triangle FXY:quad
acos(120^circ-theta)&=1
tag2label2
,
endalign
beginalign
acos(120^circ-theta)&=
frac asqrt32,sintheta
-frac a2costheta
\
&=
frac asqrt32,sintheta
-1
,
endalign
so the system eqref1,eqref2
changes to
beginalign
acostheta&=2
tag3label3
,\
a,sintheta
&=frac4sqrt33
tag4label4
,
endalign
which can be easily solved for $a$:
beginalign
a^2cos^2theta+
a^2sin^2theta
&=2^2+left(frac4sqrt33right)^2
,\
a^2&=frac 283
,\
a&=frac 23,sqrt21
.
endalign
answered Mar 31 at 17:43
g.kovg.kov
6,4971819
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the reference system with the origin in $Z$ in which $DC$ is the real axis, and let $EZ=a$. Then we have $Y=a+2i$ and:
$$e^pi i/6(a+2i) = X $$
so:
$$Im left[left(frac12+fracsqrt32iright)cdotleft(a+2iright)right]=3, $$
or:
$$ 1+fracsqrt32a = 3 $$
so $a = frac4sqrt3$, and by the Pythagorean theorem:
$$ ZY^2 = a^2 + 4 = frac163+4 = frac283 $$
so the side length is $2sqrtfrac73$.
answered Nov 21 '14 at 17:14
Jack D'AurizioJack D'Aurizio
292k33284673
$endgroup$
add a comment |
$begingroup$
Consider the reference system with the origin in $Z$ in which $DC$ is the real axis, and let $EZ=a$. Then we have $Y=a+2i$ and:
$$e^pi i/6(a+2i) = X $$
so:
$$Im left[left(frac12+fracsqrt32iright)cdotleft(a+2iright)right]=3, $$
or:
$$ 1+fracsqrt32a = 3 $$
so $a = frac4sqrt3$, and by the Pythagorean theorem:
$$ ZY^2 = a^2 + 4 = frac163+4 = frac283 $$
so the side length is $2sqrtfrac73$.
answered Nov 21 '14 at 17:14
Jack D'AurizioJack D'Aurizio
292k33284673
$endgroup$
add a comment |
$begingroup$
Consider the reference system with the origin in $Z$ in which $DC$ is the real axis, and let $EZ=a$. Then we have $Y=a+2i$ and:
$$e^pi i/6(a+2i) = X $$
so:
$$Im left[left(frac12+fracsqrt32iright)cdotleft(a+2iright)right]=3, $$
or:
$$ 1+fracsqrt32a = 3 $$
so $a = frac4sqrt3$, and by the Pythagorean theorem:
$$ ZY^2 = a^2 + 4 = frac163+4 = frac283 $$
so the side length is $2sqrtfrac73$.
answered Nov 21 '14 at 17:14
Jack D'AurizioJack D'Aurizio
292k33284673
$endgroup$
Consider the reference system with the origin in $Z$ in which $DC$ is the real axis, and let $EZ=a$. Then we have $Y=a+2i$ and:
$$e^pi i/6(a+2i) = X $$
so:
$$Im left[left(frac12+fracsqrt32iright)cdotleft(a+2iright)right]=3, $$
or:
$$ 1+fracsqrt32a = 3 $$
so $a = frac4sqrt3$, and by the Pythagorean theorem:
$$ ZY^2 = a^2 + 4 = frac163+4 = frac283 $$
so the side length is $2sqrtfrac73$.
answered Nov 21 '14 at 17:14
Jack D'AurizioJack D'Aurizio
292k33284673
answered Nov 21 '14 at 17:14
Jack D'AurizioJack D'Aurizio
292k33284673
answered Nov 21 '14 at 17:14
Jack D'AurizioJack D'Aurizio
292k33284673
answered Nov 21 '14 at 17:14
Jack D'AurizioJack D'Aurizio
292k33284673
292k33284673
add a comment |
add a comment |
$begingroup$
Hint: Extend $EY$ to meet $AB$ at $F$. Drop a vertical line from $X$, hitting $CD$ at $G$. You now have three right triangles with the hypotenuse being the side of the equilateral triangle.
answered Nov 21 '14 at 17:14
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
So you obtain three equations by pythagora formula and two other equations FY+YE=3 and GZ+ZE=XF. The system can be solved uniquely.
$endgroup$
– Idris
Nov 21 '14 at 17:32
add a comment |
$begingroup$
Hint: Extend $EY$ to meet $AB$ at $F$. Drop a vertical line from $X$, hitting $CD$ at $G$. You now have three right triangles with the hypotenuse being the side of the equilateral triangle.
answered Nov 21 '14 at 17:14
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
So you obtain three equations by pythagora formula and two other equations FY+YE=3 and GZ+ZE=XF. The system can be solved uniquely.
$endgroup$
– Idris
Nov 21 '14 at 17:32
add a comment |
$begingroup$
Hint: Extend $EY$ to meet $AB$ at $F$. Drop a vertical line from $X$, hitting $CD$ at $G$. You now have three right triangles with the hypotenuse being the side of the equilateral triangle.
answered Nov 21 '14 at 17:14
Ross MillikanRoss Millikan
301k24200375
$endgroup$
Hint: Extend $EY$ to meet $AB$ at $F$. Drop a vertical line from $X$, hitting $CD$ at $G$. You now have three right triangles with the hypotenuse being the side of the equilateral triangle.
answered Nov 21 '14 at 17:14
Ross MillikanRoss Millikan
301k24200375
answered Nov 21 '14 at 17:14
Ross MillikanRoss Millikan
301k24200375
answered Nov 21 '14 at 17:14
Ross MillikanRoss Millikan
301k24200375
answered Nov 21 '14 at 17:14
Ross MillikanRoss Millikan
301k24200375
301k24200375
$begingroup$
So you obtain three equations by pythagora formula and two other equations FY+YE=3 and GZ+ZE=XF. The system can be solved uniquely.
$endgroup$
– Idris
Nov 21 '14 at 17:32
add a comment |
$begingroup$
So you obtain three equations by pythagora formula and two other equations FY+YE=3 and GZ+ZE=XF. The system can be solved uniquely.
$endgroup$
– Idris
Nov 21 '14 at 17:32
$begingroup$
So you obtain three equations by pythagora formula and two other equations FY+YE=3 and GZ+ZE=XF. The system can be solved uniquely.
$endgroup$
– Idris
Nov 21 '14 at 17:32
$begingroup$
So you obtain three equations by pythagora formula and two other equations FY+YE=3 and GZ+ZE=XF. The system can be solved uniquely.
$endgroup$
– Idris
Nov 21 '14 at 17:32
add a comment |
$begingroup$
Let $|XY|=|YZ|=|ZX|=a$,
$angle EYZ=theta$, $|FY|=1$.
Then $angle XYF=120^circ-theta$,
beginalign
triangle EYZ:quad
acostheta&=2
tag1label1
,\
triangle FXY:quad
acos(120^circ-theta)&=1
tag2label2
,
endalign
beginalign
acos(120^circ-theta)&=
frac asqrt32,sintheta
-frac a2costheta
\
&=
frac asqrt32,sintheta
-1
,
endalign
so the system eqref1,eqref2
changes to
beginalign
acostheta&=2
tag3label3
,\
a,sintheta
&=frac4sqrt33
tag4label4
,
endalign
which can be easily solved for $a$:
beginalign
a^2cos^2theta+
a^2sin^2theta
&=2^2+left(frac4sqrt33right)^2
,\
a^2&=frac 283
,\
a&=frac 23,sqrt21
.
endalign
answered Mar 31 at 17:43
g.kovg.kov
6,4971819
$endgroup$
add a comment |
$begingroup$
Let $|XY|=|YZ|=|ZX|=a$,
$angle EYZ=theta$, $|FY|=1$.
Then $angle XYF=120^circ-theta$,
beginalign
triangle EYZ:quad
acostheta&=2
tag1label1
,\
triangle FXY:quad
acos(120^circ-theta)&=1
tag2label2
,
endalign
beginalign
acos(120^circ-theta)&=
frac asqrt32,sintheta
-frac a2costheta
\
&=
frac asqrt32,sintheta
-1
,
endalign
so the system eqref1,eqref2
changes to
beginalign
acostheta&=2
tag3label3
,\
a,sintheta
&=frac4sqrt33
tag4label4
,
endalign
which can be easily solved for $a$:
beginalign
a^2cos^2theta+
a^2sin^2theta
&=2^2+left(frac4sqrt33right)^2
,\
a^2&=frac 283
,\
a&=frac 23,sqrt21
.
endalign
answered Mar 31 at 17:43
g.kovg.kov
6,4971819
$endgroup$
add a comment |
$begingroup$
Let $|XY|=|YZ|=|ZX|=a$,
$angle EYZ=theta$, $|FY|=1$.
Then $angle XYF=120^circ-theta$,
beginalign
triangle EYZ:quad
acostheta&=2
tag1label1
,\
triangle FXY:quad
acos(120^circ-theta)&=1
tag2label2
,
endalign
beginalign
acos(120^circ-theta)&=
frac asqrt32,sintheta
-frac a2costheta
\
&=
frac asqrt32,sintheta
-1
,
endalign
so the system eqref1,eqref2
changes to
beginalign
acostheta&=2
tag3label3
,\
a,sintheta
&=frac4sqrt33
tag4label4
,
endalign
which can be easily solved for $a$:
beginalign
a^2cos^2theta+
a^2sin^2theta
&=2^2+left(frac4sqrt33right)^2
,\
a^2&=frac 283
,\
a&=frac 23,sqrt21
.
endalign
answered Mar 31 at 17:43
g.kovg.kov
6,4971819
$endgroup$
Let $|XY|=|YZ|=|ZX|=a$,
$angle EYZ=theta$, $|FY|=1$.
Then $angle XYF=120^circ-theta$,
beginalign
triangle EYZ:quad
acostheta&=2
tag1label1
,\
triangle FXY:quad
acos(120^circ-theta)&=1
tag2label2
,
endalign
beginalign
acos(120^circ-theta)&=
frac asqrt32,sintheta
-frac a2costheta
\
&=
frac asqrt32,sintheta
-1
,
endalign
so the system eqref1,eqref2
changes to
beginalign
acostheta&=2
tag3label3
,\
a,sintheta
&=frac4sqrt33
tag4label4
,
endalign
which can be easily solved for $a$:
beginalign
a^2cos^2theta+
a^2sin^2theta
&=2^2+left(frac4sqrt33right)^2
,\
a^2&=frac 283
,\
a&=frac 23,sqrt21
.
endalign
answered Mar 31 at 17:43
g.kovg.kov
6,4971819
answered Mar 31 at 17:43
g.kovg.kov
6,4971819
answered Mar 31 at 17:43
g.kovg.kov
6,4971819
answered Mar 31 at 17:43
g.kovg.kov
6,4971819
6,4971819
add a comment |
add a comment |
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$begingroup$
@ProgramFOX: Good edit.
$endgroup$
– André Nicolas
Nov 21 '14 at 17:07