![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
As shown in the figure, increase the radius of a small circular site by 5m to get a large circular site. If the site area is doubled, the radius of the small circular site will be obtained=______ .
If the radius of the small circle is XM, then the radius of the big circle is (x + 5) M,
According to the meaning of the title: π (x + 5) 2 = 2 π x2,
X = 5 + 5
2 or x = 5-5
If you don't agree with the question, give it up
So the answer is: (5 + 5
2）m．
As shown in the figure, the side length of the square ABCD is 1cm. If we make another square with the diagonal AC as the side length, then the area of the square ACEF is______ cm2．
In RT △ ABC, ab = BC = 1cm,
Then AC=
AB2+BC2=
2cm，
The area of square ACEF is ac2 = 2cm2,
So the answer is 2
As shown in the figure, the square ABCD with a side length of 4cm is first shifted upward by 2cm, and then by 1cm to the right to obtain a'b'c'd '. At this time, the area of the shadow part is______ cm2．
∵ the square ABCD with a side length of 4cm is first shifted upward by 2cm,
The width of the shadow part is 4-2 = 2cm,
∵ 1 cm to the right,
The length of the shadow part is 4-1 = 3cm,
The shadow area is 3 × 2 = 6cm2
So the answer is: 6
As shown in the figure, the side length of square ABCD is 1cm, e and F are the midpoint of BC and CD respectively. Connecting BF and De, the shadow area in the figure is () cm2 A. 4 Five B. 2 Three C. 5 Six D. 3 Four
As shown in the figure, ∵ the side length of the square ABCD is 1cm, e and F are the midpoint of BC and CD respectively, ≌≌△ CBF, easy to get, △ Bge ≌ △ DGF, so s △ Bge = s △ EGC, s △ DGF = s △ CGF, so s △ Bge = s △ EGC = s △ DGF = s △ CGF, and because s △ BFC = 1 × 12 × 12 = 14cm2, so
If the side length of the square ABCD is 1, e and F are the midpoint of BC and CD respectively, connecting BF and DF, then what is the area of shadow in the figure? To process, complete, to use the knowledge of grade two to solve
If BF and DF intersect at O, then the congruence of BCF and DCE can be proved by the edges and corners, and the congruence of BOE and DOF of triangles can be proved from the angles and edges. According to the equal areas of triangles with equal base and equal height, the areas of COF, DOF, Coe and BOE of triangles are equal, and the area of triangle BCF is equal to 1 / 4, Therefore, the area of the triangle COF, DOF, CoE and BOE is (1 / 4) / 3 = 1 / 12, and that of the quadrilateral adob is 1-1 / 12 * 4 = 2 / 3
Because there is no figure, can only explain the train of thought
As shown in the figure, the side length of the square ABCD is 4cm, e and F are the midpoint of BC and CD respectively, connecting BF and De, then the area of shadow in the figure is
As shown in the figure, GC is connected. Because the area of triangle BFC and triangle CDE are equal, and they will be equal after subtracting quadrilateral ecfg. Therefore, the area of triangle bef is equal to that of triangle DFG. Because e and F are the midpoint of BC and CD respectively, the area of triangle Berg, area of triangle EGC, area of triangle CFG and area of triangle DFG are equal, The area of triangle BFC is a quarter of the square, which is 4 * 4 / 4 = 4 square centimeter, the area of blank part is 4 / 3 * 4 = 16 / 3, and the area of shadow part is 4 * 4-16 / 3 = 32 / 3
As shown in the figure, in the quadrilateral ABCD, ad ∥ BC, and ad ﹥ BC, BC = 6cm, points P and Q start from the positions of a and C at the same time, point P moves from point a to point d at the speed of 1cm / s, and point Q moves from point C to point B at the speed of 2cm / s?
∵ the movement time is x seconds
∴AP=x，QC=2x
∵ a quadrilateral abqp is a parallelogram
∴AP=BQ
∴x=6-2x
∴x=2
A: two seconds later, the quadrilateral abqp is a parallelogram
As shown in the figure, in the quadrilateral ABCD, ad parallel BC, ad vertical DC, AB vertical AC, angle B = 60 degrees, CD = 1cm, calculate the length of BC
∵ ad parallel BC
Ψ ACB = ∠ DAC = 30 degrees
∴AC=2
In the right triangle ABC, ∠ B = 60 ° so BC = 4 / √ 3
As shown in the figure, rectangle ABCD rotates the rectangle 90 ° to the right around vertex a to find the shadow area swept by the edge of CD. (unit: cm)
3.14×（82+62）×1
4-3.14×82×1
4，
=3.14×100×1
4-3.14×64×1
4，
=3.14×25-3.14×16，
=3.14×（25-16），
=3.14×9，
=26 (square centimeter);
A: the shadow area is 28.26 square centimeters
As shown in the figure, the side length of the square is 2. Draw an arc with two vertices as the center and the side length as the radius to calculate the perimeter of the shadow part
In the first case, the triangle ABC is a regular triangle, so the perimeter of the shadow part is the arc length corresponding to the AB side of the square plus two center angles of 60 degrees. Therefore, the perimeter of the shadow C = 2 + 2x3.14x2x60 / 360x2 = 6.19
RELATED INFORMATIONS
- 1. Take the three vertices of the equilateral triangle ABC as the center of the circle and 2cm long as the radius, draw three arcs in the triangle, and the total length of the three arcs is cm
- 2. As shown in the figure, the inner angles of the hexagon abgdef are all equal, and ∠ DAB is equal to 60 degrees. What is the relationship between AB and de? Is there such a relationship between BC and ef?
- 3. Denominator rationalization: (cubic root 4 + cubic root 6 + cubic root 9) 1 / 2
- 4. Let a and B be rational numbers, and a and B satisfy the equation a squared + 2B + B times root sign 2 = 17-4 times root sign 2, and find the value of a + B
- 5. Let A.B be a rational number and A.B satisfy the equation a + 2B + a root sign 5 = 8 + 2 radical sign 5. Find the square of 3A + B
- 6. Mean inequality: let a be greater than or equal to 0, b greater than or equal to 0, a + B / 2 = 1 a times the maximum value of 1 + B under the root sign
- 7. Ad is perpendicular to BC, CE is perpendicular to AB, ad is 8, CE is 7, the sum of AB and BC is 45, the area of triangle ABC
- 8. As shown in the figure, in △ ABC, ∠ 1 + ∠ 2 = 230 ° and ad bisect ∠ BAC
- 9. As shown in the figure, △ ABC is an equilateral triangle, and points D, e and F are points on the segments AB, BC and Ca respectively, (1) If ad = be = CF, is △ def an equilateral triangle? Try to prove your conclusion; (2) If △ DEF is an equilateral triangle, is ad = be = CF true? Try to prove your conclusion
- 10. Someone plays tennis from the origin o, and the flight path of the tennis ball is a parabola, which can be represented by the image of the quadratic function y = 4x-1 / 2x 2 To find the distance between the highest flying point B and the ground and the horizontal distance between the landing point a and the point o
- 11. As shown in Figure AB is the diameter of circle O, CD is the two points on circle O and C is the midpoint of ad arc. If ∠ bad = 20 °, calculate the degree of ∠ ACO
- 12. Additional question: as shown in the figure, it is known that △ ABC is inscribed in ⊙ o, AB is the diameter, ∠ CAE = ∠ B Verification: AE and ⊙ o are tangent to point a
- 13. In the triangle ABC, ad is perpendicular to D, angle B = 2, angle C. It is proved that ab + BD = CD
- 14. In circle O, radius r = 5cm, chord ad// BC.AD=6cm BC = 8cm (1) what kind of special quadrilateral ABCD is? Prove your conjecture In circle O, radius r = 5cm, chord ad// BC.AD=6cm BC=8cm (1) What kind of special quadrilateral is quadrilateral ABCD? Prove your conjecture; (2) Find the area of quadrilateral ABCD
- 15. The diameter of a circle is 10, and a chord AB is 8. P on AB, how many p points are there when OP is an integer?
- 16. It is known that: in △ AOB and △ cod, OA = ob, OC = OD, ∠ AOB = ∠ cod = 90 ° It is known that: in △ AOB and △ cod, OA = ob, OC = OD, ∠ AOB = ∠ cod = 90 °. (1) as shown in Figure 1, points c and D are respectively on the sides OA and ob, connecting AD and BC, and point m is the midpoint of line BC and connecting OM, then the quantitative relationship between line AD and OM is, and the positional relationship is; (2) as shown in Fig. 2, rotate △ COD in Fig. 1 anticlockwise around point o with rotation angle of α (0 ° < α < 90 °), Point m is the midpoint of line BC, connecting OM. Please judge whether the two conclusions in (1) still hold. If yes, please prove; if not, please explain the reasons; (3) as shown in Figure 3, rotate △ COD in Figure 1 anticlockwise around point O to make the OD of △ cod and the edge OA of △ AOB exactly on the same line, point C falls on ob, Point m is the midpoint of line BC. Please judge whether the quantitative relationship between line AD and OM in (1) has changed, write your conjecture and prove it
- 17. Given that in a circle, the chord AB = 1 / 3 of the root sign of 2 times, and the chord center distance is 1, what is the radius of the circle?
- 18. AB, CD are the two chords in the circle, AB is perpendicular to e, AE = 2, EB = 6, ed = 4, then the diameter of the circle is The distance between the tangent points of isosceles triangle with waist 10 and base length 6 is 0
- 19. As shown in the figure, AB is the diameter of circle O, CB is the chord, OD ⊥ CB is in E, cross arc CB is in D, and ac. (1) please write down two different types of As shown in the figure, AB is the diameter of circle O, CB is the chord, OD ⊥ CB is in E, cross arc CB is in D, and AC is connected (1) Please write two different types of correct conclusions; (2) If CB = 8, ed = 2, find the radius of circle o
- 20. As shown in the figure, we know that in ⊙ o, the diameter Mn = 20, the four vertices of square ABCD are respectively on the radius OM, OP and ⊙ o, and ᙽ POM = 45 ° to find the length of ab