Surface Area Quiz - Answers

SURFACE AREA QUIZ Find — the surface area of... 1. ANSWERS a right prism: base is a triangle w ith sides 12cm, 10cm, 10cm prism height is 20 cm . 10cm The SA of a prism is alw ays the area of the tw o identical polygonal ends, plus the area of the lateral or sides. 10cm 6 cm 2. 20 cm 12 cm 2 C Area of one triangular End + Areas of lateral Sides ( ) 2 C (½) 12 cm C 8 cm + 20 cm (10 cm + 12 cm + 10 cm) 2 96 cm + 20 cm (32 cm) = 736 cm2 Find the surface area of a right pyramid, base 10 m square, height 12 m c 12 The surface area of a pyramid or cone is alw ays the area of it s base plus the area of it s triangular lateral faces. Area of square base + 4 (area of one triangular face) 10 m C 10 m + 4 ( ½ ) (10 m) (height of lateral side) 100 m2 + 4 ( ½ ) (10 m) (13 m [w ork ú]) 100 m2 + 260 m 2 = 360 m2 Lat eral Fac e c c 10 m 3. Im port ant ! The height of t he lat eral t riangular FA CE is slant ed (leaning in t ow ards t he apex ), & is LONGER t han t he height of t he py ram id. W ork ö 12 2 + 5 2 = 144 + 25 = 169 = 13 = c2 c2 c2 c 12m ²Height of py ramid 5 m a right prism, given the height is 20cm, and the base has a perimeter of 78cm, and area 26 cm2 . The surface area of a prism is the area of the tw o identical polygonal ends, plus the area of the lateral sides. For a right prism, the area of the lateral sides adds up to the (sum of all the base lengths)(height) = (Perimeter of base) (height) So the area of THIS prism is: ³ TOP Area of top + Area of bottom + Total area of lateral sides 26 cm2 + 26 cm 2 + ( 78 cm )(10 cm) 2 832 cm 4. ³bottom ...the object illustrated at right. Assume all angles that appear to be right angles are right, 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 circle cut out... Area of “ roof” : (35m)(24m) – B(10m)2 . The tw o together total (35m)(24m) = 840m 2 (t he w hole “ roof ” !) The bottom of the object is another 840m 2 The front and back are 12m C 35m rectangles, the sides, 12m C 24m rectangles . 20m The lateral sides of the cylindrical “ dome” consist of a rectangular strip, 6m high by B(20m) long. Total surface area: 6mº 12m Roof&bottom + Front & Back + 2 Sides + Cylindrical w all = [ 2(35)(24) + 2(35)(12) + 2 (2 4 ) (1 2 ) + B(6)(20) ] m2 = 1680 + 840 + 576 + 120 B m2 = ( 3096 + 120 B) m 2 * (There are no spherical surfaces in this structure.) 20m 6mº Front View 35 m Side View 24 m VOLUME PROBLEMS We can much MORE EASILY find the VOLUMES of these objects! They are: (1) V of prism = (Area of Base)(Height) = (area of prism’ s base [¡ •!]) @ (height of prism) = (½ )( 12cm)(8cm) C (20cm) = 960 cm3 (2) V of pyramid = one-third of the corresponding prism = (1/3) (area of base) C (height) = (a)( Area of square base)(height of pyramid) = (1/3) C (10m)(10m) C 12m = 400 m3 (3) V of prism (4) V = = (Area of Base)(Height) = (26cm 2 )(10cm) = 260 cm3 V of cylinder + V of box = (Area of base of cy linder ) C (height of cy linder ) + (Area of base of box ) C (height of box ) = (Area of cylinder’ s circular base) C (cylinder’ s height) + (Area of box base) C (box height) = ð (10m)2 C(6m) + (12m)(35m)C(24m) = ( 600ð + 10080 ) m3