1. The area of a circle is 1250π. Find the perimeter of the circle in simplified radical and
pi
Area = π r 2 = 1250π
r 2 = 1250
r = 1250
r = 25 2
The circumference (“perimeter”) is:
(
)
Circumference = 2π r = 2π 25 2 = 50π 2
2. Find the area of the circle below where d= 5 sqrt 6
(
πd2 π 5 6
Area =
=
4
4
)
2
=
π ( 25* 6 ) 75π
=
4
2
3. Find the area of the square with side 25 sqrt 2 ft.
(
)(
)
Area = 25 2 25 2 = ( 25* 25)
(
)
2 * 2 = 625 ( 2 ) = 1250 ft 2
4. Find the area of a rectangle with length (5+sqrt 6) and width (8-sqrt 8) in simplified
form.
(
)(
)
Area = 5 + 6 8 − 8 = ( 5) (8) + 8
Area = 40 + 8 6 − 5 8 − 48
Area = 40 + 8 6 −10 2 − 4 3
( 6 ) − 5( 8 ) − ( 6 ) ( 8 )
5. Before modern navigational tools, old sailing ships would have a small platform on top
of the front mast called a crow’s nest. Sailors in the crow’s nest could see land or other
3h
ships that were much farther away than the sailors on deck. The equation d =
2
can be used to find the distance d in miles a person h feet high above the water can see. If
the deck was 20 feet above the water and the crow’s nest was another 32 feet above the
deck, about how much farther could sailors in the crow’s nest see than those on deck?
Round to the nearest tenth of a mile
Distance seen from deck (h = 20):
d=
3 ( 20 )
3h
60
=
=
= 30 = 5.4772
2
2
2
Distance seen from crow’s nest (h = 20 + 32 = 52)
d=
3 ( 52 )
3h
156
=
=
= 78 = 8.8318
2
2
2
The difference is 8.8318 – 5.4772 = 3.3546
Rounded to the nearest tenth, the difference is 3.4 miles.
The sailor in the crow’s nest can see 3.4 miles farther than the sailor’s on deck.