As shown in Figure 1, in the quadrilateral ABCD, AB is perpendicular to ad, CD is perpendicular to ad, and BC is rotated 90 ° counterclockwise around point B to get the line segment be, connecting AE and Ce (1) As shown in Figure 1, in the quadrilateral ABCD, AB is perpendicular to AD and CD is perpendicular to AD. rotate BC 90 ° counterclockwise around point B to get the line segment be and connect AE and Ce (1). If AB = 2cm and CD = 3cm, calculate the area of triangle Abe

As shown in Figure 1, in the quadrilateral ABCD, AB is perpendicular to ad, CD is perpendicular to ad, and BC is rotated 90 ° counterclockwise around point B to get the line segment be, connecting AE and Ce (1) As shown in Figure 1, in the quadrilateral ABCD, AB is perpendicular to AD and CD is perpendicular to AD. rotate BC 90 ° counterclockwise around point B to get the line segment be and connect AE and Ce (1). If AB = 2cm and CD = 3cm, calculate the area of triangle Abe

Make ef ⊥ AB, intersect the extension line of AB at point G, and make BF ⊥ CD at point F
Then ∠ FBG = ∠ CBE = 90 °
∴∠CBF=∠FBG
∵BC=BE
∴△CBF≌△EBG
∴EG=CF=4-2=2
∴S△ABE=1/2*BA*EG=1/2*2*2=2

It is known that in the plane rectangular coordinate system as shown in the figure, ab ∥ OC, ab = 10, OC = 22, BC = 15, the moving point m starts from point a and moves along AB to point B at the speed of one unit length per second, while the moving point n starts from point C and moves along CO to point o at the speed of two units length per second. When one of the moving points moves to the end, both moving points stop moving. (1) find B (2) let the motion time be T seconds; (1) when the value of T is, the area of the quadrilateral oamn is half of the area of the trapezoidal oabc; (2) when the value of T is, the area of the quadrilateral oamn is the smallest, and the minimum area is calculated; (3) if there is another moving point P, it also moves along Ao from point m and N. under the condition of (2), the length of PM + PN is just the smallest, and the velocity of the moving point P is calculated Degree

(1) If BD ⊥ OC is given to D, then the quadrilateral oabd is a rectangle, ⊥ od = AB = 10, ⊥ CD = oc-od = 12, ⊥ OA = BD = bc2-cd2 = 9, ⊥ B (10,9); (2) from the title, we know that am = t, on = oc-cn = 22-2t, ⊥ the area of quadrilateral oamn is half of that of trapezoidal oabc, ⊥ 12 (T + 22-2t) × 9 = 12 × 12 (10 + 22) × 9, ⊥ t = 6, and (2) let the area of quadrilateral oamn be s, then s = 12 (T + 22-2t) × 9 = - 92t + 99, ⊥ 0 & Lt ; When t = 10, am = t = 10 = AB, on = 22-2t = 2, ∧ m (10,9), n (2,0), ∧ n '(- 2,0); let the function of the line Mn' be y = KX + B, then 10 K + B = 9-2k + B = 0, the solution is k = 34B = 32, P (0, 32), AP = oa-op = 152, and the velocity of the moving point P is 152 △ 10 = 34 unit length / s

It is known that in the plane rectangular coordinate system as shown in the figure, ab ∥ OC, ab = 10, OC = 22, BC = 15, the moving point m starts from point a and moves along AB to point B at the speed of one unit length per second, while the moving point n starts from point C and moves along CO to point o at the speed of two units length per second. When one of the moving points moves to the end, both moving points stop moving. (1) find B (2) let the motion time be T seconds; (1) when the value of T is, the area of the quadrilateral oamn is half of the area of the trapezoidal oabc; (2) when the value of T is, the area of the quadrilateral oamn is the smallest, and the minimum area is calculated; (3) if there is another moving point P, it also moves along Ao from point m and N. under the condition of (2), the length of PM + PN is just the smallest, and the velocity of the moving point P is calculated Degree

(1) If BD ⊥ OC is given to D, then the quadrilateral oabd is a rectangle, ⊥ od = AB = 10, ⊥ CD = oc-od = 12, ⊥ OA = BD = bc2-cd2 = 9, ⊥ B (10,9); (2) from the meaning of the title: am = t, on = oc-cn = 22-2t, ⊥ the area of quadrilateral oamn is half of that of trapezoidal oabc, ⊥ 12 (T + 22-2t) × 9 = 12 × 1

As shown in the figure, the position of right angled trapezoid oabc in the plane rectangular coordinate system is as follows: ab ∥ OC, B (5,3), ab = BC. D (x, 0) is the moving point on OC (not coincident with O and C), DP ⊥ X axis intersects PC at point P, connecting Pb
(1) The coordinates of point C are -
(2) (1) as shown in figure (1), use the algebraic expression of X to represent the coordinates of point P, and find the value of X when △ PBC becomes a right triangle
② When point P is the intersection of trapezoidal diagonal AC and Bo (as shown in Figure 2), find the value of X

C (9,0). Right angled trapezoid, the coordinate of a is known, and (0,3) B is perpendicular to X axis. Pythagorean theorem, the length of C is 5 + 4