① In △ ABC and △ def, AB / de = BC / EF = AC / DF = 4, find the perimeter ratio of △ def, M: n = 1 / 7:1 / 6, N: T = 1 / 4:1 / 3, then M: n: T =? ① In △ ABC and △ def, AB / de = BC / EF = AC / DF = 4, find the perimeter ratio of △ def ② M: n = 1 / 7:1 / 6, N: T = 1 / 4:1 / 3, then M: n: T =?

① In △ ABC and △ def, AB / de = BC / EF = AC / DF = 4, find the perimeter ratio of △ def, M: n = 1 / 7:1 / 6, N: T = 1 / 4:1 / 3, then M: n: T =? ① In △ ABC and △ def, AB / de = BC / EF = AC / DF = 4, find the perimeter ratio of △ def ② M: n = 1 / 7:1 / 6, N: T = 1 / 4:1 / 3, then M: n: T =?


① In △ ABC and △ def, AB / de = BC / EF = AC / DF = 4, find the ratio of perimeter of △ ABC and △ def
Because AB / de = BC / EF = AC / DF
Therefore, △ ABC is similar to △ def. The circumference ratio is equal to the similarity ratio
Ratio of circumference of △ ABC to △ def = 4
② M: n = 1 / 7:1 / 6, N: T = 1 / 4:1 / 3, then M: n: T =?
Because M: n = 1 / 7:1 / 6, M = (6 / 7) n
Because n: T = 1 / 4:1 / 3, t = (4 / 3) n
So: M: n: T = (6 / 7) n: n: (4 / 3) n = 6 / 7: 1: 4 / 3



If △ ABC is equal to △ def, and the circumference of △ ABC is 20cm, the length of AB is 6cm, then the length of DF + EF is - cm


I'm glad to answer your question
First, we know that two triangles are congruent
The corresponding edge is ab = ed, BC = EF, AC = DF
We know AB = 6, so Ed = 6
DF + EF = △ def perimeter - ed
Because the two congruent, so the triangle def circumference = 20
So the length of DF + EF is 20-6 = 14cm



Triangle ABC is similar to triangle def. Vertex a, B and C of triangle def correspond to def respectively. Their girths are 30 and 36, BC = 6 and EF = 5. Find the length of AC and DF


The title is wrong,
Here BC and EF are the corresponding edges, and the length does not conform to the perimeter ratio
Here five to six is not six to five



Given the vertex coordinates a (3,1) B (- 2,5) of triangle ABC, if the midpoint m and N on both sides of AC BC are on the two coordinate axes respectively,
The detailed process of finding the coordinates of point C


Let the coordinate of C be (x.y), then the midpoint m of line AC [(x + 3) / 2, (y + 1) / 2], and the midpoint n of line BC [(X-2) / 2, (y + 5) / 2]. ① when m is on the X-axis and N is on the x-axis, there is (y + 1) / 2 = (y + 5) / 2 = 0, which is impossible. ② when m is on the x-axis and N is on the y-axis, there is (y + 5) / 2 = 0, (X-2) / 2 = 0, x = 2, y = - 5