# Trapezium vector problem

• Jun 29th 2010, 09:27 PM
arze
Trapezium vector problem
ABCD is a trapezium, AB and CD are parallel sides, M is the midpoint of AD. Prove that the area of the triangle BMC is half the area of the trapezium.

I let A be the origin, then the vectors of AB, AC, and AD respectively are b, c, d. So vector AM= $\frac{1}{2}d$
Area of the trapezium= $\frac{1}{2}|b\times c|+\frac{1}{2}|c\times d|$
vector CM= $\frac{1}{2}d-c$ vector CB= $b-c$
Area BMC= $\frac{1}{2}|(b-c)\times (\frac{1}{2}d-c)|$
My problem is I am stuck here, I think it is just some rearrangement of the vectors, but I am no good at dot and cross-products.
Thanks
• Jun 29th 2010, 11:30 PM
earboth
Quote:

Originally Posted by arze
ABCD is a trapezium, AB and CD are parallel sides, M is the midpoint of AD. Prove that the area of the triangle BMC is half the area of the trapezium.

I let A be the origin, then the vectors of AB, AC, and AD respectively are b, c, d. So vector AM= $\frac{1}{2}d$
Area of the trapezium= $\frac{1}{2}|b\times c|+\frac{1}{2}|c\times d|$
vector CM= $\frac{1}{2}d-c$ vector CB= $b-c$
Area BMC= $\frac{1}{2}|(b-c)\times (\frac{1}{2}d-c)|$
My problem is I am stuck here, I think it is just some rearrangement of the vectors, but I am no good at dot and cross-products.
Thanks

Is it necessary to use vectors? If not:

1. The area of a trapezium is calculated by:

$A = \frac12 \cdot (a+c) \cdot h$

2. With $h_a = h_c = \frac12 h$ you can calculate the areas of the triangles with the bases a or c respectively. Subtract these areas from the area of the trapezium and you'll get the area of the triangle BMC:

3. $A_{\Delta(BMC)}= \frac12 \cdot (a+c) \cdot h - \frac12 \cdot a \cdot h_a - \frac12 \cdot c \cdot h_c$

$A_{\Delta(BMC)}= \frac12 \cdot (a+c) \cdot h - \frac12 \cdot a \cdot \frac12 \cdot h - \frac12 \cdot c \cdot \frac12 \cdot h$

4. Expand and collect like terms. You'll get $A_{\Delta(BMC)}= \frac14 \cdot (a+c) \cdot h$ which is half of the trapezium's area.
• Jun 29th 2010, 11:31 PM
arze
Yes, I am supposed to use vectors.
• Jun 30th 2010, 06:02 AM
sa-ri-ga-ma
Refer the attached image in#2. Draw MN parallel to AB.

$MN = \frac{a+c}{2}$

Let the vector CB be b.

Now the area of the triangle MNC = $\frac{1}{2}(\frac{a+c}{2}\times\frac{b}{2})$

Area of the triangle BMC = 2*MNC = $\frac{1}{2}(\frac{a+c}{2}\times{b})$

= $\frac{1}{2}(\frac{{a}\times{b}}{2} + \frac{{c}\times{b}}{2})$

= 1/2*[ares(ABC) + ares(DBC)]