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30
2003
1. In PQL and RPM,LPQ = MRP [Given]LQP = RPM [Given]
PQL ~ RPM [By AA axiom of similarity](i) Since, PQL ~ RPM
QLPM
=PLRM
QL × RM = PL × PM
(ii) In PQL and RQP, we haveQ = Q [Common angle]
QPL = PRM [Given] PQL ~ RQP [By AA axiom of similarity]
Then,PQQR =
QLPQ
PQ2 = QRQLHence proved
22. From the given figure,
ABD = ACD = 30º [Angle in same segment]Also, ADB = 90º [Angle in a semi-circle is 90º]In ABD, we have
DAB + ABD + ADB = 180º [Sum of all angles in a triangle is 180º]
x + 90º + 30º = 180º x + 120º = 180º x = 180º – 120º = 60º
Hence, the value of x is 60º
MT EDUCARE LTD.ICSE X
GeometrySTEP UP ANSWERSHEET
SUBJECT : MATHEMATICS
A
D
B
C
30º
xE
RQ L
P
M
MT EDUCARE LTD. X - ICSE (Maths)
31
2004
1. Given : In a figure, BAC = 30ºTo prove : BC = BDProof :
ACB = 90º [Angle in a semicircle]In ACB,ABC + ACB + BAC = 180º
ABC + 90º + 30º = 180º [ AB is a diameter of a circle and ACB is in semi-circle, so ACB = 90º]
ABC = 180º – 120º = 60ºBCD = BAC [Angles in alternate segment]
BCD = 30ºIn BCD,ABC is an exterior angle
ABC = BCD +BDC 60º = 30º + BDC BDC = 60 – 30
= 30º ... (ii)In BDC,
BDC = BCD = 30º [From equation (i) and (ii)] BC = BD [ sides opposite to equal angles
are equal]Hence proved
2. Given : DE BC
(i) To prove : ADE~ABCProof : In ADE and ABC, we have
DAE = BAC [Common angles]ADE = ABC [Corresponding angles as DE BC]
ADE ~ ABC [By AA axiom of similarity]Hence proved.
(ii) Since, ADE ~ ABC
ADAB
=DEBC
[If two triangles are similar, then
their corresponding sides areproportional]
ABAD
=BCDE
AD + BD
AD=
BCDE
[ AB = AD + BD]
30ºA B D
C
O
A
D E
B C
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MT EDUCARE LTD.X - ICSE (Maths)
1 +BDAD
=BCDE
1 +BD
1 BD2
=BCDE
1AD = BD, given2
1 +112
=BCDE
1 + 2 =4.5DE
DE =4.53
= 1.5 cmADE ~ ABC
Ar(ΔADE)Ar(ΔABC) =
2
2
ADAB
=2AD
AB
=21
3
= 1 : 9Let Ar(ADE) = kand Ar(ABC) = 9k
Ar(ΔADE)Ar(trp. BCED) =
Ar(ΔADE)Ar(ΔABC) – Ar(ADE)
=k
9k – k =
k8 k
=18
= 1 : 8
2005
B = E = 90º [ AB BC and DE BC, given]and ACB = DCE [Common angle] ABC ~ DEC [By AA axiom of similarity]
ABDE
=ACDC
[ If two triangles are similar, then their corresponding sides are proportional]
B EC
D
A
1. Given, AB = 9 cm DE = 3 cmand AC = 24 cmIn ABC and DEC,
MT EDUCARE LTD. X - ICSE (Maths)
33
93
=24DC
DC =24 3
9
= 8 cm
Now, AD = AC – DCAD = 24 – 8AD = 16 cm
2. Let 0 be the centre and r be the radius of acircle. Join OM, ON and OS.We know that the tangent at any point of acircle and the radius through the point ofcontact are perpendicular to each other. OMQ = ONQ = 90ºAlso, NQ = QM [Tangents drawn from external
point to a circle are equal in length] NQ = QM= OM = ON = r [Say]Thus, OMQN is a square.
We know that, tangents drawn from an external point are equal in lenngths. PN = PS PQ – QN = PS 3 – r = PS ... (i)and MR = SR QR – QM = SR 4 – r = SR
On adding equation (i) and (ii), we get3 – r + 4 – r = PS + SR
7 – 2r = PR = 5 [In right PQR, PR = 2 2PQ + QR(Pythagoras theorem)
= 2 23 4 = 9 16
= 25 = 5 cm] 7 – 5 = 2r 2r = 2 r = 1 cm [Radius of incircle]
3. Given : In the given figure, PA = AMTo prove :(i) PMB is isosceles.(ii) PA . PB = MB2
Proof :(i) Since PA = AM [Given] AMP = APM ... (i)
[Angles opposite to equal sides are equal]
Since, angle between the tangent and chord is equal to the angle subtended
BAP
M
P
Q
N3 c
m
M R
S
Or
4 cm
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MT EDUCARE LTD.X - ICSE (Maths)
by same chord in alternate segment. AMP = ABM = PBM APM = PBM
or BPM = PBM PM = MB
Hence, PMB is an isosceles triangle.(ii) Since, PM is tangent to the circle and PAB is a secant.
PAPB = PM2
PAPB = MB2 [PM = MB]Hence proved
4. To prove : BD = 2OCConstruction : Join CO and DB.
Proof : In OCA and BDA,OCA = BDA = 90º [ Angle in semi-circle is right
angle]and OAC = BAD [Common angle]
OAC ~ BAD [By AA axiom of similarity]
Then,OABA
=OCBD
[In similar triangles sides are
proportional]
OA2OA
=OCBD
[ BA = 2AO]
BD = 2OCHence proved
5.
On measuring, we get BCP = 30º
PA
C
3.5 cm
M
B
X
O120º
6 cm
N
A
C
B
D
O
MT EDUCARE LTD. X - ICSE (Maths)
35
6 Angle at centre is twice the angle at circumference.AOC = 2 × ADC
AOC = 2xBut AOC = 160º [Given]
160º = 2x
x =160º
2= 80º
Also, xº + yº = 180º 80º + y = 180º y = 180º – 80º yº = 100º
Now, 3y –2x = 3 × 100º – 2 × 80º= 300º – 160º= 140º
Hence proved
2006
1. Given, BAD = 65º, ABD = 70º and BDC = 45º(i) Since, ABCD is a cyclic quadrilateral.
So, the sum of opposite angles of a cyclicquadrilateral is 180º.
BCD + BAD = 180º 65º + BCD = 180º BCD = 180º – 65º = 115º(ii) In ABD,
DAB + ABD + ADB = 180º 65º + 70º + ADB = 180º ADB = 180º – 135º = 45º
ADC =BDA + BDC = 45º + 45º = 90ºi.e. ADC is in semi-circle.So AC is a diameter.
2. (i) Given, AB is a diameter of circle. ACB = 90º [ Angle in a semi-circle is 90º]
In ABC,ACB + CBA + BAC = 180º
90º + CBA + 34º = 180º CBA = 180º – 124º
CBA = 56º(ii) Since, CQ is a tangent to the circle. BCQ = BAC = 34º [Alternate segment property]
Since, AQ is a straight line. ABC + CBQ = 180º
AB
CD
65º 70º
45º
34ºA BQ
C
y
A C
x
160ºO
B
D
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MT EDUCARE LTD.X - ICSE (Maths)
56º + CBQ = 180º CBQ = 180º – 56º = 124º
Now in CBQ,CBQ + CQB + BCQ = 180º
124º + CQB + 34º = 180º CQB = CQA = 22º
3. Since, PRT is a tangent at R and QR is a chord.
QRP = RSQ = yº [Alternate segment property]Since, SQ is a diameter of a circle. QRS = 90º [Angle in semi-circle]
In PRS,SPR + PRS + RSP = 180º [Sum of all angles of a triangle is
180º] xº + (yº + 90º) + yº = 180º xº + 2yº = 180º – 90º xº + 2yº = 90º
Hence proved
4. Given, PO = 6 cm, QO = 9 cm and ar(POB) = 120 cm2
In AOQ and BOP, we haveQAO = OBP = 90º [ PB AB and QA AB]
and AOQ = POB [Vertically opposite angles] AOQ ~ POB [By AA axiom of similarity]
Then,ar( AOQ)ar( BOP) =
2
2
(OQ)(PO)
ar(AOQ) =29
6
× 120
=81 120
36
= 270 cm2
Hence, the area of AOQ is 270 cm2
xº yº
O
Q
P
S
RT
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37
5.
2007
1.
Radius = 1.5 cm
2. Given, AT = 16 cm and AB = 12 cmWe know that, chord AB and tangent at P meet at point T, then
TA × TB = PT2
16 × (AT – AB) = PT2
16 × (16 – 12) = PT2
PT2 = 16 × 4 PT2 = 64 PT = 8 cm
3. Given PBA = 45ºCalculate the value of PQBSince, AB is a diameter of circle,
A
B60º
X
D C
O
2 cm
5 cm
2 cm
4.2 cm
5.8 cm
C
O
E
A
B
X
6.4 cm
60º
P
T
BA
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MT EDUCARE LTD.X - ICSE (Maths)
So APB is in semi-circle. APB = 90º
In right APB,APB + PBA + BAP = 180º
90º + 45º + BAP = 180º 135º + BAP = 180º BAP = 180º – 135º = 45º PQB = BAP = 45º [Angles in the same segment are
equal]Hence, the value of PQB is 45º
4. In ACE,AEC + EAC + ACE = 180º [ Sum of all the angles of a
triangle is 180º] AEC + 62º + 43º = 180º AEC = 180º – 105º = 75ºSince, points A, B, D and E are the pointson a circle, so quadrilateral ABDE is acyclic quadrilateral.We know that, sum of opposite angles is 180º
ABD + AED = 180º a + 75º = 180º [ AEC = AED = 75º] a = 105º
EDF = BAE [Exterior angle property] c = 62ºIn ABF,
ABF + AFB + BAF = 180º [By angle sum property of a triangle]
105º + b + 62º = 180º b = 180º – (105º + 62º)
= 180º – (167º) = 13ºHence, a = 105º, b = 13º and c = 62º
5. Given : DE BC andADDB
=32
(i) Now,ADDB
=32
AD
AD DB=
33 2
ADAB =
35
In ADE and ABC, we getDAE = BAC [Common angle]
A E F
BD
C
b
ac
62º
43º
D E
B C
A
F
A B
Q
P
O45º
MT EDUCARE LTD. X - ICSE (Maths)
39
and ADE = ABC [Corresponding angles as DE BC] ADE ~ ABC [By AA axiom of similarity]
Then,DEBC
=ADAB
[ If two triangles are similar, then
their corresponding sides areproportional]
DEBC =
35 ... (i)
(ii) To prove : DEF ~ CBFProof : In DEF and CBF, we have
FED = FBC [Alternate interior angles as DE BC]and DFE = BFC [Vertically opposite angles] DEF ~ CBF [By AA axiom of similarity]
Then,DEBC
=FEFB
[ If two triangles are similar, then
FEFB
=35
[From equation (i)]
(iii)Since, DEF ~ CBF
ar( DEF)ar( CFB) =
2
2
DEBC
=23
5
=925
[From equation (i)]
Hence, ar(DEF) : ar(CFB) = 9 : 25
6.
(ii) 9.4 cm
AQ
D
P
C
R X
B
4 cm
45º
M
5 cm
their corresponding sides are proportional]
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MT EDUCARE LTD.X - ICSE (Maths)
2008
1. (i) area APO : ABC = 4 : 25(ii) area APO : CQO = 4 : 9
2. (i) QOP = 112º(ii) QCP = 68º
3. 12 cm
4.
AXBY is a square
2009
1. (i) 75º(ii) 15º(iii)105º
2. (i) 24 m(ii) 1000 cm3
C DX
P
Q
BA
E FY
8 cm
MT EDUCARE LTD. X - ICSE (Maths)
41
3.
4. Given : AB = 7 cm and BC = 9 cm
(i) To prove : ACD ~DCBProof : In ACD and DCB,
C = C [Common angles]CDB = BAD [Anlges in alternate segment]
ACD ~ DCB [By AA axiom of similarity](ii) Since, chord AB and tangent at D intersect each other at point C. AC × BC = CD2
16 × 9 = CD2
CD2 = 144 = (12)2
CD = 12 cm
5. Given : BA CE and AF : AC = 5 : 8
(i) To prove : ADF ~ CEFProof : In ADF and CEF,
AFD = CFE [Vertically opposite angles]DAF = FCE [Alternate angles, since BA CE]
ADF ~ CEF [By AA axiom of similarity](ii) Given : CE = 6 cm
Since, ADF ~ CEF
ADCE
=AFCF
AF 5 AC 8 AC 8–1 –1AC 8 AF 5 AF 5
AC – AF 8 – 5 CF 3 AF 5AF 5 AF 5 CF 3
A B
D
C7 cm 9 cm
CB
ED F
A
5.5cm
3.4
cm4.9 cm
A
OY
C
B
X
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MT EDUCARE LTD.X - ICSE (Maths)
AD6
=53
AD =5 6
3
AD = 5 × 2AD = 10 cm
(iii)Given, DF BC ADF = ABC [Corresponding angles]
and DAF = BAC [Common angle] ADF ~ ABC
Then,ar( ADF)ar( ABC) =
2
2
(AF)(AC) =
258
AF 5AC 8
=2564
ar(ADF) : ar(ABC) = 25 : 64
2010
1. BAO = 40º
2. (i) 12 cm(ii) Area of ABC = 36 cm2
3.
(iv) 5.2 cm
CD
P
60ºB
A
6 cm
3.5
cm
MT EDUCARE LTD. X - ICSE (Maths)
43
4.
2011
1. (i) In ADB and CAB,ADB = CAB [both 90º]ABD = CBA [common angle]
ABC DBA [AA similarity criterion]In ADC and BAC,ADC = BAC [both 90º]ACD = ACB [common angle]
DAC ABC [AA similarity criterion]If two triangles are similar to one triangle, then the two triangles aresimilar to each other.
DAC DBA or CDA ADB
(ii) Since the corresponding sides of similar triangles are proportional.
CDAD
=DADB
AD2 = DB x CD AD2 = 18 x 8
AD = 12 cm
(iii) The ratio of the areas of two similar triangles is equal to the ratio of thesquares of their corresponding sides.
So,Ar( ADB)Ar( CDA)
=2
2
ADCD
=14464 =
94
Thus, the required ratio is 9:4 .
C
A B
D8 cm
18 cm
4 cm
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MT EDUCARE LTD.X - ICSE (Maths)
2.
PQ = 4.8 cm
3. In AOC, ACO = 30º, [Given]OAC = 90º [radius is prependicular to the
tangent at the point of contact]
By angle sum property,ACO + OAC + AOC = 1800
AOC = 1800 – (900 + 300) = 600
Consider AOC and BOCAO = BO [radii]AC = BC [tangents to a circle from an
external point are equal in length]OC = OC [Common]
AOC BOC.(i) BCO = ACO = 30º(ii) AOC = BOC = 60º
AOB = AOC + BOC = 120º(iii) AOB = 2APB
So, APB =12
AOB
=12
× 120º
= 60º
4. Let the radii of the circles with A, B, and Cas centres be r1, r2 and r3 respectively.
According to the given information,AB = 10 cm = r1 + r2 ...(1)
OP
Q
R
M6 cm 3.5 cm
C
A
B
30º
PO
MT EDUCARE LTD. X - ICSE (Maths)
45
B P A
Q
C
R
BC = 8 cm = r2 + r3 ...(2)CA = 6 cm = r1 + r3 ...(3)
Adding equations (1), (2) and (3),2(r1 + r2 + r3) = 24
r1 + r2 + r3 = 12 ...(4)Subtracting (1) from (4),
r3 = 12 –10 = 2 cmSubtracting (2) from (4),
r1 = 12 – 8 = 4 cmSubtracting (3) from (4),
r2 = 12 – 6 = 6 cmThus, the radii of the three circles with centres A, B and C are 4 cm, 6 cmand 2 cm respectively.
2012
1. 11.25 cm.
2. (i) 50º(ii) 40º
3. In ABC and AMP(i) ABC = AMP [each 900]
BAC = PAM [common] ABC AMP [By AA similarity]
Since the triangles are similar, we have
(ii)ABAM
=BCMP
=ACAP
ABAM
=BC12
=1015
Taking,BC12
=1015
BC =12 10
8cm15
Now using Pythangoras theorem in triangle ABC,AB2 + BC2 – AC2
AB2 = 102 – 82 = 36AB = 6 cm
Hence, AB = 6 cm and BC = 8 cm
BA P
C
M
46
MT EDUCARE LTD.X - ICSE (Maths)
4.
5. 3 cm
2013
1. (i) In ABC,DAB + ABD + ADB = 1800
650 + 700 + ADB = 1800
ADB = 1800 – 700 – 650
= 450
Now, ADC = ADB + BDC= 450 + 450
= 900
ADC is the angle of semi-circle so AC is a diameter of the circle.(ii) ACB = ADB (angle subtended by the same segment) ACB = 450
2. (i) ADC = 80º(ii) DCT = 60º
A B
D
C450
650
700
C
B
A
D40º
100º O
T
MT EDUCARE LTD. X - ICSE (Maths)
47
3.
On measuring BCP = 30º
4. (i) In ABC and DEC,ABC = DEC = 90º (perpendiculars to BC)ACB = DCE (Common)ABC DEC (AA criterion)
(ii) Since ABC DEC,
ABDE
=ACCD
64 =
15CD
6 × CD = 60
CD =606 = 10 cm
(iii) ABC ~ DECar(ABC) : ar(DEC) = AB2 : DE2
= 62 : 42
= 36 : 16= 9 : 4
2014
1. (i) 32º(ii) 148º(iii) 32º
2. Consider the given triangleA
B D C
A
D
CB E
6 cm120º
3.5 cm
X
A
B
P
C
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MT EDUCARE LTD.X - ICSE (Maths)
Given that ABC DAC
(i) Consider the triangles ABC and DAC
ABC = DAC [given]
C = C [common]
By AA Similarity, ABC DAC
(ii) Hence the corresponding sides are proportional
ABDA
=ACDC
=BCAC
85
=4
DC
DC =4 5
8
DC =52
cm
= 2.5 cm
ABDA
=BCAC
85
=4
BC
BC =8 4
5
BC =325
cm
= 6.4 cm(iii)We need to find the ratios of the area of the triangles ABC and DAC.
Since the triangles ABC and DAC are similar triangles, we have
Thus,Area(ΔACD)Area(ΔABC) =
2
2
DAAB
So,Area( ACD)Area(ΔABC)
=
2
2
58
Area(ΔACD)Area(ΔABC) =
2564
= 25 : 64
3. (i) 12 cm(ii) 8 cm
MT EDUCARE LTD. X - ICSE (Maths)
49
4.
Radius = 1.5 cm
2015
1. (i) x = 25º(ii) y = 50º(iii) z = 40º
2.
3. Construction : Join AD and CB.
In APD and CPBA = C [Angles in the same segment]
C
A
5 cm
B N
O5.5
cm
6.5 cm
A
xF O
5 cm
B
E D
C
x
A C
P
BD
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MT EDUCARE LTD.X - ICSE (Maths)
D = B [Angles in the same segment] APD CPB [By AA Postulate]
APCP
=PDPB
[Corresponding sides of similar
triangles] AP × PB = CP × PD
4. (i) Consider ADE and ACBA = A [Common]
mB = mE = 900
ACB ADE. [AA postulate]
(ii) ConsiderADE ~ ACB
AEAB =
ADAC =
DEBC ... (i)[Corresponding sides of similar
triangles]
Consider ABCBy applying Pythagoras Theorem, we have
AB2 + BC2 = AC2
AB2 + 52 = 132
AB = 12 cmFrom equation (1), we have
412 =
AD13 =
DE5
13 =
AD13
AD =133 cm
Also,4
12 =DE5
DE =20 512 3 cm
(iii) We need to find the area of ADE and quadritateral BCED.
Area of ADE =1
× AE × DE2
A
B C
D
E
C
4 cm
D
B
A
E
9 cm13 cm
12 cm
5 cm
MT EDUCARE LTD. X - ICSE (Maths)
51
=1
× 4 ×2
53
= 210cm
=1 10
×BC × AB2
–3
=1 10
×5 ×122
-3
=10
303
=90 10
3
= 2803
cm
Thus ratio of areas of ADE to quadrilateral BCED =
1013
803
8
5.
2016
1. (i) 32º(ii) 64º(iii) 58º
C
M
AB
D
5.5 cm
PT
6 cm
NQ
R105º
3Area of quad.BCED = Area of ABC Area of ADE
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MT EDUCARE LTD.X - ICSE (Maths)
2. (i)
(ii) The figure is an irregular hexagon (or arrow head).
3.
4. (i) Since PQRS is a cyclic quadrilateral,
RSP = RQT ... (i)[Exterior angle property]In TPS and TRQ,
PTS RTQ [Common angle]RSP = RQT [From (i)]
TPS ~ TRQ [AA similarity criterion](ii) Since TPS ~ TRQ,
SPQR =
TPTR
A(–4, 4)
C(0,–3)
B(–3, 0)
S
R
Q TP
C
B
O
A Y F
M
E
N
DX
5 cm
L
MT EDUCARE LTD. X - ICSE (Maths)
53
SP4
=186
SP =18 4
6
SP = 12 m(iii) Since TPS ~ TRQ,
ar( TPS)ar( TRQ) =
2
2
SPRQ
27
ar( TRQ) =212
4
ar(TRQ) =279
ar(TRQ) = 3 cm2
ar(PQRS) = ar(TPS) – ar(TRQ) ar(PQRS) = 27 – 3 ar(PQRS) = 24 cm2
5.
6. (i) 600 m(ii) 2 m2
(iii) 175500000 m3
7.
(iv) XY = 5 cm
P
O
A
Q
C B
M2.5
cm 2.5 cm
3 cm 5 cm
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MT EDUCARE LTD.X - ICSE (Maths)
2017
1. Steps of construction :(i) Draw line AB = 7 cm and
CAB = 60º. Cut off AC = 5 cm.Join BC. ABC is the requiredtriangle.
(ii) Draw angle bisectors of Aand B.
(iii) Bisector of B meets AC at Mand bisector of A meets BCat N.
(iv) P is the point which isequidistant from AB, BC and AC.
(v) Draw a perpendicular frompoint P to AB and let itintersect at point D
(vi) With DP as the radius, draw a circle touching the three sides of thetriangle (incircle.)
2. (i) BAQ = 30ºSince AB is the bisector of CAQ
CAB = BAQ = 30ºAD is the bisector of CAP and P – A – Q
DAP + CAD + CAQ = 180º CAD + CAD + 60º = 180º 2CAD = 180º – 60º CAD = 60º
So, CAD + CAB = 60º + 30º= 90º
So, BD is the diameter of the circle.
(ii) BAQ = ACB (Alternate segment theorem) ACB = 30ºIn ABC,
CAB = 30º (AB is the bisector of CAQ) ACB = CAB AB = BC (Sides opposite to equal angles
are equal) ABC is an isosceles triangle.
3. (i) DAE = 70ºSince ABCD is cyclic quadrilateral
BCD = DAE[Exterior angle property]
BCD = 70º
C
D
M
A
N
B
P
60º
5 cm
7 cm
A
B A
D
E
O
P QA
C
B
D
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(ii) BOD = 2BCD [Angle at centre is twice theangle at the circumference]
BOD = 2(70º) BOD = 140º
(iii) In OBD, OB = OD OBD = ODBBy Angle Sum property,OBD + ODB + BOD = 180º
2OBD + 140º = 180º 2OBD = 40º OBD = 20º
4. (i) In PQR and SPR, we haveQPR = PSR (Given)PRQ = PRS (Common)
PQR ~ SPR (A.A. Axiom)
PQSP
=QRPR
=PRSR
... (i)
(ii) Now,QRPR
=PRSR
... [From (i)]
QR6
=63
QR =6 6
3
= 12 cm
Also,PQSP
=PRSR
...[From (i)]
8
SP=
63
8
SP= 2
SP =82
= 4 cm
(iii)Area of PQRArea of SPR
=2
2
PQSP
=2
2
84
=6416
= 4
Q
P
RS 3 cm
8 cm 6 cm