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What is the length of $x$ in this pentagon diagram?
The Next CEO of Stack OverflowHow to find the interior angle of an irregular pentagon or polygon?Find ratio of areas of triangle to pentagon?polygon angle namesProbability of Choosing Supplementary AnglesIn the figure: prove that $overline AM perp overline DE$ if $M$ is the middle point on $overline BC$.Find an angle in the figure defined by a equilateral triangle and a regular pentagonShow that if a convex pentagon is equilateral, and two of its adjacent inner angles measure 108 degree, then the pentagon is equiangular.Regular Pentagon sides in terms of interior pentagon and segments connecting verticesArea Of PentagonHow to prove that infinite number of pentagons exist satisfying the given requirements
$begingroup$
ABCDE is a regular pentagon. $angle AFD = angle EKC$
$|FH|=1$ cm; $|AH|=3$ cm
What is $|DK|?$
I know that triangles $EFA$ and $DEK$ are similar and that $|EK|=4$ cm. Also because this is a regular pentagon each one of the interior angles are $108^o$. Naming similar angles inside the pentagon, I tried to find an isosceles triangle, but I couldn't. I can't progress any further from here.
How can I solve this problem?
geometry euclidean-geometry polygons
edited Mar 24 at 11:14
user21820
39.9k544158
asked Mar 24 at 8:11
Eldar RahimliEldar Rahimli
42010
$endgroup$
add a comment |
$begingroup$
ABCDE is a regular pentagon. $angle AFD = angle EKC$
$|FH|=1$ cm; $|AH|=3$ cm
What is $|DK|?$
I know that triangles $EFA$ and $DEK$ are similar and that $|EK|=4$ cm. Also because this is a regular pentagon each one of the interior angles are $108^o$. Naming similar angles inside the pentagon, I tried to find an isosceles triangle, but I couldn't. I can't progress any further from here.
How can I solve this problem?
geometry euclidean-geometry polygons
edited Mar 24 at 11:14
user21820
39.9k544158
asked Mar 24 at 8:11
Eldar RahimliEldar Rahimli
42010
$endgroup$
add a comment |
$begingroup$
ABCDE is a regular pentagon. $angle AFD = angle EKC$
$|FH|=1$ cm; $|AH|=3$ cm
What is $|DK|?$
I know that triangles $EFA$ and $DEK$ are similar and that $|EK|=4$ cm. Also because this is a regular pentagon each one of the interior angles are $108^o$. Naming similar angles inside the pentagon, I tried to find an isosceles triangle, but I couldn't. I can't progress any further from here.
How can I solve this problem?
geometry euclidean-geometry polygons
edited Mar 24 at 11:14
user21820
39.9k544158
asked Mar 24 at 8:11
Eldar RahimliEldar Rahimli
42010
$endgroup$
ABCDE is a regular pentagon. $angle AFD = angle EKC$
$|FH|=1$ cm; $|AH|=3$ cm
What is $|DK|?$
I know that triangles $EFA$ and $DEK$ are similar and that $|EK|=4$ cm. Also because this is a regular pentagon each one of the interior angles are $108^o$. Naming similar angles inside the pentagon, I tried to find an isosceles triangle, but I couldn't. I can't progress any further from here.
How can I solve this problem?
geometry euclidean-geometry polygons
geometry euclidean-geometry polygons
edited Mar 24 at 11:14
user21820
39.9k544158
asked Mar 24 at 8:11
Eldar RahimliEldar Rahimli
42010
edited Mar 24 at 11:14
user21820
39.9k544158
asked Mar 24 at 8:11
Eldar RahimliEldar Rahimli
42010
edited Mar 24 at 11:14
user21820
39.9k544158
edited Mar 24 at 11:14
user21820
39.9k544158
edited Mar 24 at 11:14
user21820
39.9k544158
39.9k544158
asked Mar 24 at 8:11
Eldar RahimliEldar Rahimli
42010
asked Mar 24 at 8:11
Eldar RahimliEldar Rahimli
42010
asked Mar 24 at 8:11
Eldar RahimliEldar Rahimli
42010
42010
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Answer: $x=2$.
Since $angle EFA=angle DKE$, $angle AEF=angle EDK$ and $AE=ED$ we obtain $triangle AEF=triangle EDK$ (they are similar and corresponding sides are equal). Hence, $EF=DK=x$ and $angle FEH=angle DEK=angle EAF$. Therefore, trinagles $triangle FEH$ and $triangle FAE$ are similar, so
$$
fracFEFH=fracFAFE.
$$
It means that $x^2=FE^2=FAcdot FH=4cdot 1=4$. Thus, $x=2$.
answered Mar 24 at 8:23
richrowrichrow
38819
$endgroup$
add a comment |
$begingroup$
Let $measuredangle FEH=measuredangle EAF=alpha.$
Thus, by your work and by law of sines we obtain:
$$fracxsinalpha=frac4sin108^circ$$ and
$$fracxsin108^circ=frac1sinalpha,$$ which gives $$x^2=4$$ and $$x=2.$$
answered Mar 24 at 9:02
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Answer: $x=2$.
Since $angle EFA=angle DKE$, $angle AEF=angle EDK$ and $AE=ED$ we obtain $triangle AEF=triangle EDK$ (they are similar and corresponding sides are equal). Hence, $EF=DK=x$ and $angle FEH=angle DEK=angle EAF$. Therefore, trinagles $triangle FEH$ and $triangle FAE$ are similar, so
$$
fracFEFH=fracFAFE.
$$
It means that $x^2=FE^2=FAcdot FH=4cdot 1=4$. Thus, $x=2$.
answered Mar 24 at 8:23
richrowrichrow
38819
$endgroup$
add a comment |
$begingroup$
Answer: $x=2$.
Since $angle EFA=angle DKE$, $angle AEF=angle EDK$ and $AE=ED$ we obtain $triangle AEF=triangle EDK$ (they are similar and corresponding sides are equal). Hence, $EF=DK=x$ and $angle FEH=angle DEK=angle EAF$. Therefore, trinagles $triangle FEH$ and $triangle FAE$ are similar, so
$$
fracFEFH=fracFAFE.
$$
It means that $x^2=FE^2=FAcdot FH=4cdot 1=4$. Thus, $x=2$.
answered Mar 24 at 8:23
richrowrichrow
38819
$endgroup$
add a comment |
$begingroup$
Answer: $x=2$.
Since $angle EFA=angle DKE$, $angle AEF=angle EDK$ and $AE=ED$ we obtain $triangle AEF=triangle EDK$ (they are similar and corresponding sides are equal). Hence, $EF=DK=x$ and $angle FEH=angle DEK=angle EAF$. Therefore, trinagles $triangle FEH$ and $triangle FAE$ are similar, so
$$
fracFEFH=fracFAFE.
$$
It means that $x^2=FE^2=FAcdot FH=4cdot 1=4$. Thus, $x=2$.
answered Mar 24 at 8:23
richrowrichrow
38819
$endgroup$
Answer: $x=2$.
Since $angle EFA=angle DKE$, $angle AEF=angle EDK$ and $AE=ED$ we obtain $triangle AEF=triangle EDK$ (they are similar and corresponding sides are equal). Hence, $EF=DK=x$ and $angle FEH=angle DEK=angle EAF$. Therefore, trinagles $triangle FEH$ and $triangle FAE$ are similar, so
$$
fracFEFH=fracFAFE.
$$
It means that $x^2=FE^2=FAcdot FH=4cdot 1=4$. Thus, $x=2$.
answered Mar 24 at 8:23
richrowrichrow
38819
answered Mar 24 at 8:23
richrowrichrow
38819
answered Mar 24 at 8:23
richrowrichrow
38819
answered Mar 24 at 8:23
richrowrichrow
38819
38819
add a comment |
add a comment |
$begingroup$
Let $measuredangle FEH=measuredangle EAF=alpha.$
Thus, by your work and by law of sines we obtain:
$$fracxsinalpha=frac4sin108^circ$$ and
$$fracxsin108^circ=frac1sinalpha,$$ which gives $$x^2=4$$ and $$x=2.$$
answered Mar 24 at 9:02
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
Let $measuredangle FEH=measuredangle EAF=alpha.$
Thus, by your work and by law of sines we obtain:
$$fracxsinalpha=frac4sin108^circ$$ and
$$fracxsin108^circ=frac1sinalpha,$$ which gives $$x^2=4$$ and $$x=2.$$
answered Mar 24 at 9:02
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
Let $measuredangle FEH=measuredangle EAF=alpha.$
Thus, by your work and by law of sines we obtain:
$$fracxsinalpha=frac4sin108^circ$$ and
$$fracxsin108^circ=frac1sinalpha,$$ which gives $$x^2=4$$ and $$x=2.$$
answered Mar 24 at 9:02
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
Let $measuredangle FEH=measuredangle EAF=alpha.$
Thus, by your work and by law of sines we obtain:
$$fracxsinalpha=frac4sin108^circ$$ and
$$fracxsin108^circ=frac1sinalpha,$$ which gives $$x^2=4$$ and $$x=2.$$
answered Mar 24 at 9:02
Michael RozenbergMichael Rozenberg
109k1896201
answered Mar 24 at 9:02
Michael RozenbergMichael Rozenberg
109k1896201
answered Mar 24 at 9:02
Michael RozenbergMichael Rozenberg
109k1896201
answered Mar 24 at 9:02
Michael RozenbergMichael Rozenberg
109k1896201
109k1896201
add a comment |
add a comment |
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