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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
As shown in the figure, in the quadrilateral ABCD, ad ∥ BC, and ad > BC, BC = 6cm, P, Q start from a, C at the same time, P is 1cm/ As shown in the figure, in the quadrilateral ABCD, ad ∥ BC, and ad ∥ BC, BC = 6cm, P, Q start from a and C at the same time, P starts from C to B at the speed of 1cm / s, and Q moves from C to B at the speed of 2cm / S. after a few seconds, the quadrilateral abqp is a parallelogram?
Let ABCP be a parallelogram after x seconds
According to a set of parallelogram whose opposite sides are parallel and equal
1X＝6－2X X＝2s
A: two seconds later, the quadrilateral ABCP is a parallelogram
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, 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 rectangle ABCD is rotated 90 ° to the right along the vertex a to find the area of the shadow swept by the edge of CD
Rectangular ABCD,
You know: the length of AD, the length of DC, then the length of AC can be calculated (AC ^ 2 = ad ^ 2 + CD ^ 2)
Area of shadow = area of sector acc1 - area of triangle ad1c1 - area of sector aed1 + (area of triangle ADC - face of sector ADE)
=Area of sector acc1 -- area of sector ADD1
=1 / 4 * area of circle with radius AC --- 1 / 4 * area of circle with radius ad
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, 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, with the three vertices of the triangle as the center of the circle and the radius of 2 cm, draw an arc in the triangle, then the area of the shadow part in the figure is_______ square centimetre (the angles of the three angles of the triangle are 80 degrees, 40 degrees and 60 degrees respectively) 2. The following statement is correct () A. The circumference of a circle is 3.14 times its diameter B. A semicircle is a sector C. If the circumference of two circles is equal, their areas are equal
As shown in the figure, with the three vertices of the triangle as the center of the circle and the radius of 2 cm, draw an arc in the triangle, then the area of the shadow part in the figure is___ 2π____ square centimetre
2. The following statement is correct (c)
A. The circumference of a circle is 3.14 times its diameter
B. A semicircle is a sector
C. If the circumference of two circles is equal, their areas are also the same
The side length of the square is a, with the four vertices as the center and the side length as the radius, an arc is drawn in the square. The perimeter and area of the shadow part surrounded by the four arcs is
As shown in the figure: point E is the intersection point of B and a circle with radius a taking C as the center, so △ BEC is an equilateral triangle ∠ ECB = 60 degrees ∠ ECD = 9
Take the three vertices of the equilateral △ ABC as the center of the circle and 2cm long as the radius, draw three arcs in the triangle, and calculate the perimeter of the shadow part? The shadow part is the three corners of the triangle
The sum of the angles inside the triangle is 180 degrees, so the sum of the arc lengths of the three arcs is the arc length of the semicircle with a radius of 2 cm is 2 π
Then the perimeter of the shadow part C = 2 × 6 + 2 π = 12 + 2 π
RELATED INFORMATIONS
- 1. The following is a triangle. Take each vertex of the triangle as the center of the circle and draw an arc with a radius of 2 cm to calculate the shadow area? The shadow is the three corners
- 2. In order to know that the sum of the inner angles of the hexagon ABCDEF is equal, please judge the size of AB plus BC and de plus EF
- 3. As shown in the figure, in the hexagon ABCDEF, AF ∥ CD, ab ∥ ed, ∠ a = 140 °, B = 100 °, e = 90 °. Calculate the degrees of ∠ C, ∠ D and ∠ F
- 4. Root 15 + root 10 - root 5 / root 6 - root 2 + 2 denominator rationalization Is there any simple way to really trouble you
- 5. Radical addition, subtraction, multiplication and division! 1 ﹙√6－√2﹚²= 2 4√12÷﹙﹣√1／3﹚ 3 √18÷√8×√27／2 4 ﹙﹣7／4√24﹚﹙﹣2／7√6﹚ 5 √6﹙√24－√2﹚ 6 ﹙√7－3﹚²－﹙√7＋3﹚² Note that "1 / 3" is a third of the time, and "pro" - is a negative sign and "- is a minus sign
- 6. How to add, subtract, multiply and divide with root sign?
- 7. Given that the circumference of the right triangle ABC is 4 + 4, the root sign 2, and the length of the center line on the hypotenuse is 2, then the area of the triangle is?
- 8. In the triangle ABC, if the angle BAC = 120 degrees, ab = 2, AC = 1, D is a point on the edge BC, DC = 2bd, then the vector ad points multiply the vector BC =?
- 9. In the triangle ABC, we know that D is a point on the edge of ab. if vector ad = 2 vector dB, vector CD = 1 / 3 vector Ca + X vector CB, then x is equal to A.2/3 B.1/3 C.-1/3 D.-2/3
- 10. Let D, e, f be the midpoint of BC, AC and ab of the triangle ABC respectively How to do such a topic,
- 11. It is known that, as shown in the figure, the chords AB and CD of circle O intersect at p. it is proved that PA * Pb = PC * PD
- 12. As shown in the figure, in the parallelogram ABCD, the bisector of angle ab. C intersects with AD at point P. it is proved that PD + CD = BC
- 13. It is known that in the circle O, AB is the diameter of circle O, CD is a chord, and CD is perpendicular to point ab. connect BC, ad= PA.PB
- 14. As shown in the figure, in the circle O, the chord AB intersects OC, OD at m, n respectively. If AMC = Bnd, it is proved that am = BN
- 15. As shown in the figure, AB is the diameter of ⊙ o, CD is the chord, CE ⊥ CD intersects AB with E, DF ⊥ CD intersects AB with F, proving that AE = BF
- 16. It is known that: as shown in the figure, the diameter EF of ⊙ o intersects the chords AB, CD at the points g and h respectively, and Ag = BG, ch = DH
- 17. 1. The radius of circle O is 6cm, the chord AB = 10cm, the chord CD = 8cm, and ab is perpendicular to CD and P
- 18. Given that AB and CD are two parallel chords of circle O, and ab = 48, CD = 40, the distance between the two parallel chords is 22, find the radius of circle o
- 19. As shown in the figure, it is known that in ⊙ o, the chord ab ⊥ CD is e, AE = 2, EB = 8, and the degree of ⊙ CAD is 120 °, then the radius of ⊙ o is______ .
- 20. It is known that in the circle O, if the chord AB / / CD, ab = 2 roots, 11 cm, CD = 4 roots, 5cm, and the distance between the chord AB and CD is 1cm, then the radius of circle O is?