Question

In the adjoining figure, BD = CD and P is a point on AC such that area (ΔAPD) : area (ΔАBD) = 2 : 3. 

Find:

1. AP : PC. 
2. area (ΔDCP) : area (ΔАВС).

Answer 1

Given:

In triangle ABC, BD = CD, so D is midpoint of BC.

P lies on AC and area (ΔAPD) : area (ΔABD) = 2 : 3.

Step-wise calculation:

1. Put coordinates for convenience:

B(0, 0), C(2, 0)

So, D(1, 0). 

Let A(0, h). 

Let `(AP)/(AC) = λ`,

So, P = (2λ, h(1 – λ)).

2. Compute areas using the `1/2` |cross-product| formula or base × height/2.

`"Area" (ΔAPD) = 1/2 xx λ xx h`

`"Area" (ΔABD) = 1/2 xx h`

Hence, area (ΔAPD) : area (ΔABD) = λ : 1.

3. Given λ : 1 = 2 : 3 

⇒ `λ = 2/3` 

Therefore, `(AP)/(AC) = 2/3`.

So, AP : PC

= `2/3 : 1/3`

= 2 : 1

4. To find area (ΔDCP) : area (ΔABC):

With `λ = 2/3` we have `P = (4/3, h/3)`

DC = 1 (From x = 1 to x = 2),

Height of P above `BC = h/3`. 

So, area (ΔDCP)

= `1/2 xx 1 xx (h/3)`

= `h/6`

Area(ΔABC)

= `1/2` × BC × height 

= `1/2` × 2 × h

= h

So, area (ΔDCP) : area (ΔABC)

= `(h/6) : h`

= 1 : 6