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Maharashtra SSC Geometry (Mathematics Part II) Board Paper June 2026

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  1. Q.1 (B)(ii) In the following figure, seg AE \(\perp\) seg BC, seg DF \(\perp\) line BC, AE = 4, DF = 6, then find \(\dfrac{A(\Delta ABC)}{A(\Delta DBC)}\).

    Figure: B, E, C, F are points in that order on a line; A is above E with AE \(\perp\) BC and D is above F with DF \(\perp\) line BC; segments AB, AC, DB, DC are drawn; right-angle marks at E and F.

    1 mark · See worked answer →

  2. Q.2 (B)(i) In the following figure, in \(\Delta PQR\), seg RS bisects \(\angle R\). If PR = 15, QR = 20, PS = 12, then find QS.

    Figure: \(\Delta PQR\) with P at the lower left, Q at the lower right and R at the top; S is on side PQ; seg RS is drawn and the two equal parts of \(\angle PRQ\) are marked.

    2 marks · See worked answer →

  3. Q.3 (A)(i) In the following figure, in \(\Delta ABC\) ray BD bisects \(\angle ABC\), A – D – C, seg ED \(\parallel\) side BC, A – E – B, then to prove that \(\dfrac{AB}{BC} = \dfrac{AE}{EB}\), complete the following activity.

    Figure: \(\Delta ABC\) with A at the top, B at the lower left and C at the lower right; D is on side AC and ray BD is drawn and extended beyond D with an arrow; E is on side AB; seg ED is drawn parallel to BC; the two equal parts of \(\angle ABC\) are marked at B.

    Activity : In \(\Delta ABC\), ray BD bisects \(\angle B\).

    \(\therefore \dfrac{AB}{BC} = \dfrac{\text{____}}{\text{____}}\) .............. (I) (Angle bisector theorem)

    In \(\Delta ABC\), seg ED \(\parallel\) side BC.

    \(\therefore \dfrac{AE}{EB} = \dfrac{\text{____}}{DC}\) .............. (II) ____

    \(\therefore \dfrac{AB}{\text{____}} = \dfrac{\text{____}}{EB}\) .............. from (I) and (II)

    3 marks · See worked answer →

  4. Q.4 (i) In the following figure, line \(l\) intersects side AB and side AC of \(\Delta ABC\) in points P and Q respectively. Show that \(\dfrac{A(\Delta APQ)}{A(\Delta ABC)} = \dfrac{AP \times AQ}{AB \times AC}\).

    Figure: \(\Delta ABC\) with A at the top, B at the lower left and C at the lower right; a line \(l\) cuts side AB at P and side AC at Q and is drawn with arrows on both sides; seg PQ is drawn.

    4 marks · See worked answer →

Maharashtra SSC Geometry (Mathematics Part II) Board Paper March 2026

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  1. Q.1 (B)(1) The ratio of corresponding sides of similar triangles is 3 : 5, then find the ratio of their areas.

    1 mark · See worked answer →

  2. Q.2 (B)(1) Find QP using given information in the figure :

    Figure: Δ MNP with N at the lower left, M at the top and P at the lower right. Seg NQ is drawn to side MP, and the two angles at N are marked equal, so NQ bisects ∠MNP. MN = 25, MQ = 14, NP = 40.

    2 marks · See worked answer →

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

    Figure: Δ ABC with A at the top, B at the lower left and C at the lower right; X lies on side AB and Y lies on side BC, and seg XY is parallel to side AC.

    Activity :

    2AX = 3BX

    ∴ \(\dfrac{AX}{BX} = \dfrac{3}{2}\)

    ∴ \(\dfrac{AX + BX}{BX} = \dfrac{____ + ____}{2}\) .......... (By componendo)

    \(\dfrac{AB}{BX} = \dfrac{____}{2}\)

    Δ BCA ~ Δ BYX ............. (____ Test of similarity)

    ∴ \(\dfrac{BA}{BX} = \dfrac{AC}{XY}\) .............. (Corresponding sides of similar triangles)

    ∴ \(\dfrac{5}{____} = \dfrac{AC}{9}\)

    ∴ AC = ____

    3 marks · See worked answer →

  4. Q.4 (3) In Δ ABC, A-X-B and A-Y-C is such that segment XY || side BC. Segment XY divides Δ ABC in two equal areas, then find \(\dfrac{BX}{AB}\).

    4 marks · See worked answer →

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

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  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.

    1 mark · See worked answer →

  2. 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 worked answer →

  3. 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 worked answer →

  4. 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 worked answer →

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

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  1. Q.1 (A)(1) If \(\Delta ABC \sim \Delta DEF\), \(m\angle B = 60^\circ\), then \(m\angle E\) = ....................

    1 mark · See worked answer →

  2. Q.1 (B)(2) The ratio of corresponding sides of similar triangles is 3 : 5, then find the ratio of their areas.

    1 mark · See worked answer →

  3. Q.2 (A)(1) In the given figure, AR ⊥ BC, AR ⊥ PQ, then complete the activity for finding \(\dfrac{A(\Delta ABC)}{A(\Delta APQ)}\).

    Figure: Δ ABC with A at the top; B, P, R, Q, C lie on the base line in that order; seg AR is drawn from A to the base with a right-angle mark at R on BC.

    Activity :

    \(\dfrac{A(\Delta ABC)}{A(\Delta APQ)} = \dfrac{____ \times AR}{PQ \times ____}\)

    \(\therefore \dfrac{A(\Delta ABC)}{A(\Delta APQ)} = \dfrac{____}{____}\)

    2 marks · See worked answer →

  4. Q.2 (B)(1) In the following figure, in \(\Delta LMN\), ray MT bisects \(\angle LMN\).

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

    Figure: Δ LMN with M at the lower left; ray MT from M meets side LN at T and is produced beyond T; LM = 6, MN = 10, TN = 8, with equal angles marked at M.

    2 marks · See worked answer →

  5. Q.4 (1) Δ ABC has sides of length 4 cm, 5 cm and 6 cm while Δ PQR has perimeter of 90 cm. If Δ ABC is similar to Δ PQR, then find the length of corresponding sides of Δ PQR.

    4 marks · See worked answer →

Maharashtra SSC Geometry (Mathematics Part II) Board Paper July 2024

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  1. Q.1 (A)(1) In \(\Delta ABC\) and \(\Delta PQR\), in a one to one correspondence of vertices, \(\dfrac{AB}{QR} = \dfrac{BC}{PR} = \dfrac{CA}{PQ}\), then ....................... statement is true.

    Figure: a small triangle ABC with A at the top left, B below it and C to the right of B; a larger similarly shaped triangle with Q at the top, R below Q and P to the right of R, so that A corresponds to Q, B to R and C to P.

    1 mark · See worked answer →

  2. Q.2 (B)(2) In \(\Delta PQR\), seg RS bisects \(\angle PRQ\). If PR = 15, RQ = 20, PS = 12, then find SQ.

    Figure: \(\Delta PQR\) with S on side PQ; seg RS is drawn and marks the two equal angles at R.

    2 marks · See worked answer →

  3. Q.3 (A)(1) In the given figure, AB || CD || EF. If AC = 12, CE = 9, BD = 8, then complete the following activity to find DF.

    Figure: three parallel lines with two transversals cutting them at A, C, E on one transversal and B, D, F on the other.

    Activity : AB || CD || EF

    \(\therefore \dfrac{AC}{\text{____}} = \dfrac{\text{____}}{DF}\) ........... (Property of three parallel lines and their transversal)

    \(\therefore \dfrac{12}{\text{____}} = \dfrac{\text{____}}{DF}\)

    \(\therefore DF = \dfrac{8 \times 9}{\text{____}}\)

    \(\therefore DF =\) ____

    3 marks · See worked answer →

  4. Q.3 (B)(4) If \(\Delta PQR \sim \Delta LMN\), \(9 \times A(\Delta PQR) = 16 \times A(\Delta LMN)\) and QR = 20, then find MN.

    3 marks · See worked answer →

Maharashtra SSC Geometry (Mathematics Part II) Board Paper March 2024

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  1. Q.1 (B)(1) If \(\Delta ABC \sim \Delta PQR\) and \(AB : PQ = 2 : 3\), then find the value of \(\dfrac{A(\Delta ABC)}{A(\Delta PQR)}\).

    1 mark · See worked answer →

  2. Q.3 (A)(1)

    Figure: Δ ABC with D on side AC (A – D – C); ray BD bisects ∠ABC; E is on side AB (A – E – B) and seg DE || side BC.

    In Δ ABC, ray BD bisects ∠ABC, A – D – C, seg DE || side BC, A – E – B, then for showing \(\dfrac{AB}{BC} = \dfrac{AE}{EB}\), complete the following activity :

    Proof :

    In Δ ABC, ray BD bisects ∠B

    ∴ \(\dfrac{____}{BC} = \dfrac{AD}{DC}\) ............ (I) (____)

    In Δ ABC, DE || BC

    ∴ \(\dfrac{____}{EB} = \dfrac{AD}{DC}\) ............ (II) (____)

    \(\dfrac{AB}{____} = \dfrac{____}{EB}\) ............ [from (I) and (II)]

    3 marks · See worked answer →

  3. Q.3 (B)(4) Prove that, ‘If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the side in the same proportion’.

    3 marks · See worked answer →

  4. Q.5 (1) □ ABCD is a trapezium, AB || CD, diagonals of trapezium intersects in point P.

    Write the answers of the following questions :

    (a) Draw the figure using given information.

    (b) Write any one pair of alternate angles and opposite angles.

    (c) Write the names of similar triangles with test of similarity.

    3 marks · See worked answer →