Question

In the adjoining figure; AN = NB and DM = MC in the parallelogram ABCD. If area (ABCD) = 44 cm²,

1. find the area of ΔВСЕ. 
2. find the area of parallelogram BNMC.

Answer 1

Given: From the figure: AN = NB (N is midpoint of AB) and DM = MC (M is midpoint of DC) in parallelogram ABCD and area (ABCD) = 44 cm².

Step-wise calculation:

1. Set up vectors

Put A at the origin, let AB = b and AD = d.

Then B = b, D = d, C = b + d.

Area (ABCD) = |b × d| = 44.

2. Area of parallelogram BNMC

N is midpoint of AB

`N = b/2`.

M is midpoint of DC

`M = (D + C)/2` 

= `(d + (b + d))/2`

= `b/2 + d`

The four points B, N, M, C give

`BN = N - B = -b/2` 

And NM = M – N = d

Area (BNMC) = |(BN) × (NM)| 

= `|-b/2 xx d|` 

= `1/2 |b xx d|` 

= `1/2` area (ABCD)

So area (BNMC)

= `1/2 xx 44` 

= 22 cm²

3. Location of E from the figure and area of ΔBCE

Line BM:

`B + t(M - B) = b + t(d - b/2)`

Line AD through A and D: s × d.

Solve `s xx d = b + t(d - b/2)`. 

Writing both sides in the basis {b, d} gives `1 - t/2 = 0` and s – t = 0.

So, t = 2 and s = 2.

Hence, E = s × d = 2d. That is, E lies on AD produced and AE = 2 × AD

So, DE = AD.

Now compute area (ΔBCE):

B – C = b – (b + d) 

= –d

E – C = 2d – (b + d)

= d – b

(B – C) × (E – C) = (–d) × (d – b)

= d × b

= b × d   ...(up to sign) 

So, area (ΔBCE) = `1/2` |b × d|

= `1/2` area (ABCD)

Thus, area (ΔBCE)

= `1/2 xx 44`

= 22 cm²