Maharashtra 10th Std Model Question Paper Mathematics Part II Geometry

Maharashtra State Board Sample Paper 2026

N 697 Seat No.

2025 VII 02 1100 – N 697– MATHEMATICS (71) GEOMETRY—PART II (E)

(REVISED COURSE)
Time : 2 Hours (Pages 11) Max. Marks : 40

Note :—

(i) All questions are compulsory.

(ii) Use of 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) If ABC ~ DEF, mB = 60º, then mE = .................

(A) 30º

(B) 60º

(C) 90º

(D) 45º

P.T.O.

2/N 697
(2) Two circles of radii 5.5 cm and 4.2 cm touch each other externally
then distance between their centres is ...................

(A) 9.7 cm

(B) 1.3 cm

(C) 5.5 cm

(D) 4.2 cm

(3) A line makes an angle of 45º with the positive direction of
X-axis. So the slope of the line is .......................

1
(A)
2
3
(B)
2
(C) 1

(D) 3

(4) The volume of a cube of side 2 cm is .....................

(A) 4 cm3

(B) 2 cm3

(C) 6 cm3

(D) 8 cm3

(B) Solve the following subquestions : 4

(1) Find the diagonal of square whose side is 10 cm.

(2) The ratio of corresponding sides of similar triangles is 3 : 5, then
find the ratio of their areas.

3/N 697
(3) Find the slope of the line passing through the points A(2, 3)
and B(4, 7).

7
(4) If sin   , then find the value of cosec  .
25

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

(1) In the given figure, AR  BC , AR  PQ , then complete the
A(ABC)
activity for finding .
A(APQ)
A

B P R Q C

Activity :

A(ABC)  AR
=
A(APQ) PQ ×

A(ABC)
 =
A(APQ)

P.T.O.

4/N 697
(2) In the following figure, seg PS is a tangent segment, line PR

is a secant. If PQ = 3.6, QR = 6.4, then find PS by completing
the following activity.

P
Q

R

S

Activity :

PS2 = PQ ×

(tangent secant segments theorem)

 
PS2 = PQ × PQ  
 
PS2 = 3.6 × 3.6  
 PS2 = 3.6 × 10

 PS2 = 36

 PS =

5/N 697
(3) Measure of an arc of a circle is 90º and its radius is 14 cm.
Complete the following activity to find the length of an arc.
Activity :

Length of an arc =  ....... (Formula)
360
90 22
= 2 
360 7
1
= ×
4

Length of an arc = cm

(B) Solve the following subquestions (any four) : 8

(1) In the following figure

In LMN, ray MT bisects LMN.

If LM = 6, MN = 10, TN = 8, then find LT.

L

T
6

8

M 10 N

P.T.O.

6/N 697
(2) Find surface area of sphere of radius 7 cm.

(3) In the following figure, m(arc NS) = 125º, m(arc EF) = 37º. Find

mNMS .

N
F

M
E

S

(4) Find the co-ordinates of midpoint of the segment joining the points

P(22, 20) and Q(0, 16).

(5) Find the volume of a cone if radius of its base is 7 cm and its

perpendicular height is 15 cm.

3. (A) Complete the following activities and rewrite it (any one) : 3

sin   cos 
(1) If tan  = 1, then find the value of by completing
sec   cosec 
the following activity.

7/N 697
Activity :

tan   1 ........ (given)

but tan 1

  =

sin   cos  sin 45º  cos 45º
 = sec 45º  cosec 45º
sec   cosec 

1 1

2
=
2

2
= 2

sin   cos  1
=
sec   cosec 

(2) In the following figure, point O is the centre of the circle and

length of chord AB is equal to the radius of the circle. Find the

measures of :

(i) AOB

(ii) arc AB

(iii) ACB

P.T.O.

8/N 697
by completing the activity.
C

O

A
B
Activity :
In AOB,
AO = OB = AB

 AOB is an triangle.

 mAOB =

mAOB = m(arc AB) = (definition of

measure of an arc)
1
mACB = ×
2

.................

1
= × 60º
2
mACB =

9/N 697
(B) Solve the following subquestions (any two) : 6

(1) Find co-ordinates of point P, if P divides the line segment joining

the points A(–1, 7) and B(4, –3) in the ratio 2 : 3.

(2) Draw a circle with centre O of radius 3.4 cm. Draw a chord MN

of length 5.7 cm in it. Construct tangents at point M and N to

the circle.

(3) A storm broke a tree and the treetop rested 20 m from the base

of the tree, making an angle of 60º with the horizontal. Find

the height of the tree.

(4) Prove that, ‘In a right-angled triangle, the square of the

hypotenuse is equal to the sum of the squares of remaining two

sides’.

4. Solve the following subquestions (any two) : 8

(1) ABC has sides of length 4 cm, 5 cm and 6 cm while PQR has perimeter

of 90 cm. If ABC is similar to PQR, then find the length of

corresponding sides of PQR.

BC 5
(2) ABC ~ PBR, BC = 8 cm, AC = 10 cm, B = 90º,  , then
BR 4
construct PBR.

P.T.O.

10/N 697
(3) In the following figure ABC is an isosceles triangle with perimeter

44 cm. The base BC is of length 12 cm. Side AB and AC are congruent.

A circle touches the three sides of triangle as shown.

Find the length of tangent segment from A to circle.

A

P Q

B R C

5. Solve the following subquestions (any one) : 3

(1) Draw right-angled ABC of lengths of sides are 3 cm, 4 cm and 5 cm.

Draw median on the hypotenuse of ABC.

Then :

(i) Measure the length of median and write it.

(ii) By obeserving lengths of median and hypotenuse write your

observations.

11/N 697
(2) Observe the given figure and answer the following questions :

P

A B
O

(i) How many surfaces does a solid cone have ?

(ii) What are the names of slant height and perpendicular height

in the given figure ?

(iii) If slant height of solid cone is 10 cm and perpendicular height

is 8 cm, then find diameter of base of solid cone ?

P.T.O.

Maharashtra State Board Sample Paper 2026

2025 VII 02 - 1100
N 698 Seat No.

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

(REVISED COURSE)
Pages - 11 Total Marks : 40

(i)

(ii)

(iii)

(iv) [ 1(A)]

(v)

(vi)

(vii)

1. (A) 4

(1) ABC ~ DEF mB = 60º, mE =

(A) 30º

(B) 60º

(C) 90º

(D) 45º

P.T.O.

2/N 698
(2) 5.5 4.2

(A) 9.7

(B) 1.3

(C) 5.5

(D) 4.2

(3) X- 45º
=
1
(A)
2
3
(B)
2
(C) 1

(D) 3

(4) 2

(A) 4

(B) 2

(C) 6

(D) 8

(B) 4

(1) 10

(2) 3:5

3/N 698
(3) A(2, 3) B(4, 7)

7
(4) sin   , cosec 
25

2. (A) 4

(1) AR  BC, AR  PQ,

A(ABC)
A(APQ)

A

B P R Q C

A(ABC)  AR
=
A(APQ) PQ ×

A(ABC)
 =
A(APQ)

P.T.O.

4/N 698
(2) PS PR

PQ = 3.6, QR = 6.4, PS

P
Q

R

S

:

PS2 = PQ × .......

..........

 
PS2 = PQ × PQ  
 
PS2 = 3.6 × 3.6  
 PS2 = 3.6 × 10

 PS2 = 36

 PS =

5/N 698
(3) 90º 14

:

=  .......
360
90 22
= 2 
360 7
1
= ×
4

=

(B) 8

(1) LMN MT LMN
LM = 6, MN = 10, TN = 8, LT

L

T
6

8

M 10 N

P.T.O.

6/N 698
(2) 7

(3) m( NS) = 125º, m ( EF) = 37º, NMS

N
F

M
E

S

(4) P(22, 20) Q(0, 16)

(5) 7 15

3. (A) 3

sin   cos 
(1) tan  = 1,
sec   cosec 

7/N 698
:

tan   1 ........ ( )

tan 1

  =

sin   cos  sin 45º  cos 45º
 = sec 45º  cosec 45º
sec   cosec 

1 1

2
=
2

2
= 2

sin   cos  1
=
sec   cosec 

(2) O AB

(i) AOB

(ii) AB

(iii) ACB

P.T.O.

8/N 698

C

O

A
B

AOB

AO = OB = AB

 AOB

 mAOB =

mAOB = m( AB) = ...........

.............

1
mACB = ×
2
.................
1
= × 60º
2

mACB =

9/N 698
(B) 6

(1) P A(–1, 7) B(4, –3)

2 : 3 P

(2) O 3.4 5.7

MN M N

(3)

60º

20

(4)

4. 8

(1) ABC 4 5 6 PQR

90 ABC PQR PQR

BC 5
(2) ABC ~ PBR, BC = 8 AC = 10 B = 90º,  ,
BR 4
PBR

P.T.O.

10/N 698
(3) ABC 44

BC 12 AB AC

A

A

P Q

B R C

5. 3

(1) ABC 3 4 5 ABC

(i)

(ii)

11/N 698
(2)

P

A B
O

(i)

(ii)

(iii) 10 8

P.T.O.