As shown in the figure, D and E are the points on △ ABC sides AB and AC respectively, de ‖ BC. If ad = BD, find the ratio of the area of triangle ade to the area of trapezoid bced

As shown in the figure, D and E are the points on △ ABC sides AB and AC respectively, de ‖ BC. If ad = BD, find the ratio of the area of triangle ade to the area of trapezoid bced

∵ D, e are the points on △ ABC edge AB, AC respectively, de ∥ BC., ad = BD
∴DE=1/2BC
The area of ∧ ade = 1 / 4 of ∧ ABC
The area of trapezoidal bced = 3 / 4 △ ABC
Area of triangle ade: area of trapezoid bced = 1:3
A: the area ratio of triangle ade to trapezoid bced is 1:3

As shown in the figure, in △ ABC, De is parallel to BC, ad = 2cm, BD = 3cm, and the ratio of De to BC is calculated. If the area of trapezoidal bced is 63cm & sup2;, then the area of △ ade

Ade is similar to ABC, ad: ab = de: BC = 2:5
Ade ABC area ratio = (2:5) ^ 2 = 4:25
Ade area = 63 / (1-4 / 25) * 4 / 25 = 12

As shown in the figure, De is the median line of △ ABC and FG is the median line of trapezoidal bced. If de = 4, FG is equal to ()
A. 6B. 8C. 10D. 12

∵ De is the median line of △ ABC, ∵ BC = 2DE = 8, ∵ FG is the median line of trapezoidal bced, ∵ FG = BC + de2 = 6

Triangle ABC, AC = 8, BC = 6, AC is perpendicular to AB, according to AC, AB make square ACDE, bcfg, connect EF, find the area of triangle FDB
It's aced, bgfc, AC, BC

Yes, it should be AC perpendicular to BC, C is right angle. Make a square according to AC, BC
Well, I'm too lazy to upload a picture. Let's compare it with myself. N-many right triangles and Pythagorean theorem
In RT triangle BCF, by Pythagorean theorem, BF = (BC ^ 2 + CF ^ 2) ^ (1 / 2) = 6 √ 2,
In the RB triangle bed, by Pythagorean theorem, BD = (be ^ 2 + de ^ 2) ^ (1 / 2) = (14 ^ 2 + 8 ^ 2) ^ (1 / 2) = 2 √ 65,
In the R T triangle fad, by Pythagorean theorem, FD = (AF ^ 2 + ad ^ 2) ^ (1 / 2) = (14 ^ 2 + 8 ^ 2) ^ (1 / 2) = 2 √ 65,
So, triangle FDB is isosceles, BD = FD
Then the height of BF is [BD ^ 2 - (BF / 2) ^ 2] ^ (1 / 2) = [4 * 65 - 9 * 2] ^ (1 / 2) = 11 √ 2,
Therefore, the final area is 6 √ 2 * 11 √ 2 / 2 = 66