In ladder shaped ABCD, the area of triangle ABO is 6 square meters, and the area of triangle ADO is 15. Calculate the area of ladder shaped ABCD. Triangle ABO = triangle doc In ladder ABCD, the area of triangle ABO is 6 square meters, and the area of triangle ADO is 15 square meters. Find the area of ladder ABCD. Triangle ABO = triangle doc

In ladder shaped ABCD, the area of triangle ABO is 6 square meters, and the area of triangle ADO is 15. Calculate the area of ladder shaped ABCD. Triangle ABO = triangle doc In ladder ABCD, the area of triangle ABO is 6 square meters, and the area of triangle ADO is 15 square meters. Find the area of ladder ABCD. Triangle ABO = triangle doc

ABO=DOC
Trapezoidal ABCD = ABO + ADO + doc = 15 + 6 + 6 = 27
∵S△AOC：S△AOB=AO：OD=S△COD：S△AOD=6：15
∴S△AOC=36/15=12/5=2.4
S ladder ABCD = s △ AOC + s △ AOB + s △ cod + s △ AOD = 2.4 + 6 + 6 + 15 = 29.4

Trapezoid ABCD, AC intersect BC with O, ab = 1.5cd, triangle ABO + CDO = BCO + ADO, BCO = 12cm 2, calculate the area of triangle ADO

According to the question, s △ AOB / s △ BOC = AO / OC = 10 / 15 = 2 / 3
AO/OC=BO/OD=2/3
Area of s △ BOC / s △ doc = Bo / OD = 2 / 3
S△DOC =15/2/3=22.5
S△AOB/S△AOD=OB/OD=2/3
So s △ AOD = 10 / 2 / 3 = 15

In known rectangle ABCD, diagonal lines AC and BD intersect at O, AE ⊥ BD, perpendicular foot is e, ∠ DAE: ∠ BAE = 3:1, then ∠ EAC=______ ．

As shown in the figure ∵ DAE: ∵ BAE = 3:1, ∵ BAE = 22.5 °, ∵ Abe = 67.5 °, ∵ quadrilateral ABCD is a rectangle, ∵ AC = BD, Ao = Co, Bo = do ∵ OA = ob, ∵ OAB = ∵ Abe = 67.5 °∵ EAC = ∵ OAB - ∵ BAE = 67.5 ° - 22.5 ° = 45 °. So the answer is: 45 °

In rectangular ABCD, AE is perpendicular to BD and E, DAE = 3 BAE, and the degree of EAC is calculated

Because ∠ DAE = 3 ∠ BAE ∠ DAE + ∠ BAE = 90 °
So 4 ∠ BAE = 90 °∠ BAE = 90 ° / 4
Because ∠ Abe = ∠ DBA ∠ bad = ∠ AED = 90 ° because ABCD is rectangular and AC is diagonal, △ Abe ∽ abd ∽ ACD
Therefore, BAE = ADB = CAD, EAC = 90 ° - BAE - CAD = 90 ° - 2 ∠ BAE = 90 ° / 2 = 45 °