Draw the largest circle in a square. What percentage of the square area does the circle occupy? How to find it?

Draw the largest circle in a square. What percentage of the square area does the circle occupy? How to find it?

Draw the largest circle in a square. The diameter of the circle is equal to the side length of the square
Let the radius of the circle be a and the side length of the square be 2A
The area of a circle is π a 2, and the area of a square is (2a) 2 = 4A
The area of circle accounts for 78.5% of square area

Draw the largest circle in a square with an area of 9 square centimeters. The circumference of the circle is () and the area is ()

Draw the largest circle in a square with an area of 9 square centimeters. The circumference of the circle is (9.42 cm) and the area is (7.065 square centimeters)

Given that the circumference of the big circle is 50.24 cm, find the area of the small circle Request method is we can understand!

50.24 △ 3.14 = 16 cm - diameter of large circle
Divide the square area into two triangles: 16 × 8 △ 2 × 2 = 128 square centimeters
Square of radius of small circle × 4 = square area, 128 △ 4 = 32 square centimeter -- square of radius of small circle
32 × 3.14 = 100.48 square centimeter -- small circle area

If AF = 3 BC = 1 CD = de AB = CD + 1, find the circumference of the hexagon As above

If the sum of the internal angles of a hexagon is 720 degrees, then all the internal angles are equal, and each angle is equal to 120 degrees
Because the six angles of the hexagon ABCDEF are all 120 degrees, the degree of each external angle of the hexagon ABCDEF is 60 degrees. Therefore, the triangle APF, triangle BGC, triangle DHE and triangle GHP are equilateral triangles
Let CD = x, according to GH = GP, 3 + X + 1 + 1 = 1 + X + x = 4
Similarly, EF = 2
AB=5, BC=1, CD=4, DE=4, EF=2, AF=3, the circumference is 19

As shown in the figure, in the parallelogram ABCD, ab = 6cm, ad = AC = 5cm. Starting from C, the point P moves uniformly along the direction of CA with a speed of 1cm / S; meanwhile, the line EF starts from ab At the same time, the line EF starts from AB and moves uniformly along the ad direction with the speed of 1cm / s, crosses AC to Q and connects PE.PF Let the exercise time be t (s) (0 〈 t 〈 5) (1) What is the value of CD? (2) Try to judge the shape of triangle PEF and explain the reason; (3) When 0 < T < 2.5, ① Whether the area of the Pentagon abfpe is a fixed value during the above motion? If so, write the area of the Pentagon abfpe directly; if not, explain the reason; ② Try to find the value range of the area of triangle PEQ

(1) Firstly, the length of AE, CP and AP is expressed by T. If PE ∥ CD, then △ ape ∽ ACD, then the value of t can be obtained according to the proportional line segment obtained from the similar triangle
(2) Since ad = AC and QE ‖ CD, △ AQE is also isosceles triangle, namely AQ = AE. From the velocity of P and Q, we can know that CP = AE = AQ, and then CQ = AP. Similarly, it can be proved that △ CFQ is also an isosceles triangle, that is CF = CQ. Thus, CF = AP, AE = PC, and ∠ DAC = ∠ FCP, can be proved to be △ FCP ≌ △ PAE, and PF = PE, that is, △ PEF is isosceles triangle
(3) (1) according to the congruent triangle of (2), the areas of △ AEP and △ EPC are equal, so the area of Pentagon can be transformed into the area of △ ABC, so the area of Pentagon is a fixed value;
② From the similar triangles of (1), it is easy to get the expression of QE. The vertical lines CG and pH of AB and EF are made by passing through C and P respectively. According to the property of three lines in an isosceles triangle, the values of Ag and BG can be easily obtained, and then the cosine value of ∠ ACG (i.e. ∠ Eph) can be obtained. The value of high pH on QE side can be expressed according to the length of PQ, The functional relation between the area of △ PQE and t can be obtained. According to the properties of the function, the maximum area of △ PQE can be obtained, and the range of its area can be obtained
(1) According to the meaning of the title, AE = BF = CP = t, AP = 5-t,
In ▱ ABCD, ad = BC = AC = 5, ab = EF = CD = 6,
When PE ‖ CD, △ ape ∽ ACD,
∴ T/5=5-T/5,
∴t=2.5．
(2) It's an isosceles triangle
It is proved that in ▱ ABCD, ad = BC = AC = 5, ab = EF = CD = 6, ▱ cab = ∠ CBA,
∵AB∥EF,∴∠CQF=∠CAB,∠CFQ=∠CBA,
∴∠CFQ=∠CQF,∴CF=CQ,
∴AQ=BF=AE,∴AP=CQ=CF,
∵AD∥BC,∴∠PAE=∠FCP,
∴△PAE≌△FCP（SAS）,∴PE=PF．
(3) 1 is a constant value, which is 12
Reason: from the congruent triangle of (2): s △ AEP = s △ PCF, that is, s Pentagon bfpea = s △ ABC;
CG ⊥ AB to G through C,
In isosceles △ ACB, if Ag = BG = 3, AC = BC = 5, then CG = 4;
The s Pentagon bfpea = s △ ABC = × 6 × 4 = 12
② From (1) we know: △ ape ∽ ACD,
﹣ OE / CD = AE / AD, that is, OE / 6 = t / 5, QE = 6t / 5;
PH ⊥ EF to h through P,
It is easy to get: cos ∠ APH = cos ∠ ACG = 4 / 5,
Therefore, pH = 4 / 5pq = 4 / 5 (5-2t);
Let the area of △ PEQ be y, then y = - 24 / 25 (T-5 / 4) squared + 3 / 2,
When t = 5 / 4, y max = 3 / 2,
∴ 0>sPEO

In this paper, the trajectory equation of the midpoint of chord is obtained by the hyperbola x ^ 2 - (y ^ 2) / 4 = 1 which passes through the fixed point (0,1)

A (x1, Y1) B (X2, Y2) is a (x1, Y1) B (X2, Y2) chord midpoint P (x, y) X1 + x2 = 2xy1 + y2 = 2Y (y1-y2) / (x1-x2) = (Y-1) / (x-0) X1 ^ 2 - (Y1 ^ 2) / 4 = 1x2 ^ 2 - (Y2 ^ 2) / 4 = 1 = 1 two formula subtraction (x1-x2) (x1 + x2) (x1 + x2) - (y1-y2) (Y2 + Y1) / 4 = 0 (x1 + x2) - (Y1 + Y2) (Y2) (Y2 + Y1) / 4 (x1-x1 + x2) - (Y1 + Y2) (y1-y2) / 4 (x1-y1-y2) / 4 (x1-x1-y2) / 4 (x1-y1 x2) = 02x-2y (Y-1) / 4x = 04x ^ 2

Given the hyperbola x ^ 2 / 4 + y ^ 2 = 1 (1), find the equation of the line where the chord with the point (- 1,1 / 2) as the midpoint (2) find the midpoint trajectory equation of the chord passing through the point (- 1,1 / 2)

Given the ellipse x ^ 2 / 4 + y ^ 2 = 1 (1), find the equation of the line where the chord with point P (- 1,1 / 2) is located. (2) find the midpoint trajectory equation of the chord passing through the point (- 1,1 / 2)

The equation of the circle (x-3) 2 + (y + 1) 2 = 1 about the symmetric circle of the straight line x + 2y-3 = 0 is______ .

Let the coordinates of the center of a circle (3, - 1) with respect to the symmetric point of the straight line x + 2y-3 = 0 are (a, b), then B + 1a-3 × (- 12) = - 13 + A2 + 2 × b-12-3 = 0, so a = 195, B = 35. Therefore, the coordinates of the center (3, - 1) of the circle with respect to the symmetric point of the straight line x + 2y-3 = 0 are (195, 35)

Let the symmetric point of a point a (2,3) on the circle about the straight line x + 2Y = 0 is still on the circle, and the chord length of the intersection between the circle and the straight line X-Y + 1 = 0 is twice the root sign 2, and the equation of the circle is obtained The symmetry point of point a about the straight line I calculated is (- 6 / 5, - 17 / 5), and then the distance from the center of the circle to the straight line d ^ 2 + chord length ^ 2 = R ^ 2. In the next step, we can not calculate the symmetry point of a into the standard equation of circle (circle is set)

（1） It is easy to know that the center of the circle O is on the straight line x + 2Y = 0, because the straight lines passing through the center of the circle are the symmetry axes of the circle. The center of the circle O (- 2T, t) can be set. Then the radius r = | OA | = (2t + 2) 2 + (T-3) 2]; and the distance from the center of the circle to the line X-Y + 1 = 0 is d = | 3t-1 / / √ 2. 9 according to the problem and "theorem of vertical diameter", we can get: [(3t-1) 2] / 2] + 2 = (2t + 2) 2 + (T-3) 2) 2) 2) 2) 2) 2) 2) 2) 2) 2) 2) 2) 2) 2) when t = 3 or T = 7. (2) when t = 3, the center of the circle (- 6,3), The radius r = 8.when the equation of the circle is (x + 6) 2 + (Y-3) 2 = 64. (3) t = 7, the center of the circle (- 14,7) and the radius r = 4 √ 17. The equation of the circle is (x + 14) Ω + (Y-7) 2 = 272

It is known that the symmetric point of the circle point a2,3) about the line x + 2Y = 0 is still on the circle, and the dark length obtained from the line X-Y + 1 = 0 of the circle is 2 radical sign 2, and the equation of the circle is obtained

Let the center of the circle be (- 2A, a), (x + 2a) ^ 2 + (Y-A) ^ 2 = R ^ 2 chord center distance d = │ - 2a-a + 1 │ / radical 2, so that (2 + 2a) ^ 2 + (3-A) ^ 2 = R ^ 2, then (a + 3) (a + 7) is obtained