Sample Problem 1. Find the area of a triangle with perimeter p whose inscribed circle has a radius of r.
Just like most geometry problems, we will start by drawing a diagram.
Seeing that we have the length of altitudes and having information about the lengths of the sides which the altitudes are perpendicular to, we think about trying to split our big triangle into smaller triangles with the altitudes that we have already draw. We can easily do so by connecting the incenter to the vertices of the triangle. We will also let the sides of the triangle have lengths of a, b, and c,n such that a+b+c=p.
From this diagram, it should be clear that the lines from the incenter to the vertices of the triangle split our original triangle up into 3 smaller ones. Using the formula Area=\(\frac{1}{2}bh\), we get the area of our smaller triangles to be \(\frac{ar}{2}\), \(\frac{br}{2}\), and \(\frac{cr}{2}\). Thus, our original triangle has an area of \(\frac{ar}{2}\)+\(\frac{br}{2}\)+\(\frac{cr}{2}\)=\(\frac{(a+b+c)r}{2}\)=\(\frac{pr}{2}\). You will often seen this formula referred to by letting s, the semi perimeter, be equal to \(\frac{p}{2}\), and thus achieving \(Area=rs\).
The formula for the area of a triangle given above is just one of the many helpful formulas. Below, we will provide a list of formulas to know and a brief idea of how to prove each. We will let [ABC] be the area of the triangle, r be its inradius, s be its semiperimeter, R be its circumradius, and a, b, and c be the lengths of its sides. We will also let the triangle have angles with measures of A, B, and C where angle A is opposite side BC, angle B is opposite side AC, and angle C is opposite side AB.
Formulas:
Formulas:
- [ABC]=\(\frac{1}{2}bh\) (draw a rectangle around the triangle and look at the two parts the altitude splits the triangle into separately)
- [ABC]=rs (proved above)
- [ABC]=\(\frac{1}{2}ab\sin{C}\) (will be proved in trigonometry lesson)
- Note that this formula implies the if two triangles have a congruent angle, then the ratio of the areas of the triangles is equal to the ratio of the product of the length of the sides of the triangle that form the congruent angle. This can be easily proved with the definition of a sine, or the knowledge that the sine of an angle is always the same (the definition of basic trigonometric functions will be given in the trigonometry lesson).
- [ABC]=\(\frac{abc}{4R}\) (proved easily using the law of sines, motivation being that we see the circumradius in the formula)
- [ABC]=\(\sqrt{s(s-a)(s-b)(s-c)}\) (called Heron's Formula, proved by dropping an altitude, using Pythagorean Theorem, and lots of algebra)
Sample Problem 2. Find the length of the altitude to the longest side of a triangle whose side lengths are are 4, 13, and 15.
Because we know that Area=\(\frac{bh}{2}\), we could solve for the length of the altitude to the longest side if we knew the area of the triangle (b=15, h is the length of the altitude we want to solve for). However, using Heron's Formula, we can find the area of the triangle to be \(\sqrt{16\cdot 12\cdot 3\cdot 1}=24\). Thus, we get \(24=\frac{15h}{2}\), giving \(h=\boxed{\frac{16}{5}}\).
This problem introduced the important concept of finding the area of a triangle as an intermediate step, then using the area and other knowledge to find another value. This is where knowing all of the formulas can be helpful because you will often need to use two formulas--one to find the area and one to use the area you found, and you need to know the right formulas for the information you are given.
This problem introduced the important concept of finding the area of a triangle as an intermediate step, then using the area and other knowledge to find another value. This is where knowing all of the formulas can be helpful because you will often need to use two formulas--one to find the area and one to use the area you found, and you need to know the right formulas for the information you are given.
Exercises
- Find the ratio of the inradius of a triangle to the inradius of the triangle formed by connecting that triangle's midpoints.
Solution 1
From the equation A=rs, we have r=A/s. Using similar triangles, we can find that the triangle formed by connecting the larger triangle's midpoints has 1/4 the area and 1/2 the semiperimeter (each side of the triangle can be easily count parallel to the side of the triangle that wasn't used in forming that side, and from there we can easily find the inner triangle similar to the outside one with a side ratio of 1/2). Thus, the inradius of the smaller triangle is divided by 4 and multiples by 2, leaving a ratio of the inradius of the outer triangle to the inradius of the inner triangle to be 2 to 1.