dao 2011 solved paper 2 mensuration with key

18
(4) oAD/599 A tdangle can have (1) fivo right angles / (2) tlvo abtuse angles a (3) each angle less thao 60'> t f Lwo aarLe ar'gles _,ll , , ") (1n e'-.'' - " \:4 1. In a ^,4-BC ml1= 720' . If 41 is rlhe foot of the perpendicular liom A on BC , then the points are in the order (]) B-M -C A. ]n (2) M-B-C rl )r" t\+ JEl M-C B \:.C - O --. I 14) C M-B t, BM is an altitude. similar to (1) ^MBC (2) L ACB (3) I\CAM &) ^,43C -A Then A AMB is The sum of Dentagon is (1) 360' (2) 180' ,6 540' (.4) 270" In AABC , angle is (1) l4 In a A,4BC, !l is a right angle and the interior angles qL4_ tz"-'j)'2. -:4r'7' -,& : 'iv !.1 D ,<i Tle angles r,f a lriangle ar- 21 35 3' -5 and 5( 50 dpgrpcs. Then r = L v-\ t\ : l (1) 10' 3', - < qz - fll 6 zo' "-',=*'- tua -'ro :) (3) 30' L^. (4) 40. 'n ' *. 1\0 '^ ' Y7' ,t. l:Lz 4= o' + "', th"t' the-lEht tc ln L: )-B /N z1 i\ n ._T-*o o"saa p 4b I lI the areles A.B.C ol a l-rienqle cr" in the ratio.2 : A : 4, then alg{p]s .tf ao' (2) 30" (3) 40' (4) 90" 6a , Thrce anele- of a ouadrilaleral arts rpspe.trvclr equal to I l0', 40' and 5u then the fourth angle is i1) 70' (2) 140' ,t21' 160" (4) 90' c l: ' th"' " 5p .)+ .6 (3) \4) orlC appsc group.blogspot.com

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Page 1: DAO 2011 solved paper 2 mensuration with KEY

(4) oAD/599

A tdangle can have

(1) fivo right angles /(2) tlvo abtuse angles a(3) each angle less thao 60'>

t f Lwo aarLe ar'gles

_,ll , , ")(1n e'-.''

- "\:4

1. In a ^,4-BC

ml1= 720' . If 41 is rlhe

foot of the perpendicular liom A on

BC , then the points are in the order(]) B-M -C A.

]n(2) M-B-C rl )r"t\+JEl M-C B \:.C - O

--. I14) C M-B

t,

BM is an altitude.

similar to

(1) ^MBC

(2) L ACB

(3) I\CAM

&) ^,43C -A

Then A AMB is

The sum ofDentagon is

(1) 360'

(2) 180'

,6 540'

(.4) 270"

In AABC ,

angle is

(1) l4

In a A,4BC, !l is a right angle and

the interior angles qL4_

tz"-'j)'2.

-:4r'7'-,&

: 'iv !.1

D,<i

Tle angles r,f a lriangle ar-21 35 3' -5 and 5( 50 dpgrpcs.

Then r =L v-\ t\ : l(1) 10' 3', - <qz - fll6 zo' "-',=*'-tua -'ro :)(3) 30' L^.

(4) 40. 'n ' *.1\0

'^ ' Y7'

,t. l:Lz

4= o' + "', th"t' the-lEht

tclnL:

)-B/N

z1 i\n ._T-*o

o"saap 4b I

lI the areles A.B.C ol a l-rienqle cr"

in the ratio.2 : A : 4, then alg{p]s.tf ao'

(2) 30"

(3) 40'

(4) 90"6a ,

Thrce anele- of a ouadrilaleral arts

rpspe.trvclr equal to I l0', 40' and 5u

then the fourth angle is

i1) 70'

(2) 140'

,t21' 160"

(4) 90'

c

l: ' th"'

"5p .)+ .6(3)

\4) orlC

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Page 2: DAO 2011 solved paper 2 mensuration with KEY

\o' ?'a

:|rv?A\otv ->" zt72.l "7

9. In \ABa, rhc s.des are 6. I0 and 8

then AABC is a

right angled trjangle

acute angled t angle

obtuse angled trra gle

equilateral triangle

The total surface area ofa cylindel i-s10.

(2)

(3)

(4)

(7) 2rr1.2)

"sr(41

t2)

(3)

2tzrh

2nr\r+h)

2n\r+h)

1'--)tlG't/r;'4+ , " "t'\'7tlt (t:+')

11. A qgtal p+9 iq !f lqlong, Th14I'.91drqmeter of a cross secl io-L!:-il cm

ThlLthe il]ner,cur:ved-sudacq 4t194 is

968 cm'

484 cm'?

102 cm':

986 cm'

- 1,4 _l,-/2> \ tgt>/12

'?1

#{-

/ tz. m. curvcd surf.rce area ol c right\ -.' circular rvljnder ol beiqhl 14 .m <

88 cm'. Then the diameter of the base

of the cylinder is

(1) l cm

5y' z.^(3) 4 cm

(4) 0.5 cl]rI7l

{

27/A)

,'*

(6)

Lq

(1) 748 m't-A/ 7 48 m')

(3) 74.8 m?

\.4) 7 480 n2 ) '

::y,;*"r@oAD/599

+;$.,ta' I

1 t3., To rncke a rlo.ed ct intlrical tank, o[\ -/ r."i*tl r 'o

*"JG"" ";--, '.. U! "- -

from , m"tal shpet, hc sheel reqJirpd

-n'to ,'i l2n.- r' rt a

Y

,14. A Jukers ccp i-in r'lclorm ol a righl\ - crrculcr conc of bcso r.'dius/ / , m and

n.igl^t 24 cm. Th"n ihp^ ar6a of th-shaet rcqLrrTed lo mdRp ru rru.n caps ls

A\ et'"/ t\

.7'n\. t0 (.-na! +'tI6 ,)

2-3aJ {,{ i

(1) 500 cm'

(2) ?00 cm'?

yfl 55oo crr''

(4) 550 cm,

15. The radius and slant height of a coneare in the ratio 4 ; ?. If its clunedsq1!4qe qlpq.. E-.ry--l!+':, ther theradius is

(1) 10 cm

(.2) 22 cm

(3) 20 cm

4gF tz"

Il h" radiru-,a nd vertrcal he.ghL of a

cone are 5 cm and 12 cm r:espectivelv,ihe culved suriace area is

,4...

/ lYv'h{! )t.ffi t11\

'va-t'/-r "I .,n/"'

%t- ,?

fit .l/ , 172-)- -t?- ,-

7

:,. '

16.

(1) 20{) cm'z

\'-."1 \'@<

(3)

(4)

2A4 I crr'27

200? cm'?7

204 cm'

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Page 3: DAO 2011 solved paper 2 mensuration with KEY

t{'(8)

6)

1:* '''

9

2 tz '2-

of tbe sphere of

rt: I -,x''.-._ .f:r '?

,.@ .A

,,"1 ^,'/

mqoil- lsof theratio of

of the roorn is

(1) 60 Ctr

(2) ,15 dn

(.3) 72 d.l\ /(4) 30 dn *-

.!,\n,

'],'.lc 6"r., 'l

19. If the diameter of theappror.inately one fourthdicmeicr of fhc crrlh, ihetherr sur'faqe preasjs

16

€J. 1v'

!s.-.1" !-/{

"\'1" "'

23. The inner di3:x:tel of a cvlindrical

'lvooden pipe is 2'4 cn and its otil.er

diameler is 28 cm. 'I\: leeg!! of ihe

pipe is 35 cm. The volume oi the rvood

used is

(1) 5700 crn3

X s120 entj

(3) 5000 cmr

(4) 502i1 nnl

spher'e witb sudace

A\l"' .,.,

,-ti'-r-^

17, In two cones, if the cgllql-sql]face.rc. of one i.-I.yli!! 'hat ol thc othetand th" :lcttf l.rq\1 o' LIlo r"1 Pr s

r\\'rre rhrl of " iot^"i, rhe rctro o[their radii is

F(.2)

(3)

,1 :L

4:5

3:5

The surface arearadius 7 cm is

(1) 394 cml

(.2)

(3)

*4

2464 cm'?

(rl

(.2)

"F1.4r

20. The ladius of aarea 154 clr]'z. rs

(1) 7 cm

p1' 35crn

(3) 14 cm

(4) 49 cm

3:1

616 ctL') d rt%"

t\r- = - tl

@

A dght ci..'cular cylindcr juit encl!:ers

a sohcr" ofr-o,'r- , Th.n lre rcliu ni'

-urlers arn, ', i.-.1rr"* cld iliv-cdsurface area oi t-ie c1-liridir is

yf, t,tt2) 1:2

(3)

\+)

Thts lpng'n. broadth ard ' orgnr oI e/\

roo1l1:rre in tire ratio(3./: 2 : 1 I!-$"volume is 929'14 dm3, then the len9lh

r .1"_.) 'rLl-.i

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Page 4: DAO 2011 solved paper 2 mensuration with KEY

-ir /z'

E/ ,r. Thc inner radius of a pipe is 2 cm.

Horv much water ca:r 14 n ofthis pipehold?

(1) 0.176 mi ,----l,-/(2) 176 n:l , .v

t \ nt ..t/a^(3r 1?6m l/'A-itfr., .a

)4t u.urtolt7' I

A solid cylinder has a total lurfac-qare" ol 462 rm- Its -curvpd slrdacealea is one.thlrd oflts tolcl sur{acearea. Then the radius of the cylinder

26. T\vo cy-indrlcal cans havp bascs oI Llresdme sr7c. The diamcler of eacn is--- ^ , ,ir{.c}D,une oTTne cans ls tu cm lenglnand the other is 20 cm. Then the ratioof tlleir capa€ities is

(l) 'l:5gf r,z(3) 3:4(.4) 4:7

27. I'he volume of a circulag eq4g is

(1)

(y I ",3

2 rnz h

!). 7 cm

(2) 3.5 cm

(3) 14 cm

(4) 21 cm

lt l r iro-)\4 .,,n',."'s-

, ')'q , ^'/

;':,"{30$

rr_:--l fl,/^\q6

t>2'

(2)

l4) 15

(10)

,n 6,.";?

rlh- t:l oAD/599

28. If the height of a cone is 15 cm and itsvolume is 770 cm3, then the radius ofthe base is

(1) 5 cm

(:J)

,/ \ j\t6L -,j1tt" r i .,1 , '"-tulf-7

29. The volume oI a right circular cone is

9!59-em;. If thc diameter of the base

i, 28 cm. thel rhe heieht oI the cone -.

,{2,,

11'h ' 1? -h

- '--

(4.J

'1

\(3) 21crrL tt

7cm

10 cm

12 cm

6 +a"

e) 32 cm 3

(4) 12 cm

The base

ratio

is

(1)

""/a s

(,r) 9

radii o{ two right circular

the Ea4rc igight are jn the

3 : 5. The ratio of lheir volumes

73

'?'n''''"

(.2)

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Page 5: DAO 2011 solved paper 2 mensuration with KEY

31. In the triangle

then cos A =

11) #,?#(3) 1!

9

(4r *

(1) 3: 1

3

2

'r@a''.\,

*rf-'4

ABC , if sinA= 9.15

it

32. The diameter of a ci4lecqct ruqsc4 qt48:a.sgttarg_is 10 cm. Itsside u.ill be

(1) 3.5 cnlD .5'E cm

(3) 6 cm

(1) 4rli crr

t.2) 2

(3) 3

(4L. 1

Th" cr". o tl)p.l-rdco region in rhefigule if ,4,BCD is a square of lengthol each side 14 cm

llv 12 sq.cr,1

(2) 100 sq.cm

13) 15.1sq.cm(4) 196 sq.cm

Gi .-,,-.-tr- .r\[}

If the volumes of two hemisphe-res arein the ratio 1:27, then the ratio of

'ff:ri.

(12) oAD/599

A sP. or i- ' L|ro,D r c ., le o -ao Ur21 cm. The angle of i,he sector is 150"Then the length ofthe arc is

(2) 45 ctn

(3) 48 cIn

(4) 55 cm

The distance befiveen the centres ofthp rrvo cirr.es o ruJii ,, "nd r i" r/'fhel will rourlr pach orreln'prna.l.

;--)\1- -/'-7--,..I ni-J

,an . r-i\L--'l?$*

or'.1,,, , .-)

(-

35.

:@.. .,t'r

j

'4

if(.7). d=r. r"

(2) d=\+12

(3) .t = J\;(1) d =2(\ + r,)

An arc makes an angle of 72" atcentre of a circle of radius 10 crn.length will be

theIts

u'L

(7) 2

(2) 4

(3) 6

(4) 8 :t4- I -

7'/ ''>

T\vo r ir, lc" hrve. romrrun. rord .,!9rI r ne radlJs u' thp ( .,p rro aqLJl 1013 cm ald rhe dis enre L,eruecrl rh.centres is 41 cm, then the length ofthe chord is

( 1),\ 5 cn1

(2) 72 cn(3) 25 cm

(4),.. 10 cm

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Page 6: DAO 2011 solved paper 2 mensuration with KEY

Ifthe chords A,B and CD intersect atO as shown in the ligure, then "9

tro"

0g

40. The slope ofthe line

(.7) AC.AB= BD 'CD(2J AO OD=CO OB

(3) . AO OB =CO O1)

(4) CO AO =OD'

De

(1) 3

(2) O

(3) 5

t4J -3

4t. Two straighf lines areslopes are

(1) zero

(2) equai

(3) not equal

(4) unde{ined

(r) r

(2) 1

(3) 0

parallel if their

42. If L$ro lines intersect at right aEgles,then the product ol their slopes is

j--

(14)

-.1iq9'2 t|' ,r'.v -\,^" I-l ^^^,--v [7 oAD'ses

l):re slope of a line perpendicular tothe iine s:r|j 2y+4=0 is

(r)

(2)

(3)

\.4 )

_!5

5t2

(7) 2,7

(2)- 6,2(3) 5,2

(4) 3,5

(1) (3, - 2)

(2) ( 3,2)

(3) (2,3)

I (3, 2)

b

-2

2

5

lf A=14,2), B=(1,y) and AB=5,then the possible values of l are

^aPo (!'J) = "

' o:1)'" '"

''l - '1 z'

@45. One end ofla@glqelqglq=cqrgle is

(3,2) and the cenhe is (0,0). Then thecoordinates of the other end of th,e

diameter is ._ {.lt " ,)

c

- ),'"

46. If R divides the joj-n ef P (3.9J--and

A(6,-f) in the ratio 1 : 2, then the

coordinates ofthe point R is

(1) 0,6)

(2) (1,2)

(> (4,1)

(4) (2,3)

I

0,"') \

tfc t 2rt

>( ("1. t)

-vT3

Lta1

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Page 7: DAO 2011 solved paper 2 mensuration with KEY

E47.

48.

49. A rexagesimal measure of 72o is equalin circular measure to

The equation of a line with the slope

I and passing through a point (2, 3)J

1S - f n. ,(1) 2r+3y ?=0\2) ic 3] 13 = 0

,: , .t *.yn-L)'3- 2r-3y l3-0 ,..-1 '

-a a t(4) r. +2t =9 " !,r ., ..lif'.':,.-

Find the equation of a line passinglhrouqh rhe poinr t4. 3J whose sum ofrhe nrerc"pls on the coordinale a-\es

& q-rr3+t'','r'- 2'I -- - )-t ,y

--.f V_ ,i

-.r't i

G) i

@

rs -I(7Y r 2y+2=0(.2) r.+2y-2=0(3) y-2x+3=0(.4) 2y+x 2-0

3'a'a*'t"

(2)

(3)

.'-/'

3

--rTxlL

50. In a right angled A. BC, !! is

acute, Lq=90" and tanA =5

. tnen12'

(1)

l2)

5

13

t25

l312

12

BU

(3) 1j.:(41 (4)

(16)

and a, is

oAD/599

acute, then

?*{=v ,-.3z

51. If sin d =95

(1) ::3

(s) +

.,_.. +'r+,

I "'1

Qt 2/

5t4)4

2I'he value of cos60'

cos 30'

,y r+2Ji2

(2)

(3)

2

1+16

t'{)6 4

.(a .+T.), 61' .!- .',,.-' '.

- ..t'').4::?+ sin 60" + 2 sin 30"

zr)--

,.9.

(4) + 9*

53. The dnglejr+2y 6=0

(1) 45.

between the linesand2r+4y-6=0is

z

(.2)

(31

60'

s0"&

D"/

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Page 8: DAO 2011 solved paper 2 mensuration with KEY

54. A laddcr 6 $1g r" plcr.pd ag.r.nst a

verlical $ail .-o rhar || .naqes al rngleof 60" with the souna. 4!:Sb3!,!9fCbtabove the ground does the laddet

Itlr Em

e) 6",6,n

(3) s"li m./(4) .J5 n

1h" lengh ofthe shadow ofa pr'lar rsE-i3 rmps irs heighr. TJ^o angJe uf

elevation of the source of light is

,./,

AtL)a"e. , \

4.lj .l7 ,<:.

) J-l

(1) 20.

\2) 30" -/(3) 45'

(4) 60'

If x is acute and sin 2J. = cos rc , then

0) nE z'1

(3) -.,5

,.i-'/..''t'I

)' r'" ' '1-- t,'- 3 a

(18) oAD/599

If ir = sec d+ tan d andy = sex9 lan 0 , then the relationbetween ic and y by elirninating d is

(1) ry=\E(.2) r+-r,=1 o'l' I

(3) q, =1.,. c,i.]'

(4) 1= 1 , k(',' ',

| 58)

(1)

t2)

(3)

(41

i{2

0/1

,

+{,

., {! 'r,,,,t

+ sin355'=

- 8., 5

a<,<

5 + {n ' t)"

- srn'd cos' d1ne vatue oI 1s1-cosd 1+sind

rrJYub(1) sin d cos l/

(2) sn 0 + cos 0z/

(3) cos a - sin d

(4) 2si.n0-cos0

a, va

\'.')j:.,'

\'',itit

60. If the lines 2r + 3y + 9 = 0 and24x + hy 1.3 = A are perpendicular,then the value of A is

(1) -16/(2) 1E

(3) -8@) 2I

7.z.iz)-- *1?R

I< r - lc-

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Page 9: DAO 2011 solved paper 2 mensuration with KEY

61. Thc ratio of the volurnes of tt'o conesis 4 : 5 and the ratio of the-r4dij' oftheir bases is 2 : 3. Thc ralio of theirvertical heights is

(1) 4: 5

(2) 5,4(3) 3:5(4.) 9:5 .'

63. The volume of the largest ghtr r, ulrr , onp tlrJ Ca.l bp r.rr out nl e

cube with 9 cm edge, is

26',13

1.1

26'.13

't

729 cm3

891 cmi

-Att' - '!x !)?- 1 n-,>

-.

62. If thc circumference of the bas-e.of a9 rrr hiqh woqde4 solid cone is 44 rn,rlre'r rFc v"lLme of rh" conc i"

( 1) 440 mr /1,, "9 / "\.)^ l\,2, 4600 n ' ?-\

..' '3 r42Bm "rt.

u" ,1.t' ' ./u-,, ?. 4 4c2 n:-/ */: " "-4

(t)

t2)

(3)

(.,1)

liT\ o'"/-.--1 1 r,

:"4't;'or"/

If a cylinder and a conc hull"gg.+!radij 9{ thgir bases and equal hcightstheD the ralio oltheir l'olu1nes is

(1) 1 :2(.2) 4:11.3) 3:7414) 2: I

(20) oAD/599

C rl,'es o A-BI- hr\:rA edge" l8 crza cm a;d--50 -cm cre melted undrecr"r inlo .1 np$ cubc lhn pdgp oi lhp

"a.i' "i"6; ih..i r;-;-d is

(1) 26 cm

(2) 36 cm.'(3) 32 cm

(4) 38 cm

rrrL6 n'r'n+l+q ^f -llL in :

ncrr r"phcrieuJ bowl oI dran"t.r10.5 cm is (unto three decimals) / tj6) 0.303 1t s r!2

| {-l(2) 0.102lt \..-si3) 0.2131t --?)

".1^.h>,'14) 0 243 lt ? r'z't! -3

If the nrrrrb"r' of jqlrare (pn.r.meLers

on the sur:face of a sphere is equal tothe number olcubt( ccrlrrrelF"s in it.volume, then the diametcr of the

66.

sphere is

(1) 10 cm

(2) B cm

(3) 6 cn /a(4) 4 cm

,tF= +1:

r^,i7

..r3

68. If the rndius of the sphere is do-u!1e{,then thc ratio o{ t}re volume of thefirsl spbere to bhat ofthc second is

(1) 1 :4(2) 1 :3(3) 1: 5

(4) 1:8r'

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Page 10: DAO 2011 solved paper 2 mensuration with KEY

69. Ifthe surface area of a splS:re is eqqalto the area of a cfucle of diameter5.6 cm, then-the rid;us oT-tEe-iFhGre

.^v' nl'r- .,''x(1) 1cm .l; , '.../..\(2) 2 cm -1

(.3) 7.4 cm/(4) 23cm

The raLio of thp volump of a cJb. to{hdljl a lplterg,I. hrch wrll c\aclly lur4sidc t!r-e, cube is a3.

6 ',1

The longest chord of a circle is the

(1) radius

(2) tangent

B) dianetet'/

72. The number of circles aliawn throughthree non-colLinear points in a plane rs

(7) 1 -',"

(.2) 2

(3) 3

(4) 0

rq; ?nei^,),.' ud;.

\r L.43

1.2) 6.tt /

(3) r:a(.1) B:r

73.

oAD/599

Th" ccpacrly oi a c\lindrica' rark i.1540 m . If the radius ol its base i-7 m, the depth of the tanh is

sE:-.- 1-(1) 14m LJ,n'1^'..',/

'1 ^'

74.

(3) 13 m

(,1) 12 m

(1)

(3)

(4)

60.

90'22

120'

180'

The angle subtended

at the center is

' i'j 10

2i l'a t-..'/

b5' a semicircle

of the

cyclj!

(2)

75. The sum of the either

opposite angles of

quadrilateral is

parr

(1)

(3)

(.2)

90"

780" "'

270'

(4) 360'

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Page 11: DAO 2011 solved paper 2 mensuration with KEY

lr a AABa. il tnc eircle drcwr on

BC as diamcter passes through A,then A ABC is

(1) an acute anglcd tliangle(2) an equilateral triangle

(3) an obtuse angled triangle

(.1) ./a dght angled triangly'f I -:-\

'.I-- ,-t*

78.

77. In the figure, DCt=

(1)

t2)

(3)

(4)

t20'

110'

70" 1

130'

\ 1{ h of lhp foJ,o"ing . 6 ,r.1 ,

quadrilateral?

(1) a trapezlum

(2) a parallelogram

(3) a rhombus

(4) a rectangle /

?9. The tangents at the ends of a diameterof a circle

(1) ar-e perpendicular

(2) are parallcl ,7

(3) bisects

(4) intcrsects at the center

t2L\ OAD.5::

80. The length of the tangent drayltq acrrcle u. h raoius r rm from a poir':p which is d cm arvay from the

center:, is

I

(1)

(2J

(3)

81. If ti'o circles touch internally, thenthe number of their common tangents

1S

(.r) 2

t2) 1..'(3) 4

(4) 3

-^.i' -'l

I

(1) congruent,/

(2) oniy similar but not congruent

(3) not congruent

(4) neither congruent nor similar:

J,I*,

TVo crrcles with equal radii alg82.

83. If trro circles ol radii 3

touch internally, Ihenbet\ieen their q€nters is

(1) 3 cm

(2) 8 cm

(3) 2 ct:t,'

(4) 15 cm

a)cm and 5€the distance

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Page 12: DAO 2011 solved paper 2 mensuration with KEY

84. I'he angle between a

circle and the radiuspoint ol contact is

tangeni 1:o a

drawn at the

circle from

(3)

The tangents drawn

an external Doint are

(1) 60'

t.2) 30'

45"

(1) so" -.'

(1)

e) 2-.

(1)

(3)

toa

(3)

\4)

86. l" rwo crrclp" are radii5 c_m and I2 cm

touch e>r:te_r,nally, --tbgL!-he -distenrebetween their: center:s iL

t2)

21 cm

7cm

1'7 crn,/

2cm(4)

(26) oAD/599

Two circles are of radii 3 cnr a4{1-rmgnd the distance beiwaen their centersi;-5 cnf Ean-nia:L"tgE si- !Fi'tiiiiiwe$p sogrlqeal4g€ent is

88.

( 1) 3 cm/- .-,,<' i., ^ "

jY,2, ,rcm \ /'

]

^-'h- I13) 5cm 2'* -l"t(4) 8cm t a

I

How many g941palq, gaq! of lenelh1,1 m and diameter 2 cm can be made 1it"-]olifof 0.g8 cubic -eters ol iron?

(1) 140 j. o.ov Q ?=- t|

,2,200./ )11 ,".,'"f,r" 7

o,Kt ' f,3. 280 -' ,. 4'F

(4) 320

89. If the length of the chor4gf q circle iseSld,lojb radius, then the anglesubtended by the chord at tbglgqUqIS

(1) 30'

(2) 45"

$) 60" .,.'

(4) 90"

90. lf two concentdc circles grq &Illcdwith radii 7 cm and 14 crrr, thc area'+.-+between tne two clTcles rs

(1) 154 sq. cm

(2) 250 sq. cm

(3) 462 sq. cm I

(4) 49? sq. cm

n'

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Page 13: DAO 2011 solved paper 2 mensuration with KEY

lii t,' n t"s.,lo. polj'gon, ifthe sum 01 thei / irrt.rlo. angles is twice tbe sum of the

extedor angles. then the nurlrbFr ofsides olthe legulal polvgon is f9| ..v.

':' "-, ,,'f) 'rn.i.2.t 6.

, ... ,, , ,€(3) 8 l'' *

"r'',' '/'(.1) t{}

' '.,,, :.'t-. .,, ooo^*

@' ,:' i.'iu a.nl A(-=4tm'' i '

tbe[ the radius of the circumcircle olthe tdaDgle is \(1) 5cm -l\L2) 10 crn -l \(31 2.5 cnr.' e

14) 15 clll

If AD is p:uallel to BE, C- i1 tbcniclfloinf of Bt ar.rd the arca of

I\.BD-E=40cm', thcn the area ol

96.

,\ ,c,E is(1) 30 cm:L2) 60 cml(3) 40 ctti(4) 20 ctrr:r

/{/- 7,bc'-

of

ot

A

93.

^ 94'

98.

99.

.! ___ I)DLI=

^ DEF -!

4

' o' ,l l\'" '(3) a''

e:t tb+f)/ "

95. If A and E are two congllLcul i.gure;,then

11) area ol .4 = area of B r'

(2) ared ol A

13) ] area of A = area of B

1,1) arc'a of A = ll alea of B

A A'C(3) area of A .DEF =

A AB'](4) arca of A ABC -

\ D]]F

1 "-

i. ,. .*,, r jr '),1.c qrc"riL orrl i

rrhirh AB AD B-! =('D 'tne",1r I!B.=\LBA *_l;

I pBC = 2lcBp

i Apts-=l pBCl2')

j pBC = 2wD/

rnd -\/l = DF , ttren XF =(1) t ctr12) 16 cm(3) 4 cm

(4) 10 cmr' ,c

In the figure, ABCD is a ryrtangleand BDE is an isosceleslS$-qgglcl .

triangle. The {tlca of the figue t--'

( 1,'t cab e " F ...{,e

97.

(1)

(2)

A ABC

area of

4

2

.J

100. ll ,1. dicgo rrls-.At . .,1, 1 9? "1 '

" I r-z u n ABa lr ::lt AB I' D(-

intcrsect eacb other at O , tlrgn tbearea of r\ AOD =|' dtts o \BOC/

(21 ] areaofjBoC ;" o '-:;,'2r

,3, | -" o rts.)C r4(4) 2 area of ABOC

If the area oI :r rhombus is 2-l!q41' and,rp of i., i,Sor d F :s o lPrel\ 18.n_then the length of one of its sjdes is

\,o| 12 art2r 15 cm,/ )rt 1.' --;1,lr 13 (n1 - -),-" ,/a( 1) 20 cnL

lf D,-U and a a1e lcsPectively the :nidpoiDts of the sij]es -BC, CA arLd. ,'AB ol'a triangle ABC. then

IL

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Page 14: DAO 2011 solved paper 2 mensuration with KEY

101. The arca ofthe rhombus is(1) th€ ploduct of the lengths oI the

diagonals(2) trvice thc product of the lentghs

of the diagonal-.1(3) : the nrodrrct of ihe lengths ofl2'the diagonalsl(4) i the product of the lengths of3-

the diagonals

102. Iffto chods Al) anrl CD intersect atO. AO=1cn, OB=5cm andDO = 2 cn, lhen CO =,..---{(1) 4cm i' I(2) 6cm ^ \ :. 1,q )a,J b cn(4) 10 cmr

.^-103. II p"ch dirsoral ol'a ou.rdrilarprlll' separates io iwo tr-ii;-1es;T-;qualarea, then the quadrilateral is a(1) parallelogranr(2) square(3) rhombus(4) rectangle

104, If the parallelosram lAC,Q._gl4racl rneTe r1tsEF are on tl-e "enc br.pAB and lave -9qqa1 arcas, thgL-t_llepedmeter: of the-parell-ellgggis(1) less than that ofthe rectangleX(2), {reatcr. than that of the /,

recrangte(3) equal to ihat olthe rectangle

(4) ] .f thc Derimeter of the2'rectangle

lO5, l' AIlc.D r" r prlrl.elogrur. X .rnoY a-r ,l n In.dpoinrs oi BC ard eDrespectively, lhen the area of theAAxI =

(1)

t2)

(3)

(.1)

L area. of ABCD tII alea of ABCD ^I area of ABCD8

! aye,a ol AB(ID2

(30) oAD/599

106. In a t4?4g1e_ A!-C.' , right angled at B,BD fAq, BD =9 cmand.4D = 3 cm,then the value of AC is(1) 12 cm

(2) 18 cm(3) 30 cm,\4t Zt trn .l .

107. if the area of a triaritgle i.s 12 crn, andir. hssF i" 6.r..:hen il".orr.slondrnt

)rBt.h. -'v'

108. qlhg !g4lg!tc19! 4q!{apqzjlqn 196 cm, 8 cm and the distance belwecnthem is ,1 cm, then its area is -!t.l+

!.f

/ 1' .

4- t '"'

I

1^'2i 'L+':

altitude is(1) 12 cm

(2) 6 cm

(iJ) 8 cm

(4) 4cnt

l1) 28 crn2z

(2) 48 cm,

i3) 32 cm'z

(.4) 24 cn'

order is(.1) 2:5(2) :J :'l(.il) \0:21t(4) 14 :15

'.t'

109. If the ratio of the bases of trvorflcrlgre: rs z : J anLl -c rat.o ol thc rcorresponding altitudes is 5 : 7, thenthe ratio of their areas in '.he same

z/< ?/:*1A 2 t

110. 1f the area of a rectangle i-c 2,4 crn, a.lldits lengrhis 6 qg. lhen its pelimeter is(1) 10 ct,\2) 20 cn, | )-. b

,3) JU cmb'I/l) t4,,rr!

t !J] '\

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Page 15: DAO 2011 solved paper 2 mensuration with KEY

111. If the qleq- -gl. j!. tariuglltsljs7!S!t\ n,n and

'' are the midpoints

of the sides BC, CA and ABl'cspectively, then the area of the,..nstc DFF i"-. ;'-i, ern^ .,1 \,,'.*2l 4 cnr 1_1 \(3) 3.5 cm'A) 1.5 cm2/

trt-2 " "

112. If the ratio of the basq! !r lIorri,,ngl4- is-o:.1,"d rh" i.',;;;i"",,"aqLrl in arca. rnen the rdtro oflhFrrcorrespondlng sll,rtudcs rs{1) o:b L.^\2) b d.(.3) a:11) 2a: b

Il3. Tn r 'riangl" {Bl rlrc b."er !o4 of-/JAC intersect .BC in D, then

Bt)DC

(3)

(4)

(1) AB.AC'ACABAD

ACAD

114. If in IABC, AC'= AI'_iB-C,, then

IABC=

(2) B'a)a-

(1) 180.(2) 90"/(3) 46.(4) 50'

115. 1he lengths of thc sides !f, cq+arntdangles are gt"cn b?Iui Which ofthem is a dght angled t4g4C1e1(1) 6cm Icnl, 11c r(2) 5 cm, I cm, 10 cm(3) 8 cm, 15 cm, 17 cm/(.1) 7 clr',23 crr,,24 cm

e. L-

&

116. If ABCtriaDgle,AB, =

oAD/599

is an isosceles right angledright anglcd at C, then

(1) AC"(.2) 2AC,/

{

(4)

(3) !.ac,2

1AC'

I17. A "ote uf 2.2 rn ,ong h.. b€ct p!t_!-lt_tbg_f-orrgaf 4 circle The area of thecu cle so lo1.Ined ls(1) 4400 cm: , fl f(2) 2620 cm2(3) 2560 cm, 'L'".a('1) 3850 cm':1

..7- .i/a

?.-) '-..r

",'t ,t ''7

PR:AC IBC:QR

t?

'n.ti)

rr9. If AB:ZY = BC : XYAABC i6 similar to

then

<_vr2o. rf G;ta:^pet) 44_!.":r",

PQ=2.4 cm and PR=5.4 cm. thcn

;A

h(1) ^XYZ /\,/\t.2) t zxY / \

(3) LZYX. t it4)

^YZX

(1) 3.6 cm(2) 5.4 cn:r

{3) 4.5 cm(4) 8.1 crr' ,

2:

L

z\5-.1

l,^':: -::n* ' :nen 4!' PQ =

(3)

?c

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Page 16: DAO 2011 solved paper 2 mensuration with KEY

121. The sum o{ thequadrilateral is(1) 90'(2) 180.1.3) 270'(4) 360e,

interior angles of a

(1) 5

(2) 1.0 /(3) 12(4) 8

@123. In a triangle AtsC, il BC= AB ard

122. Tf 'he e\tcrtor arglF ot a rcgu rrpolygon is 36', then the nurallel orsidcs of Lhe polyeon is

124. In a triangie 44c_,,jdrt ergl"d "t -B ,BD IAC , BD =9cm and,4r=3cm,the value of AC is(1.) 18 cms(.2) 27 cni's(3) 30 cmsT P'

| ,-Dr1160"ar" ix\ --s#. c-

"9.

t1t 12 rms 1' " lt *dK

/tzr25. If 4-is lbqrqi{1!ar&qllh.9.}ip,o-tS.+}se

AC of a right angled triangle ,48C,

l! = B0' , then L{ =

(1) 30.(2) 60'(.3) 20" .(4) 90.

then BD =

o) +AC /

(2)

(3)

!ac2

!dn2

IAD2

.r') (,'7.

(34)

angle of an

50'

60" /70"

90'

oAD/599

"qg!!914-,t4q€19

a triangle q{9 tllhethen the angles of the

l2^, a \ 6o

,.r t,0 ;

t26. Eachis

(1)

l2)

(3)

\4)

727. In LABC ,

which is the

(1) AB

(.2) BC /(3) AC

if !l =!l= a5", then

(4) All sides are equal

128. The angles ofratio 3:4.: 5,r.rangle are l(t7) 45",60",75. t(2) 40",50.,90.(3) 45., 45., 90"

(4) ,10., 60., 80.

129. Tn a parallelogam thp diagonal"

(1) are equal

(2) bisect each other /(3) intersect at dght angles

(4) ere parallel to each other

130. If ali sides of a quaddlateralequal, then it is a

(t) rhombus I(2) rectangle

(ts) squarc

(4) parallelogt am

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Page 17: DAO 2011 solved paper 2 mensuration with KEY

11 a triangle and a parailelogramon the sane base and bet\rcensarne parallels, thcn the area of

thethe

-r-

60" B

1V120"/160"

133. Tf rhc -un ofrrro rlgl.s orr triansleiq 8u'. th"u ih-ther ansi; ; ,;;Lrr snacuteanqle,..-?or2 r u nsht ansle ''J(3) an obtuse angle,/(4) a leflex angle

T e a rriang c ABa . lf o ;" anl pinr\'rtlnn it such tbat AlJ, + Br{ + CA1 = 3

\oA' + oB, + oC, ) ihen O is thc

i 13.+

(1) excentre12) incentre

Pf , eirtiimcentre(4) 1 centroid

\\1r, r o, rp tollusrng r. a .onpi.lsLatcnent?rr I t\\'o II]soglcs are :rlrvays srmtlcrZ iro.qudrF- rr. cl$-y"..tD.'J-./rrlr iio r or'.bu.cs ar'e rlurv-

slmrlal'r I luu rc,.rane1.- tue ,l$_,.

f-lL\!)

131.

132. Irl the lbllorviug figure,

(1)

l2)\.;1)

(1)

tliangle is equal to(ll areaofparailelogranl

(3)

(41

i.2)1

ar-ea ol the pr rullclo{Tnmy'

1"rr"er (]1 Lbe puatlelngrJrn

area ol the parallclogranl

(36)

Thc leng1h of the mediaoIigirt angled triaDgle ii

oAD/599

BD in rhe

regular peDtagon

1:.. I , .'i -,' ,

" ,\i" ,.,,."o

136.

I39. Tl e arcb of an eqgilgr"ra| r.r.lnglc\. hr.h ii 'brn cd orr r t-p srde "t usou-"eu h8ctnsda. " :n.; ;(1) 1216

t.2) 76nE/.

(3) 18 a6(4) 14./5

140. II on" Jnete of rr rri*ellc_r" obrust.I nFn rt c 01 ie,' t\.o dlgle. m, isLbc(1) acute angles I(2) iess thau 45" each(3) straight angles

(4) one riglrt angle and the other-acute angle

6' j;{.iI ,q\1

(l) 12 cm .h(2) 6 cn ../ ll ,..i3) 1J cm --.- n,r c'ru a.s cm2 f;$ .

137. The perineter of a\r-ith side 7 cn is(1) 35 cm7t.2) 2E cm(3) 56 crn\4) 42 crn

(1) 1000.(2) 1280.(3) 1080'./(4) 1800"

138. The sum of the interior a+gles of arcg1llar octagon is

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Page 18: DAO 2011 solved paper 2 mensuration with KEY

141,

142. ln a parallelogram PQRS, X '" the

midpoint of P@ and I is themidpoint of ,RS. Then XY divides theparallelogram into lwo parts as

In a tnanel" A19a. DE ..- drawr._ ,lD Aparallel to Be. Then

"" = n,t\

.. BC DHEDEatl\2) :-1/ ts't-Ea ;d" i"

B) l! f otb. BD/&t 49^

AE/

(l) 1:2(2) 3:1(.3) 7 . 1/(4) 2:7

143. The interiornanogone is(1) 110'(2) r2o'(3) 130'(4) 740'/

angle o{

@

--t

L/_1

and the other is

a regtrlar

triangle the other144. In a right angledtwo angles must be(1) acute angles 7(2) obtuse angles(3) equal to 15'(4) one is acute

obtuse

145. Intersecting point of angular bisectorsof a triangle is tbe.,(1) circuncentre(2) incentre;(3) centroid(4) orthocenhe

(38) oAD/599

If the sides of a triangle ale in the

ratio 1: rE: 1, then the triangle is(1) equilateral \,\ {u(2) isosceles 'eL t,,..',(3) righi angled l..i(4) .,.right angled isosceles "

the tdangle

@ +(BD" + AD')

(4) BC' + CD2

. . ..!at,- . -_ltlDr.v/

t ,,.,76 11

146.

t47- Il AD is the rnedian of.4BC , then AB2 + AC2 ts

0) z(ao'+ tn") zq 2lBC'+cD')

trian?e, ihen its a

ia'- (-a,

148. Th"_gioe- of q tr a rg1",ere 4!)3m41 cm and 9 cm. Then the arca of thetdangle is(1) 120 cm?(2) 320 cm'(3) 180 cm'z./(4) 200 cm:

149. If (6, 2) ar'd (2,6) are $re--e-4{s of ah]'potenuse of al:tgb! argled isosceles

in square units

(1) 8 /.

13) 2

Q) "E

If A(0.0) and B(6,0) are trvo vertices

ol J nghr anpled rrrar gre ABr- "rghtangled at A and "s(3,4) is the

circumcenlre, then its tlrird ve ex is(1) (6, 8)(2) (8, 6)(3) (0,8)/(4) (4,3J

,YL

@)€,t)

150.

r-?,b') .r, fr,?""'

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