Surface Area & Volume Quiz Solutions

SURFACE AREA & VOLUME QUIZ solutions 1. The surface area of a right prism, show n, [w ith base an equilateral triangle w ith side 12 cm, height 10cm] IS: 10 cm The SURFA CE A REA OF A PRISM is alw ays the area of the tw o identical polygonal ends, plus the area of the lateral or sides– 2 2 2 2 2. C C C C 12 cm Area of one triangular End + 3 C area of one lateral Side ( ) (area of equilateral > w . side 12cm) + 3 C (area of 10cm x 12 cm rectangle) ( ½ ) (base length) (altitude of ª )u + 3 C (length ) (height) (½ ) ( 12cm) (12cm(%& 3 )/2) + 3 C (12cm) (10 cm) u See discussion of height of ( 36 %& 3 + 360 ) sq cm equilateral below ! Find the surface area of a hexagonal based pyramid, given the base is a regular hexagon w ith side 6cm, and the height of the pyramid is 3 cm. The SURFA CE A REA OF A PY RA M ID OR CONE is alw ays the area of it s base plus the area of it s triangular lateral faces. BASE: ...to get height of equilateral 6 cm hex agon base can 3 cm be seen as Area of 6C ( ½ ) 6C ( ½ ) ( 54%& 3 h 3 cm In an equilateral triangle, the (h) is alw ays (%& 3 '2 )CSIDE ...because h2 + ( s'2 )2 = s2 HEIGHT h2 + 3 2 = 6 2 h2 = 27 ... So h = 3 %& 3 cm SIX hexagonal base + 6 (area of one triangular lateral face) ( 6cm) (6cm(%& 3 )/2) + 6 ( ½ ) (6 cm) (height of lateral face) ( 6cm) (6cm(%& 3 )/2) + 6 ( ½ ) (6 cm) (6 cm [see w ork below ]) + 108 ) cm2 c 2 = 9 cm2 + 27cm2 c 3 cm 3(%& 3 )cm 3. c 2 = 36 cm c = 6 cm 2 Find the surface area of a prism, given that the height is 10cm, and the base has a perimeter of 78cm, and area 26 cm2 . The SURFA CE A REA OF A PRISM is alw ays the area of the tw o identical polygonal ends, plus the area of the lateral sides– the area of the lateral sides adds up to the (sum of all the base lengths)(height) = (Perimeter of base) (height) ³ top So the area of THIS prism is: Area of top + Area of bottom + Total area of lateral sides 26 cm2 + 26 cm2 + ( 78 cm )(10 cm) 2 832 cm Top ³bottom The prism may have any sort of polygonal base... the ones show n are just tw o of many possibilities. 4. Find the surface area of the object illustrated at right. Assume all angles that appear to be right angles are so, all arcs that appear circular or semicircular are, all surfaces that appear flat are so, and all surfaces that appear spherical* are so. The small circular disc at the very top has area B(10m)2 On the next level dow n is a rectangle w ith a circular disc missing... Area (35m)(24m) – B(10m)2 . The tw o together total (35m)(24m) = 840m 2 The bottom of the object is another 840m 2 The front and back are 12m x 35m rectangles, the sides, 12m x 24m rectangles . The lateral surface of the cylindrical “ dome” consists of a rectangular strip, 6m x B(20m) long. 20m 6mº 20m 6mº Front View Total surface area: [ 2(35)(24) + 2(35)(12) + Side View 12m 2 (2 4 ) (1 2 ) + B(6)(20) ] m2 35 m 24 m * (There are no spheric al surf ac es in t his st ruct ure.) Finding the Volumes of these objects is much easier. They are: (1) V of prism = (area of base [¡ •!])(height) = ( ½ )( 12cm)(12cm(%& 3 )/2) C(10cm) 3 = 1440 %& 3 cm (The key here is to realize the triangular “ end” is the base.) (2) V of pyramid The volume of a pyramid or cone is one-third that of the corresponding prism or cylinder. = (1/3)( Area of hexagonal base)(height of pyramid) = (1/3) C6C(½)(6cm)(6cm(%& 3 )/2) C(3cm) 3 = 54 %& 3 cm (3) V of prism The volume of any cylinder or prism is (Area of Base)(Height) = (Area of Base)(Height) = (26cm 2 )(10cm) = 260 cm 3 (4 ) V = V of cylinder + V of box The KEY: realize the volume of the w hole is the sum of volumes of the parts. = (Area of cylinder’ s base)C(height of cylinder) + (Area of prism base)C(Height of prism) = ð (10m)2 (6m) + (12m)(35m)(24m) = 600ð m 3 + 10080 m 3 = ( 600ð + 10080 ) m 3