# Quadrilateral with area and 2 sides only

Printable View

• Feb 5th 2011, 10:38 AM
Neverquit
Quadrilateral with area and 2 sides only
Hi

I have 2 triangles when joined by a common side form a quadrilateral:

1. right angled triangle with the opposite side about 17 metres
2. non-right angled triangle with one side of about 21 metres.

Joined together their area is about 200m squared.

The adjacent side of the right angled triangle joins the non-right angled triangle so they share a side.

I'm trying to find the length of the shared side and the unknown side of the non-right angled triangle.

I'm thinking something like 200=x+y
Where x is area of right triangle and y is area of non right triangle

x=0.5(17)(shared side)
y=0.5(shared side)h(somehow?).

I was thinking of Herons rule but I dont have enough information and it would be a lot of working.

Any ideas?

Regards,
Neverquit
• Feb 5th 2011, 10:43 PM
Wilmer
Quote:

Originally Posted by Neverquit
Any ideas?

Not unless you provide a diagram.
Plus the EXACT wording of the original problem.

The way you got it worded, with approximates, means we can theoretically
make the 2nd triangle also a right triangle, work it out, then change a side
of the 2nd triangle by .000000001 or so, so it is no longer "quite right!";
get my drift?
• Feb 6th 2011, 02:39 AM
Neverquit
---------------------------------------
|
|
|
• Feb 6th 2011, 03:04 AM
Neverquit
picture . this is the best I can do
Hi I have done a rough drawing attached. This is the best I can do.

I was thinking that the triangle bcd could be solved with sine rule for area. I am trying to work out length of bd and cd..

Thanks,
Neverquit
• Feb 6th 2011, 03:49 AM
Unknown008
I don't think that you have enough information.

You could very well have different values for the shared side and different values for the height of triangle BCD, while keeping the area as 200 m^2, meaning that there is no exact solution to your problem.
• Feb 6th 2011, 04:04 AM
Neverquit
Hi,

how would you go about finding the diffferent values for the lengths bd and cd? Use excel trial and error?

Regards,
Neverquit
• Feb 6th 2011, 04:21 AM
Unknown008
Ok, let BD = 4 m, x = 34 m^2, y = 166 m^2, h = 83 m

Let BD = 8 m, x = 68 m^2, y = 132 m^2, h = 33 m

Need I say more? [I used even numbers for BD so that I have whole numbers]

The area of BCD does not depend on the side BC which is 21 m because no angle is specified and the vertex can be adjusted to get BC = 21 m in all cases.
• Feb 6th 2011, 08:02 AM
Wilmer
WHY are you considering it solved?

Seems like an "interesting" problem IF given sufficient information.

Why don't you "experiment"; say take an actual integer-sided right triangle,
say 12-16-20, then construct a non-right triangle using side=12 ...
• Feb 6th 2011, 08:39 AM
Neverquit
Hi okay,

I tried the integers 12-16-20 which gives triangle ABC area of 96m^2
200-96=104.
104=0.512h
h=17.3333

(BA-x)^2=21^2-17.33^2=140.6711^2. sqrt(140.6711)=11.8557
x=12-11.8557=0.1443

other triangle right triangle with side CD:
CD^2=17.3333^2+0.1443^2=300.4641. sqrt(300.4641)=17.3339.

Its the same as h to 3 decimal places? So that would indicate that triangle BCD is a right triangle too!

That is interesting...........
• Feb 6th 2011, 11:20 AM
Wilmer
I kinda meant set up something ENTIRELY with integers; like:
Code:

```                  B                   E          D A                  C```
Right triangle ABC: AB=50, AC=48, BC=14 ; area = 336

Now tack on non-right triangle BCD: BD=13, CD=15 ; area 84
Height DE=12; BE=5, CE=9

Now your original problem would be worded:
given right triangle ABC, AC=48;
non-right triangle BCD added, CD=15;
total area of resulting quadrilateral ABDC = 420

It will now be easier for you to "see" that more info is required.
Suggestion: pretend height DE=12 is given instead of CD=15;
see how far you get with that...
• Feb 6th 2011, 10:35 PM
Neverquit
Hi
it becomes 2 right angled triangles. share side, corresponding angles are the same. then triangles are congruent.
So they have equal area. 11.77 works best for shared side.