Isosceles Trapezoid, given sides, find the circumcircle area [on hold]
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
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BPP
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weareallinweareallin
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put on hold as off-topic by Lord Shark the Unknown, Abcd, Ali Caglayan, Davide Giraudo, José Carlos Santos 2 days ago
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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 2 days ago
BPP
2,149927
asked 2 days ago
weareallinweareallin
51
put on hold as off-topic by Lord Shark the Unknown, Abcd, Ali Caglayan, Davide Giraudo, José Carlos Santos 2 days ago
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 |
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 2 days ago
BPP
2,149927
asked 2 days ago
weareallinweareallin
51
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 2 days ago
BPP
2,149927
asked 2 days ago
weareallinweareallin
51
edited 2 days ago
BPP
2,149927
asked 2 days ago
weareallinweareallin
51
edited 2 days ago
BPP
2,149927
edited 2 days ago
BPP
2,149927
edited 2 days ago
BPP
2,149927
2,149927
asked 2 days ago
weareallinweareallin
51
asked 2 days ago
weareallinweareallin
51
asked 2 days ago
weareallinweareallin
51
51
put on hold as off-topic by Lord Shark the Unknown, Abcd, Ali Caglayan, Davide Giraudo, José Carlos Santos 2 days ago
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.
put on hold as off-topic by Lord Shark the Unknown, Abcd, Ali Caglayan, Davide Giraudo, José Carlos Santos 2 days ago
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
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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 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
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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 2 days ago
BPPBPP
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
add a comment |
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 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
add a comment |
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 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
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 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
answered 2 days ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.5k42865
73.5k42865
add a comment |
add a comment |
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 2 days ago
BPPBPP
2,149927
add a comment |
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 2 days ago
BPPBPP
2,149927
add a comment |
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 2 days ago
BPPBPP
2,149927
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 2 days ago
BPPBPP
2,149927
answered 2 days ago
BPPBPP
2,149927
answered 2 days ago
BPPBPP
2,149927
answered 2 days ago
BPPBPP
2,149927
2,149927
add a comment |
add a comment |
