Semi-circle folded into a cone with a circular base
$begingroup$
From my 7th-grade math book:
The semicircle shown is folded to form a right circular cone so that the arc PQ becomes the circumference of the base. Find the diameter of the base,
Let $text{circumference of cone base}=C$ and $text{diameter}=d$.
I think the diameter should be $frac{2C}{pi}=frac{2 cdot 5cm}{pi}approx 3.183cm$. But my book's answer key says $d=2.5cm$, which seems to be half the circumference ($frac{5cm}{2}$). But I don't get the reasoning behind it. The semicircle's diameter is the cone base's circumference. And if that's the circumference ($frac{pi d}{2}$), then the diameter should be $2C/pi$, shouldn't it?
Why? I'm very confused. I feel the amount of information provided is insufficient somehow.
The question is unique and original. This question seems quite similar, with an answer which is appropriate in my case too. But the problem is the answer only states (facts I've already deduced), but doesn't explain.
solid-geometry
asked Jul 27 '17 at 3:43
Soha Farhin PineSoha Farhin Pine
369319
$endgroup$
add a comment |
$begingroup$
From my 7th-grade math book:
The semicircle shown is folded to form a right circular cone so that the arc PQ becomes the circumference of the base. Find the diameter of the base,
Let $text{circumference of cone base}=C$ and $text{diameter}=d$.
I think the diameter should be $frac{2C}{pi}=frac{2 cdot 5cm}{pi}approx 3.183cm$. But my book's answer key says $d=2.5cm$, which seems to be half the circumference ($frac{5cm}{2}$). But I don't get the reasoning behind it. The semicircle's diameter is the cone base's circumference. And if that's the circumference ($frac{pi d}{2}$), then the diameter should be $2C/pi$, shouldn't it?
Why? I'm very confused. I feel the amount of information provided is insufficient somehow.
The question is unique and original. This question seems quite similar, with an answer which is appropriate in my case too. But the problem is the answer only states (facts I've already deduced), but doesn't explain.
solid-geometry
asked Jul 27 '17 at 3:43
Soha Farhin PineSoha Farhin Pine
369319
$endgroup$
$begingroup$
You forgot to divide by 2.
$endgroup$
– 伽罗瓦
Jul 27 '17 at 3:47
$begingroup$
The circumference of the circle the arc forms is not 5. Right approach but not the correct circumference.
$endgroup$
– 伽罗瓦
Jul 27 '17 at 4:01
$begingroup$
@ntntnt Okay. Edited the question.
$endgroup$
– Soha Farhin Pine
Jul 27 '17 at 4:06
$begingroup$
Circumference of the circle the arc forms is the length of the arc...which is $5pi/2.$ I think you can calculate the diameter from there. Remember the circumference is $2pi r.$
$endgroup$
– 伽罗瓦
Jul 27 '17 at 4:15
add a comment |
$begingroup$
From my 7th-grade math book:
The semicircle shown is folded to form a right circular cone so that the arc PQ becomes the circumference of the base. Find the diameter of the base,
Let $text{circumference of cone base}=C$ and $text{diameter}=d$.
I think the diameter should be $frac{2C}{pi}=frac{2 cdot 5cm}{pi}approx 3.183cm$. But my book's answer key says $d=2.5cm$, which seems to be half the circumference ($frac{5cm}{2}$). But I don't get the reasoning behind it. The semicircle's diameter is the cone base's circumference. And if that's the circumference ($frac{pi d}{2}$), then the diameter should be $2C/pi$, shouldn't it?
Why? I'm very confused. I feel the amount of information provided is insufficient somehow.
The question is unique and original. This question seems quite similar, with an answer which is appropriate in my case too. But the problem is the answer only states (facts I've already deduced), but doesn't explain.
solid-geometry
asked Jul 27 '17 at 3:43
Soha Farhin PineSoha Farhin Pine
369319
$endgroup$
From my 7th-grade math book:
The semicircle shown is folded to form a right circular cone so that the arc PQ becomes the circumference of the base. Find the diameter of the base,
Let $text{circumference of cone base}=C$ and $text{diameter}=d$.
I think the diameter should be $frac{2C}{pi}=frac{2 cdot 5cm}{pi}approx 3.183cm$. But my book's answer key says $d=2.5cm$, which seems to be half the circumference ($frac{5cm}{2}$). But I don't get the reasoning behind it. The semicircle's diameter is the cone base's circumference. And if that's the circumference ($frac{pi d}{2}$), then the diameter should be $2C/pi$, shouldn't it?
Why? I'm very confused. I feel the amount of information provided is insufficient somehow.
The question is unique and original. This question seems quite similar, with an answer which is appropriate in my case too. But the problem is the answer only states (facts I've already deduced), but doesn't explain.
solid-geometry
solid-geometry
asked Jul 27 '17 at 3:43
Soha Farhin PineSoha Farhin Pine
369319
asked Jul 27 '17 at 3:43
Soha Farhin PineSoha Farhin Pine
369319
edited Jul 27 '17 at 4:05
Soha Farhin Pine
asked Jul 27 '17 at 3:43
Soha Farhin PineSoha Farhin Pine
369319
asked Jul 27 '17 at 3:43
Soha Farhin PineSoha Farhin Pine
369319
asked Jul 27 '17 at 3:43
Soha Farhin PineSoha Farhin Pine
369319
369319
$begingroup$
You forgot to divide by 2.
$endgroup$
– 伽罗瓦
Jul 27 '17 at 3:47
$begingroup$
The circumference of the circle the arc forms is not 5. Right approach but not the correct circumference.
$endgroup$
– 伽罗瓦
Jul 27 '17 at 4:01
$begingroup$
@ntntnt Okay. Edited the question.
$endgroup$
– Soha Farhin Pine
Jul 27 '17 at 4:06
$begingroup$
Circumference of the circle the arc forms is the length of the arc...which is $5pi/2.$ I think you can calculate the diameter from there. Remember the circumference is $2pi r.$
$endgroup$
– 伽罗瓦
Jul 27 '17 at 4:15
add a comment |
$begingroup$
You forgot to divide by 2.
$endgroup$
– 伽罗瓦
Jul 27 '17 at 3:47
$begingroup$
The circumference of the circle the arc forms is not 5. Right approach but not the correct circumference.
$endgroup$
– 伽罗瓦
Jul 27 '17 at 4:01
$begingroup$
@ntntnt Okay. Edited the question.
$endgroup$
– Soha Farhin Pine
Jul 27 '17 at 4:06
$begingroup$
Circumference of the circle the arc forms is the length of the arc...which is $5pi/2.$ I think you can calculate the diameter from there. Remember the circumference is $2pi r.$
$endgroup$
– 伽罗瓦
Jul 27 '17 at 4:15
$begingroup$
You forgot to divide by 2.
$endgroup$
– 伽罗瓦
Jul 27 '17 at 3:47
$begingroup$
You forgot to divide by 2.
$endgroup$
– 伽罗瓦
Jul 27 '17 at 3:47
$begingroup$
The circumference of the circle the arc forms is not 5. Right approach but not the correct circumference.
$endgroup$
– 伽罗瓦
Jul 27 '17 at 4:01
$begingroup$
The circumference of the circle the arc forms is not 5. Right approach but not the correct circumference.
$endgroup$
– 伽罗瓦
Jul 27 '17 at 4:01
$begingroup$
@ntntnt Okay. Edited the question.
$endgroup$
– Soha Farhin Pine
Jul 27 '17 at 4:06
$begingroup$
@ntntnt Okay. Edited the question.
$endgroup$
– Soha Farhin Pine
Jul 27 '17 at 4:06
$begingroup$
Circumference of the circle the arc forms is the length of the arc...which is $5pi/2.$ I think you can calculate the diameter from there. Remember the circumference is $2pi r.$
$endgroup$
– 伽罗瓦
Jul 27 '17 at 4:15
$begingroup$
Circumference of the circle the arc forms is the length of the arc...which is $5pi/2.$ I think you can calculate the diameter from there. Remember the circumference is $2pi r.$
$endgroup$
– 伽罗瓦
Jul 27 '17 at 4:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
So for the first part, you must find the circumference of the semicircle that is pictured (not including the straight side), which is $$c=0.5dpi=2.5pi$$
When the cone is formed, this circumference of the semicircle becomes the new circumference of the base of the cone, which is $$c=Dpi=2.5pi$$ It can be implied that the diameter must be 2.5 centimeters.
answered Aug 17 '17 at 13:56
Griffin ModjeskiGriffin Modjeski
262
$endgroup$
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%2f2373128%2fsemi-circle-folded-into-a-cone-with-a-circular-base%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
So for the first part, you must find the circumference of the semicircle that is pictured (not including the straight side), which is $$c=0.5dpi=2.5pi$$
When the cone is formed, this circumference of the semicircle becomes the new circumference of the base of the cone, which is $$c=Dpi=2.5pi$$ It can be implied that the diameter must be 2.5 centimeters.
answered Aug 17 '17 at 13:56
Griffin ModjeskiGriffin Modjeski
262
$endgroup$
add a comment |
$begingroup$
So for the first part, you must find the circumference of the semicircle that is pictured (not including the straight side), which is $$c=0.5dpi=2.5pi$$
When the cone is formed, this circumference of the semicircle becomes the new circumference of the base of the cone, which is $$c=Dpi=2.5pi$$ It can be implied that the diameter must be 2.5 centimeters.
answered Aug 17 '17 at 13:56
Griffin ModjeskiGriffin Modjeski
262
$endgroup$
add a comment |
$begingroup$
So for the first part, you must find the circumference of the semicircle that is pictured (not including the straight side), which is $$c=0.5dpi=2.5pi$$
When the cone is formed, this circumference of the semicircle becomes the new circumference of the base of the cone, which is $$c=Dpi=2.5pi$$ It can be implied that the diameter must be 2.5 centimeters.
answered Aug 17 '17 at 13:56
Griffin ModjeskiGriffin Modjeski
262
$endgroup$
So for the first part, you must find the circumference of the semicircle that is pictured (not including the straight side), which is $$c=0.5dpi=2.5pi$$
When the cone is formed, this circumference of the semicircle becomes the new circumference of the base of the cone, which is $$c=Dpi=2.5pi$$ It can be implied that the diameter must be 2.5 centimeters.
answered Aug 17 '17 at 13:56
Griffin ModjeskiGriffin Modjeski
262
edited Aug 17 '17 at 14:02
answered Aug 17 '17 at 13:56
Griffin ModjeskiGriffin Modjeski
262
answered Aug 17 '17 at 13:56
Griffin ModjeskiGriffin Modjeski
262
answered Aug 17 '17 at 13:56
Griffin ModjeskiGriffin Modjeski
262
262
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%2f2373128%2fsemi-circle-folded-into-a-cone-with-a-circular-base%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
$begingroup$
You forgot to divide by 2.
$endgroup$
– 伽罗瓦
Jul 27 '17 at 3:47
$begingroup$
The circumference of the circle the arc forms is not 5. Right approach but not the correct circumference.
$endgroup$
– 伽罗瓦
Jul 27 '17 at 4:01
$begingroup$
@ntntnt Okay. Edited the question.
$endgroup$
– Soha Farhin Pine
Jul 27 '17 at 4:06
$begingroup$
Circumference of the circle the arc forms is the length of the arc...which is $5pi/2.$ I think you can calculate the diameter from there. Remember the circumference is $2pi r.$
$endgroup$
– 伽罗瓦
Jul 27 '17 at 4:15