Hands-on math!

Washington State Math Olympiad
Hints and Solutions
2009 Grade 8 Geometry
Problem
Hint
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 = ____ 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 = ______ sq. in.

  3. The side length of the inner square is _______________ sq. in.

  4. That makes the radius of the inscribed circle __________

  5. So the area of the circle is:



    Area = _________ 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 = ______________ sq. ft.

  2. Area of an equilateral triangle =
    a2(√ 3) / 4 where a is the side length.
  3. Area of 1 equilateral triangle = _____________ sq. ft.

  4. Area of 3 triangles = ___________ sq. ft.
  5. Area covered by the white rocks =



    ______ sq. ft.

Problem
Hint
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


  2. The surface area of the larger box (doubling the width and height) is


  3. Multiply the first equation by a factor that will eliminate the LW and LH parameters when added to the second:




    Therefore WH = _________ 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 = _____ degrees
  3. Therefore angle PMT = ________ degrees
  4. Angle RMT = ______ degrees
  5. Therefore, angle PMR = _____
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. The translation to get P to the origin is (___,___). Therefore, the translation to get it back to it's original position is (___,___)
  2. Translate point R by this same translation: = (___,___)
  3. Rotate it 90 degrees clockwise around the origin:
    Rotating point (x,y) 90 degrees clockwise makes it (y,-x), so
    Point R is now (___,___)
  4. Now translate it back (using the inverse translation above): R = (___,___)