Isosceles Trapezoid, given sides, find the circumcircle area [closed]
$begingroup$
Isosceles trapezoid $ABCD$ has side lengths $AB = 10$ cm, and $BC = CD = DA
= 6$ cm. A circle passes through points $A$, $B$, $C$, and $D$. What is the area of this circle? Express your answer in terms of $pi$.
geometry euclidean-geometry
edited Jan 5 at 19:50
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weareallinweareallin
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closed as off-topic by Lord Shark the Unknown, Abcd, Ali Caglayan, Davide Giraudo, José Carlos Santos Jan 5 at 19:51
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$begingroup$
Isosceles trapezoid $ABCD$ has side lengths $AB = 10$ cm, and $BC = CD = DA
= 6$ cm. A circle passes through points $A$, $B$, $C$, and $D$. What is the area of this circle? Express your answer in terms of $pi$.
geometry euclidean-geometry
edited Jan 5 at 19:50
BPP
2,149927
asked Jan 5 at 18:07
weareallinweareallin
51
$endgroup$
closed as off-topic by Lord Shark the Unknown, Abcd, Ali Caglayan, Davide Giraudo, José Carlos Santos Jan 5 at 19:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lord Shark the Unknown, Abcd, Ali Caglayan, Davide Giraudo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Isosceles trapezoid $ABCD$ has side lengths $AB = 10$ cm, and $BC = CD = DA
= 6$ cm. A circle passes through points $A$, $B$, $C$, and $D$. What is the area of this circle? Express your answer in terms of $pi$.
geometry euclidean-geometry
edited Jan 5 at 19:50
BPP
2,149927
asked Jan 5 at 18:07
weareallinweareallin
51
$endgroup$
Isosceles trapezoid $ABCD$ has side lengths $AB = 10$ cm, and $BC = CD = DA
= 6$ cm. A circle passes through points $A$, $B$, $C$, and $D$. What is the area of this circle? Express your answer in terms of $pi$.
geometry euclidean-geometry
geometry euclidean-geometry
edited Jan 5 at 19:50
BPP
2,149927
asked Jan 5 at 18:07
weareallinweareallin
51
edited Jan 5 at 19:50
BPP
2,149927
asked Jan 5 at 18:07
weareallinweareallin
51
edited Jan 5 at 19:50
BPP
2,149927
edited Jan 5 at 19:50
BPP
2,149927
edited Jan 5 at 19:50
BPP
2,149927
2,149927
asked Jan 5 at 18:07
weareallinweareallin
51
asked Jan 5 at 18:07
weareallinweareallin
51
asked Jan 5 at 18:07
weareallinweareallin
51
51
closed as off-topic by Lord Shark the Unknown, Abcd, Ali Caglayan, Davide Giraudo, José Carlos Santos Jan 5 at 19:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lord Shark the Unknown, Abcd, Ali Caglayan, Davide Giraudo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Lord Shark the Unknown, Abcd, Ali Caglayan, Davide Giraudo, José Carlos Santos Jan 5 at 19:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lord Shark the Unknown, Abcd, Ali Caglayan, Davide Giraudo, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
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2 Answers
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Let $M$ be the middle point of our circle.The three triangles $$ADM,MDC,BMC$$ are congruent. And let $$angle AMD=alpha$$ so we get by the theorem of cosines we get
$$36=2R^2(1-cos(alpha))$$
$$100=2R^2(1-cos(2pi-3alpha))$$
Now we use that $$cos(2pi-3alpha)=4cos^3(alpha)-3cos(alpha)$$
Now you will get one equation for $$alpha$$ and you can compute $$R$$
answered Jan 5 at 19:03
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
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Let $H$ be the orthogonal projection of $C$ onto $AB$. By symmetry $BH=2$ so $CH=sqrt{BC^2-BH^2}=sqrt{36-4}=4sqrt{2}$ and $AC=sqrt{AH^2+CH^2}=4sqrt{6}$. The radius of the circumcircle of $ABC$ is given by:
$$R=dfrac{abc}{sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}=3sqrt{3}$$
by WA.
answered Jan 5 at 19:47
BPPBPP
2,149927
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $M$ be the middle point of our circle.The three triangles $$ADM,MDC,BMC$$ are congruent. And let $$angle AMD=alpha$$ so we get by the theorem of cosines we get
$$36=2R^2(1-cos(alpha))$$
$$100=2R^2(1-cos(2pi-3alpha))$$
Now we use that $$cos(2pi-3alpha)=4cos^3(alpha)-3cos(alpha)$$
Now you will get one equation for $$alpha$$ and you can compute $$R$$
answered Jan 5 at 19:03
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
add a comment |
$begingroup$
Let $M$ be the middle point of our circle.The three triangles $$ADM,MDC,BMC$$ are congruent. And let $$angle AMD=alpha$$ so we get by the theorem of cosines we get
$$36=2R^2(1-cos(alpha))$$
$$100=2R^2(1-cos(2pi-3alpha))$$
Now we use that $$cos(2pi-3alpha)=4cos^3(alpha)-3cos(alpha)$$
Now you will get one equation for $$alpha$$ and you can compute $$R$$
answered Jan 5 at 19:03
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
add a comment |
$begingroup$
Let $M$ be the middle point of our circle.The three triangles $$ADM,MDC,BMC$$ are congruent. And let $$angle AMD=alpha$$ so we get by the theorem of cosines we get
$$36=2R^2(1-cos(alpha))$$
$$100=2R^2(1-cos(2pi-3alpha))$$
Now we use that $$cos(2pi-3alpha)=4cos^3(alpha)-3cos(alpha)$$
Now you will get one equation for $$alpha$$ and you can compute $$R$$
answered Jan 5 at 19:03
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
Let $M$ be the middle point of our circle.The three triangles $$ADM,MDC,BMC$$ are congruent. And let $$angle AMD=alpha$$ so we get by the theorem of cosines we get
$$36=2R^2(1-cos(alpha))$$
$$100=2R^2(1-cos(2pi-3alpha))$$
Now we use that $$cos(2pi-3alpha)=4cos^3(alpha)-3cos(alpha)$$
Now you will get one equation for $$alpha$$ and you can compute $$R$$
answered Jan 5 at 19:03
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Jan 5 at 19:03
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Jan 5 at 19:03
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Jan 5 at 19:03
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
73.6k42864
add a comment |
add a comment |
$begingroup$
Let $H$ be the orthogonal projection of $C$ onto $AB$. By symmetry $BH=2$ so $CH=sqrt{BC^2-BH^2}=sqrt{36-4}=4sqrt{2}$ and $AC=sqrt{AH^2+CH^2}=4sqrt{6}$. The radius of the circumcircle of $ABC$ is given by:
$$R=dfrac{abc}{sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}=3sqrt{3}$$
by WA.
answered Jan 5 at 19:47
BPPBPP
2,149927
$endgroup$
add a comment |
$begingroup$
Let $H$ be the orthogonal projection of $C$ onto $AB$. By symmetry $BH=2$ so $CH=sqrt{BC^2-BH^2}=sqrt{36-4}=4sqrt{2}$ and $AC=sqrt{AH^2+CH^2}=4sqrt{6}$. The radius of the circumcircle of $ABC$ is given by:
$$R=dfrac{abc}{sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}=3sqrt{3}$$
by WA.
answered Jan 5 at 19:47
BPPBPP
2,149927
$endgroup$
add a comment |
$begingroup$
Let $H$ be the orthogonal projection of $C$ onto $AB$. By symmetry $BH=2$ so $CH=sqrt{BC^2-BH^2}=sqrt{36-4}=4sqrt{2}$ and $AC=sqrt{AH^2+CH^2}=4sqrt{6}$. The radius of the circumcircle of $ABC$ is given by:
$$R=dfrac{abc}{sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}=3sqrt{3}$$
by WA.
answered Jan 5 at 19:47
BPPBPP
2,149927
$endgroup$
Let $H$ be the orthogonal projection of $C$ onto $AB$. By symmetry $BH=2$ so $CH=sqrt{BC^2-BH^2}=sqrt{36-4}=4sqrt{2}$ and $AC=sqrt{AH^2+CH^2}=4sqrt{6}$. The radius of the circumcircle of $ABC$ is given by:
$$R=dfrac{abc}{sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}=3sqrt{3}$$
by WA.
answered Jan 5 at 19:47
BPPBPP
2,149927
answered Jan 5 at 19:47
BPPBPP
2,149927
answered Jan 5 at 19:47
BPPBPP
2,149927
answered Jan 5 at 19:47
BPPBPP
2,149927
2,149927
add a comment |
add a comment |
