similarity and pythagoras' theorem · geometry (std. x) 1 similarity and pythagoras'...

GeometryGeometryGeometryGeometry (Std.(Std.(Std.(Std. X)X)X)X)1111

Sim ilarity And Pythagoras ' Theorem1 . A rig h t trian g le h as h yp oten u se of len g th p cm an d on e s id e len g th q cm . If p q = 1 , fin d th e

len g th of th e th ird s id e of th e trian g le .

2 . ABC is an isos ce les trian g le rig h t- an g led a t B. Sim ila r trian g les ACD an d ABE are con s tru cted on

s id es AC an d AB. Fin d th e ra t io b e tween th e areas of ABE an d ACD.

3 . P an d Q are p o in ts on th e s id es CA an d CB resp ective ly of ABC rig h t an g led a t C.

Prove th a t AQ2 + BP2 = AB2 + PQ2.

4 . In ABC, If AD BC an d AD2 = BD DC, Prove th a t BAC = 9 0 .

5 . ABCD is a p ara lle log ram , P is a p o in t on s id e BC an d DP wh en p rod u ct m eets AB p rod u ced at L.

Prove th a t

i)BLBLBLBLDCDCDCDC

PLPLPLPLDPDPDPDP

ii)DCDCDCDCALALALAL

DPDPDPDPDLDLDLDL

.

6 . ABCD is a q u ad rila te ra l in wh ich AB = AD. Th e b is ecto r of BAC an d CAD in tersect th e s id es BC

an d CD at th e p o in ts E an d F res p ective ly. Prove th a t EF | | BD.

7 . ABC is a trian g le in wh ich AB = AC an d D is a p o in t on AC su ch th a t BC2 = AC CD. Prove th a t BD

= BC.

8 . In trap ez iu m ABCD, AB | | DC an d DC = 2 AB. EF d rawn p ara lle l to AB cu ts AD in F an d BC in E su ch

th a tECECECEC

BEBEBEBE =44443333 . Diag on al DB in tersects EF at G. Prove th a t 7 FE = 1 0 AB.

9 . Th rou g h th e m id - p o in t M of th e s id e CD of a p ara lle log ram ABCD, th e lin e BM is d rawn in tersect in g

AC in L an d AD p rod u ced in E. Prove th a t EL = 2 BL.

1 0 . Prove th a t th ree t im es th e su m of th e sq u are s of th e s id es of a trian g le is eq u al to fou r t im es th e

su m of th e sq u are s of th e m ed ian s of th e trian g le .

1 1 . P an d Q are th e m id – p oin t of th e s id es CA an d CB resp ective ly of a ABC, rig h t an g led a t C. Prove

th a t : i) 4 AQ2 = 4 AC2 + BC2 ii) 4 BP2 = 4 BC2 + AC2

iii) 4 (AQ2 + BP2) = 5 AB2

1 2 . A p oin t O in th e in te rio r of a rectan g le ABCD is jo in ed with each of th e vert ice s A,B, C an d D. Prove

th a t OB2 + OD2 = OC2 + OA2

1 2

A

CB D

GeometryGeometryGeometryGeometry (Std.(Std.(Std.(Std. X)X)X)X)2222

1 3 . ABC is a rig h t trian g le rig h t- an g led a t C an d AC = 3 BC. Prove th a t ABC = 6 0 .

1 4 . In ABC, D an d E trisect BC an d B- D- E- C. Prove th a t 8 AE2 = 3 AC2 + 5 AD2.

1 5 . In fig ., two ch ord s AB an d CD in tersect each o th er a t th e p o in t P. Prove th a t :

i) APC ~ DPB ii) AP.PB = CP.DP

1 6 . In a rig h t trian g le ABC rig h t- an g led a t C, P an d Q are th e p o in ts on th e s id es CA an d CB

resp ective ly, wh ich d ivid e th ese s id es in th e ra t io 2 :1 . Prove th a t

(i) 9 AQ2 = 9 AC2 + 4 BC2 (ii) 9 BP2 = 9 BC2 + 4 AC2 (iii) 9 (AQ2 + BP2) = 1 3 AB2 .

1 7 . If A b e th e area of a rig h t trian g le an d b e on e of th e s id es con ta in in g th e rig h t an g le , p rove th a t

th e len g th of th e a lt itu d e on th e h yp oten u se is22224444 AAAA4444bbbb

AbAbAbAb2222

.

1 8 . In an eq u ila te ra l trian g le ABC th e s id e BC is trisected a t D. Prove th a t 9 AD2 = 7 AB2 .

1 9 . D is a p o in t on s id e BC of ABC su ch th a tCDCDCDCDBDBDBDBD =

ACACACACABABABAB . Prove th a t AD is th e b is ecto r of BAC.

2 0 . If two s id es an d a m ed ian b isect in g th e th ird s id e of a trian g le are res p ective ly p rop ort ion a l to th e

corresp on d in g s id es an d th e m ed ian of an o th er trian g le , th en p rove th a t th e two trian g les are

s im ila r.

2 1 . Th e areas of two s im ila r trian g les are 3 6 cm 2 an d 2 5 cm 2 res p ective ly. If th e on e s id e of th e secon dtrian g le is 3 cm lon g , fin d th e len g th of th e corresp on d in g s id e of th e firs t t rian g le .

2 2 . If th e areas of two s im ila r trian g les are eq u al, p rove th a t th ey are con g ru en t .

2 3 . In th e g iven fig u re , DE | | AB an d EF | | BD, p rove th a t CD2 = AC × CF.

2 4 . ABC is a trian g le . PQ is a lin e seg m en t in te rsect in g AB in P an d AC in Q su ch th a t PQ | | BC an dd ivid es ABC in to two p arts eq u al in area . Fin d

ABBP .

2 5 . Prove th a t th e area of eq u ila te ra l trian g le d es crib ed on th e s id e of a sq u are is h a lf th e area of th eeq u ila te ra l trian g le d es crib ed on its d iag on al.

2 6 . In th e g iven fig u re , DE | | BC an d32

DBAD

. Ca lcu la te : (i))ABC(A)ADE(A (ii)

)ABC(A)DECB(A

D

B

A

C

P

GeometryGeometryGeometryGeometry (Std.(Std.(Std.(Std. X)X)X)X)3333

2 7 . In th e g iven fig u re , ABC is a trian g le with EDB = ACB. Prove th a t ABC ~ EBD. If BE = 6 cm ,EC = 4 cm , BD = 5 cm an d area of BED = 9 cm 2 , ca lcu la te :(i) Len g th of AB. (ii) Area of ABC.

2 8 . In th e g iven fig u re , DE | | BC an d CD | | EF. Prove th a t AD2 = AB × AF

2 9 . Let X b e an y p oin t on th e s id e BC of a ABC. XM, XN are d rawn p ara lle l to BA an d CA m eetin g CA, BAin M, N res p ective ly; MN m ee ts BC p rod u ced in T. Prove th a t TX2 = TB × TC

3 0 . In th e g iven fig u re , AD an d CE are m ed ian s of ABC. DF is d rawn p ara lle l to CE. Prove th a t(i) EF = FB(ii) AG : GD = 2 : 1

3 1 . In th e g iven fig u re , ABD = CDB = PQB = 9 0 °. If AB = x u n its , CD = y u n its an d PQ = z u n its ,

p rove th a t z =yx

xy

.

GeometryGeometryGeometryGeometry (Std.(Std.(Std.(Std. X)X)X)X)4444

3 2 . ABCD is a trap ez iu m in wh ich AB | | CD an d AB = 2 CD. Th e d iag on als AC an d BD in tersect each o th era t O. If a r( AOB) = 8 4 cm 2, fin d ar( COD).

3 3 . Eq u ila te ra l trian g les are d rawn on th e s id es of a rig h tan g led trian g le . Sh ow th at th e area of th etrian g le on th e h yp oten u se is eq u al to su m of areas of o th er two trian g les d rawn on th e rem ain in gtwo s id es .

�������

Circle

1 . In fig ., XP an d XQ are tan g en ts from X to th e circle with cen tre O. R is a p o in t on th e circle . Proveth a t , XA + AR = XB + BR.

2 . In ., fig ., th e in circle of ABC tou ch es th e s id es BC, CA an d AB at D, E an d F res p ective ly. Sh ow th at

AF + BD + CE = AE + BF + CD =2222

1111 (Perim eter of ABC)

3 . ABCD is a q u ad rila te ra l su ch th a t D = 9 0 . A circle C(O, r) tou ch es th e s id es AB, BC, CD an d DA at P,

Q, R an d S res p ective ly. If BC = 3 8 cm , CD = 2 5 cm an d BP = 2 7 cm , fin d r.

Q

RX

A

B

P

B

E

D

F

C

A

S

R

O

P

C

Q

D

B

A

GeometryGeometryGeometryGeometry (Std.(Std.(Std.(Std. X)X)X)X)5555

4 . In fig ., l an d m are two p ara lle l tan g en ts a t A an d B. Th e tan g en t a t C m akes an in te rcep t DE b etween l

an d m . Prove th a t DFE = 9 0 .

5. Th e rad iu s of th e in circle of a trian g le is 4 cm an d th e seg m en ts in to wh ich on e s id e is d ivid ed b y th e

p oin t o f con tact a re 6 cm an d 8 cm . Determ in e th e o th er two s id es of th e trian g le .

6 . PQ is a ch ord of len g th 8 cm of a circle of rad iu s 5 cm . Th e tan g en ts a t P an d Q in tersect a t a p o in t T.

fin d th e len g th TP.

7 . In th e g iven fig u re , O is th e cen tre of th e circle . Fin d th e valu e of x.

D

m

l

C

E

F

B

A

4 cm

y

5 cm

x

R

P

O

Q

T4 cm

5 cm

C

N

A

M

L

I

B

6 cm

8 cmX cm

X cm

GeometryGeometryGeometryGeometry (Std.(Std.(Std.(Std. X)X)X)X)6666

8 . In th e g iven fig u re , ACE = 4 3 ° an d CAF = 6 2 °. If AEC = x, fin d th e va lu es of a, b an d c.

9 . In th e g iven fig u re , ABCD is a cyclic q u ad rila te ra l. If fin d th e va lu e of x, y an d z.

1 0 . In a cyclic q u ad rila te ra l, if on e p air o f op p o s ite s id es is eq u al, th e o th er p a ir is p ara lle l. Prove it .

1 1 . ABCD is a cyclic q u ad rila te ra l in wh ich DAC = 2 7 °, DBA = 5 0 ° an d ADB = 3 3 °. Calcu la te :(i) DBC (ii) DCB (iii) CAB

1 2 . In th e g iven fig u re , PQ an d RS are two s tra ig h t lin es th rou g h th e cen tre O of a circle . If POR = 8 0 °an d RSK = 4 0 °, fin d th e n u m b er of d eg rees in

(i) SRK (ii) PQR

GeometryGeometryGeometryGeometry (Std.(Std.(Std.(Std. X)X)X)X)7777

1 3 . In th e g iven fig u re , O is th e cen tre of th e circle , BAD = 7 5 ° an d ch ord BC = ch ord CD.Fin d : (i) BCD (ii) OBD (iii) BOC

1 4 . In th e g iven fig u re , AB = AC = CD. If ADC = 3 8 °, ca lcu la te :(i) ABC (ii) BCE

1 5 . In trian g le PQR, PQ = 2 4 cm , QR = 7 cm an d PQR = 9 0 °.Fin d th e rad iu s of th e in scrib ed circle .

1 6 . In th e g iven fig u re , q u ad rila te ra l ABCD is circum s crib ed an d AD DC. Fin d x, if rad iu s of th e in circleis 1 0 cm .

GeometryGeometryGeometryGeometry (Std.(Std.(Std.(Std. X)X)X)X)8888

1 7 . In th e g iven fig u re , two circles tou ch each oth er ex te rn a lly a t a p o in t P. AB is th e d irect com m ontan g en t of th ese circles . Prove th a t :

(i) APB = 9 0 ° (ii) tan g en t a t p o in t P b is ects AB

1 8 . In th e g iven fig u re , p rove th a t : AP + BQ + CR = BP + CQ + AR

1 9 . Th ree circles tou ch each oth er ex te rn a lly. A trian g le is fo rm ed b y jo in in g th e cen tres of th ese th ree

circles . Fin d th e rad ii o f th e circles , if th e s id es of th e trian g le form ed are 6 cm , 8 cm an d 9 cm .

�������

Co-OrdinateCo-OrdinateCo-OrdinateCo-Ordinate GeometryGeometryGeometryGeometry1 . Wh at p o in t on th e x- ax is is eq u id is tan t from (7 , 6 ) an d (– 3 , 4 )?

2 . Th e cen tre of a circle is (2 – 1 , 7 ) an d it p ass es th rou g h th e p oin t (– 3 , –1 ). If th e d iam ete r of th ecircle is 2 0 u n its , th en fin d th e valu e of .

3 . Th e p oin t R d ivid es th e lin e seg m en t AB wh ere A(–4 , 0 ), B(0 , 6 ) are su ch th a t 4 AR = 3 AB. Fin d th ecoord in a te s of R.

4 . If A(4 , – 8 ), B(3 , 6 ) an d C(5 , –4 ) are th e vert ices of ABC, D is th e m id p oin t o f BC an d P is a p o in t onAD jo in ed su ch th a t AP = 2 PD, fin d th e coord in a tes of P.

5 . If C is a p o in t lyin g on th e lin e seg m en t AB jo in in g A(1 , 1 ) an d B(2 , – 3 ) su ch th a t 3 AC = CB, th enfin d th e coord in a tes of C.

6 . If two vert ices of a p ara lle log ram are (3 , 2 ), (– 1 , 0 ) an d th e d iag on als cu t a t (2 , – 5 ), find th e o th ervert ice s of th e p ara lle log ram .

7 . In wh at ra t io d oes th e lin e x – y – 2 = 0 d ivid e th e lin e seg m en t jo in in g (3 , –1 ) an d (8 , 9 )?

8 . Th e p oin ts (p, q); (m, n) an d (p – m, q – n) a re co llin ear, sh ow th a t pn = qm.

9 . If th e p o in t (x, y) b e eq u id is tan t from th e p oin ts (a + b, b – a) an d (a – b, a + b), p rove th a t bx = ay.

1 0 .Fin d th e len g th s of th e m ed ian s of ABC h avin g vert ices a t A(5 , 1 ), B(1 , 5 ) an d C(– 3 , – 1 ).

1 1 .If R(x, y) is a p o in t on th e lin e seg m en t jo in in g th e p o in ts P(a, b) an d Q(b, a), th en p rove th a t x + y= a + b.

GeometryGeometryGeometryGeometry (Std.(Std.(Std.(Std. X)X)X)X)9999

Trigonom etry

1 . If tan (A + B) = 3 an d tan (A B) =3

1 ; 0 < A + B 9 0 ; A > B, fin d A an d B.

2 . If cos (4 0 + x) = s in 3 0 , Fin d x .3 . If tan 2 A = cot(A 1 8 ), wh ere 2 A is an acu te an g le , fin d th e va lu e of A.4 . If tan A = cot B, Prove th a t A + B = 9 0 .

5 . If l (AD) = 1 0 cm , fin d l (CD)

6 . If A, B, C are in te rn a l an g les of a trian g le ABC, th en sh ow th at s in

2

CB = cos2A .

7 . Prove : 0Bs inAs inBcosAcos

BcosAcosBs inAs in

8 . If s in + cos = p an d sec + cosec = q , sh ow th a t q (p 2 – 1 ) = 2 p .

9 . Prove th a t : 9 sec2 5 tan 2 = 5 + 4 sec 2

1 0 . Prove th a t : (s in + cosec )2 + (cos + sec )2 = 7 + tan 2 + cot 2 .

1 1 . Prove th a t : If cos + s in = 2 cos , p rove th a t cos s in = 2 s in .

1 2 . In an acu te an g led trian g le ABC, if tan (A + B C) = 1 an d , sec (B + C A) = 2 , fin d th e va lu e of A,

B, an d C.

1 3 . Prove th a t :1sectan1sectan

=

coss in1

1 4 . If

coscos = m an d

s incos = n sh ow th at (m 2 + n 2) cos 2 = n 2

1 5 . If x = r s in A cos C, y = r s in A s in C an d z = r cos A, p rove th a t r2 = x 2 + y2 + z 2 .

1 6 . If tan A = n tan B an d s in A = m sin B, p rove th a t cos 2 A =1n1m

2

2

1 7 . Prove th a t : 2 (s in 6 + cos 6 ) 3 (s in 4 + cos 4 ) + 1 = 0

1 8 . Prove th a t : s in 6 + cos 6 + 3 s in 2 cos 2 = 1 .

1 9 . Prove th a t : sec 2 (1 s in 4) 2 tan 2 = 1 .

2 0 . If sec = x +x4

1 , p rove th a t : sec + tan = 2 x or,x2

1 .

4 5 DB

3 0

A

C

GeometryGeometryGeometryGeometry (Std.(Std.(Std.(Std. X)X)X)X)10101010

2 1 . Prove :AsinAsin

1Acot

11Atan

11 4222

2 2 . Prove : AcotAeccos.AsecAcos1

AtanAcos1

Asin

2 3 . Prove : cos 4 A – cos 2 A = sin 4 A – s in 2 A

2 4 . Prove: If x = a sec , y = b tan , p rove th a t 1yb

xa

2

2

2

2

.

2 5 . If a cos – b s in = x an d a s in + b cos = y. Prove th a t a2 + b2 = x2 + y2 .

2 6 . Prove: Asec2AcosAsin1

Asin1Acos

2 7 . Prove:Acos1

AsinAsinAcos1

2 8 . Prove:AcotAeccos

1Asin1

Asin1

AcotAeccos1

2 9 . Prove: Acot.Aeccos21Asec

11Asec

1

3 0 . If tan + s in = m an d tan – s in = n, sh ow th a t m2 – n2 = mn4

3 1 . If sec + tan = p, p rove th a t s in =1p1p

2

2

3 2 . Prove :

sec2

sin1sin1

sin1sin1

3 3 . If sec + tan = p, fin d th e va lu e of cosec .

3 4 . A vert ica l tower s tan d s on a h oriz on ta l p lan e an d is su rm ou n ted b y a vert ica l flag – s ta ff o f h e ig h t

h . At a p o in t on th e p lan e , th e an g les of e leva tion of th e b o ttom an d th e top of th e flag – s ta ff a re

an d res p ective ly. Prove th a t th e h e ig h t of th e tower is

tantantantantantantantan

tantantantanhhhh .

3 5 . Th e an g les of e leva tion of th e top of a tower from two p oin ts a t d is tan ces a an d b m etres from th e

b ase an d in th e sam e s tra ig h t lin e with it a re com p lem en tary. Prove th a t th e h e ig h t of th e tower is

ab m eters .

3 6 . Two s ta t io n s d u e sou th of a lean in g tower wh ich lean s toward s th e n orth are a t d is tan ces a an d b

from its foo t . If , b e th e e leva tion s of th e top of th e tower from th ese s ta t io n s , p rove th a t its

in clin a t ion to th e h oriz on ta l is g iven b y co t =ab

cotaco tb

GeometryGeometryGeometryGeometry (Std.(Std.(Std.(Std. X)X)X)X)11111111

3 7 . Th e an g le of e leva tion of a je t p lan e from a p oin t A on th e g rou n d is 6 0 . Afte r a flig h t of 1 5

secon d s , th e an g le of e leva tio n ch an g es to 3 0 . If th e je t p lan e is flyin g a t a con s tan t h e ig h t of

1 5 0 0 3 m . fin d th e sp eed of th e je t p lan e .

3 8 . A rou n d b alloo n of rad iu s r su b ten d s an an g le a t th e eye of th e ob se rver wh ile th e an g le of

e leva tion of it ’s cen tre is . Prove th a t th e h e ig h t of th e cen tre of th e b a lloon is r s in cosec / 2 .

3 9 . Th e an g le of e leva tion of a cliff from a fixed p oin t is . Afte r g o in g u p a d is tan ce of k m e ters

toward s th e top th e cliff a t an an g le of , it is fou n d th a t th e an g le of e leva tion is . Sh ow th a t th e

h e ig h t of th e cliff is

cotcotcotcotcotcotcotcotcotcotcotcotsinsinsinsincoscoscoscoskkkk m etres .

4 0 . Th e an g le of e leva tion of th e top of a tower a t a d is tan ce of 1 2 0 m from a p oin t A on th e g rou n d is4 5 °. If th e an g le of e leva tio n of th e top of a flag s ta ff fixed a t th e top of th e tower, from A is 6 0 °,th en fin d th e h eig h t of th e flag s ta ff. [Use = 1 .7 3 ]

4 1 . Th e an g le of e leva tion of a je t fig h ter from a p oin t A on th e g rou n d is 6 0 °. Afte r a flig h t of 1 5secon d s , th e an g le of e leva tio n ch an g es to 3 0 °. If th e je t is flyin g a t a sp eed of 7 2 0 km / h r, find th econ s tan t h e ig h t . ( 3 = 1 .7 3 2 ).

4 2 . From th e top of a b u ild in g 6 0 m h ig h , th e an g les of d ep res s io n of th e top an d b otto m of a vert ica llam p p os t are ob s erved to b e 3 0 ° an d 6 0 ° res p ective ly. Fin d(i) Th e h oriz on ta l d is tan ce b etween th e b u ild in g an d th e lam p p os t .(ii) Th e h eig h t of th e lam p p os t , 3 = 1 .7 3 2

4 3 . Th e an g le of e leva tion of an aerop lan e from a p oin t A on th e g rou n d is 6 0 °. Afte r a flig h t of 3 0secon d s , th e an g le of e leva tio n ch an g es to 3 0 °. If th e p lan e is flyin g a t a con s tan t h e ig h t of3 6 0 0 3m , fin d th e sp eed in km / h r of th e p lan e .

4 4 . Th e an g les of e leva tion an d d ep res s io n of th e top an d b otto m of a lig h t- h ou se from th e top of a 6 0m h ig h b u ild in g are 3 0 ° an d 6 0 ° resp ective ly. Fin d(i) th e d iffe ren ce b etween th e h eig h ts of th e lig h t- h ou se an d th e b u ild in g .(ii) th e d is tan ce b etween th e lig h t- h ou se an d th e b u ild in g .

4 5 . Fro m th e to p of a b u ild in g 1 5 m h ig h , th e an g le of e leva t ion of th e to p of a to wer is fou n d to b e 3 0 °.From th e b ottom of th e sam e b u ild in g , th e an g le of e leva tio n of th e top of th e tower is fou n d to b e4 5 °. Dete rm in e th e h e ig h t of th e tower an d th e d is tan ce b etween th e tower an d th e b u ild in g .

4 6 . At a p o in t A, 2 0 m e tres ab ove th e leve l of water in a lake , th e an g le of e leva tion of a clou d is 3 0 °.Th e an g le of d ep ress ion of th e reflect ion of th e clou d in th e lake , a t A is 6 0 °. Fin d th e d is tan ce ofth e clou d from A.

4 7 . Two p ole s of eq u al h e ig h ts are s tan d in g op p os ite to each o th er on eith er s id e of th e road wh ich is8 0 m wid e . From a p oin t P b etween th em on th e road , th e an g le of e leva tion of th e top of a p o le is6 0 ° an d th e an g le of d ep ress ion from th e top of an o th er p o le a t p o in t P is 3 0 °. Fin d th e h eig h ts ofth e p o le s an d th e d is tan ce of th e p o in t P from th e p ole s .

GeometryGeometryGeometryGeometry (Std.(Std.(Std.(Std. X)X)X)X)12121212

Mensura tion1 . A p en d u lu m swin g s th rou g h an an g le of 3 0 an d d es crib es an arc 8 .8 cm in len g th . Fin d th e len g th

of th e p en d u lu m . [Use = 2 2 / 7 ].2 . Th e m in u te h an d of a clock is 1 0 cm lon g . Fin d th e area of th e face of th e clock d es crib ed b y th e

m in u te h an d b etween 9 A.M. an d 9 .3 5 A.M.

3 . A car h as wip ers wh ich d o n ot overlap . Each wip e r h as a b lad e of len g th 2 5 cm sweep in g th rou g h anan g le of 1 1 5 . Fin d th e to ta l a rea clean ed at each sweep of th e b lad es .

4 . To warm sh ip s for u n d erwate r rocks , a lig h t h ou se th rows a red co lou red lig h t over a secto r of 8 0 an g le to a d is tan ce of 1 6 .5 km . Fin d th e area of th e sea over wh ich th e sh ip s area warm ed .[Use = 3 .1 4 ]

5 . Determ in e th e ra t io of th e vo lu m e of a cu b e th a t o f sp h e re wh ich will exactly fit in s id e th e cu b e .6 . Fin d th e m axim u m volu m e of a con e th a t can b e carved ou t of a so lid h em isp h e re of rad iu s r.7 . Th e circu m feren ce of a circle exceed s th e d iam ete r b y 1 6 .8 cm . Fin d th e rad iu s of th e circle .8 . A wire is loop ed in th e form of a circle of rad iu s 2 8 cm . It is re - b en t in to a sq u are form . Dete rm in e

th e len g th of th e s id e of th e sq u are .9 . A race track is in th e form of a rin g wh ose in n er circu m feren ce is 3 5 2 m , an d th e ou ter

circum feren ce is 3 9 6 m . Fin d th e wid th of th e track.1 0 . A wh ee l h as d iam e te r 8 4 cm . Fin d h ow m an y com p le te revo lu t io n s m u s t it take to cover 7 9 2 m ete rs .1 1 . Fin d th e areas of th e sh ad ed reg ion in th e fig .

1 2 . Th e rad ii o f th e b ases of two rig h t circu lar so lid con es of sam e h eig h t are r1 an d r2 resp ective ly. Th econ es are m elted an d recas t in to a so lid sp h ere of rad iu s R. Sh ow th at th e h e ig h ts of each con e is

g iven b e h = 22

21

3

rrR4

.

1 3 . Mayan k m ad e a b ird - b a th for h is g ard en in th e sh ap e of a cylin d er with a h em isp h e rica l d ip re ss ion

a t on e en d as sh own in fig . Th e h eig h t of th e cylin d er is 1 .4 5 m an d its rad iu s is 3 0 cm . Fin d th e

to ta l su rface area of th e b ird - b a th . (Take = 2 2 / 7 ).

1 4 . Two circle tou ch ex tern a lly. Th e su m of th e ir a reas is 1 3 0 sq . cm . an d th e d is tan ce b etween th e ir

cen tres is 1 4 cm . Fin d th e rad ii o f th e circles .

1 5 . A cop p e r wire , wh en b en t in th e form of a sq u are , en closes an area of

1 .4 5 M

3 0 cm

GeometryGeometryGeometryGeometry (Std.(Std.(Std.(Std. X)X)X)X)13131313

4 8 4 cm 2 . If th e sam e wire is b en t in th e form of a circle , fin d th e are en closed b y it .

(Use = 2 2 / 7 ).

1 6 . A car h as wh eels wh ich are 8 0 cm in d iam ete r. How m an y com p le te revo lu t io n s d oes each wh eel

m ake in 1 0 m in u tes wh en th e car is trave lin g a t a sp eed of 6 6 km p er h ou r?

1 7 . In Fig ., d ep icts an arch ery targ e t m arke t with its five scorin g areas from th e cen tre ou tward s as

Gold , Red , Blu e Black an d Wh ite . Th e d iam ete r of th e reg ion rep re sen tin g Gold score is 2 1 cm an d

each of th e o th er b an d s is 1 0 .5 cm wid e . Fin d th e area of each of th e five scorin g reg ion s .

1 8 . PQRS is a d iam ete r of a circle of rad iu s 6 cm . Th e len g th s PQ, QR an d RS are eq u al. Sem i- circles are

d rawn on PQ an d QS as d iam ete rs as sh own in fig ., Fin d th e p erim e ter an d area of th e sh ad ed

reg ion .

1 9 . On a circu lar tab le cover of rad iu s 3 2 cm , a d es ig n is fo rm ed leavin g an eq u ila te ra l trian g le ABC in

th e m id d le as sh own in fig . Fin d th e area of th e d es ig n (sh ad ed reg ion ).

2 0 . If th e d iam ete r of cross - sect ion of a wire is d ecreased b y 5 % h ow m u ch p ercen t will th e len g th b e

in creased so th a t th e vo lu m e rem ain s th e sam e?

2 1 . A cylin d rica l p ip e h as in n er d iam ete r of 7 cm an d water flows th rou g h it a t 1 9 2 .5 lit res p er m in u te .

Fin d th e ra te of flow in kilom e ters p er h ou r.

2 2 . Th e rad ii o f th e in te rn a l an d ex te rn a l su rface of a m e ta llic sp h erica l sh e ll a re 3 cm an d 5 cm

resp ective ly. It is m e lted an d recas t in to a so lid rig h t circu lar cylin d er of h e ig h t321 0 cm . Fin d th e

d iam ete r of th e b ase of th e cylin d er.

2 3 . A con ica l ves se l o f rad iu s 6 cm an d h eig h t 8 cm is com p le te ly filled with water. A sp h e re is

lowered in to th e water an d its s iz e is su ch th a t wh en it tou ch es th e s id es , it is ju s t im m ersed as

sh own in fig ., Wh at fract ion of water over flows?

GeometryGeometryGeometryGeometry (Std.(Std.(Std.(Std. X)X)X)X)14141414

2 4 . ABCD is a fie ld in th e sh ap e of a trap ez iu m . AB | | DC an d ABC = 9 0 , DAB 6 0 . Fou r secto r are

form ed with cen tres A, B, C an d D. Th e rad iu s of each secto r is 1 7 .5 m . Fin d th e

(i) to ta l a rea of th e fou r secto rs .

(ii) a rea of rem ain in g p ort io n g iven th a t AB = 7 5 m

an d CD = 5 0 m .

2 5 . How m an y sp h e rica l b u lle ts can b e m ad e ou t of a so lid cu b e of len d wh ose ed g e m easu res 4 4 cm ,

each b u lle t b e in g 4 cm in d iam ete r.

2 6 . A so lid iron rectan g u lar b lock of d im en s ion s 4 .4 m , 2 .6 m an d 1 m is cas t in to a h o llow cylin d rica l

p ip e of in te rn a l rad iu s 3 0 cm an d th ickness 5 cm . Fin d th e len g th of th e p ip e .

2 7 . A well, wh ose d iam e ter is 7 m , h as b een d u g 2 2 .5 m d eep an d th e earth d u g ou t is u sed to form an

em b an km en t arou n d it . If th e h e ig h t of th e em b an km en t is 1 .5 m , fin d th e wid th of th e

em b an km en t.

2 8 . Water is flowin g at th e ra te of 7 m e tres p er secon d th rou g h a circle p ip e wh ose in te rn a l d iam ete r is

2 cm in to a cylin d rica l tan k th e rad iu s of wh ose b ase is 4 0 cm . Determ in e th e in crease in th e water

leve l in ½ h ou r.

2 9 . A rig h t trian g le , wh ose s id es are 1 5 cm an d 2 0 cm , is m ad e to revo lve ab ou t its h yp oten u se . Fin d

th e Volu m e an d su rface area of th e d ou b le con e so form ed . (Use = 3 .1 4 )

3 0 . An iron p illa r h as som e p art in th e form of a rig h t circu lar cylin d er an d rem ain in g in th e form of a

rig h t circu lar con e . Th e rad iu s of th e b ase of each of con e an d cylin d er is 8 cm . Th e cylin d rica l p art

is 2 4 0 m h ig h an d th e con ica l p art is 3 6 cm h ig h . Fin d th e weig h t of th e p illa r if on e cu b ic cm of

iron weig h s 7 .8 g ram s .

3 1 . A lead p en cil con s is ts o f a cylin d er of wood with a so lid cylin d er of g rap h ite filled in to it . Th ed iam ete r of th e p en cil is 7 m m , th e d iam ete r of th e g rap h ite is 1 m m an d th e len g th of th e p en cil is1 0 cm . Calcu la te th e weig h t of th e wh ole p en cil, if th e sp ecific g ravity of th e wood is 0 .7 g m / cm 3

an d th a t o f th e g rap h ite is 2 .1 g m / cm 3.�������