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 yesterday
user21820
39.7k544158
asked yesterday
Eldar RahimliEldar Rahimli
41810
$endgroup$
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 yesterday
user21820
39.7k544158
asked yesterday
Eldar RahimliEldar Rahimli
41810
$endgroup$
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 yesterday
user21820
39.7k544158
asked yesterday
Eldar RahimliEldar Rahimli
41810
$endgroup$
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 yesterday
user21820
39.7k544158
asked yesterday
Eldar RahimliEldar Rahimli
41810
edited yesterday
user21820
39.7k544158
asked yesterday
Eldar RahimliEldar Rahimli
41810
edited yesterday
user21820
39.7k544158
edited yesterday
user21820
39.7k544158
edited yesterday
user21820
39.7k544158
39.7k544158
asked yesterday
Eldar RahimliEldar Rahimli
41810
asked yesterday
Eldar RahimliEldar Rahimli
41810
asked yesterday
Eldar RahimliEldar Rahimli
41810
41810
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 yesterday
richrowrichrow
30319
$endgroup$
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 yesterday
Michael RozenbergMichael Rozenberg
109k1896200
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3160269%2fwhat-is-the-length-of-x-in-this-pentagon-diagram%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
$$
frac{FE}{FH}=frac{FA}{FE}.
$$
It means that $x^2=FE^2=FAcdot FH=4cdot 1=4$. Thus, $x=2$.
answered yesterday
richrowrichrow
30319
$endgroup$
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 yesterday
richrowrichrow
30319
$endgroup$
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 yesterday
richrowrichrow
30319
$endgroup$
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 yesterday
richrowrichrow
30319
answered yesterday
richrowrichrow
30319
answered yesterday
richrowrichrow
30319
answered yesterday
richrowrichrow
30319
30319
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 yesterday
Michael RozenbergMichael Rozenberg
109k1896200
$endgroup$
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 yesterday
Michael RozenbergMichael Rozenberg
109k1896200
$endgroup$
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 yesterday
Michael RozenbergMichael Rozenberg
109k1896200
$endgroup$
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 yesterday
Michael RozenbergMichael Rozenberg
109k1896200
answered yesterday
Michael RozenbergMichael Rozenberg
109k1896200
answered yesterday
Michael RozenbergMichael Rozenberg
109k1896200
answered yesterday
Michael RozenbergMichael Rozenberg
109k1896200
109k1896200
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3160269%2fwhat-is-the-length-of-x-in-this-pentagon-diagram%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
