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It is known that the four vertices of the trapezoid ABCD are all on the circle O, ad is parallel to BC, known ad = 12cm, BC = 16cm, the diameter of circle O is 20cm, calculate the area of trapezoid Urgent I don't have a picture. Do you think it is
There are two kinds of heights. AD and BC are on one side of the dot or on both sides. Connect OA, ob, OC, OD through o to ef ⊥ ad, cross ad to e, cross BC to F, then E and F are the midpoint of AD and BC respectively. OE = root number (OD ^ - ed ^) = root number (100-36) = 8, of = root number (OC ^ - FC ^) = root number (100-64) = 6
In the isosceles trapezoid ABCD, ab ‖ CD, ad = BC = 6, ab = 6, ab = 12, CD = 4, then the area of trapezoid ABCD is
The height of trapezoid is 2 × heel sign 5. Area s = 1 / 2 × (AB + CD) × height = 1 / 2 × 16 × 2 × heel 5 = 16 × heel 5
Given that the center of the circle passes through the vertices a and B of the square ABCD and is tangent to the edge CD, if the side length of the square is two, what is the radius of the center of the circle?
The midpoint of AB is set as E
The distance OE from the center of a circle to the line AB is OE = 2-r
In the right triangle AOE, AE = 2 / 2 = 1, OE = (2-r), and the hypotenuse Ao = R
So 1 ^ 2 + (2-r) ^ 2 = R ^ 2
Then r = 4 / 5
As shown in the figure, each vertex of ABCD with side length of 2 is on circle O, AC is diagonal, P is the midpoint of edge CD, and extension AP intersects circle at point E (1)∠E=___ . (2) Write a pair of unequal triangles in the diagram and explain the reason (3) Find the length of the chord De De is linked. It should be a congruent similar triangle.
∵ = ∵, ∵∵? = ∵) (∵) the diameter of ∵ PDAC is not equal to that of ∵ PAC ∵ 2, but the diameter of ﹤ PDAC is not equal to that of ∵ PAC
In a square ABCD with side length of 3, circle O is tangent to AB and ad, and circle O 'is tangent to BC and CD and circumscribed to circle O. the sum of the radii of these two circles is obtained
Is it 6-3 times root 2
If you draw a picture, you can see that when two circles are circumscribed, the diagonal line AC is related to the radius of the two circles. Making a vertical line from the center of the two circles as the tangent edge, they both form two small squares
The length of AC = R + R + root 2 * r + root 2 * r = 3 times root 2
After putting forward root number 2, we can get (1 + root 2) * (R + R) = 3 times root 2
R = 3
If the side length of the square is 2, what is the radius of the circle AB2 point inscribed on circle 'BC2 point circumscribed on circle
Let AB be the x-axis and a be at the origin. Let the circle equation be (x-a) ^ + (y-b) ^ = R ^. Because the circle O passes through the fixed point a'B of square ABCD and is tangent to the edge of CD, if the side length of the square is 2, then
A ^ + B ^ = R ^; (2-A) ^ + B ^ = R ^; (1-A) ^ + (2-B) ^ = R ^
a=1;b=3/4;r=5/4
So the radius is 5 / 4
The square ABCD with side length of 2 has an inscribed circle, and the equilateral triangle EFG is inscribed with circle O. the side length of triangle EFG is proved
The radius of the inscribed circle of a square is half of the side length of the square, that is, r = 2 / 2 = 1,
The center point of an equilateral triangle inside a circle is the outer center and also the center of gravity. Therefore, two thirds of the length of the center line is equal to the radius of the circle, that is, the length of the center line of the regular triangle is 1 / (2 / 3) = 3 / 2, then the side length of the regular triangle EFG = (3 / 2) / cos30 ° = (3 / 2) / (√ 3 / 2) = √ 3
As shown in the figure, the square ABCD is inscribed in circle O, e is the midpoint of DC, and the line be intersects circle O at point F. if the radius of circle O is root 2, find the distance from point O to be
Let's take a look at △ BOE. According to the radius of the circle is √ 2, it is easy to get OE = 1, OB = √ 2, and we can know ∠ BOE = 135 ° so s △ BOE = (1 / 2) × 1 ×√ 2 × sin135 ° = (1 / 2) × 1 × √ 2 × (√ 2 / 2) = 1 / 2, we can also get BC = 2, CE = 1
As shown in the figure, it is known that ⊙ o passes through the vertices a and B of the square ABCD and is tangent to the edge of CD. If the side length of the square is 2, the radius of the circle is () A. 4 Three B. 5 Four C. Five Two D. 1
Pass through point o as OE ⊥ AB, cross AB to point E, and connect ob,
Let the radius of ⊙ o be r, ∵ the side length of square is 2, and CD is tangent to ⊙ o,
∴OF=R，
∴OE=2-R，
In RT △ OBE,
Oe2 + Eb2 = ob2, i.e. (2-r) 2 + 12 = R2, r = 5
4．
Therefore, B
As shown in the figure, we know that in the semicircle o, the diameter Mn = 10, the four vertices of the square ABCD are on the radius OM, OP and circle O respectively, and the angle POM = = 45 ° is positive
∠DOC=45 ∠DCO=90 CO=DC
Connecting Ao (BC + CO) squared + AB squared = Ao squared, if BC is equal to X (AO radius), then 5 squared = 4 (xsquare) + (xsquare)
X = radical 5
RELATED INFORMATIONS
- 1. As shown in the figure, we know that in ⊙ o, the diameter Mn = 10, the four vertices of square ABCD are on the radius OM, OP and ⊙ o respectively, and ᙽ POM = 45 °, then the length of AB is () A. 5 B. 4 C. 3 D. Five
- 2. 0
- 3. 0
- 4. 0
- 5. As shown in the figure, in RT △ ABC, ∠ C = 90 °, B = 30 °, BC = 4cm, with point C as the center of the circle and 2cm long as the radius as the circle, then the position relationship between ⊙ C and ab is () A. Separation B. Tangency C. Intersection D. Tangency or intersection
- 6. In the triangle ABC, (isosceles triangle), ab = AC, angle BAC = 120 degrees, ad is perpendicular to AC, if the square of AB = 3, then BC The figure is a large isosceles triangle. The large triangle has a small isosceles triangle. A is the vertex. D is a point on BC
- 7. As shown in Figure 1, △ ABC is inscribed in ⊙ o, ad bisects BAC, the intersection line BC is at point E, and the intersection ⊙ o is at point D (1) It is proved that Mn is the tangent line of ⊙ o; (2) The results showed that ab · AC = ad · AE; (3) As shown in Figure 2, AE bisects the outer angle of BAC ∠ fac, the extension line of BC intersects at point E, and the extension line of EA intersects ⊙ o at point D. conclusion whether AB? AC = ad? AE is still valid? If yes, please write the proof process; if not, please explain the reasons
- 8. Ad, AE and CB are all tangent lines, and the tangent points are D, e, F, ad = 4, so we can find the circumference of the triangle ABC
- 9. As shown in the figure, given the triangle ABC, the angle a = 90 degrees, take AB as the diameter to make the semicircle intersect BC at point D, and cross point D as the tangent of circle O to intersect AC at point P. it is proved that PA = PC
- 10. As shown in Fig. 1, ⊙ o is the circumscribed circle of ⊙ ABC, and ∠ B = ∠ CAD
- 11. As shown in the figure, in the rectangle ABCD, e is the midpoint of BC, and de ⊥ AC, then CD: ad=______ .
- 12. The area of a regular hexagon The bottom surface of this part is hexagonal, with a floor area of 120 square centimeters, a height of 5 centimeters and a volume of () cubic centimeters
- 13. The triangle whose largest angle is an acute angle must be an acute triangle () in a triangle, at least two angles are acute angles
- 14. Trigonometric function relation of the same acute angle sin^2 x + cos^2 x tan A * cot A
- 15. If cos alpha = 3 / 5, alpha belongs to (- π, 0), then cos (π / 4-alpha)
- 16. Why when n tends to infinity, the limit of one part of n to open an n-th root is one
- 17. What is the sum of the factorials from 0 to positive infinity n
- 18. Is it fifty-five or fifty-four sixty-one?
- 19. If a is not divisible by 2 It is proved that a squared minus 1 can be divisible by 8
- 20. +6! How to calculate, seek specific steps