The diagram shows a trapezium ABCD. AB is parallel to DC. AB=6a, DC=2a and DA=3b. AC and BD intersect at P such that AP:PC=3:1. Express in terms of a and/ or b

Question

AC

Answer 1

Since AP:PC = 3:1, we can express AP as 3x and PC as x for some value of x.

From triangle ADC, we can use the triangle proportionality theorem to find that AC/CD = AP/PD. Since AB is parallel to CD, triangle ADC is similar to triangle ABC. Therefore, we can replace AD with AB in the proportion.

We get AC/2a = 3x/(3x + 6a).

Cross multiplying, we have AC(3x + 6a) = 2a * 3x.

Expanding and simplifying, we have 3ACx + 6ACa = 6ax.

Dividing both sides by 3 and factoring out an x, we have ACx + 2ACa = 2ax.

Dividing both sides by x, we have AC + 2aAC/x = 2a.

Since AP is 3x, we can replace x with AP/3 to get AC + 2aAC/(AP/3) = 2a.

Multiplying both sides by 3/1, we have 3AC + 6aAC/AP = 6a.

Dividing both sides by 3, we have AC + 2aAC/AP = 2a.

Multiplying both sides by AP, we have AC(AP) + 2aAC = 2a(AP).

Using the fact that AP:PC = 3:1, we can express AP as 3(PC).

Therefore, we have AC(3PC) + 2aAC = 2a(3PC).

Expanding, we have 3ACPC + 2aAC = 6aPC.

Rearranging, we have 3ACPC = 6aPC - 2aAC.

Factoring out a 2, we have 3ACPC = 2(3aPC - aAC).

Dividing both sides by 3a, we have ACPC/a = 2(3PC - AC).

Therefore, AC = a(2(3PC - AC)/PC).