Question

In the adjoining figure; ABCD is a parallelogram whose area is 324 cm². If P is a point on AB such that AP : PB = 1 : 2, find:

1. Area of ΔAPD. 
2. QP : QD.

Answer 1

Given: ABCD is a parallelogram with area 324 cm².

P lies on AB with AP : PB = 1 : 2.

So, `AP = 1/3 AB`. 

DP meets diagonal AC at Q.

Find

1. Area of ΔAPD
2. QP : QD

Step-wise calculation:

1. Let h be the perpendicular distance (height) between the pair of parallel lines AB and CD same as between AD and BC.

The area of parallelogram ABCD

= AB × h   ...(Using that triangle area with base on AB uses the same height h)

= 324

So, AB × h = 324.

2. Area of ΔAPD = `1/2` × base AP × h. 

Since `AP = 1/3 AB` 

Area (ΔAPD) = `1/2` × AP × h 

= `1/2 xx 1/3 AB xx h` 

= `1/6 xx AB xx h`

3. Substitute AB × h = 324:

Area (ΔAPD)

= `1/6 xx 324` 

= 54 cm²

4. To find QP : QD use coordinates (convenient and exact).

Put A = (0, 0), B = (3, 0) 

So, AP : PB = 1 : 2 gives P = (1, 0).

Let D = (0, h) and C = (3, h).

So, AB × h = 3h

= 324

⇒ h = 108, though h cancels in the ratio.

Parametrize AC: (3t, 108t). 

Parametrize DP: D + s(P – D) = (s, 108 – 108s) where s = 0 at D and s = 1 at P. 

Solve intersection AC = DP:

3t = s and 108t = 108 – 108s

⇒ `t = s/3` and t = 1 – s.

So, `s/3 = 1 - s`

⇒ `4/3 s = 1`

⇒ `s = 3/4`

5. `s = 3/4` means Q is `3/4` of the way from D to P.

So, `DQ = 3/4 xx DP` and `QP = 1/4 xx DP`. 

Therefore, QP : QD

= `1/4 : 3/4`

= 1 : 3