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# Give a vector based proof of Heron’s formula for computing the area of a triangle

Let denote the area of a triangle with sides of lengths . Heron’s formula states that

We prove this formula using vectors via the following steps. Assume the triangle has vertices , and with

1. Using the identities

prove the formula

2. Simplify the formula in part (a) to obtain the formula

and use this to deduce Heron’s formula.

1. Proof. We know the area of the triangle (in terms of the vectors and ) is

Therefore,

2. Proof. We simplify the formula in part (a),

# Find a point satisfying conditions if given three vertices of a parallelogram

Let the three points

be three vertices of a parallelogram.

1. Find the points which are permissible fourth points of the parallelogram.
2. Find the area of the triangle .

1. There are three possible points,

2. The area of the triangle is given by

# Use the cross product to compute the area of triangles with given vertices

For each of the following sets of points use the cross product to compute the area of the triangle with vertices .

1. ;
2. ;
3. .

1. The area of the triangle with vertices is given by

2. The area of the triangle with vertices is given by

3. The area of the triangle with vertices is given by

# Find a function whose graph has given properties

Let be a function such that the points and are on the graph of . Assume that for every point on the graph of , the graph is above the line segment joining to . Further, assume that the area of the region bounded by the graph of and the line segment is . Find a formula for the function .

Incomplete.

# Find a function whose graph has given properties

Let be a function on the interval that is nonnegative and differentiable, and such that . If for each in the open interval the line given by the equation divides the ordinate set of into regions and where denotes the leftmost region. If the areas of the two regions obey the equation

where is a constant that does not depend on , find and .

Incomplete.

# Find a function whose ordinate set satisfies a given property

Find a nonnegative function whose ordinate set on an interval has area proportional to the product of the function values at the endpoints of the interval.

With reference to this previous exercise (Section 8.24, Exercise #18) we write

But

As in the previous exercise (Section 8.24, Exercise #19) this implies

# Find a function whose ordinate set satisfies a given property

Find a nonnegative function whose ordinate set on an interval has area proportional to the sum of the function values at the endpoints of the interval.

With reference to the previous two exercises here and here we write

But,

If then for all ; hence, . If , then for all since was arbitrary. Thus,

# Find a function whose ordinate set on an interval has a given property

Find a nonnegative function whose ordinate set on an interval has area proportional to the difference of the function values at the endpoints of the interval.

With reference to the previous exercise (Section 8.24, Exercise #17) we can write the area as

Therefore,

Thus,

# Find a function whose ordinate set over an interval has area proportional to the length of the interval

Find a nonnegative function whose ordinate set over any interval has area proportional to the length of the interval.

The area of the ordinate set of over an interval is given by . So, if it is proportional to the length of the interval then we have

# Compute the area of regions enclosed by a circular sector and tangent lines

In the figure below let be the area of the shaded region and the area of the triangle . Compute

1. ;
2. ;
3. .

4. We have the following diagram, with some additional markings from the diagram in Apostol,

1. First, to calculate the area of the triangle we calculate the area of the polygon and subtract the area of the triangle . Since the polygon is two congruent triangles (since the line from to bisects the polygon into two equal sized triangles) we have

To calculate the area of the triangle we use the auxiliary point . The line from to is the height of the triangle and it has length . Therefore,

Hence,

2. The area of the shaded region is the area of circular sector minus the area of the triangle (which we calculated in part (a) as ). The area of the sector is given by

Hence,

3. For this we take the limit using our expressions for and obtained in parts (a) and (b) and use L’Hopital’s,

Then, we note that (from the trig identity ). Therefore,