Find the area of the quadrilateral

[chwala]

Homework Statement

See attached

Relevant Equations

sine rule

I was looking at this problem today, and i was trying to figure out its area with the given dimensions shown. First, is this even possible?...i later looked at the problem in detail and realized that i had missed out on some dimension that was given on the text.
Having said that, i would like to try and see if its possible to find the area of the quadrilateral with the given dimensions as indicated...i would nevertheless like to know whether this is possible. I will try look at it and share my working later...i will make use of sine, cosine rule and the angle property of parallelogram...

 

[PeroK]

There is not enough to determine where point $D$ is. The area of triangle $ACD$ could be anything.

 

[chwala]

There is not enough to determine where point $D$ is. The area of triangle $ACD$ could be anything.

I was thinking of angle $B$+$D$=$180^0$... this has to fix the point $D$ at some point...

 

[PeroK]

I was thinking of angle $B$+$D$=$180^0$... this has to fix the point $D$ at some point...

You can put D wherever you like. BCD is any triangle with a base of 5m.

 

[Steve4Physics]

I was thinking of angle $B$+$D$=$180^0$... this has to fix the point $D$ at some point...

Angles B and D would only add to 180 degrees for a cyclic quadrilateral.

 

[chwala]

Let me try and crack this tomorrow...sorry i have been a bit busy with work...let's see where i reach on this...i will try and find the dimensions of the quadrilateral...of course by use of the parallelogram property...then i may agree with you if its impossible...thanks Perok and Steve4Physics man!

 

[chwala]

Angles B and D would only add to 180 degrees for a cyclic quadrilateral.

Aaargh that's true...i overlooked that! let's see how far I go with this...

 

[sysprog]

@chwala, what part of @PeroK's point escaped you? You don't need trig functions and such for this; it's 5th grade geometry:

$\mathrm{area~of}~\triangle = \frac 12(\mathrm{base}\times\mathrm{height})$.

In the illustration, increasing the lengths of CD and AD would increase the height. Since we don't know those lengths, we can't find the area of triangle ACD, and without knowing that area, we can't know the area of quadrilateral ABCD.

 

[chwala]

@chwala, what part of @PeroK's point escaped you? You don't need trig functions and such for this; it's 5th grade geometry:

$\mathrm{area~of}~\triangle = \frac 12(\mathrm{base}\times\mathrm{height})$.

In the illustration, increasing the lengths of CD and AD would increase the height. Since we don't know those lengths, we can't find the area of triangle ACD, and without knowing that area, we can't know the area of quadrilateral ABCD.

True, it's not possible to find the area...I just tried looking at it...cheers mate...