# area of shaded figure herons formula

I am trying to figure out how to go about solving the following figures. Find the area to two decimal places. Finding the area of a triangle is fairly simple, but when it comes to these figures I am stuck. I tried making them into triangles and using law of sin/cos but it doesnt seem to work.

I tried the following, but its wrong:

S = 1/2 (5 + 6 + 8 + 7)
S = 13

A = sqrt( 13(13-5)(13-6)(13-8)(13-7) )
A = 147.78


here is a link to the figures http://i.stack.imgur.com/9HCcS.png

Hint: Heron's formula for quadrilaterals, (also known as Brahmagupta's formula) is $$A=\sqrt{(s-a)(s-b)(s-c)(s-d)}.$$ Note that in your computation, you have an extra factor of $s$ inside the square root. However, one must be careful since this formula only holds if the quadrilateral is cyclic. If it is not cyclic, you will need to cut it into triangles and use the cosine law to find the other side. (Extra hint: Your quadrilateral is in fact not cyclic)

For the next question with the triangle, notice that in the big triangle we have two sides of length $4$ with an angle of $60$ between then. This means they form an equilateral triangle with sides $4,4,4$. It then follows that the area is the area of the equilateral triangle with sides $4,4,4$ minus the area of the triangle with sides $3,3,4$.

You cannot apply Heron's formula to a quadrilateral. As mentioned by Eric Naslund, you may use Brahmagupta's formula instead. If you insists on using Heron's formula, here is a way: For the quadrilateral in fig. 33, cut it into two triangles, an upper one and a lower one. Let the unknown common side length of the two triangles be $c$. The area of the upper triangle is then $A=\frac12(5)(6)\sin100^\circ$. However, by Heron's formula, $A=\sqrt{s(s-5)(s-6)(s-c)}$, where $s=(5+6+c)/2$. So you may solve for $c$. Once it is solved, you may calculate the area of the lower triangle by using Heron's formula again.

For the quadrilateral in fig. 34, join an edge between the two upper vertices and let its length be $c$. The area of the quadrilateral is then the difference of the area of two triangles. Then you may proceed like the above. However, as André Nicolas has pointed out, this part of the question can be solved by other easier methods.

Take for example the second bird-like picture. This is an isosceles triangle with two of the sides equal to $4$, with angle $60^\circ$ between them, minus a certain isosceles triangle with two sides equal to $3$.

The area of the "big" triangle can be found in various ways, it is equilateral. You should get $4\sqrt{3}$. If you wish, Heron's formula can be used, but there are several more elementary ways.

The triangle to be subtracted is isosceles with sides $3,3,4$. One can now use Heron's formula. More simply, split it in two right-angled triangles, hypotenuse $3$, one leg equal to $2$. By the Pythagorean Theorem, the remaining leg is $\sqrt{5}$, and the triangle to be removed has base $4$ and height $\sqrt{5}$, and therefore area $2\sqrt{5}$.