MH 10th Question Paper 2025 Maths Part 2 Geometry

Maharashtra Board

Question Paper
2025
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QUESTION PAPER
Presented By

N 832 Seat No.

2025 III 07 1100 – N 832– MATHEMATICS (71) GEOMETRY—PART II (E)

(REVISED COURSE)
Time : 2 Hours (Pages 12) Max. Marks : 40

Note :—

(i) All questions are compulsory.

(ii) Use of a calculator is not allowed.

(iii) The numbers to the right of the questions indicate full marks.

(iv) In case of MCQs [Q. No. 1(A)] only the first attempt will be evaluated

and will be given credit.

(v) Draw proper figures wherever necessary.

(vi) The marks of construction should be clear. Do not erase them.

(vii) Diagram is essential for writing the proof of the theorem.

1. (A) Choose the correct alternative from given : 4

(1) Out of the following which is a Pythagorean triplet ?

(A) (1, 5, 10)

(B) (3, 4, 5)

(C) (2, 2, 2)

(D) (5, 5, 2)

P.T.O.

2/N 832
(2) ACB is inscribed angle in a circle with centre O. If ACB

= 65º, then what is measure of its intercepted arc AXB ?

(A) 65º

(B) 230º

(C) 295º

(D) 130º

(3) Distance of point (3, 4) from the origin is .................... .

(A) 7

(B) 1

(C) 5

(D) –5

(4) If radius of cone is 5 cm and its perpendicular height is 12 cm,

then the slant height is ..................... .

(A) 17 cm

(B) 4 cm

(C) 13 cm

(D) 60 cm

3/N 832
(B) Solve the following sub-questions : 4
(1) In the following figure ABC, B – D – C and BD = 7, BC = 20,
A( ABD)
then find A ( ABC) .

(2) In the following figure MNP = 90º, seg NQ seg MP , MQ = 9,
QP = 4, find NQ.

(3) Angle made by a line with the positive direction of X-axis is 30º.
Find slope of that line.
(4) In cyclic quadrilateral ABCD m A 100º , then find m C .

P.T.O.

4/N 832
2. (A) Complete the following activities and rewrite it (any two) : 4

(1) The radius of a circle with centre ‘P’ is 10 cm. If chord AB of

the circle subtends a right angle at P, find area of minor sector

by using the following activity. ( = 3.14)

Activity :
r = 10 cm, = 90º, = 3.14.

A(P–AXB) = ×
360

= 3.14 102
360
1
= ×
4

A (P–AXB) = sq. cm.

5/N 832
(2) In the following figure chord MN and chord RS intersect at

point D. If RD = 15, DS = 4, MD = 8, find DN by completing

the following activity :

Activity :

MD × DN = × DS .............. .

................. (Theorem of internal division of chords)

× DN = 15 × 4

DN =
8

DN =

P.T.O.

6/N 832
(3) An observer at a distance of 10 m from tree looks at the top
of the tree, the angle of elevation is 60º. To find the height of
tree complete the activity. ( 3 1.73 )

Activity :
In the figure given above, AB = h = height of tree, BC = 10 m,
distance of the observer from the tree.
Angle of elevation ( ) = BCA = 60º

tan = .................. (I)
BC

tan 60º = .................. (II)

AB
= 3 (From (I) and (II))
BC

AB = BC × 3 = 10 3

AB = 10 × 1.73 =

height of the tree is m.

7/N 832
(B) Solve the following sub-questions (any four) : 8

(1) In ABC, DE BC. If DB = 5.4 cm, AD = 1.8 cm, EC = 7.2 cm,

then find AE.

(2) In the figure given below, find RS and PS using the information

given in PSR.

P.T.O.

8/N 832
(3) In the following figure, circle with centre D touches the sides
of ACB at A and B. If ACB = 52º, find measure of ADB.

(4) Verify, whether points, A(1, –3), B(2, –5) and C(–4, 7) are collinear
or not.

11
(5) If sin , find the values of cos using trigonometric
61
identity.
3. (A) Complete the following activities and rewrite it (any one) : 3
(1) In the following figure, XY seg AC. If 2AX = 3BX and XY = 9.
Complete the activity to find the value of AC.

9/N 832
Activity :

2AX = 3BX (Given)

AX 3
=
BX
AX BX 3 2
= .......... (by componendo)
BX 2
5
= ................ (I)
BX 2

Now BCA ~ BYX ............. ( test of similarity)
BA AC
= .............. (corresponding sides of similar
BX XY
triangles)

AC
= .............. from (I)
9

AC =

(2) Complete the following activity to prove that the sum of squares
of diagonals of a rhombus is equal to the sum of the squares
of the sides.

P.T.O.

10/N 832
Given :

PQRS is a rhombus. Diagonals PR and SQ intersect each other
at point T.

To prove : PS2 + SR2 + QR2 + PQ2 = PR2 + QS2

Activity :

Diagonals of a rhombus bisect each other.

In PQS, PT is the median and in QRS, RT is the
median.

by Apollonius theorem,

PQ2 + PS2 = + 2QT2 ................. (I)

QR2 + SR2 = + 2QT2 ................. (II)

adding (I) and (II),

PQ2 + PS2 + QR2 + SR2 = 2 (PT2 + ) + 4QT2

= 2 (PT2 + ) + 4QT2

.............. (RT = PT)

= 4PT2 + 4QT2

2
= ( ) + (2QT)2

PQ2 + PS2 + QR2 + SR2 = PR2 + .

11/N 832
(B) Solve the following sub-questions (any two) : 6

(1) Show that points P(1, –2), Q(5, 2), R(3, –1), S(–1, –5) are the
vertices of a parallelogram.

(2) Prove that tangent segments drawn from an external point to
a circle are congruent.

(3) Draw a circle with radius 4.1 cm. Construct tangents to the circle
from a point at a distance 7.3 cm from the centre.

(4) How many solid cylinders of radius 10 cm and height 6 cm can
be made by melting a solid sphere of radius 30 cm ?

4. Solve the following sub-questions (any two) : 8

(1) In the following figure DE BC , then :

(i) If DE = 4 cm, BC = 8 cm, A ( ADE) = 25 cm 2 , find
A( ABC).

(ii) If DE : BC = 3 : 5, then find A ( ADE) : A ( DBCE).

P.T.O.

12/N 832
(2) ABC ~ PQR. In ABC, AB = 3.6 cm, BC = 4 cm and AC = 4.2 cm.

The corresponding sides of ABC and PQR are in the ratio 2 : 3,

construct ABC and PQR.

(3) The radii of the circular ends of a frustum of a cone are 14 cm and

8 cm. If the height of the frustum is 8 cm, find : ( = 3.14)

(i) Curved surface area of frustum.

(ii) Total surface area of the frustum.

(iii) Volume of the frustum.

5. Solve the following sub-questions (any one) : 3

(1) ABCD is a rectangle. Taking AD as a diameter, a semicircle AXD

is drawn which intersects the diagonal BD at X. If AB = 12 cm,

AD = 9 cm, then find the values of BD and BX.

(2) Taking = 30º to verify the following Trigonometric identities :

(i) sin 2 cos2 1

(ii) 1 tan 2 sec 2

(iii) 1 cot 2 cosec 2 .

2025 III 07 - 1100
N 833 Seat No.

Time : 2 Hours MATHEMATICS (71) GEOMETRY—PART II (M)

(REVISED COURSE)
Pages - 12 Total Marks : 40

(i)

(ii)

(iii)

(iv) 1(A)]

(v)

(vi)

(vii)

1. (A) 4

(1)

(A) (1, 5, 10)

(B) (3, 4, 5)

(C) (2, 2, 2)

(D) (5, 5, 2)

P.T.O.

2/N 833
(2) O ACB 65º

AXB

(A) 65º

(B) 230º

(C) 295º

(D) 130º

(3) (3, 4)

(A) 7

(B) 1

(C) 5

(D) –5

(4) 5 12

(A) 17

(B) 4

(C) 13

(D) 60

3/N 833
(B) 4
A( ABD)
(1) ABC B–D–C BD = 7, BC = 20, A ( ABC)
=

(2) MNP = 90º, NQ MP, MQ = 9,
QP = 4, NQ

(3) X- 30º

(4) ABCD m A 100º , m C

P.T.O.

4/N 833
2. (A) 4

(1) P 10 AB

( = 3.14)

r = 10 = 90º, = 3.14.

A(P–AXB) = ×
360

= 3.14 102
360

1
= ×
4

A (P–AXB) =

5/N 833
(2) MN RS D

RD = 15, DS = 4, MD = 8, DN

MD × DN = × DS .............. .

× DN = 15 × 4

DN =
8

DN =

P.T.O.

6/N 833
(3) 10
60º
( 3 1.73 )

C AB
AB = h = BC = 10
( ) = BCA = 60º.

tan = .................. (I)
BC
tan 60º = .................. (II)
AB
= 3 ((I) (II) )
BC
AB = BC × 3 = 10 3

AB = 10 × 1.73 =

7/N 833
(B) 8

(1) ABC DE BC. DB = 5.4 AD = 1.8

EC = 7.2 AE

(2) PSR RS PS

P.T.O.

8/N 833
(3) D ACB
A B ACB = 52º, ADB

(4) A(1, –3), B(2, –5) C(–4, 7)

11
(5) sin , cos
61

3. (A) 3
(1) XY AC. 2AX = 3BX XY = 9,
AC

9/N 833
2AX = 3BX

AX 3
=
BX
AX BX 3 2
= .........
BX 2
5
= ................ (I)
BX 2

BCA ~ BYX .............

BA AC
= ..............
BX XY

AC
= .............. (I)
9

AC =

(2)

P.T.O.

10/N 833

PQRS PR SQ T

PS2 + SR2 + QR2 + PQ2 = PR2 + QS2

PQS PT QRS RT

PQ2 + PS2 = + 2QT2 ................. (I)

QR2 + SR2 = + 2QT2 ................. (II)

(I) (II)

PQ2 + PS2 + QR2 + SR2 = 2 (PT2 + ) + 4QT2

= 2 (PT2 + ) + 4QT2

.............. (RT = PT)

= 4PT2 + 4QT2

2
= ( ) + (2QT)2

PQ2 + PS2 + QR2 + SR2 = PR2 + .

11/N 833
(B) 6

(1) P(1, –2), Q(5, 2), R(3, –1), S(–1, –5)

(2)

(3) 4.1 7.3

(4) 30 10 6

4. 8

(1) DE BC

(i) DE = 4 BC = 8 A ( ADE) = 25 2

A ( ABC)

(ii) DE : BC = 3 : 5, A ( ADE) : A ( DBCE)

P.T.O.

12/N 833
(2) ABC ~ PQR, ABC AB = 3.6 BC = 4

AC = 4.2 ABC PQR

2 : 3 ABC PQR

(3) 14 8

8 ( = 3.14)

(i)

(ii)

(iii)

5. 3

(1) ABCD AD BD X

AXD AB = 12 AD = 9 BD BX

(2) = 30º

(i) sin 2 cos2 1

(ii) 1 tan 2 sec 2

(iii) 1 cot 2 cosec 2 .

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