Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2009 Grade 8 Geometry
Problem
Solution
1) The area of the outside square is 64 square inches. Points A, B, C, and D are midpoints on their respective line segments. What is the area of the shaded circle? Give an exact answer in terms of
  1. If the area of the outside square is 64 sq. in., then the side length = 8 in.
  2. Connected midpoints of the sides of a square make an inner square whose area is 1⁄2 of the outer square's area = 32 sq. in.
  3. The side length of the inner square is √32
  4. That makes the radius of the inscribed circle √32 / 2
  5. So the area of the circle is:
    (√32 / 2 )2 =
    (32/4) = 8 sq. in.
2) A garden is made up of three shaded equilateral triangles inside a circle as shown. The triangles have side length 10 feet. The parts of the circle that are not part of the garden are filled with white decorative rocks and are shown as the unshaded regions. What is the area covered by the white rocks? Give your answer as an exact answer OR rounded to the nearest tenth. You may use = 3.142
  1. Area of the circle = 102 = 100 sq. ft.
  2. Area of an equilateral triangle =
    a2(√ 3) / 4 where a is the side length.
  3. Area of 1 equilateral triangle = 25√ 3 sq. ft.
  4. Area of 3 triangles = 75√ 3 sq. ft.
  5. Area covered by the white rocks =
    100 - 75√ 3 sq. ft.
Note: You should memorize the area of an equilateral triangle. It comes up often. But just in case you haven't, here is it's derivation:
To compute the area of an equilateral triangle,
you must find its height:
To find this height, you must use the pythagorean theorem:
The sides are a and the height (h) divides the base into halves (a/2) So, by the Pythagorean theorem:
h = √ a 2 - (a/2) 2  =
√ a 2 - a 2/4  = √ 3a 2/4 = a (√ 3 )/2
Now, the area of 1/2 of the equilateral triangle = (a/2) h/2 = (a h)/4 Two of them, which equals the area of the equilateral triangle =
(a h)/2 = a (a√ 3 )/4 = a2(√ 3  ) / 4

Problem
Solution
3) The surface area of a rectangular prism box is 288 square inches. Another box that has twice the width and twice the height has surface area 672 square inches. What is area of one side panel of the smaller box given by the product of its width and height?

 Let L = the length of the box, H the height and W the width.
  1. The surface area of the smaller box is
    2WH + 2LW + 2LH = 288 sq. in.
  2. The surface area of the larger box is
    2(2W)(2H) + 2L(2W) + 2L(2H) = 672 sq. in. so:
    8WH + 4LW + 4LH = 672 sq. in.
  3. Double the first equation and subtract it from the second:
      8WH + 4LW + 4LH =  672
    - 4WH - 4LW -  4LH = -576
       4WH = 96 sq. in.
    Therefore WH = 96/4 = 24 sq. in.
4) Paula is playing around with sketches of a partially opened laptop. The top and the bottom of the laptop are both represented by parallelograms. If angle NPM is 45 degrees and angle MRS is 20 degrees, what is angle PMR?
  1. Because the opposite sides of a parallelogram are parallel PN is parallel to MT and RS.
  2. Angle MTN is equal to angle NPM = 45 degrees
  3. Therefore angle PMT = (360 - 45x2) / 2 = 135 degrees
  4. Angle RMT = (360 - 20x2) / 2 = 160 degrees
  5. Therefore, angle PMR = 360 - 135 - 160 = 65 degrees
5) If the rhombus in the figure is rotated clockwise around point P by 90 degrees, what will be the coordinates of the image of point R? Consider sketching the entire image rhombus to help you think. Clockwise is this way: ↷
  1. Translate point P to the origin: P(5-5,4-4) = (0,0). The translation to get P to the origin is (-5,-4). Therefore, the translation to get it back to it's original position is (5,4)
  2. Translate point R by this same translation:
    (1-5,8-4) = (-4,4)
  3. Rotate it 90 degrees clockwise around the origin:
    Rotating point (x,y) 90 degrees clockwise makes it (y,-x), so
    (-4,4) becomes (4,4)
  4. Now translate it back: (4+5,4+4) = (9,8)