Question

In the adjoining figure; ABCD is a parallelogram. If AP = BP and CP meets the diagonal BD at Q and area of ΔBPQ = 20 cm², find:

1. PQ : QC.
2. Area of ΔPBC.
3. Area of parallelogram ABCD.

Answer 1

Given: ABCD is a parallelogram. P lies on AB with AP = BP, so P is the midpoint of AB. CP meets diagonal BD at Q. Area (ΔBPQ) = 20 cm².

Step-wise calculation:

1. Put coordinates (convenient choice):

A = (0, 0), B = (2, 0)

So, P = (1, 0).

Let D = (u, v)

So, C = B + AD = (2 + u, v).

Parametrize BD:

B + s(D – B) = (2 + s(u – 2), sv).

Parametrize CP:

C + t(P – C) = (2 + u – t(1 + u), v(1 – t)).

Equate y-coordinates:

sv = v(1 – t)

⇒ s = 1 – t

Equate x-coordinates and substitute s = 1 – t.

Solve for t:

t(2 – u) = 2 – t(1 + u) 

⇒ 3t = 2

⇒ `t = 2/3`

So, `s = 1/3`.

Thus, Q divides CP with CQ : QP

= t : (1 – t) 

= `2/3 : 1/3` 

= 2 : 1 

So, PQ : QC = 1 : 2.

2. Areas on the same altitude:

ΔBPQ and ΔBPC share the same altitude from B to line PC.

So, `("Area"(ΔBPQ))/("Area"(ΔBPC))`

= `(PQ)/(PC)` 

= `1/3`

Therefore, Area (ΔBPC) 

= 3 × Area (ΔBPQ) 

= 3 × 20

= 60 cm²

3. Relation to parallelogram area:

With P the midpoint of AB, vector argument or coordinate calc gives Area (ΔBPC) = `1/4` × Area (parallelogram ABCD).

Hence, Area (ABCD) 

= 4 × Area (ΔBPC)

= 4 × 60

= 240 cm²