Question

AC = 25 cm $\bullet$ PT || AB and SR || BC

(i) AS:SB = 3:2 (ii) $\frac{CT}{TB}$ = 2:1

Answer 1

AB = 20 cm, BC = 15 cm

Answer 2

We begin by placing triangle ABC in a coordinate system with the right angle at B. Let

• B = (0, 0)
• A = (a, 0)
• C = (0, c)

Since AC = 25, we have
  √(a² + c²) = 25.                       (1)

Step 1. Locate S on AB.

It is given that AS:SB = 3:2. Write S = (s, 0) with S between A and B. Then   AS = (a – s)  and  SB = s. So,
  (a – s)/s = 3/2 ⇒ 2(a – s) = 3s ⇒ 2a = 5s ⇒ s = 2a/5. Thus, S = (2a/5, 0).

Step 2. Locate T on BC.

Given CT:TB = 2:1. Since B = (0, 0) and C = (0, c), let T = (0, t). Write TB = t and CT = (c – t). Then   (c – t)/t = 2/1 ⇒ c – t = 2t ⇒ c = 3t ⇒ t = c/3. Thus, T = (0, c/3).

Step 3. Locate P on AC and determine its parameter.

Let P lie on AC. Writing P in the form   P = (a(1 – λ), cλ), with 0 ≤ λ ≤ 1. A line through P parallel to AB is horizontal. Since PT ∥ AB and T is on BC, the horizontal line through P must have the same y‐coordinate as T. That is,   cλ = c/3 ⇒ λ = 1/3. So,
  P = (a(1 – 1/3), c/3) = (2a/3, c/3).

Step 4. Locate Q and use QT = 8 cm.

We have drawn two lines:  • PT is horizontal (through P at y = c/3).
 • SR is vertical (since SR ∥ BC) and passes through S at x = 2a/5.

Their intersection is
  Q = (2a/5, c/3).

The problem tells us that QT = 8 cm. Now, T = (0, c/3) and Q = (2a/5, c/3) are horizontal-separated so that   QT = 2a/5 = 8 ⇒ a = (8 × 5)/2 = 20.

Step 5. Find c using (1).

Using a = 20 in equation (1):   20² + c² = 25² ⇒ 400 + c² = 625 ⇒ c² = 225 ⇒ c = 15.

Therefore, AB = 20 cm and BC = 15 cm.