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As shown in the figure, De is the median line of △ ABC, M is the midpoint of De, and the extension line of CM intersects AB at point n, then s △ DMN: s quadrilateral anme equals () A. 1：5B. 1：4C. 2：5D. 2：7
∵ De is the median line of △ ABC, ∵ de ∥ BC, de = 12bc, if the area of △ ABC is 1, according to de ∥ BC, ∥ ade ∥ ABC, ∥ s ∥ ade = 14, am is connected. According to the title, s ∥ ADM = 12S ∥ ade = 18S ∥ ABC = 18, ∥ de ∥ BC, DM = 14bc, ∥ DN = 14bn, ∥ DN = 13bd = 13ad. ∥ s ∥ DNM = 13s ∥ ADM = 124, ∥ s quadrilateral anme = 14 − 124 = 524, ∥ s ∥ DMN: s quadrilateral anme = 124:524 = 1 So a
As shown in the figure, De is the median line of △ ABC, M is the midpoint of De, and the extension line of CM intersects AB at point n, then s △ DMN: s quadrilateral anme equals ()
A. 1：5B. 1：4C. 2：5D. 2：7
∫ De is the median line of △ ABC, ∥ de ∥ BC, de = 12bc. If the area of △ ABC is 1, according to de ∥ BC, ∥ s ∥ ade = 14, am is connected. According to the title, s ∥ ADM = 12S ∥ ade = 18S ∥ ABC = 18, ∥ de ∥ BC, DM = 14bc, ∥ DN = 14bn, ∥ DN = 13bd = 13ad. ∥ s
It is known that the three vertices of △ ABC are a (- 1,4), B (- 2, - 1), C (2,3). 1
First of all, the coordinate of point D is (0,1). Let the ad linear equation be y = KX + B. since it passes through two points a and D, B = 1, - K + B = 4 and K = - 3 are obtained, so the linear equation is y = - 3x + 1. As for the coordinate of D, it is the midpoint of BC, so its abscissa is (- 2 + 2) divided by 2 = 0, and its ordinate is (- 1 + 3) divided by 2 = 1
Given that the three vertices of △ ABC are a (- 5,0) B (3, - 3) C (0,2), try to find the point oblique equation of the line with the height on the edge of BC
The equation of the line BC is
(y+3)/(2+3)=(x-3)/(0-3)
That is y = - 5 / 3x + 2
So the slope of the high on the BC side is 3 / 5
Therefore, the point oblique equation is y = 3 / 5 (x + 5), that is, y = 3 / 5x + 3
RELATED INFORMATIONS
- 1. It is known that the coordinates of the three vertices of △ ABC are a (- 5,0) B (3, - 3) C (0,2), and the linear equations of height on the BC edge are obtained respectively
- 2. As shown in the figure, in the trapezoidal ABCD, ab ∥ CD, ab = 3CD, diagonal lines AC and BD intersect at point O, and median lines EF intersect with AC and BD at two points m and N respectively, then the area of the shadow part in the figure is () A. 12B. 13C. 14D. 47
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- 8. Given that y in the first-order function y = KX + 2K decreases with the increase of X, then the inequality KX + 2K
- 9. Given the function f (x) = (M + 1) x ^ 2 - (m-1) x + M-1, if M belongs to (- 1,1) inequality and the solution set of F (x) > 0 is r, the value range of X is obtained Please give the detailed process, thank you, this kind of question is a little dizzy, usually is to find the scope of M, thank you!
- 10. It is known that function FX is an odd function defined on R, and f (x + 2) is equal to negative FX. If F1 is equal to 1, then f (3) - f (4)=
- 11. As shown in the figure, ⊙ O's chord AB and AC have an included angle of 50 °, Mn is the midpoint of arc AB and arc AC respectively, and OM and on intersect AB and AC at points E and f respectively, then the degree of ∠ mon is () A. 110°B. 120°C. 130°D. 100°
- 12. It is known that in ⊙ o, n is the midpoint of the chord AB, on intersects the arc AB in M, if AB = 2, radical 3, Mn = 1, the radius of ⊙ o is obtained
- 13. For the two strings AB and Cd in the circle O, AB > CD, the distances from the center O to AB and CD are OM and on respectively, then om on is filled with ＞＜ =
- 14. It is known that the circle O and the circle K are the big circle and the small circle of the ball o, and their common chord length is equal to the radius of the ball o, OK = 3 / 2, and the angle between the circle O and the plane of the circle K is 60 degrees, Then the surface area of ball 0 is?
- 15. Let m.n be two points on the radius op of the ball 0 and NP = Mn = OM. Make three circles that are perpendicular to the surface of OP through m.n.o. respectively, then the area ratio of these circles is
- 16. As shown in the figure, ⊙ o, if the chord ad ∥ BC, Da = DC, ∠ AOC = 160 °, then ∠ BCO is equal to______ Degree
- 17. In triangle ABC, ab = AC, EF is the median line of triangle ABC, extend AB to point D, make BD = AB, connect CD, what do you think is the relationship between CE and CD
- 18. In the triangle ABC, ab = AC, EF is the median line of the triangle ABC, intersecting AB, AC at e, f respectively, extending AB to D, making BD = AB, connecting CD
- 19. In the isosceles trapezoid ABCD, the angle B is 60 degrees, ad is parallel to BC, and the angle BAC is equal to 90 degrees. The median line length of the isosceles trapezoid is 12cm. Find the circumference of the isosceles trapezoid
- 20. As shown in the figure, in the isosceles trapezoid ABCD, ad is parallel to BC, angle B = 45 degrees, AE is perpendicular to BC, and point E, AE = ad = 2cm