to", \~~!' LSI'r-=- 5 b H 2-l

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Math 10 Name_\< Ei_' _ WORKSHEET 2.2 - Surface Area Part 1 - Find SA of different objects Usepage 3 of your data booklet to find the formulae for SAof objects! 1. Find the SA of the square pyramid below. The pyramid has a square base measuring 4 ft by 4ft, and a slant height of 5ft. 2-.::sR -=- d- b~ -\: '0 j~;;r-:, - ('J,., 1-\,(5) ~ ( L\- ) 2- / /----~1_-~\--\ - u 0 -t \ l.o I, I.. "'/ ~ to", \~~!' LSI'r-=- 5 b H 2-l ~b~ 2. Find the SA of this right rectangular pyramid below. Use the formula for a 'general right pyramid'. S f\ -=-.5\-\.rn 0\ 0\'\ \ a. c. e S. =- ( q m 'f,. 6m J t ( ~ t'r\ y..'t.it))) -\:(5m 'I- bm ':=. 5 L\ \'(\1- 't L\ d.:~ rn'2- + ''3o rf\:l =~d-\o,~ rna. J 1

3. Find the SA of the cone below to the nearest square millimetre. Formula for SA = Substitute r = 3..5..- and s = -rc.y2. --\\C, S q0 3ft ~6-c"X3 5) 2-4 (TC~3 5 1- ~ 0) SR :=- 3<'64'0 \q -t q <3 q fa, 0 L.s R-" =- \ 3 Il-t L\- l-\- "" (Y\ 2-. I 4. Find the SA of the cone below to the nearest square foot. Formula for SA = 1C y-":l- -\ \C r S Substitute r = --=:L and s = a 0 5rt:::-(TC ~92-) --t (TC~qj(20) ~ d-s 11-.4-i- -\-5la S I =\ «, \~ _'\ b t\-2> \ '-\ ~ 5. Find the SA of the sphere below to the nearest square inch. 'L Formula for SA= 4-1v {' SA =- L-\- ~ TC ~ \(:,2- - L\ -,.\\..; y. ~ 56 6. Find the SA of the sphere below to the nearest square centimetre. 2- Formula for SA= -rr,~... 5'A- -=- lc X \1- L -R'l-d~~ ls~ - qo-=\, q CI'II 2- ~ 2

@-d~il\-cm 7. A solid cork ball is covered in gold plating. The diameter of the ball is 14 cm. What is the area of the gold plating? 3Pr - TCd\:L -=-.\c ~\Y. 2-8. A cylindrical can, shown below, has a radius of 6 cm and a height of 20 cm. What is the surface area of the can? n 2. n h Formula for SA = 0-.IT''1 "T <7'- ll-r SA -=- (~1'1C"b 2-) -t- ( ~ ~ \C'f. 6 '(:20) -- ~?lo, ~ -t 153,q 'b \ Sf1- - q 'Do ';).C \l'i 2- \ 9. Find the surface area of the cylinder shown below: 2... n... L Formula for SA = ~ \\\ "'t ~ 1vr f) OOcfn r= \f> era Sf\ = \?-><rr'< 15"-)""\ (;;t" IT~ Ii? Y- '10) ~ ~o35 < 7> -r \ C)\l-~..S ~'d-~\t.\-.\o c-\'(\2- \ 10. Find the SA of this cube: /1 Ị. 5; 'ern :.... _- - -7./!'}crn 5cfl'l ~ 5o..rn...e- las 0c\-. f r\' ~ rn Formula for SA = g. t wh -\...Q.u) -\- J..h) w= 5,1= 5,h=---"'S"----_ Sf\;,. g @\<S) -'rls». s) + (51'5)] - (}...(9-.S -\'d- S -;-'J.S) -?- ("1-5)

fj ill. 1. A sphere has a SA of 3567 m2. What is the radius of the sphere to the nearest tenth of a metre? I \ -TT""' c- 2.. Formula for SA = a.;- \ L-, ------- Rearrange... s~'<la (~ foot-:: <~ ~ ~~ SA -=- ~'(2-4--ir ~ 3ft t.ttl 2- - ( - y/ r3sb~ (-::: 1L..t 1T ( ~ A' 2.f)3, ~s 1 (" -=- \ b.is5 rf'\ 2. A tennis ball has a SA of 706.86 ern". What is the diameter of the ball to the nearest em? \ "..12- C~he(~) Formula for SA = _\_L_U\ _ Rearrange... S A "TC:.-, -R 0\ 2- yc SR rgr r{sa 4 ~16(". ~~

3. A sphere has a circumference of 62.83 mm. What is the surface area of the sphere? (You will need to solve for r first!) Formula for C = 9. \'C r Rearrange... c.. '= ~ T\ r ~\\ 9-1\ C, - r ~'TC ~-a.~)~_ q,qq~1- ::.r (~ \C) [\ ~~f \ 4. Challenge Question** The cylinder shown below has radius of 7 inches. The SA of the cylinder is 1011.6 in2. What is the height (h) of the cylinder to the nearest inch? Formula for SA = ~"T\('2- -t ~ \\, h Substitute in all the values you know...then solve for h: ~ ~ ( ) ~t\~ \O\\\~ = rtf(i-)') -+ \Q.~TC'kl- y; h \0 \\ I b --= 3D -=l \ 1'1- -\ 4-3~ct ~ \) \0\\,6-3()1\~--=tlo3~l~ - L\-3.Q13 h '-t '3, cr::o h \)\~\~i 'co\.n :5 \ do 'oj if?>.o,'(, : 5 -"1 tf~ Ii?:> h 4-31qf> \ \Co.Din"'- ~ \\

Right Triangles (Pythagorean theorem) and Composite Objects Recall: The Pytha.gorean Theorem ln 3 ig t trian.gle wit )"pol!.e IJse length cad leg lenqths a and b. (2=3 2 b 1 1. Hndthe value of r, to the nearest tenth of a centlrnetre. '13 ern 2. The cone below has a height of 6 ft from the base to the vertex. The radius is 2 ft. What is the slant height (s)? Use the Pythagorean theorem to solve. 'd.~b <5 '2- ~ qo -Va ~\6,3 ~t 1. 3. Find the surface area of this composite object: 6

Remember... A-=-J.,,\.0 :~ 1 11:;;rn -=- h,l _ /" tn crn -:::.W 1-: 10 [;111 SApyramid = SA rect prism = \;Jh -t w'h + J. vj.-\ -t..q.n + ~h no \-op Total SA = ~ b S -\- Wh""," Wh -\- J( vj -+ 1'-'-\-.i~ ~ ( \0 '1- 'q') -\-('\CJ j.. \ \) "T l\0 ~ \\") t (\0.,..\ 0) + ( \o."\\') ~( 10r-1~ \ 'Do -t \\ 0 -\ \\0 -t \ ()0 t \\ 0 -t \\0 ~Sf\ =,20 Cxn2. J 4. Find the surface area of this composite object: in /2.. - I ov--- SAcone = (" T I L(S. (,\v.. ~\)~'\- ~{\~ ro bctl<..- '0 Il.<J- \) R SYtCOM ~ ('~) z,-t (IT~3"5) Q-{\b- ~(\b1( 6 em ::: q t \S 1\: 0\- Ckj \ :.~~~..~;~ J ern ~ S~ \ \ TotalSA= I.,\ 5'0.\ -t \ 4-\. 1" 1- SAcylinder ~y ("2.. -t d--if,h 1. DD~' ju~4 bottom, Sl\ C'-f\' ~ 1L (3")2. -t( ~.l( -= ~1C +36lC = \1-\Ll-\- \\1t3,,~

5. A tent has the shape of a square pyramid on top of a cube as shown. How ~many square feet of material is needed for the outside of the tent? (Find the surface area in square feet). Remember, that the shaded regions will not be included in your calculation! 7ft. 7 fl. f~n:.he::a oi t e exposed 5 rfaces of each sopa awqbj~ n Q-\-c,f' -,~~(:'\~=~::-:Il I f \ '- t I 7 l '":..n I ft.. i n" 'po..~.l------ ----- ( bl..') Ill-=- vj f\ C) bam. lj.)cw} SAsQuare pyramid = ~ 'b S -\ SAcube = ::A. ( LV" ~ / -\- JL h ) - a '0 -S ) -= ~ (8- '1-1-) -\- (~ ~ ~ )).. - ~(~ )(3 ':: 9- ( L\q -\ 4-~ ') usl p~~go;'an~h~;e~~,~~: - ("L\, <t ] ~Sl\ ~ \ '\ ~j.., ;;..- 2- S.- =. 3 -\-3. '::J S:L =: q + t"j. ~ 5 S2-~a\.~s = ~ ~\. ~5 ~ i-\-." Total SA;;;; ~4.4 -\-\q6-f?~b,t-t-\-2...1 8

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