As shown in the figure, the radius of the sector is 12 cm and the perimeter is 50 cm Where does half come from

As shown in the figure, the radius of the sector is 12 cm and the perimeter is 50 cm Where does half come from

C sector = (degree of center angle / 180 degrees) * 3.14 * sector radius s sector = (degree of center angle / 360 degrees) * 3.14 * square of sector radius, because sector = two radii + arc length. If radius is R and the center angle of the opposite sector is n °, then sector perimeter: C = 2R + n π R △ 180 in a circle with half diameter R, because 360

As shown in the figure, it is known that in ⊙ o, the diameter AB is 13cm, the chord AC is 5cm, the bisector of ⊙ ACB intersects ⊙ o at D, and the lengths of BC, ad and BD are calculated

The diameter of ∵ o is 13cm, the diameter of ∵ ACB is 90 °, the chord of AC is 5cm, the bisector of ∵ BC is 132 − 52 = 12cm, the bisector of ∵ ACB intersects with D, ad = BD, ad = BD, in RT △ ADB, ad2 + BD2 = AB2, ∵ AB = 13cm, ad = BD = 1322cm

To draw two circumscribed circles with radius less than 4cm and 1cm on a rectangular piece of paper, what is the minimum area of the rectangle?
First, draw two tangent circles and draw a straight line which is tangent to both circles and has different tangent points. Make a straight line which is tangent to the big circle and perpendicular to the above straight line, a straight line which is tangent to the small circle and perpendicular to the first straight line, a straight line which is tangent to the big circle and parallel to the first straight line and perpendicular to the second and third straight lines. The rectangle formed by the intersection of the four straight lines is the rectangle with the smallest area
Connecting the centers of two circles, making a straight line a perpendicular to the length of the rectangle through the center of a small circle, and making a straight line perpendicular to the width of the rectangle through the center of a large circle, a right triangle is formed with the line segment connecting the center of the circle as the hypotenuse. It is easy to know that the short right side is 3. (if the center of a small circle is perpendicular to the width of the circle, the short right side is 4-1 = 3)
So the area is 9 × 8 = 72
I don't quite understand the answer. Do you have a more detailed answer?

First draw two tangent circles and draw a straight line which is tangent to both circles and has different tangent points. Make a straight line which is tangent to the big circle and perpendicular to the above straight line, a straight line which is tangent to the small circle and perpendicular to the first straight line, a straight line which is tangent to the big circle and parallel to the first straight line and perpendicular to the second and third straight lines

To draw two circumscribed circles with radius of 4cm and 1cm on a rectangular piece of paper, the minimum area of the rectangular piece of paper is______ cm2．

As shown in the figure, if WG ⊥ SC is made, then the quadrilateral wdcg is a rectangle, ∵ two circles are tangent, ∵ WS = SC + WD = 1 + 4 = 5, ∵ SG = sc-gc = 4-1 = 3, ∵ WG = 4, ∵ the length ab of rectangular QHB = AD + CD + CB = 1 + 4 + 4 = 9, the width BH = 4 + 4 = 8, ∵ the minimum area of rectangular paper = 8 × 9 = 72cm2

As shown in the figure, take each vertex of the regular hexagon as the center and draw a circle with a radius of 1cm, then the area of the shadow part in the figure is______ Cm 2. (result retention π)

The sum of inner angles of hexagon = (6-2) × 180 ° = 720 ° and shadow area = 720360 π × 12 = 2 π

As shown in the figure, the section radius of the horizontal circular cylindrical water pipe is 0.6m, and the water surface is 0.3m high. Calculate the area of the section with water (the result retains π)

Connect OA and ob, make OD ⊥ AB through O, intersect AB at point E, ∵ od = 0.6m, de = 0.3m, ∵ OE = od-de = 0.6-0.3 = 0.3m, ∵ cos ∠ AOE = oeoa = 0.30.6 = 12, ∵ AOE = 60 °∵ AE = OA · sin ∠ AOE = 0.6 × 32 = 3310, ab = 2ae = 335 ∵ AOB = 2 ∵ AOE = 2 × 60 ° = 120 °, s shadow =

As shown in the figure, the section radius of the horizontal cylindrical drainage pipe is 6dm, in which the water surface is 3DM high, then the area of the section with water is () DM2

12π-9√3.

As shown in the figure, the section radius of the horizontal cylindrical drainage pipe is 0.5m, and the water surface width AB is 0.6m, then the maximum depth of water is 0.5m______ m．

According to the Pythagorean theorem, if the chord center distance is 0.4, the maximum depth of water is 0.4 + 0.5 = 0.9m

As shown in the figure, the section radius of the horizontal cylindrical drainage pipe is 0.5m, and the water surface width AB is 0.6m, then the maximum depth of water is 0.5m______ m．

According to the Pythagorean theorem, if the chord center distance is 0.4, the maximum depth of water is 0.4 + 0.5 = 0.9m

As shown in the figure, the section radius of the horizontal cylindrical drainage pipe is 0.5m, and the water surface width AB is 0.6m, then the maximum depth of water is 0.5m______ m．

According to the Pythagorean theorem, if the chord center distance is 0.4, the maximum depth of water is 0.4 + 0.5 = 0.9m