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 180 o . 1 (iii) In ABC, ABC = 90 o B A C D Seg BD is the median on hypotenuse AC. BD = 1 2 AC (Median drawn to the hypotenuse is half of it) 7 = 1 2 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 tan 40 cot 50 = o o tan 40 tan(90 50) - = o o tan 40 tan 40 = 1 1
14
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B MT - xssc.maheshtutorials.com/images/SSC_Testpapers/school/STB...8 / MT - x SET - B = ½ = tan + cot ½ = R.H.S. sec2 q+cosec2 q = tan + cot A.4. Solve ANY THREE of the following
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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)