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Maharashtra SSC Geometry (Mathematics Part II) Board Paper March 2025

Written board papers, 40 marks and 2 hours each: Science & Technology Part 1 and 2, Algebra, Geometry and more.

2025, March 120 minutes 40 marks Save WhatsApp

Sections

Q.1 (A) Choose the correct alternative (4 marks) 4 questions
Q.1 (B) Solve the following sub-questions (4 marks) 4 questions
Q.2 (A) Complete the activities, any two (4 marks) 3 questions
Q.2 (B) Solve any four sub-questions (8 marks) 5 questions
Q.3 (A) Complete the activities, any one (3 marks) 2 questions
Q.3 (B) Solve any two sub-questions (6 marks) 4 questions
Q.4 Solve any two sub-questions (8 marks) 3 questions
Q.5 Solve any one sub-question (3 marks) 2 questions

March 2025 SSC Geometry board paper

This is the Mathematics Part II (Geometry) paper from the Maharashtra Board Class 10 exam held on 7 March 2025 (English medium, paper code N 832). You get 2 hours for 40 marks, with no calculator. Figures must be drawn where needed and construction marks must not be erased. The paper follows the usual pattern of five questions: four MCQs and four short questions in Q.1, then activities and sub-questions with internal choice in Q.2 to Q.5. Figures from the original paper are described in words.

All seven chapters are tested. Some examples: Pythagorean triplets and the geometric mean in a right triangle, intersecting chords and a cyclic quadrilateral, the slope of a line at 30°, collinearity and showing that four points form a parallelogram, the angle of elevation of a tree, DE ∥ BC and similar triangles, constructing tangents from an external point and a triangle similar to a given one, recasting a solid into cylinders, a frustum of a cone, and verifying identities for θ = 30°.

How to practise with it

Treat it as a real exam. Use a compass and ruler for the constructions, write proofs with figures, and stop at 2 hours. Then check each answer against the model solution and see where marks were lost: in the steps, the figure or the final value. Revise weak chapters with the Geometry full subject test and the solved questions on the Geometry subject page.

FAQs

Is this paper free?

Yes, the paper and solutions are free.

Which syllabus does it follow?

The current Maharashtra Board (Balbharati) Class 10 Geometry syllabus. All SSC papers are listed on the exam hub.

Chapters covered

Instructions

Solve on paper with a timer, then check against the solutions.

Q.1 (A) Choose the correct alternative (4 marks)

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

    1. (A) (1, 5, 10)
    2. (B) (3, 4, 5)
    3. (C) (2, 2, 2)
    4. (D) (5, 5, 2)
  2. Q.1 (A)(2) ∠ACB is inscribed angle in a circle with centre O. If m∠ACB = 65º, then what is measure of its intercepted arc AXB ?

    1. (A) 65º
    2. (B) 230º
    3. (C) 295º
    4. (D) 130º
  3. Q.1 (A)(3) Distance of point (3, 4) from the origin is .................... .

    1. (A) 7
    2. (B) 1
    3. (C) 5
    4. (D) –5
  4. Q.1 (A)(4) If radius of cone is 5 cm and its perpendicular height is 12 cm, then the slant height is ..................... .

    1. (A) 17 cm
    2. (B) 4 cm
    3. (C) 13 cm
    4. (D) 60 cm

Q.1 (B) Solve the following sub-questions (4 marks)

  1. Q.1 (B)(1) In the following figure Δ ABC, B – D – C and BD = 7, BC = 20, then find \(\dfrac{A(\Delta ABD)}{A(\Delta ABC)}\).

    [Figure: Δ ABC with point D on side BC; segment AD drawn.]

  2. Q.1 (B)(2) In the following figure ∠MNP = 90º, seg NQ ⊥ seg MP, MQ = 9, QP = 4, find NQ.

    [Figure: right-angled Δ MNP with the right angle at N; Q on hypotenuse MP with NQ ⊥ MP.]

  3. Q.1 (B)(3) Angle made by a line with the positive direction of X-axis is 30º. Find slope of that line.

  4. Q.1 (B)(4) In cyclic quadrilateral ABCD m∠A = 100º, then find m∠C.

Q.2 (A) Complete the activities, any two (4 marks)

  1. Q.2 (A)(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. (π = 3.14)

    [Figure: circle with centre P, chord AB, ∠APB = 90º, X on minor arc AB.]

    Activity :

    r = 10 cm, θ = 90º, π = 3.14.
    A(P–AXB) = \(\dfrac{\theta}{360}\) × ____
    = \(\dfrac{\text{____}}{360}\) × 3.14 × 102
    = \(\dfrac{1}{4}\) × ____
    A(P–AXB) = ____ sq. cm.

    2 marks See answer →
  2. Q.2 (A)(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 :

    [Figure: circle with chords MN and RS intersecting at D inside the circle.]

    Activity :

    ∴ MD × DN = ____ × DS .............. (Theorem of internal division of chords)
    ∴ ____ × DN = 15 × 4
    ∴ DN = ____ / 8
    ∴ DN = ____

    2 marks See answer →
  3. Q.2 (A)(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. (\(\sqrt3\) = 1.73)

    [Figure: right-angled Δ ABC, right angle at B; AB is the tree, BC = 10 m, ∠BCA = 60º.]

    Activity :

    In the figure given above, AB = h = height of tree, BC = 10 m, distance of the observer from the tree.
    Angle of elevation (θ) = ∠BCA = 60º
    tan θ = ____ / BC .................. (I)
    tan 60º = ____ .................. (II)
    \(\dfrac{AB}{BC} = \sqrt3\) (From (I) and (II))
    AB = BC × \(\sqrt3\) = \(10\sqrt3\)
    AB = 10 × 1.73 = ____
    ∴ height of the tree is ____ m.

    2 marks See answer →

Q.2 (B) Solve any four sub-questions (8 marks)

  1. Q.2 (B)(1) In Δ ABC, DE || BC. If DB = 5.4 cm, AD = 1.8 cm, EC = 7.2 cm, then find AE.

    [Figure: Δ ABC with D on AB and E on AC, DE parallel to BC.]

    2 marks See answer →
  2. Q.2 (B)(2) In the figure given below, find RS and PS using the information given in Δ PSR.

    [Figure: Δ PSR with ∠S = 90º, ∠P = 30º and hypotenuse RP = 12.]

    2 marks See answer →
  3. Q.2 (B)(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.

    [Figure: circle with centre D; CA and CB are tangents touching the circle at A and B; radii DA and DB drawn.]

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

    2 marks See answer →
  5. Q.2 (B)(5) If \(\sin\theta = \dfrac{11}{61}\), find the values of cos θ using trigonometric identity.

    2 marks See answer →

Q.3 (A) Complete the activities, any one (3 marks)

  1. Q.3 (A)(1) In the following figure, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC.

    [Figure: Δ ABC with X on side AB and Y on side BC, XY parallel to AC.]

    Activity :

    2AX = 3BX (Given)
    ∴ \(\dfrac{AX}{BX} = \dfrac{3}{\text{____}}\)
    \(\dfrac{AX + BX}{BX} = \dfrac{3 + 2}{2}\) .......... (by componendo)
    \(\dfrac{\text{____}}{BX} = \dfrac{5}{2}\) ................ (I)
    Now Δ BCA ~ Δ BYX ............. (____ test of similarity)
    ∴ \(\dfrac{BA}{BX} = \dfrac{AC}{XY}\) .............. (corresponding sides of similar triangles)
    ∴ \(\dfrac{\text{____}}{\text{____}} = \dfrac{AC}{9}\) .............. from (I)
    ∴ AC = ____

    3 marks See answer →
  2. Q.3 (A)(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.

    [Figure: rhombus PQRS with diagonals PR and QS intersecting at T.]

    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 Δ PQS, PT is the median and in Δ 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 + ____ .

    3 marks See answer →

Q.3 (B) Solve any two sub-questions (6 marks)

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

    3 marks See answer →
  2. Q.3 (B)(2) Prove that tangent segments drawn from an external point to a circle are congruent.

    3 marks See answer →
  3. Q.3 (B)(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.

    3 marks See answer →
  4. Q.3 (B)(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 ?

    3 marks See answer →

Q.4 Solve any two sub-questions (8 marks)

  1. Q.4 (1) In the following figure DE || BC, then :

    [Figure: Δ ABC with D on side AB and E on side AC, DE parallel to BC.]

    (i) If DE = 4 cm, BC = 8 cm, A(Δ ADE) = 25 cm2, find A(Δ ABC).
    (ii) If DE : BC = 3 : 5, then find A(Δ ADE) : A(□ DBCE).

    4 marks See answer →
  2. Q.4 (2) Δ ABC ~ Δ PQR. In Δ ABC, AB = 3.6 cm, BC = 4 cm and AC = 4.2 cm. The corresponding sides of Δ ABC and Δ PQR are in the ratio 2 : 3, construct Δ ABC and Δ PQR.

    4 marks See answer →
  3. Q.4 (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 : (π = 3.14)
    (i) Curved surface area of frustum.
    (ii) Total surface area of the frustum.
    (iii) Volume of the frustum.

    4 marks See answer →

Q.5 Solve any one sub-question (3 marks)

  1. Q.5 (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.

    3 marks See answer →
  2. Q.5 (2) Taking θ = 30º to verify the following Trigonometric identities :
    (i) \(\sin^2\theta + \cos^2\theta = 1\)
    (ii) \(1 + \tan^2\theta = \sec^2\theta\)
    (iii) \(1 + \cot^2\theta = \operatorname{cosec}^2\theta\).

    3 marks See answer →