2024 ____ ____ 1100
MT Seat No.
MT - MATHEMATICS (71) Geometry - PRELIM - II - PAPER - I
Time : 2 Hours Model Answer Paper Max. Marks : 40
Q.1. (A) Choose the correct alternative for each of the following.
(i) (B) (3,4,5) 1
(ii) (A) 50° 1
(iii)
(C) (–3, 1) 1
(iv) (B) 550 cm2 1
Q.1. (B) Solve any 4 out of 4
(i) (Triangles with common base)
(ii) In ∆AMB, ∠AMB = 90o (Given)
∠B = 60o (angle of an equilateral triangle)
∠BAM = 30 (Sum of all angles of a triangle is 180o)
o
∴ ∆AMB is 30o - 60o - 90o triangle
By 30o – 60o – 90o triangle theorem,
AM = × AB (Side opposite to 60o)
∴ AM = × 2a
∴ AM = a
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1
(iii) m∠ABE =
[m(arc DC) + m(arc AE)] ½
21
\ 108° = [m(arc DC) + 95°]
2
\ 216° = m(arc DC) + 95°
\ m(arc DC) = 216° – 95°
\ m(arc DC) = 121° ½
(iv) Let A(22, 20) = (x1, y1) and
B (0, 16) = (x2, y2)
Let M (x, y) be the midpoint of seg AB.
By midpoint formula,
and
x = and y=
x = and y=
x = 11 and y = 18
\ M(11, 18)
Q.2. (A) Complete 2 activities out of 3
(i) The angle inscribed in the semicircle
is a right angle Prove the result by
completing the following activity.
Given: ∠ABC is inscribed angle in a
semicircle with center M.
To prove : ∠ABC is a right angle.
Proof: segment AC is a diameter of the circle.
∴ m(arc AXC) = 180º
Arc AXC is intercepted by the inscribed angle ∠ABC.
1
∠ABC = 2 ...(Inscribed angle theorem)
1
= × 180º
2
∴ m∠ABC = 90º
∴ ∠ABC is a right angle.
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5
(ii) − 5 cot2 θ, Complete the activity given below.
sin2 θ
5
Activity → − 5 cot2 θ
sin2 θ
1
= 5 ( − cot2 θ )
sin2 θ
1
= 5 ( cosec2 θ − cot2 θ) ......... ( = cosec2 θ )
sin2 θ
= 5 (1)
= 5
(iii) Find the volume of greatest right circular cone, which can be cut from
a cube of a side 4 cm. (Answer in terms of )
Diameter of cone = edge of the cube
\ h = 4 cm
r = 2 cm
1
Volume of cone = r2h ........... (Formula)
3
1
V = (2) 3
×4
3
16
V= cm3
3
Q.2. (B) Solve any 4 out of 5 subquestions
(i) In ∆ABC, ray BD bisects ∠ABC (Given)
∴ (Angle bisector property of a triangle)
x (x + 2) = (x + 5) (x – 2)
x2 + 2x = x2 + 5x – 2x – 10
x2 + 2x = x2 + 3x – 10
x2 + 2x – x2 – 3x = –10
–x = –10
x = 10
∴ x = 10
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(ii) In ∆AMB, ∠AMB = 90o (Given)
∠B = 60o (Angle of an equilateral triangle)
∴ ∠BAM = 30o (Sum of all angles of a trangle is 180o)
∴ ∆AMB is 30o - 60o - 90o triangle
By 30o - 60o - 90o triangle theorem,
AM = × AB (Side opposite to 60o)
∴ AB = × AM
∴ AB
∴ AB = 2 cm
∴ Perimeter of ABC = 3 × AB
= 3 × 2 = 6cm
∴ Perimeter of ∆ABC = 6 cm
(iii) Let WT = a
WX = WT + TX ...(W – T – X)
\ 25 = a + TX
\ TX = (25 – a)
YZ = YT + TZ ...(Y-T-Z)
\ 26 = 8 + TZ
\ TZ = 26 – 8
\ TZ = 18
chords XW and YZ intersect each other in the interior of the circle at point T.
WT × TX = TZ × YT
\ a(25 – a) = 18 × 8
\ 25a – a2 = 144
\ a2 – 25a + 144 = 0
\ a2 – 16a – 9a + 144 = 0
\ a(a – 16) – 9 (a – 9) = 0
\ (a – 16) (a – 9) = 0
\ a – 16 = 0 or a – 9 = 0
\ a = 16 or a = 9
\ WT = 16 units or WT = 9 units
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(iv) 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,
0
\ 0 × (m + n) = –9m + 3n
\ 0 = –9 +3n
\ 9m = 3n
\ =
\ =
\ m : n = 1 : 3
\ Y-axis divides segment joining points
A and B in the ratios 1 : 3
(v) CH represents the height of the church and
C represents its top. P is the position of the
person at a distance of 80 m from the church.
\ PH = 80 m
ÐCPH is the angle of the elevation
\ ÐCPH = 45°
In ∆ PHC, ÐCHP = 90°
CH
\ tan ÐCPH = (By definition)
PH
CH
\ tan 45° =
80
CH
\ 1 =
80
\ CH = 80 m
\ Height of the church is 80 m
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Q.3. (A) Complete 1 activity out of 2
(i) A line is parallel to one side of a triangle
which intersects remaining two sides in
two distinct points, then than line
divides the sides in same proportion.
Given : In ∆ABC, line l ║ side BC and
line l intersect side AB in P and side AC
at Q.
AP AQ
To Prove : =
PB QC
Contruction : Draw seg CP and seg BQ
Proof : ∆APQ and ∆PQB have a equal heights.
AP
A(∆APQ) ½
= ...(i) (areas in proportion of bases)
A(∆PQB) PB
A(∆APQ) AQ
= ...(ii) (areas in proportion of bases) ½
A(∆PQC) QC
In ∆PQC and ∆PQB PQ is a common base ½
seg PQ║ seg BC hence heights of ∆PQC and ∆PQB are equal
∴ A(∆PQC) = A(∆PQB) ...(iii) ½
A(∆APQ) A(∆ APQ )
∴ = [From (i), (ii) and (iii)] 1
A(∆PQB) A(∆ PQC )
AP AQ
∴ = ..... [From (i), (ii) and (iii)]
PB QC
(ii) If the point P (6, 7) divides the segment joining A (8, 9) and B(1, 2) in
some ratio. Find that ratio.
Point P divides segment AB in the ratio m : n.
A(8, 9) = (x1 , y1), B(1, 2) = (x2 , y2) and P(6, 7) = (x , y)
Using Section formula of internal division,
m( 2 ) + n(9)
∴ 7 =
m+n
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∴ 7m + 7n = 2m + 9n
∴ 7m – 2m = 9n – 7n
∴ 5m = 2n
m 2
∴ n = 5
Q.3. (B) Solve any 2 out of 4 subquestions
(i) In ΔLMN, LMN = 900.
M
seg MR side LN
MR2 = LR × RN ....(i)
[Theorem of geometric mean] L R N
In ΔLKN, LKN = 900.
seg KR side LN K
KR2 = LR × RN ....(ii)
[Theorem of geometric mean]
\ MR2 = KR2 [From (i) & (ii)]
\ MR = KR
\ R is the midpoint of seg MK
(ii) Theorem: Opposite angles of a cyclic D
quadrilateral are supplementry.
Fill in the blanks and
C
complete the following proof.
Given : □ ABCD is cyclic. A
B
To prove : (i) ∠B + ∠D = 180°
(ii) ∠A + ∠C = 180°
Proof : Arc ABC is intercepted by the inscribed angle ∠ADC.
1
\ m∠ADC = m(arc ABC) .......... (I) (Inscribed angle theorem)
2
Similarly ∠ABC is an inscribed angle. It intercepts arc ADC.
1
\ m∠ABC =
m(arc ADC) ..... (II)
2
1 1
\ m∠ADC + m∠ABC = m(arc ABC) + m(arc ADC)
2 2
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... [adding (I) & (II)]
1
= [ m(arc ABC) + m(arc ADC)]
2
1
= × 360° ... a
rc ABC and
2
arc ADC constitute a
complete circle.
= 180º
Similarly we can prove, ∠A + ∠C = 180°. 3
(iii) Analytical figure:
Line AQ and line BQ are the required tangents to the circle from point Q.
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(iv) For segment PRQ, q = ∠POQ = 90°
A (segment PRQ) = 114 cm2
sin
A (segment PQR) = r2 360 2 1
\ 114 = r2 3.14 90 sin 90
360 2
3.14 1
\ 114 = r2
4 2
1.57 1
\ 114 = r2
2
0.57
\ 114 = r2
2
114 × 2
\ = r2
0.57
114 × 2 × 100
\ r2 =
57
\ r2 = 2 × 2 × 10 × 10
\ r = 20 ...(Taking square roots)
\ Radius of the circle is 20 cm 1
Q.4. Solve any 2 out of 3 subquestions
(i) In Δ ABC, D
seg AE bisects BAC,
A
AB BE F
\ = ...(i)
AC EC
[Property of an angle bisector of a triangle]
In Δ ADC, B
E
seg AF bisects DAC, C
AD DF
\ = ...(ii)
AC FC
[Property of an angle bisector of a triangle]
Also AB = AD ...(iii) [Given]
In Δ BDC,
BE DF
= ...[From (i), (ii) and (iii)]
EC FC
\ seg EF ║ seg BD [Converse of basic proportionality theorem]
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(ii) Analytical figure:
L
A
m
5. 1c
40°
B
4.8 cm C N
B1
B2
B3
B4
B5
B6
B7
x x
DLBN is the required triangle similar to the DABC. 4
(iii) For the wire,
2
Diameter = cm
5
1
radius (r) = cm
5
length of wire = 'h'
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Radius of a sphere (R) = 9 cm
According the given information
Volume of sphere = Volume of wire drawn
4
R3 = r2h
3
4 1 1
×9×9×9 = × ×h
3 5 5
4 × 3 × 9 × 9 × 25 = h
\ h = 24300 cm
\ h = 243 m
\ Length of the wire drawn is 243 m
Q.5. Solve any 1 out of 2 subquestions
(i) Construction : Draw seg MN.
Proof : PNMR is a cyclic quadrilateral
(Definition)
∠MNQ is an exterior angle of PNMR
(Definition)
\ ∠MNQ @ ∠PRM ...(i)
(Exterior angle of cyclic quadrilateral is congruent to the angle
opposite to its adjacent interior angle)
MNQS is cyclic quadrilateral (Definition)
\ ∠MNQ + ∠MSQ = 180° ...(ii)
(Opposite angles of a cyclic quadrilateral are supplementary)
\ ∠PRM + ∠MSQ = 180° [From (i) and (ii)]
\ ∠PRS + ∠RSQ = 180° (R - M - S)
\ seg PR segQS (Interior angles test)
cot A – cos A
(ii) LHS = cot A + cos A
cos A – cos A
sin A
=
cos A
+ cos A
sin A
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1
cos A − 1
sin A
=
1
cos A +1
sin A
cosec A – 1
= cosec A + 1
LHS = RHS
vvvvv
