course Mth 158
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6. True or False: The volume of a sphere of radius r is 4/3pier^2. False.. Volume= 4/3pier^3.
The lengths of the legs of a right triangle are given. Find the hypotenuse.
9. a=10 b=24. c^2=a^2+b^2 …c^2=10^2+24^2…c^2=100+576…c^2=676…square root of 676=c…c=26
12. a= 14 b=48. c^2=14^2+48^2…c^2=196+2304…c^2=2500…square root of 2500=c…c=50
The lengths of the sides of a triangle are given. Determine which are right triangles. For those that are, identify the hypotenuse.
15. 4,5,6. 6^2=4^2+5^2…36=16+25…36 does not equal 41.Not a right triangle.
18. 10,24,26. 26^2=24^2+10^2…676=576+100…676=676, Right triangle
21.Find the area A of a rectangle with length 4 inches and width 2 inches. A=l*w…A=4*2..A=8in^2
24. Find the area A of a triangle with height 9 centimeters and base 4 centimeters. A=1/2b*h…A=1/2*4*9…A=18cm^2
27. Find the volume V and surface area S of a rectangular box with length 8 feet, width 4 feet, and height 7 feet. Volume=l*w*h…V=8*4*7…V=224 ft squared Surface area=2*l*w+2*l*h+2*w*h… S=2*8*4+2*8*7+2*4*7…S=64+112+56…S=176+56…S=232 ft squared
30. Find the volume V and surface area S of a sphere of radius 3 feet. V=4/3*pie*r^3…V=4/3*pie*3^3…V=4/3*pie*27…V=4.2*27…V=113.4 ft. squared.
S area=4*pie*r^2…S area=4*pie*3^2…S area=4*pie*9…S area=12.6*9…S area=113.4 ft squared
Find the area of the shaded region.
33. shaded area of a circle inside a square. The squares length is 2 and width is 2. Area of a circle= pie*r^2… the diameter is 2, so the radius is 1 because it is half the diameter. A=pie*1^2…A=pie*1…A=pie square units.
36. shaded area is a circle, inside the circle is a square with a length of 2 and width of 2. Area of a square=height * base..A=2*2..A=4… If the square is inside the circle, and the Area of the square is 4, then half the area =the circles radius. Area of a circle=pie*r^2..A=pie*2^2…A=pie*4…A=12.6
39.In the figure shown, ABCD is a square, with each side of length 6 feet. The width of the border (shaded portion) between the outer square EFGH and ABCD is 2 feet. Find the area of the border.
Area of the border= area of EFGH-area of ABCD=10^2-6^2=100-36=64 ft.^2
42. A circular swimming pool, 20 feet in diameter, is enclosed by a wooden deck that is 3 feet wide. What is the area of the deck? How much fence is required to enclose the deck?
20 feet in diameter plus the 3 feet of the width of the deck on both sides of the pool = 20+3+3, which =’s 26. Area of the deck =’s area of a circle, which is A=pie*r^2…A=pie*26^2…A=pie*376..A=2123.7 ft squared. To enclose the deck you use the circumference of a circle..C=pie*diameter..C=pie*26..C=81.7..81.7ft squared is the amount of fencing needed to enclose the deck.
What is the radius of the pool?
What is the radius of the circle that includes the pool and the deck?
What are the areas of these two circles?
What therefore is the area of the deck?
Use the facts that the radius of Earth is 3960 miles and 1 mile=5280 feet.
45. The deck of a destroyer is 100 feet above sea level. How far can a person see from the deck? How far can a person see from the bridge, which is 150 feet above sea level?
Use the Pythagorean theorem.. 100 feet= 100feet*1 mile/5280 feet which =’s around 0.018939 mile. D^2=(3960+0.018939)^2-3960^2, around 150 square root of 150 is around 12.247 miles
From the bridge.. also use the Pythagorean theorem..150 feet=150*1 mile/5280 feet which =’s around 0.028409.. d^2 equals around (3960+0.028409)^2-3960^2 which equals around 225..square root of 225=’s around 15 miles.
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