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To find the ratio of the side lengths of inscribed regular hexagon and circumscribed regular hexagon of a circle
Let the radius of the circle be r, for the inner solution of a regular hexagon, the side length L1 = R, because the center angle of each side pair of a regular hexagon is 60 degrees, and for the circumscribed regular hexagon, the distance from the center of the circle to each side = R, and the opposite angle is also 50 degrees. The side length L2 = 2 √ 3 / 3rl1: L2 = R: 2 √ 3 / 3R = √ 3:2
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It is known that the area of the inscribed hexagon of the circle is 3, and the length of the circumscribed square of the circle is online
If the radius of the circle is r, the circle can be divided into six equilateral triangles with equal area and side length R
According to the formula of triangle area: S = 1 / 2Sin @ * a * b = 1 / 2sin60 * r * r = root 3 / 4r2, because the area of the inscribed hexagon in the circle is 3, the above formula is equal to 3 / 6, and R ≈ 1.075 is obtained. From the length of the circumscribed square of a circle twice the radius of the circle, the side length of the circumscribed square of the circle = 2R ≈ 2.15
If the rectangle ABCD is inscribed with ⊙ o, and ad arc: CD arc = 1:2, what is ᙽ AOB equal to
Because it is inscribed, the diagonal AC goes through the center of the circle, that is, the arc AC is a semicircle
Ad arc: CD arc = 1:2, ∠ cod = 2 / 3 * 180 ° = 120 °
∠AOB=∠COD=120°
As shown in the figure, the quadrilateral ABCD is inscribed in ⊙ o, ad ∥ BC, arc AB + arc CD = arc AD + arc BC. If ad = 4, BC = 6, the area of quadrilateral ABCD is______ .
Connect OA, ob, OC, OD, make OE ⊥ ad to e, extend the intersection BC to point F, ∵ ad ∥ BC, ∵ of ⊥ BC, isosceles ⊥ AOD and isosceles △ BOC: OE ⊥ ad, of ⊥ BC, so ≁ AOE = 12 ﹣ AOD, ﹤ BOF = 12 ﹣ BOC; AE = 2, BF = 3, ∵ arc AB + arc CD = arc AD + arc BC, ? AOE + ⊥ BOF = 90 °
As shown in the figure, the square ABCD is connected to ⊙ o, and the point P is on the arc ad, then ∠ BPC=______ .
Connect OB and OC, as shown in the figure,
∵ the square ABCD is connected with ⊙ o,
ν BC arc = 1
4 circumference,
∴∠BOC=90°，
∴∠BPC=1
2∠BOC=45°．
So the answer is 45 degrees
Given that the vertices of the isosceles trapezoid ABCD are all on ⊙ o, ab ∥ CD, arc AB + arc CD = arc AD + arc BC, if AB = 4, CD = 6, calculate the area of the isosceles trapezoid ABCD
Because the arc AB + arc CD = arc AD + arc BC, so ∠ AOB + ∠ cod = ∠ BOC + ∠ AOD because ∠ AOB + ∠ cod + ∠ BOC + ∠ AOD = 360 ° so ∠ AOB + ∠ cod = 180 ° because Bo = Ao = co = do
As shown in the figure, it is known that ⊙ o passes through the vertices a and B of the square ABCD and is tangent to the edge of CD. If the side length of the square is 2, the radius of the circle is () A. 4 Three B. 5 Four C. Five Two D. 1
Pass through point o as OE ⊥ AB, cross AB to point E, and connect ob,
Let the radius of ⊙ o be r, ∵ the side length of square is 2, and CD is tangent to ⊙ o,
∴OF=R，
∴OE=2-R，
In RT △ OBE,
Oe2 + Eb2 = ob2, i.e. (2-r) 2 + 12 = R2, r = 5
4．
Therefore, B
As shown in the figure, it is known that ⊙ o passes through the vertices a and B of the square ABCD and is tangent to the edge of CD. If the side length of the square is 2, the radius of the circle is () A. 4 Three B. 5 Four C. Five Two D. 1
Pass through point o as OE ⊥ AB, cross AB to point E, and connect ob,
Let the radius of ⊙ o be r, ∵ the side length of square is 2, and CD is tangent to ⊙ o,
∴OF=R，
∴OE=2-R，
In RT △ OBE,
Oe2 + Eb2 = ob2, i.e. (2-r) 2 + 12 = R2, r = 5
4．
Therefore, B
Square ABCD and regular triangle AEF are inscribed in circle O and connected with be. Try to judge that be is the side length of regular polygon inscribed in circle o process
The center angle of chord AE is 360 ° / 3 = 120 °
The center angle of chord AB is: 360 ° / 4 = 90 °
The center angle of the chord be is 120 ° - 90 ° = 30 °
Therefore: be is the number of sides of a regular polygon inscribed in circle O: 360 ° / 30 ° = 12
That is, be is the side length of circle O inscribed with a regular 12 polygon
As shown in the figure, it is known that ⊙ o passes through the vertices a and B of the square ABCD and is tangent to the edge of CD. If the side length of the square is 2, the radius of the circle is () A. 4 Three B. 5 Four C. Five Two D. 1
Pass through point o as OE ⊥ AB, cross AB to point E, and connect ob,
Let the radius of ⊙ o be r, ∵ the side length of square is 2, and CD is tangent to ⊙ o,
∴OF=R，
∴OE=2-R，
In RT △ OBE,
Oe2 + Eb2 = ob2, i.e. (2-r) 2 + 12 = R2, r = 5
4．
Therefore, B
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