B MT - xssc.maheshtutorials.com/images/SSC_Testpapers/school/STB...8 / MT - x SET - B = ½ = tan +...
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2018 ___ ___ 1100 - MT - x - MATHEMATICS (71) Geometry - SET - B (E)
MT - x
Time : 2 Hours Preliminary Model Answer Paper Max. Marks : 40
Q.P. SET CODE
B A.1.(A) Solve ANY FOUR of the following :
(i) If diagonals of a parallelogram are congruent, then it is a rectangle. 1
(ii) If a transversal intersects two parallel lines then the sum of interior angles on the same side of the transversal is 180o . 1
(iii) In ABC, ABC = 90o B
A CD
Seg BD is the median
on hypotenuse AC.
BD =12
AC
(Median drawn to the hypotenuse is half of it)
7 =12
AC
AC = 14 cm 1
(iv) A quadrilateral is a parallelogram if a pair of opposite sides is parallel and congruent. 1
(v) Equation of the Y – axis is X = 0 1
(vi)o
o
tan40cot50
=o
o
tan40tan(90 50)-
=o
o
tan40tan40
= 1 1
SET - B2 / MT - x
A.1.(B) Solve ANY TWO of the following :
(i) For a cone Area of the base = 1386 sq. cm Height (h) = 28 cm
Volume of a cone = r2 h 1
=13
× 1386 × 28 ½
Volume of a cone = 12936 cm3 ½
(ii) A circle with centre O
O
A BM
chord AB = 24 cm
seg OM chord AB
OM = 5 cm
AM = 12
AB ½
[Perpendicular drawn from the centre to the chord, bisects the chord]
= 12
× 24
AM = 12 cm ½ In OMA, OMA = 90o
OA2 = 52 + 122 [Pythagoras theorem] ½ OA2 = 25 + 144 OA2 = 169 OA = 13 cm Radius of the circle is 13 cm ½
(iii) In FAN,
AF80o 40o
N
F = 80o
A = 40o
N = 60o [Remaining angle] ½ F > N > A ½ AN > AF > FN [In a triangle, ½
side opposite to greater angle is greater] Greatest side is AN and the smallest side is FN ½
SET - B3 / MT - x
A.2.(A) Select the correct alternative answer and write it :
(i) (c) 4 1
(ii) (b) cosec2 – sin2 = 1 1
(iii) (b) 25 cm 1
(iv) (c) 1
A.2.(B) Solve ANY TWO of the following :
(i) B(k, –5) = (x1, y1)
C(1, 2) = (x2, y2)
Slope of line BC =
½
7 = ½
7(1 – k) = 2 + 5
7(1 – k) = 7 ½
1 – k =77
1 – k = 1 1 – 1 = k k = 0 ½
(ii) Proof : D
E F
Q
P
X
R
In ∆XDE, PQ DE ...(Given)
XPPD
=XQQE
...(i) (Basic proportionality theorem) ½
In ∆XEF, seg QR side EF ... (Given)
XQQE
=XRRF
...(ii) (Basic proportionality theorem) ½
XPPD
=XRRF
...[From (i) and (ii)] ½
seg PR side DF ...(Converse of Basic Proportionality theorem) ½
SET - B4 / MT - x
(iii) mABE = [m(arc DC) + m(arc AE)] A
BE
CD
½
108 = [m(arc DC) + 95] ½
216 = m(arc DC) + 95°
m(arc DC) = 216 – 95° ½
m(arc DC) = 121° ½
A.3.(A) Carry out ANY TWO of the following activites :
(i) Given : In ABC, ABC = 90o
To prove : AC2 = AB2 + BC2
Construction : Draw seg BD hypotenuse AC. A – D – C
Proof : In ABC, ABC = 90o (Given)
seg BD hypotenuse AC (Construction)
ABC ~ ADB ~ BDC (Similarity of right angle triangle) 2
ABC ~ ADB A
B C
D
ABAB
=
AB2 = AC × AD ...(i)
ABC ~ BDC
BCDC
= ...(ii)
BC2 = AC × DC ...(ii)
Adding (i) and (ii),
AB2 + BC2 = AC × AD + AC × DC
AB2 + BC2 = AC (AD + DC)
AB2 + BC = AC AC(AD + DC)
AB2 + BC2 = AC2
SET - B5 / MT - x
(ii) Given : In ABC, seg CE bisects ACB
A B
C
D
E
To prove : =
Construction : Through B, draw a line parallel to ray CE, Extend AC to intersect it at point D. Proof : In ABD, seg EC || seg BD (Construction)
= . . .(i) By B.P.T
ray CE || ray BD and AD is transversal
ACE CDB . . .(ii) (Corresponding angles) 2
Now, BC as transversal
ECB CBD . . .(iii) (Alternate angles)
But, ACE ECB . . .(iv) (Given) In CBD,
CBD CDB [from (ii), (iii), (iv)]
seg CB seg CD . . .(v) (converse of isosceles triangle theorem)
= from (i) and (v)
(iii) Given : ABCD is cyclic
A
B
C
D
To prove : A + C = 180o
Proof :
A = 12
m (arc BCD) ...(i) } (Inscribed angletheorem)
C =12
m (arc BAD) ...(ii) 2
Adding (i) and (ii)
A + C = 12
[m (arc BCD) + m (arc BAD)]
A + C =12
× 360
A + C = 180o
SET - B6 / MT - x
A.3.(B) Solve ANY TWO of the following :
(i) Analytical fi gure: Radius = 3.3 cm (Given) Chord = 6.6 cm (Given)
Chord is twice of radius. Chord PQ is a diameter.
6.6 cm
3.3 cm
O
P
Q
6.6 cm
3.3 cm
O
P
Q
l
1 mark for drawing circle and diameter 2 marks for drawing tangents at P & Q.
SET - B7 / MT - x
(ii) A(3, 8) = (x1, y1)
B(–9, 3) = (x2, y2)
Let point P(0, a) be a point on Y-axis which divides seg AB in the ratio m : n.
P(0, a) = (x, y)
By Section formula,
½
0 ½
0 (m + n) = –9m + 3n
0 = –9m + 3n ½
9m = 3n
=
=
m : n = 1 : 3
Y-axis divides segment joining pointsA and B in the ratios 1 : 3
½
(iii) LHS = 2 2sec cosecq + q
= ½
=
=
= ½
1tancot
tan cot 1
SET - B8 / MT - x
= ½
= tan + cot ½ = R.H.S.
2 2sec cosecq + q = tan + cot
A.4. Solve ANY THREE of the following :
(i) GD is ground level.
BC is base of the ladder of
the fi re brigade van at a
height of 2 m from ground level. ½
‘T’ is top of ladder of the
fi re brigads van at the maximum height
TBC = 70° ...(Angle of elevation)
BT is the length of the ladder
BT = 20 m, BG = 2 m
BGDC is a rectangle ...(By definition) ½
BG = CD = 2 m ...(Opposite sides of a rectangle)
In BCT, BCT = 90°
sin TBC = ...(By definition) ½
sin 70° =
0.94 =
TC = 0.9420
TC = 18.80 m ½
TD = TC + CD ...(T – C – D) ½
TD = 18.80 + 2
TD = 20.80 m
Other end of the ladder can reach20.80 m above the ground ladder. ½
SET - B9 / MT - x
(ii) For cylindrical wrapper,
Diameter = 14 mm ½
Radius (R) mm = 7 mm
Height (H) = 10 cm
i.e. H = 100 mm
For cylindrical tablet, ½
Radius (r) = 7 mm, Height (h) = 5 mm
Let ‘N’ number of tablets can be wrapped in the given wrapper.
N × Volume of tablet = Volume of wrapper. ½
N × r2h = R 2H ½
N × × 7 × 7 × 5 = × 7 × 7 × 100
N = ½
N = 20
20 tablets can be packed in thegiven wrapper. ½
(iii) Analytical fi gure:
SET - B10 / MT - x
1 mark for PQR 1 mark for constructing RR3Q RR4T 1 mark for constructing RQP RTL
(iv) AR = 5AP ...(Given)
A
P Q
RS
...(i) ½
AS = 5AQ ...(Given)
...(ii) ½
In ∆ASR and ∆AQP,
...[From (i) and (ii)].
SAR QAP ...(Vertically opposite angles) 1 ∆ASR ~ ∆AQP ...(By SAS Test of similarity)
...(c.s.s.t.) ½
...[From (i)]
SR = 5 PQ ½
SET - B11 / MT - x
A.5. Solve ANY ONE of the following :
(i) Proof :
A
Q
B
C
PDDAP PAB [Ray AP bisects DAB]
Let mDAP = mPAB = xo ...(i)DCQ QCB [Ray CQ bisects DCB]
Let mDCQ = mQCB = yo ...(ii) ½ABCD is cyclic
DAB + mDCB =180o
[Opposite angles of cyclic quadrilateral are supplementary] ½
DAP + PAB + DCQ + QCB = 180o
x + x + y + y = 180 [From (i) and (ii)] 2x + 2y = 180 ...(iii)
mDAP =12
m(arc DP) [Inscribed angle theorem] ½
x = 12
m(arc DP) [From (i)]
m(arc DP) = 2x ...(iv)
mDCQ =12
m(arc DAQ) [Inscribed angle theorem] ½
y = 12
m(arc DAQ) [From (ii)]
m(arc DAQ) = 2y ...(v) ½
2x + 2y = 180 ...(iii) m(arc DP) = 2x ...(iv) m(arc DAQ) = 2y ...(v) Adding (iv) and (v),
m(arc DP) + m(arc DAQ) = 2x + 2y ½ m(arc PDQ) = 2x + 2y [Arc addition property]
m(arc PDQ) = 180o [From (iii)] ½ Arc PDQ is a semicircle
Seg PQ is the diameter of the circle ½
SET - B12 / MT - x
(ii) Construction : Draw seg AE side BC ½ such that B – D – E - C Proof : In AEB, mAEB = 90o [Construction]
AB2 = AE2 + BE2 ...(i) [By Pythagoras theorem] ½
In AED, mAED = 90o [Construction]
AD2 = AE2 + BE2 ...(ii) [By Pythagoras theorem] ½
Subtracting equation (ii) from (i),A
B CD
AB2 – AD2 = AE2 + BE2 – (AE2 + DE2) ½
AB2 – AD2 = AE2 + BE2 – AE2 – DE2
AB2 – AD2 = BE2 – DE2
AB2 – AD2 = (BE + DE) (BE – DE)
AB2 – AD2 = (BE + DE) BD ...(iii) [B – D – E] ½
In AEB and AEC,mAEB = m AEC = 90o [Construction]
Hypotenuse AB Hypotenuse AC [Given]
seg AE seg AE [Common side]
AEB AEC [Hypotenuse-side theorem] 1
seg BE seg CE …(iv) [c.s.c.t.]
AB2 – AD2 = (CE + DE) BD [from (iii) and (iv)]
AB2 – AD2 = CD BD [C – E – D] ½
A.6. Solve ANY ONE of the following :
(i) To Prove : 2AB2 = 2AC2 + BC2A
BCD
Proof :
DB = CD ...(i) (Given)
In ADB, ADB = 90o (Given)
AB2 = AD2 + DB2 (By Pythagoras theorem) ½
AB2 = AD2 + (3CD)2 [From (i)] AB2 = AD2 + 9CD2 ...(ii)
In ADC, ADC = 90o ...(Given) AC2 = AD2 + CD2 (By Pythagoras theorem) ½ AD2 = AC2 – CD2 ...(iii)
AB2 = AC2 – CD2 + 9CD2 [From (ii) and (iii) ½ AB2 = AC2 + 8CD2 ...(iv)
But BC = CD + DB ...[C - D - B]
SET - B13 / MT - x
BC = CD + 3CD ...[From (i)] BC = 4CD
CD = ...(v) ½
AB2 = AC2 + 8 ...[From (iv) and (v)]
AB2 = AC2 + 8 × ½
AB2 = AC2 + ...[From (iv) and (v)]
2AB2 = 2AC2 + BC2 (Multiplying throughout by 2) ½
(ii) In PQRPQR = 90º [Tangent Theorem]R = 30º [Given]
P = 60º [Remaing angle] PQR is 30º – 60º – 90º triangle
PQ =12
PR (30º – 60º – 90º triangle theorem)
= 12
× 12
AB
Q
P
R
PQ = 6 cm Radiam of the circle (r) = 6 cm
QR = 32
PR
= 32
× 12
QR = 6 3
A( PQR) = 12
QR × PQ
=12
6 6
= 18 = 18 1.73 = 31.14 cm2 1
Area of a sector = r2
SET - B14 / MT - x
A(P–QAB) = 31.14 62
= 3 14 6 6 = 3.14 6
= 18.84 cm2 1 Area of shaded region = A(PQR) – A(P–QAB) = 31.14 – 18.84 = 12.30 cm2 1 Area of the shaded region = 12.30 cm2