Question
I have a question my brother asked me but I need a math expert.
You have two interlocking circles and the radius of circle B goes through the center of circle A and of course the radius of circle A goes through the center of circle B. The radius of each circle is 30 feet. Now, if you draw a line C from the top center of circle A across to the top center of circle B, the line would be 60 feet long and would leave an area below it created by the line C and part of an arc of circle A and part of an arc of circle B. Please give me the area of that space in square feet or the formula for how it's worked. Thanks.
You have two interlocking circles and the radius of circle B goes through the center of circle A and of course the radius of circle A goes through the center of circle B. The radius of each circle is 30 feet. Now, if you draw a line C from the top center of circle A across to the top center of circle B, the line would be 60 feet long and would leave an area below it created by the line C and part of an arc of circle A and part of an arc of circle B. Please give me the area of that space in square feet or the formula for how it's worked. Thanks.
Answers
We will describe the problem as follows.
Two circles A,B of equal radii (r=30') are centred at P, Q, distance r apart.
Points R and S on circles A, B are such that RS form a common tangent to both circles. Hence PRSQ form a square of side r.
Arcs are drawn with centre P and Q, radius r, which intersect at point D inside the square.
Hence Δ PDQ is an equilateral triangle of side r.
The required area bounded by the side RS , arcs RD and DS will be equal to
Area of square PRSQ - Area of ΔADQ - area of sector RPD - Area of sector SQD.
Note that sectors RPD and SQD have centrai angles of (90-60)=30°.
Two circles A,B of equal radii (r=30') are centred at P, Q, distance r apart.
Points R and S on circles A, B are such that RS form a common tangent to both circles. Hence PRSQ form a square of side r.
Arcs are drawn with centre P and Q, radius r, which intersect at point D inside the square.
Hence Δ PDQ is an equilateral triangle of side r.
The required area bounded by the side RS , arcs RD and DS will be equal to
Area of square PRSQ - Area of ΔADQ - area of sector RPD - Area of sector SQD.
Note that sectors RPD and SQD have centrai angles of (90-60)=30°.
Typo corrections:
Area of square PRSQ - Area of ΔPDQ - area of sector RPD - Area of sector SQD.
Note that sectors RPD and SQD have central angles of (90-60)=30°.
Area of square PRSQ - Area of ΔPDQ - area of sector RPD - Area of sector SQD.
Note that sectors RPD and SQD have central angles of (90-60)=30°.
Many thanks.
You're welcome!
Related Questions
You are left in charge of taking care of your baby brother. Your mother asked you to prepare the mil...
Fix errors
"Who is your brother" asked Nethstel.
"Jonren he is my brother" said Lanceren "he mus...
One hundred students were asked about their favorite subject the results are as follows
Science 15...
Use the analogy to answer the question.
Watching from my room, I saw that my little brother was a...