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Maharashtra SSC March 2026 Mathematics Part II (Geometry) Question Paper with Step-by-Step Solutions

AI-written · numbers calculator-checked on 1 Nov 2026

Step-by-step working for 27 questions of the board's own exam paper “MATHEMATICS (71) GEOMETRY – PART II (E), REVISED COURSE, paper N 932” (40 marks, 2 hours), written by Shishya. Each question is in our own shorter words; open the board's paper for the exact wording, figures and instructions.

The board's paper (PDF) ↗Where the board lists it (mahahsscboard.in) ↗

How to answer this paper: All five questions are compulsory, with choices inside them: Q.2 (A) any two activities, Q.2 (B) any four, Q.3 (A) any one, Q.3 (B) any two, Q.4 any two and Q.5 any one. A calculator is not allowed; draw proper figures, keep construction marks, and a proof of a theorem needs its diagram.

How these were checked: the solutions were drafted with AI from the board's English-medium paper (read on 1 Nov 2026). On 1 Nov 2026, every final number was recomputed from the question's own numbers by a calculator check that runs before each release. They were not compared with a board answer key: none was listed beside this paper on the board's page when Shishya read it. Graphs, constructions and diagrams are described step by step, not drawn. If something looks wrong, your teacher and the board's own answer key, where one exists, are the authority.

Maths in the Maharashtra SSC is two separate 40-mark papers: Part I (Algebra) and Part II (Geometry). Both March 2026 papers are solved on Shishya. Several Geometry questions use a figure printed in the board's paper; each one is described in words here.

Free to read: the first 5 solutions are open to everyone; the other 22 open with a free Shishya account (for students 13 and above).

Q.1 (A)

  1. Q1(A)(i)

    1 markPythagoras theorem

    Which of these is a Pythagorean triplet? Options: (A) (1, 5, 10), (B) (3, 4, 5), (C) (2, 2, 2), (D) (5, 5, 2).

    1. In a Pythagorean triplet the square of the largest number equals the sum of the squares of the other two.
    2. (3, 4, 5): 3² + 4² = 9 + 16 = 25 = 5² ✓.
    3. The others fail: 1² + 5² = 26 ≠ 100, 2² + 2² = 8 ≠ 4, 2² + 5² = 29 ≠ 25.

    Answer: (3, 4, 5) — option (B).

  2. Q1(A)(ii)

    1 markCircle

    Circles of radii 4 cm and 5 cm touch internally. The distance between their centres is … Options: (A) 4 cm, (B) 5 cm, (C) 1 cm, (D) 9 cm.

    1. For circles touching internally, the distance between the centres = difference of the radii.
    2. Distance = 5 − 4 = 1 cm.

    Answer: 1 cm — option (C).

  3. Q1(A)(iii)

    1 markCo-ordinate geometry

    Which point lies on the X-axis, to the right of the origin? Options: (A) (−2, 0), (B) (0, 2), (C) (2, 3), (D) (2, 0).

    1. A point on the X-axis has y-co-ordinate 0; it is to the right of the origin when its x-co-ordinate is positive.
    2. (−2, 0) is to the left, (0, 2) is on the Y-axis and (2, 3) is not on any axis; (2, 0) has y = 0 and x = 2 > 0.

    Answer: (2, 0) — option (D).

  4. Q1(A)(iv)

    1 markMensuration

    The volume of a cube of side 5 cm. Options: (A) 10 cm³, (B) 100 cm³, (C) 125 cm³, (D) 25 cm³.

    1. Volume of a cube = (side)³.
    2. Volume = 5 × 5 × 5 = 125 cm³.

    Answer: 125 cm³ — option (C).

Q.1 (B)

  1. Q1(B)(i)

    1 markSimilarity

    Corresponding sides of two similar triangles are in the ratio 3 : 5. Find the ratio of their areas.

    1. The areas of similar triangles are in the ratio of the squares of their corresponding sides.
    2. Ratio of areas = 3² : 5² = 9 : 25.

    Answer: 9 : 25

  2. Q1(B)(ii)

    1 markPythagoras theorem

    The diagonal of a square is 10√2 cm. Find its side.

    1. Diagonal of a square = √2 × side.
    2. √2 × side = 10√2, so side = 10√2 / √2 = 10 cm.

    Answer: Side = 10 cm

    Steps and answer: open with a free account

  3. Q1(B)(iii)

    1 markCircle

    A minor arc measures 70°. Find the measure of the corresponding major arc.

    1. Measure of a major arc = 360° − measure of the corresponding minor arc.
    2. = 360° − 70° = 290°.

    Answer: 290°

    Steps and answer: open with a free account

  4. Q1(B)(iv)

    1 markCo-ordinate geometry

    A line makes an angle of 45° with the positive direction of the X-axis. Find its slope.

    1. Slope of a line = tan θ, where θ is the angle the line makes with the positive direction of the X-axis.
    2. Slope = tan 45° = 1.

    Answer: slope = 1

    Steps and answer: open with a free account

Q.2 (A)

  1. Q2(A)(i)

    2 marksCircle

    Activity: PS is a tangent segment and line PR a secant through Q (P–Q–R). PQ = 3.6 and QR = 6.4; fill the boxes to find PS.

    Uses a figure printed in the board's paper — open the paper above to see it.

    1. Tangent secant segment theorem: PS² = PQ × PR.
    2. PR = PQ + QR, so PS² = PQ × [PQ + QR] = 3.6 × [3.6 + 6.4].
    3. = 3.6 × 10 = 36.
    4. PS = √36 = 6.

    Answer: Boxes: PQ, 3.6, 10, 6 — PS = 6

    Steps and answer: open with a free account

  2. Q2(A)(ii)

    2 marksTrigonometry

    Activity: if sec θ = 25/7, fill the boxes to find tan θ.

    1. 1 + tan²θ = sec²θ.
    2. 1 + tan²θ = (25/7)² = 625/49.
    3. tan²θ = 625/49 − 1 = (625 − 49)/49 = 576/49.
    4. tan θ = √(576/49) = 24/7.

    Answer: Boxes: sec²θ, 25/7, 1, 24/7 — tan θ = 24/7

    Steps and answer: open with a free account

  3. Q2(A)(iii)

    2 marksMensuration (area of a sector)

    Activity: square ABCD has side 7 cm and sector D-AXC is drawn with centre D and radius DA. Given that the sector's area is 38.5 cm², fill the boxes to find the shaded area (the part of the square outside the sector).

    Uses a figure printed in the board's paper — open the paper above to see it.

    1. Area of square = (side)² = 7² = 49 cm².
    2. The sector D-AXC is a quarter circle of radius 7 (central angle 90°): 90/360 × 22/7 × 7 × 7 = 38.5 cm², as given.
    3. A(shaded region) = A(square ABCD) − A(sector D-AXC) = 49 − 38.5 = 10.5 cm².

    Answer: Boxes: 49; square ABCD, sector D-AXC; 10.5 — shaded area = 10.5 cm²

    Steps and answer: open with a free account

Q.2 (B)

  1. Q2(B)(i)

    2 marksSimilarity (angle bisector)

    In △MNP, NQ bisects ∠MNP with Q on MP; the figure gives MN = 25, NP = 40 and MQ = 14. Find QP.

    Uses a figure printed in the board's paper — open the paper above to see it.

    1. In △MNP, NQ bisects ∠MNP, so by the property of an angle bisector of a triangle: MN / NP = MQ / QP.
    2. 25 / 40 = 14 / QP.
    3. QP = 14 × 40 / 25 = 560 / 25 = 22.4.

    Answer: QP = 22.4

    Steps and answer: open with a free account

  2. Q2(B)(ii)

    2 marksPythagoras theorem (30°-60°-90° triangle)

    In △RST, ∠S = 90°, ∠T = 30° and RT = 12 cm. Find RS and ST.

    1. In △RST, ∠S = 90° and ∠T = 30°, so ∠R = 60°: it is a 30°-60°-90° triangle with hypotenuse RT.
    2. The side opposite 30° is half the hypotenuse: RS = 1/2 × RT = 1/2 × 12 = 6 cm.
    3. The side opposite 60° is √3/2 times the hypotenuse: ST = √3/2 × 12 = 6√3 cm.

    Answer: RS = 6 cm and ST = 6√3 cm

    Steps and answer: open with a free account

  3. Q2(B)(iii)

    2 marksCircle

    Chords MN and RS of a circle intersect at D. RD = 15, DS = 4 and MD = 8. Find DN.

    Uses a figure printed in the board's paper — open the paper above to see it.

    1. Chords MN and RS intersect inside the circle at D, so by the theorem of intersecting chords: RD × DS = MD × DN.
    2. 15 × 4 = 8 × DN, so 60 = 8 × DN.
    3. DN = 60 / 8 = 7.5.

    Answer: DN = 7.5

    Steps and answer: open with a free account

  4. Q2(B)(iv)

    2 marksCo-ordinate geometry

    Find the midpoint of the segment joining P(0, 6) and Q(12, 20).

    1. Midpoint formula: ((x₁ + x₂)/2, (y₁ + y₂)/2).
    2. x = (0 + 12)/2 = 6 and y = (6 + 20)/2 = 13.

    Answer: The midpoint is (6, 13)

    Steps and answer: open with a free account

  5. Q2(B)(v)

    2 marksTrigonometry

    Prove that sin²θ + cos²θ = 1.

    1. Draw △ABC with ∠B = 90° and let ∠A = θ.
    2. sin θ = BC/AC and cos θ = AB/AC.
    3. sin²θ + cos²θ = BC²/AC² + AB²/AC² = (BC² + AB²)/AC².
    4. By Pythagoras' theorem, AB² + BC² = AC², so sin²θ + cos²θ = AC²/AC² = 1.

    Answer: sin²θ + cos²θ = 1 (proved, using Pythagoras' theorem in a right triangle)

    Steps and answer: open with a free account

Q.3 (A)

  1. Q3(A)(i)

    3 marksSimilarity

    Activity: in △ABC, X is on AB and Y on BC with seg XY ∥ side AC. 2AX = 3BX and XY = 9. Fill the boxes to find AC.

    Uses a figure printed in the board's paper — open the paper above to see it.

    1. 2AX = 3BX, so AX/BX = 3/2.
    2. By componendo, (AX + BX)/BX = (3 + 2)/2, so AB/BX = 5/2.
    3. △BCA ~ △BYX by the AA test of similarity (∠B is common and ∠BAC = ∠BXY, corresponding angles as XY ∥ AC).
    4. So BA/BX = AC/XY (corresponding sides of similar triangles): 5/2 = AC/9.
    5. AC = 5 × 9 / 2 = 45/2 = 22.5.

    Answer: Boxes: 3, 2; 5; AA; 2; 22.5 — AC = 22.5

    Steps and answer: open with a free account

  2. Q3(A)(ii)

    3 marksPythagoras theorem (45°-45°-90° triangle)

    Activity: in right △ABC, ∠B = 90°, AB = BC (marked equal in the figure) and AC = √8. Fill the boxes to find AB and BC.

    Uses a figure printed in the board's paper — open the paper above to see it.

    1. AB = BC (given in the figure), so ∠BAC = ∠BCA; with ∠B = 90°, each is (180° − 90°)/2 = 45°: ∠BAC = 45°.
    2. In a 45°-45°-90° triangle, each side making the right angle is 1/√2 × hypotenuse: AB = BC = 1/√2 × AC.
    3. = 1/√2 × √8 = 1/√2 × 2 × √2 (since √8 = 2√2).
    4. AB = BC = 2.

    Answer: Boxes: given; 45°; 1/√2; √8; √2; 2 — ∠BAC = 45°; AB = BC = 2

    Steps and answer: open with a free account

Q.3 (B)

  1. Q3(B)(i)

    3 marksCircle (theorem)

    Prove: the tangent segments drawn to a circle from an external point are congruent.

    1. Given: a circle with centre O; from an external point P, PA and PB are tangent segments touching the circle at A and B. To prove: seg PA ≅ seg PB. Draw the figure and join OA, OB and OP.
    2. ∠OAP = ∠OBP = 90° (a tangent is perpendicular to the radius at the point of contact).
    3. In right triangles OAP and OBP: side OA ≅ side OB (radii of the same circle) and hypotenuse OP ≅ hypotenuse OP (common side).
    4. So △OAP ≅ △OBP by the hypotenuse-side test.
    5. Hence seg PA ≅ seg PB (corresponding sides of congruent triangles).

    Answer: △OAP ≅ △OBP (hypotenuse-side test), so the tangent segments PA and PB are congruent.

    Steps and answer: open with a free account

  2. Q3(B)(ii)

    3 marksGeometric constructions

    Draw a circle with centre O and radius 3.5 cm, take a point P 5.7 cm from O, and draw the tangents from P to the circle.

    1. Draw a circle with centre O and radius 3.5 cm. Mark a point P with OP = 5.7 cm and draw seg OP.
    2. Draw the perpendicular bisector of seg OP; let it meet OP at M (the midpoint of OP).
    3. With centre M and radius MO, draw a circle; it cuts the given circle at A and B.
    4. Draw lines PA and PB: these are the required tangents (∠OAP = ∠OBP = 90°, angles in semicircles).
    5. Keep all the construction marks; seg PA and seg PB measure the same.

    Answer: Rays PA and PB, drawn through the points where the circle on OP as diameter meets the given circle, are the tangents from P.

    Steps and answer: open with a free account

  3. Q3(B)(iii)

    3 marksCo-ordinate geometry (section formula)

    P divides the segment joining A(−1, 7) and B(4, −3) in the ratio 2 : 3. Find the co-ordinates of P.

    1. Section formula: x = (m x₂ + n x₁)/(m + n), y = (m y₂ + n y₁)/(m + n), with m : n = 2 : 3, A(−1, 7) and B(4, −3).
    2. x = (2 × 4 + 3 × (−1))/(2 + 3) = (8 − 3)/5 = 5/5 = 1.
    3. y = (2 × (−3) + 3 × 7)/(2 + 3) = (−6 + 21)/5 = 15/5 = 3.

    Answer: P(1, 3)

    Steps and answer: open with a free account

  4. Q3(B)(iv)

    3 marksMensuration (frustum)

    A bucket is a frustum of height 28 cm with end radii 12 cm and 15 cm. Find its capacity in litres (π = 22/7).

    1. Volume of a frustum = 1/3 × π × h × (r₁² + r₂² + r₁ × r₂), with h = 28, r₁ = 15, r₂ = 12.
    2. r₁² + r₂² + r₁ × r₂ = 225 + 144 + 180 = 549.
    3. Volume = 1/3 × 22/7 × 28 × 549 = 22 × 4 × 183 = 16104 cm³.
    4. 1 litre = 1000 cm³, so the capacity = 16104 / 1000 = 16.104 litres.

    Answer: 16,104 cm³ = 16.104 litres

    Steps and answer: open with a free account

Q.4

  1. Q4(i)

    4 marksGeometric constructions (similar triangles)

    △ABC ~ △ADE. In △ABC, AB = 6.3 cm, ∠CAB = 50°, AC = 5.6 cm and AB/AD = 7/5. Construct △ADE.

    1. Analysis: △ABC ~ △ADE with AB/AD = 7/5 = AC/AE, so AD = 6.3 × 5/7 = 4.5 cm and AE = 5.6 × 5/7 = 4 cm; ∠DAE = ∠CAB = 50°.
    2. Draw seg AB = 6.3 cm, draw ∠BAC = 50° at A and mark C on its arm with AC = 5.6 cm; join BC.
    3. Draw a ray from A making an acute angle with AB on the side away from C, and mark 7 equal parts A₁, A₂, …, A₇ on it.
    4. Join A₇B. Through A₅ draw a line parallel to A₇B, meeting AB at D (then AD = 4.5 cm).
    5. Through D draw a line parallel to BC, meeting AC at E (then AE = 4 cm). △ADE is the required triangle.

    Answer: △ADE with AD = 4.5 cm, ∠DAE = 50° and AE = 4 cm (constructed inside △ABC, DE ∥ BC)

    Steps and answer: open with a free account

  2. Q4(ii)

    4 marksMensuration (sphere and cube)

    The surface area of a sphere equals the total surface area of a cube. Show that volume of the sphere : volume of the cube = √6 : √π.

    1. Let the radius of the sphere be r and the side of the cube be a.
    2. Surface area of the sphere = total surface area of the cube: 4πr² = 6a², so a² = 4πr²/6 = 2πr²/3.
    3. Volume of sphere / volume of cube = (4/3 × π × r³) / a³.
    4. Square both sides: (V₁/V₂)² = (16/9 × π² × r⁶) / a⁶, and a⁶ = (a²)³ = 8π³r⁶/27.
    5. (V₁/V₂)² = (16π²r⁶/9) × 27/(8π³r⁶) = 432/(72π) = 6/π.
    6. V₁/V₂ = √(6/π) = √6/√π, so V₁ : V₂ = √6 : √π.

    Answer: Volume of sphere : volume of cube = √6 : √π (shown)

    Steps and answer: open with a free account

  3. Q4(iii)

    4 marksSimilarity (areas)

    In △ABC, X is on AB and Y on AC with seg XY ∥ side BC, and XY divides the triangle into two parts of equal area. Find BX/AB.

    1. XY ∥ BC, so ∠AXY = ∠ABC and ∠AYX = ∠ACB (corresponding angles): △AXY ~ △ABC by the AA test.
    2. Areas of similar triangles are in the ratio of the squares of corresponding sides: A(△AXY)/A(△ABC) = AX²/AB².
    3. XY divides △ABC into two equal areas, so A(△AXY) = 1/2 × A(△ABC): AX²/AB² = 1/2 and AX/AB = 1/√2.
    4. BX/AB = (AB − AX)/AB = 1 − AX/AB = 1 − 1/√2 = (√2 − 1)/√2.
    5. Multiplying the top and bottom by √2, this is also (2 − √2)/2.

    Answer: BX/AB = (√2 − 1)/√2, i.e. (2 − √2)/2

    Steps and answer: open with a free account

Q.5

  1. Q5(i)

    3 marksCircle (inscribed angles)

    PQ is a chord of a circle with centre O; arc PXQ is the major arc, arc PYQ the minor arc and ∠POQ a central angle. (a) Draw the figure. (b) Find ∠PXQ and ∠PYQ and add them. (c) Find ∠POQ and write how ∠PXQ and ∠POQ are related.

    1. (a) Draw a circle with centre O and a chord PQ (not a diameter). Mark X on the major arc and Y on the minor arc; join XP, XQ, YP, YQ, and OP, OQ.
    2. (b) Measure ∠PXQ and ∠PYQ with a protractor. ∠PXQ is inscribed in the major arc and ∠PYQ in the minor arc; PXQY is a cyclic quadrilateral, so m∠PXQ + m∠PYQ = 180° (opposite angles of a cyclic quadrilateral are supplementary).
    3. (c) Measure ∠POQ, the central angle on the minor arc PYQ. ∠PXQ intercepts the same arc PYQ, so by the inscribed angle theorem m∠PXQ = 1/2 × m∠POQ.

    Answer: m∠PXQ + m∠PYQ = 180°; m∠PXQ = 1/2 × m∠POQ (the measured values depend on the figure drawn)

    Steps and answer: open with a free account

  2. Q5(ii)

    3 marksTrigonometry (heights and distances)

    Two eagles fly at a height of 10√3 m. A boy on the ground sees them at angles of elevation of 60° and 30°. Find the distance between the eagles.

    1. Let the boy stand at B; the eagles are at height 10√3 m, at horizontal distances d₁ (elevation 60°) and d₂ (elevation 30°) from B, on the same side of the boy.
    2. tan 60° = 10√3 / d₁, so d₁ = 10√3 / √3 = 10 m.
    3. tan 30° = 10√3 / d₂, so d₂ = 10√3 × √3 = 30 m.
    4. Both eagles are at the same height, so the distance between them = d₂ − d₁ = 30 − 10 = 20 m.

    Answer: The eagles are 20 m apart

    Steps and answer: open with a free account

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