Category 2
Geometry
Meet #4 February 2007 – Practice #2
1) The area of this square is 121 in2. How many square inches are in
the area of the unshaded region below. Use 3.14 for  and round
your answer to the nearest whole number.
2) The large circle to the right has a diameter of 12 cm. The two inner
circles are tangent to each other and to the outer circle. How many
square centimeters are in the shaded region. Use 3.14 for  and round
your answer to the nearest whole number.
3) Inside the rectangle to the right is a circle and a semicircle. The two are
tangent to each other and the full circle is tangent to the rectangle on
three sides. The semicircle uses the height of the rectangle as its diameter
which is 10 in. How many square inches are in the shaded region.
Use 3.14 for  and express your answer as a mixed number fraction.
4) The square to the right has an area of 144 cm2. Four congruent circle are all
tangent to each other and the square. How many square centimeters are in the
shaded region. Use 3.14 for  and round your answer to the nearest
whole number.
5) Square JCMS has an area of 256 ft2. A, B, and C are midpoints of MS , SJ , JC
respectively. JB and MA are used as the radii of two quarter circles inside
the square. Circle c1 and c2 are both tangent to one side of the square and
the dotted line connecting A and C. How many square centimeters are in
the shaded region. Use 3.14 for  and round your answer to the nearest
whole number.
Category 2 - Geometry - Meet #4 February 2007 – Practice #2
1) *If the area is 121, the side of the square is 11. If 11 is the diameter of the circle, then
the radius is 5.5. The two semicircles can combine to make one full circle with
radius 5.5 and an area of 3.14(5.52) = 3.14(30.25) = 94.985.
The unshaded region has an area of 121 – 94.985 = 26.015  26
2) * If the diameter of the large circle has diameter then the inner circle each have
diameter of 6cm. The are of the large circle equals  r2 = 3.14(62) =113.04.
The inner circles each have an area of  r2 = 3.14(32) = 28.26 and a total area
of 2(28.26) = 56.52. The shaded region then has an area of 113.04 – 56.52 = 56.52
3) *The height of the rectangle is 10, so the diameter of the big circle is also 10.
The length of the rectangle is equal to the diameter of the big circle plus the
radius of the semicircle or 10 + 5 as shown. The area of the rectangle
is 15(10) = 150. The area of the full circle is  r2 = 3.14(52) = 78.5. The
area of the semicircle is half the area of the circle = .5(78.5) = 39.25.
The shaded region = 150 – (78.5 + 39.25) = 32.25
4) *Since the area of the square is 144, each side is 12, so the diameter of each
circle is half of that and equal to 6 and radius equal to 3. By drawing in the
dotted lines as shown you can see that there are 16 congruent pieces outside
10 5
 of the area outside the circles
the circles and 10 of them are shaded. So
16 8
is shaded. The area of each circle is  r2 = 3.14(32) = 28.26. The area of the four
circles would be 4(28.26) = 113.04. The area inside outside the circles =
5
5
144 – 113.04 = 30.96. We already determined that of that is shaded we get (30.96)  19.3
8
8
5) *With an area of 256, each side equals 16. AM and JC are both equal to 8
and are radii of congruent quarter circles. The two together make a semicircle
with radius 8. The area of would be (1/2)(  r2) = (1/2)(3.14)(82) = 100.48.
The two small circles have a diameter of 8 and a radius of 4. The area of
each would be  r2 = 3.14(42) = 50.24 and combined the area of the two
would be 100.48. The area of the shaded region would equal to
256 – (100.48 + 100.48) = 55.04