What is the length of $x$ in this pentagon diagram?
$begingroup$
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
40k544160
asked Mar 24 at 8:11
Eldar RahimliEldar Rahimli
42510
$endgroup$
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
40k544160
asked Mar 24 at 8:11
Eldar RahimliEldar Rahimli
42510
$endgroup$
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
40k544160
asked Mar 24 at 8:11
Eldar RahimliEldar Rahimli
42510
$endgroup$
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
40k544160
asked Mar 24 at 8:11
Eldar RahimliEldar Rahimli
42510
edited Mar 24 at 11:14
user21820
40k544160
asked Mar 24 at 8:11
Eldar RahimliEldar Rahimli
42510
edited Mar 24 at 11:14
user21820
40k544160
edited Mar 24 at 11:14
user21820
40k544160
edited Mar 24 at 11:14
user21820
40k544160
40k544160
asked Mar 24 at 8:11
Eldar RahimliEldar Rahimli
42510
asked Mar 24 at 8:11
Eldar RahimliEldar Rahimli
42510
asked Mar 24 at 8:11
Eldar RahimliEldar Rahimli
42510
42510
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
$$
frac{FE}{FH}=frac{FA}{FE}.
$$
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:
$$frac{x}{sin{alpha}}=frac{4}{sin108^{circ}}$$ and
$$frac{x}{sin108^{circ}}=frac{1}{sinalpha},$$ 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
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active
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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
$$
frac{FE}{FH}=frac{FA}{FE}.
$$
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
$$
frac{FE}{FH}=frac{FA}{FE}.
$$
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
$$
frac{FE}{FH}=frac{FA}{FE}.
$$
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
$$
frac{FE}{FH}=frac{FA}{FE}.
$$
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:
$$frac{x}{sin{alpha}}=frac{4}{sin108^{circ}}$$ and
$$frac{x}{sin108^{circ}}=frac{1}{sinalpha},$$ 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:
$$frac{x}{sin{alpha}}=frac{4}{sin108^{circ}}$$ and
$$frac{x}{sin108^{circ}}=frac{1}{sinalpha},$$ 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:
$$frac{x}{sin{alpha}}=frac{4}{sin108^{circ}}$$ and
$$frac{x}{sin108^{circ}}=frac{1}{sinalpha},$$ 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:
$$frac{x}{sin{alpha}}=frac{4}{sin108^{circ}}$$ and
$$frac{x}{sin108^{circ}}=frac{1}{sinalpha},$$ 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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