The diameter of a circle has been increased by 5cm, the circumference has been increased by () and the area has been increased () Perimeter increased by () cm, area increased by () square centimeter?

If the circumference is 3.14, then 5 times 3.14
Area: 6.25

The diameter of a circle is 4cm. Its circumference and area are equal. Is it right or wrong?

No. the units are different and cannot be compared

In a circle of radius 6, find the arcuate area enclosed by a chord of length 6 and its inferior arc

The arcuate area is equal to the sector area minus the triangle area
S=3.14×6^2/6-6×6×sin60/2=18.84-15.678=3.162

As shown in the figure, the distance from the point P in the circle with radius 2 to the center O is 1, and the chord AB passing through the point P and the minor arc form an arch, then the minimum perimeter of the arch is () A. 4π 3+2 Three B. 2π 3-2 Three C. 2π 3+2 D. 4π 3-2

When p is at the midpoint of chord AB, the arc length is the minimum
An arch is an irregular figure, which needs to be cut and mended,
According to Pythagorean theorem, the sector angle is 120 degrees,
According to the formula of arc length, the length of arc AB = 4 π
3，
In △ AOB, ab = 2
3，
The minimum value of bow circumference = 4 π
3+2
3．
Therefore, a

As shown in the figure, if the radius of ⊙ o is 5, P is a point in the circle, and the distance from point P to the center O of the circle is 4, then the minimum value of chord length passing through point P is______ .

Connect OP and extend to intersect with circle at C. pass through point P as ab ⊥ CQ, AB is the shortest chord
Because Ao = 5, Op = 4,
According to Pythagorean theorem AP=
52−42=3，
According to the vertical diameter theorem,
AB=3×2=6．

If the chord length of the bow is 4 and the height of the bow is 1, then the radius of the circle where the bow is located

Let the radius of the circle O where the arch is located is r, the vertical line od of AB is the crossing point O, the perpendicular foot is C, and the intersection of ⊙ o with D is ⊙ ACO = 90 °
∵AB=4，
∴AC=1
2AB=2．
In RT △ AOC, OA = R, OC = R-1, AC = 2,
According to Pythagorean theorem, oc2 + ac2 = oa2,
That is (R-1) 2 + 22 = R2,
The solution is: r = 2.5
Therefore, the radius of the circle where the arch is located is 2.5

How to find the chord length when the arc area and circle radius are known

0

The minor arc is 1 / 6 of the circumference, which is equal to the third of π R. the arcuate area is equal to the sector area minus the area of the regular triangle. The sector area is equal to π R 2 divided by 6. The side length of the regular triangle is r, and the height is the root of the two thirds of R

In a circle of radius 6, find the area of the bow formed by the first with length 6 and the inferior arc he is opposite

Center angle = 60
Area = sector area - triangle area = π * 6 ^ 2 / 6-radical 3 * 6 ^ 2 / 4 = 6 π - 9 root 3 = 3.26

As shown in the figure, O is a point on the diagonal of the square ABCD, with o as the center of the circle, the length of OA as the radius, O and BC are tangent to M, Verification: circle O is tangent to CD

O is the point on the diagonal line! It should be a point on the diagonal AC! Because it is a point on the diagonal AC of a square, then the distance from O to BC and DC is the same. The circle is tangent to BC, of course, to CD

The diameter of a circle has been increased by 5cm, the circumference has been increased by () and the area has been increased () Perimeter increased by () cm, area increased by () square centimeter?