## Monday, August 29, 2016

### Mathcounts Strategy: Shoestring (or Shoelace) method of finding the area of any polygon

Check out Mathcounts here, the best competition math program for middle school students.

Shoelace formula from Wikipedia

More on Shoelace

Problems: Solutions below

#1:  Find the area of a quadrilateral polygon given the four end points (3, 5), (11, 4), (7,0) and (9,8) in a Cartesian plane.

#2 2007 Chapter Target Round: A quadrilateral in the plane has vertices at (1,3),  (1,1), (2, 1) and (2006, 2007). What is the area of the quadrilateral?

#3: Find the area of a polygon with coordinates (1, 1), (3, -1),  ( 4, 4), and  (0.3)

#4: What is the number of square units in the area of the pentagon whose vertices are
(1, 1 ), ( 3, -1),  (6, 2), (5, 6), and (2, 5)?

#5: Find the area of a polygon with coordinates ( -6, 0), (0, 5), (3, -2), and (4, 7)

#6: Find the area of a polygon with coordinates (20, 0), (0, 12), (3, 0), (4, -4)

#7: Find the area of a polygon with coordinates (-8, 4), (2, 12), (3, -5), (4, -4)

#8: Find the area of a triangle with coordinate (-8, -4), (-3, 10), (5, 6)

Solution I: Draw a rectangle and use the area of the rectangle minus the four triangles to get the area of the quadrilateral polygon.

Solution II: Using shoestring method. First, plug in the four points. Second, choose one starting point and list the other points in order (either clockwise or counterclockwise)  and at the end, repeat the starting point. The answer is 33 square units.

Use this link to practice finding the area of any irregular polygon. Keep in mind that a lot of the times you don't need to use shoestring method. Be flexible!! Scroll to the middle section.

## Tuesday, August 16, 2016

### Pathfinder

From Mathcounts Mini :

Counting/Paths Along a Grid

From Art of Problem Solving

Counting Paths on a Grid

Math Principles : Paths on a Grid : Two Approaches

Question #1: How many ways to move the dominoes on a 6 by 6 checker board if you can only move the dominoes to the right or to the bottom starting from the upper left and you can't move the dominoes diagonally?

Solution :
You can move the dominoes 5 times to the right at most and 5 down to
the bottom at most, so the answer is $$\dfrac {\left( 5+5\right) !} {5! \times 5!}$$ = 252 ways

Question # 2: How many ways can you  move from A to B if you can only move downward and to right?

Solution : There are $$\dfrac {\left( 4+4\right) !} {4!\times 4!}$$ * 2 * $$\dfrac {\left( 4+4\right) !} {4!\times 4!}$$ = 9800 ways from A to B