Quadrilateral updated

[miamivince]

A quad has an area of 360 cms. It's side lengths are 28, 25 and 15 cms. Find the missing side length ? The soln is 13 cms. Don't know how to arrive @ it. There seem to be a lot of formulae and many types of quad. Thanks. Could you break part of the quad into two right angle triangles and proceed from there ?

 

[lookagain]

Remember that to calculate the area, > > 5 < < pieces of information
are required, like 4 sides plus an angle.

Denis, for a quadrilateral in general, it is not sufficient to know the "5 pieces of information." $\displaystyle \ \ \$For example, in the following case, if we know the lengths all four sides, and that a particular angle is a right angle, we cannot state what the area is, because we need the measure of another angle, length of a diagonal, etc.

Counterexample:
In quadrilateral ABCD, let the length of AB = the length of BC = 1. Let the length of CD = 2, and let the length of DA = $\displaystyle \ \ \sqrt{2}. \ \ \$ Suppose we know that the measure of angle C is 90 degrees. Does that mean that we have a quadrilateral with a unique area? No. Angle B might (also) be a right angle, making the measure of angle A 135 degrees and ABCD be a convex quadrilateral. But, quadrilateral ABCD might instead be non-convex, with angle A being a reflex angle with a measure of 225 degrees and angle B being an acute angle. The area of the non-convex quadrilateral is necessarily different (less) than that of the convex quadrilateral in my example. The inclusion of a sixth piece of information, such as knowing that the measure of angle B is, say 90 degrees, would give a quadrilateral with a unique area.

 

[DrPhil]

. . . making the measure of angle A 135 degrees and ABCD be a convex quadrilateral. But, quadrilateral ABCD might instead be non-convex, with angle A being a reflex angle with a measure of 225 degrees and angle B being an acute angle. The area of the non-convex quadrilateral is necessarily different (less) than that of the convex quadrilateral in my example. The inclusion of a sixth piece of information, such as knowing that the measure of angle B is, say 90 degrees, would give a quadrilateral with a unique area.

Is it adequate for the 6th piece of information to be "convex quadrilateral"?

 

[lookagain]

Is it adequate for the 6th piece of information to be "convex quadrilateral"?

In my example, yes.

 

[Deleted member 4993]

For a cyclic quadrilateral, Bramhagupta's (or Heron's) formula can be used:

$\displaystyle A \ = \ \sqrt{(s-a)(s-b)(s-c)(s-d)}$

where a,b,c,d are the sides of the cyclic quadrilateral and s = (a+b+c+d)/2

 

[miamivince]

The quadrilateral I mentioned in my previous question represented a lawn. In the middle of the lawn, a flag is planted , it is a proper quadrilateral, the flag is an equal distance from each corner. The other sides are 28, 24 and and 15 meters. It's area is 360 m sqd. Fourth side = ? I was working from memory. Thanks, apologies.

 

[stapel]

The quadrilateral I mentioned in my previous question ... is a proper quadrilateral

How does your book define a "proper" quadrilateral?

Thank you.

 

[HallsofIvy]

The quadrilateral I mentioned in my previous question represented a lawn. In the middle of the lawn, a flag is planted , it is a proper quadrilateral, the flag is an equal distance from each corner. The other sides are 28, 24 and and 15 meters. It's area is 360 m sqd. Fourth side = ? I was working from memory. Thanks, apologies.

Draw a line from the center of the quadrilateral (the flag) to each corner and you have 4 isosceles triangles. Dropping a perpendicular from the vertex (where the two equal sides intersect) to the base of an isosceles triangle divides it into two right triangles with hypotenuse of length x and base b/2. By the Pythagorean theorem the other leg has length $\displaystyle \sqrt{x^2- b^2/4}= (1/2)\sqrt{4x^2- b^2}$. The area of the right triangle, then, is $\displaystyle (1/2)(b/2)(1/2)\sqrt{4x^2- b^2}= (1/8)b\sqrt{4x^2- b^2}$ and so the area of the isosceles triangle is $\displaystyle (1/4)b\sqrt{4x^2- b^2}$.

You can find the area of the three triangles with the given bases as functions of x. Subtract that from 360 to find the area of the fourth triangle, with base "b". That will give you an equation in x and b. You should be able to solve for the fourth side as a function of x but I don't believe you will be able to find a specific value for b.

 

[Deleted member 4993]

Did your buddy "steal" that from Heron?

I do not for sure - but I have seen Brahmagupta's name associated with that equation.

Of course Hero came around 10 AD whereas Brahmagupta cam around ~700 AD.

Since there were no planes or Books or internet those days, Brahmagupta probably derived this independently.