Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2010 Grade 5 Geometry
Problem
Hint
1) The square shown is divided into 4 congruent rectangles. Each of them has perimeter 30 cm. What is the perimeter of the square?
 This is actually an algebra problem.
1. The short side of each rectangle is 1/4 of the long side because 4 of them fit in a square.
2. Let X = the short side length. Then:
  X + X + 4X +4X = 30 cm.
3. Solve this equation for the length of the short side, then determine the length of the long side.
4. From that, compute the perimeter of the square = _____ cm.
2) The points A (4, 0), B (6, 4) and C (2, 6) are three vertices of a rhombus. The fourth vertex, D, also lies on this coordinate grid. What are the coordinates of D?
1. Plot and label these points on the grid.
(the first coordinate is X and the second is Y)
2. Draw the lines between the points.
3. The point D connects points A and C.
4. Find the X and Y offsets from point B to get to point C.
5. Apply these offsets to point A to get the coordinates of point D.

Problem
3) In the diagram, the length of the rectangle is 8 cm, the width is 6 cm.
The area of isosceles triangle PDC is 4 cm2.
The height of isosceles triangle PAB is 50% greater than the height
of trapezoid ABCD.
What is the area of isosceles trapezoid ABCD?
Hint
There are 3 ways to solve this problem: 2 by area subtraction and one by the formula for the area of a trapezoid.
1. First, determine the height of triangle PAB.
    It is 50% greater than the height of ABCD (which is 6 cm.), so it is ____ cm. Call this H1
2. This makes the height of triangle PDC ____ - 6 cm. = _____ cm. Call this H2.
We can use one of three methods to determine the trapezoid area.
Here they are:
Method 1: Area subtraction #1.
1. The idea is to determine the areas the large triangle, PAB, and subtract the area of PDC from it. That leaves the trapezoid!
2. Triangle PAB area =
    (8 x H1) / 2 = _____
3. Triangle PDC area = 4 sq. cm.
4. Subtract the second area from the first to get the area of the trapezoid = _____ sq. cm. 		Method 2: Area subtraction #2.
1. The idea is to determine the area the large rectangle and subtract the areas of the two dotted triangles from it. That leaves the trapezoid!
2 From the area of triangle PDC (4 cm2) you can get the length DC:
    Triangle Area = base x height / 2
    4 = (H2 x DC) / 2, so
    DC = (4 x 2) / H2 = ______.
3. Area of the rectangle = _____
4. Base of one dotted triangle =
    (8 - DC) / 2 = _____.
5. Area of one dotted triangle = _____
6. Subtract 2 of these from the rectangle = _____ 		Method 3: Area of a trapezoid
1 From the area of triangle PDC
    (4 cm2) you can get the length DC:
    Triangle Area = base x height / 2
    4 = (H2 x DC) / 2, so
    DC = (4 x 2) / H2 = ______.

2. The area of a trapezoid is
    Area = (b1 + b2) H / 2 where:
    b1 is the length AB (8 cm.)
    b2 is the length DC _____
and   H = the height (6 cm.)
3. You have all the required lengths, so plug them in to the above equation to get the answer: = _____ sq. cm.

Problem
Hint
4) Quadrilateral ABCD has three congruent sides AB, AD, and DC. It has a right angle at A, an acute angle at D, and an obtuse angle at C. What is the range of values for angle at D for these things to be true? 1. In this case, congruent means AB, AD and DC are all equal in length.
2. Using the grid from problem #2 plot the points A, B and D, (point A upper left) making sure AB and AD form a 90o angle. This makes A,B, and D the corners of a square.
3. Since the angle at C is obtuse (greater than 90o and less than 180o), it cannot form a square with A, B and D. Try to draw a point for C on that grid that makes it an obtuse angle with points B and D.
4. Note that if you move it too close to A it "turns around" and is more than 180o with B and D. and is no longer obtuse.
5. So the angle at D must be less than ____ degrees because if it were that angle or greater, then C would be 90o or less.
6. It also must be greater than ___ degrees to keep the angle at C obtuse (less than 180o.
5) Dee has 9 cubes. She wants to build a large cube that takes more than 9 cubes. What is the fewest number of cubes she would need to add to be able to build it? Make a table like this:
Cube side length
(# of small cubes)		Total number of small cubes
in the larger cube
11
2 
  
  
Fill out this table until one of the entries has more than 9 small cubes. Subtract 9 from this number of cubes to get the answer = _____ cubes.