Board Previous Year Questions

Board Previous Year Questions (PYQs)

These are exam-style questions modelled on CBSE and State Board papers from recent years. Each is fully solved with the reasoning a board examiner expects.

Scoring tip: In Triangles, always state the theorem or similarity criterion (BPT, AA, SSS, SAS, Pythagoras) explicitly — naming it earns marks even before the calculation.

Work through all 26. They span 1-mark, 2-mark, 3-mark and 5-mark patterns.

PYQ 1 (1 mark): In $\triangle ABC$, $DE \parallel BC$ with $AD = 1.5$ cm, $DB = 3$ cm, $AE = 1$ cm. Find $EC$.

Solution:

1. $\dfrac{AD}{DB} = \dfrac{AE}{EC} \Rightarrow \dfrac{1.5}{3} = \dfrac{1}{EC}$.
2. $EC = \dfrac{3}{1.5} = 2$ cm.

Final Answer: $EC = 2$ cm.

PYQ 2 (1 mark): Two similar triangles have areas $36\,\text{cm}^2$ and $49\,\text{cm}^2$. Find the ratio of their corresponding sides.

Solution:

1. Side ratio $= \sqrt{\dfrac{36}{49}} = \dfrac{6}{7}$.

Final Answer: $6 : 7$.

PYQ 3 (1 mark): The sides of a triangle are 7, 24, 25. Is it a right triangle?

Solution:

1. $7^2 + 24^2 = 49 + 576 = 625 = 25^2$.
2. Equality holds ⇒ right triangle.

Final Answer: Yes, right-angled.

PYQ 4 (1 mark): $\triangle ABC \sim \triangle DEF$ and $2\,AB = DE$. If $BC = 8$ cm, find $EF$.

Solution:

1. $\dfrac{AB}{DE} = \dfrac{1}{2}$, so $\dfrac{BC}{EF} = \dfrac{1}{2}$.
2. $EF = 2 \times 8 = 16$ cm.

Final Answer: $EF = 16$ cm.

PYQ 5 (2 marks): In $\triangle ABC$, $D$ and $E$ are on $AB$ and $AC$ with $AD = 2$ cm, $BD = 3$ cm, $AE = 3$ cm, $CE = 4.5$ cm. Is $DE \parallel BC$?

Solution:

1. $\dfrac{AD}{DB} = \dfrac{2}{3}$; $\dfrac{AE}{EC} = \dfrac{3}{4.5} = \dfrac{2}{3}$.
2. Ratios equal ⇒ by converse of BPT, $DE \parallel BC$.

Final Answer: Yes, $DE \parallel BC$.

PYQ 6 (2 marks): A 15 m tall pole and a tower cast shadows of 10 m and 60 m respectively at the same time. Find the height of the tower.

Solution:

1. AA similarity: $\dfrac{15}{10} = \dfrac{h}{60}$.
2. $h = \dfrac{15 \times 60}{10} = 90$ m.

Final Answer: 90 m.

PYQ 7 (2 marks): In $\triangle ABC$, right-angled at $B$, $AB = 24$ cm, $BC = 7$ cm. Find $AC$.

Solution:

1. $AC = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25$ cm.

Final Answer: $AC = 25$ cm.

PYQ 8 (2 marks): Sides of two similar triangles are in ratio $4 : 9$. Find the ratio of their areas.

Solution:

1. Area ratio $= 4^2 : 9^2 = 16 : 81$.

Final Answer: $16 : 81$.

PYQ 9 (2 marks): In $\triangle ABC$, $DE \parallel BC$, $AD = x$, $DB = x-2$, $AE = x+2$, $EC = x-1$. Find $x$.

Solution:

1. $\dfrac{x}{x-2} = \dfrac{x+2}{x-1} \Rightarrow x(x-1) = (x+2)(x-2)$.
2. $x^2 - x = x^2 - 4 \Rightarrow x = 4$.

Final Answer: $x = 4$.

PYQ 10 (3 marks): Prove that the ratio of areas of two similar triangles equals the square of the ratio of their corresponding medians.

Solution:

1. Let $\triangle ABC \sim \triangle PQR$ with medians $AM$, $PN$.
2. Since similarity ⇒ $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = k$ and $\dfrac{BM}{QN} = k$ (medians bisect proportional sides).
3. So $\triangle ABM \sim \triangle PQN \Rightarrow \dfrac{AM}{PN} = k$.
4. $\dfrac{\text{ar}(ABC)}{\text{ar}(PQR)} = k^2 = \left(\dfrac{AM}{PN}\right)^2$.

Final Answer: Proved.

PYQ 11 (3 marks): In $\triangle ABC$, right-angled at $C$, $BD \perp AB$ produced… Instead: $CD \perp AB$ with $C$ the right angle, $AC = 6$, $BC = 8$. Find $CD$.

Solution:

1. $AB = \sqrt{6^2 + 8^2} = 10$.
2. Area: $\dfrac{1}{2}(AC)(BC) = \dfrac{1}{2}(AB)(CD) \Rightarrow 6 \times 8 = 10 \times CD$.
3. $CD = 4.8$ cm.

Final Answer: $CD = 4.8$ cm.

PYQ 12 (3 marks): State and prove the Basic Proportionality Theorem (Thales).

Solution:

1. Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, it divides them in the same ratio.
2. Given: $\triangle ABC$, $DE \parallel BC$.
3. Construction: Join $BE$, $CD$; draw $DM \perp AC$, $EN \perp AB$.
4. $\dfrac{\text{ar}(ADE)}{\text{ar}(BDE)} = \dfrac{AD}{DB}$ and $\dfrac{\text{ar}(ADE)}{\text{ar}(CDE)} = \dfrac{AE}{EC}$ (same height).
5. $\triangle BDE$ and $\triangle CDE$ have equal areas (same base $DE$, between parallels).
6. Hence $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.

Final Answer: Proved.

PYQ 13 (3 marks): In a trapezium $ABCD$ with $AB \parallel DC$, diagonals meet at $O$. Show that $\dfrac{OA}{OC} = \dfrac{OB}{OD}$.

Solution:

1. In $\triangle OAB$ and $\triangle OCD$: $\angle AOB = \angle COD$ (vertically opposite).
2. $\angle OAB = \angle OCD$ (alternate angles, $AB \parallel DC$).
3. By AA, $\triangle OAB \sim \triangle OCD$.
4. Hence $\dfrac{OA}{OC} = \dfrac{OB}{OD}$.

Final Answer: Proved.

PYQ 14 (3 marks): $\triangle ABC \sim \triangle PQR$. Areas are $25\,\text{cm}^2$ and $49\,\text{cm}^2$. If $QR = 9.8$ cm, find $BC$.

Solution:

1. $\dfrac{BC}{QR} = \sqrt{\dfrac{25}{49}} = \dfrac{5}{7}$.
2. $BC = \dfrac{5}{7} \times 9.8 = 7$ cm.

Final Answer: $BC = 7$ cm.

PYQ 15 (3 marks): A ladder 13 m long reaches a window 12 m above the ground. Find the distance of the foot of the ladder from the wall.

Solution:

1. Distance $= \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5$ m.

Final Answer: 5 m.

PYQ 16 (3 marks): In $\triangle ABC$, $DE \parallel BC$ and $\dfrac{AD}{DB} = \dfrac{3}{5}$. If $AC = 5.6$ cm, find $AE$.

Solution:

1. $\dfrac{AE}{EC} = \dfrac{3}{5}$, so $AE = \dfrac{3}{8} \times AC = \dfrac{3}{8} \times 5.6 = 2.1$ cm.

Final Answer: $AE = 2.1$ cm.

PYQ 17 (3 marks): Two poles of heights 6 m and 11 m stand on level ground. If the distance between their feet is 12 m, find the distance between their tops.

Solution:

1. Vertical difference $= 11 - 6 = 5$ m; horizontal $= 12$ m.
2. Distance between tops $= \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13$ m.

Final Answer: 13 m.

PYQ 18 (3 marks): In $\triangle PQR$, $M$ and $N$ are on $PQ$ and $PR$ with $PM = 4$, $MQ = 6$, $PN = 6$, $NR = 9$. Prove $MN \parallel QR$ and find $MN$ if $QR = 15$ cm.

Solution:

1. $\dfrac{PM}{MQ} = \dfrac{4}{6} = \dfrac{2}{3}$; $\dfrac{PN}{NR} = \dfrac{6}{9} = \dfrac{2}{3}$ ⇒ $MN \parallel QR$ (converse BPT).
2. $\dfrac{PM}{PQ} = \dfrac{4}{10} = \dfrac{2}{5}$, so $MN = \dfrac{2}{5} \times 15 = 6$ cm.

Final Answer: $MN = 6$ cm.

PYQ 19 (5 marks): State and prove the Pythagoras Theorem.

Solution:

1. Statement: In a right triangle, the square on the hypotenuse equals the sum of squares on the other two sides.
2. Given: $\triangle ABC$, right-angled at $B$. Draw $BD \perp AC$.
3. $\triangle ADB \sim \triangle ABC$ ⇒ $\dfrac{AD}{AB} = \dfrac{AB}{AC} \Rightarrow AB^2 = AD \cdot AC$.
4. $\triangle BDC \sim \triangle ABC$ ⇒ $\dfrac{CD}{BC} = \dfrac{BC}{AC} \Rightarrow BC^2 = CD \cdot AC$.
5. Add: $AB^2 + BC^2 = (AD + CD)\,AC = AC \cdot AC = AC^2$.

Final Answer: $AC^2 = AB^2 + BC^2$. Proved.

PYQ 20 (5 marks): Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Solution:

1. Let $\triangle ABC \sim \triangle PQR$. Draw altitudes $AM$ and $PN$.
2. $\dfrac{\text{ar}(ABC)}{\text{ar}(PQR)} = \dfrac{\frac{1}{2} \cdot BC \cdot AM}{\frac{1}{2} \cdot QR \cdot PN} = \dfrac{BC}{QR} \cdot \dfrac{AM}{PN}$.
3. $\triangle ABM \sim \triangle PQN$ (AA) ⇒ $\dfrac{AM}{PN} = \dfrac{AB}{PQ} = \dfrac{BC}{QR}$.
4. So ratio $= \dfrac{BC}{QR} \cdot \dfrac{BC}{QR} = \left(\dfrac{BC}{QR}\right)^2$.

Final Answer: Proved.

PYQ 21 (5 marks): In an equilateral triangle $ABC$, $D$ is a point on $BC$ such that $BD = \dfrac{1}{3}BC$. Prove that $9\,AD^2 = 7\,AB^2$.

Solution:

1. Let side $= a$. Draw $AE \perp BC$, so $BE = \dfrac{a}{2}$.
2. $BD = \dfrac{a}{3}$, so $DE = BE - BD = \dfrac{a}{2} - \dfrac{a}{3} = \dfrac{a}{6}$.
3. $AE^2 = a^2 - \dfrac{a^2}{4} = \dfrac{3a^2}{4}$.
4. $AD^2 = AE^2 + DE^2 = \dfrac{3a^2}{4} + \dfrac{a^2}{36} = \dfrac{27a^2 + a^2}{36} = \dfrac{28a^2}{36} = \dfrac{7a^2}{9}$.
5. So $9\,AD^2 = 7a^2 = 7\,AB^2$.

Final Answer: Proved.

PYQ 22 (5 marks): $\triangle ABC$ has $DE \parallel BC$, $AD = 6$, $DB = 9$, $AE = 8$. Find $EC$ and the ratio of areas of $\triangle ADE$ to $\triangle ABC$.

Solution:

1. $\dfrac{AD}{DB} = \dfrac{AE}{EC} \Rightarrow \dfrac{6}{9} = \dfrac{8}{EC} \Rightarrow EC = 12$.
2. $\dfrac{AD}{AB} = \dfrac{6}{15} = \dfrac{2}{5}$.
3. Area ratio $= \left(\dfrac{2}{5}\right)^2 = \dfrac{4}{25}$.

Final Answer: $EC = 12$; area ratio $4 : 25$.

PYQ 23 (3 marks): $O$ is a point inside rectangle $ABCD$… Instead: In $\triangle ABC$, $AD$ is the bisector of $\angle A$ meeting $BC$ at $D$. If $AB = 6$, $AC = 8$, $BC = 7$, find $BD$.

Solution:

1. Angle bisector divides opposite side in ratio of adjacent sides: $\dfrac{BD}{DC} = \dfrac{AB}{AC} = \dfrac{6}{8} = \dfrac{3}{4}$.
2. $BD = \dfrac{3}{7} \times 7 = 3$ cm.

Final Answer: $BD = 3$ cm.

PYQ 24 (3 marks): The diagonal $BD$ of a parallelogram $ABCD$ intersects segment $AE$ at $F$, where $E$ is any point on $BC$. Prove that $DF \cdot EF = FB \cdot FA$.

Solution:

1. In $\triangle AFD$ and $\triangle EFB$: $\angle AFD = \angle EFB$ (vertically opposite).
2. $\angle DAF = \angle BEF$ (alternate angles, $AD \parallel BC$).
3. By AA, $\triangle AFD \sim \triangle EFB$.
4. $\dfrac{DF}{BF} = \dfrac{FA}{FE} \Rightarrow DF \cdot EF = FB \cdot FA$.

Final Answer: Proved.

PYQ 25 (3 marks): Two similar triangles have perimeters 30 cm and 20 cm. If one side of the first is 12 cm, find the corresponding side of the second.

Solution:

1. Side ratio = perimeter ratio $= \dfrac{30}{20} = \dfrac{3}{2}$.
2. $\dfrac{12}{x} = \dfrac{3}{2} \Rightarrow x = 8$ cm.

Final Answer: 8 cm.

PYQ 26 (5 marks): In $\triangle ABC$, right-angled at $B$, $D$ is the midpoint of $BC$. Prove that $AC^2 = 4\,AD^2 - 3\,AB^2$.

Solution:

1. Let $BD = DC = x$, so $BC = 2x$.
2. $AD^2 = AB^2 + BD^2 = AB^2 + x^2$ ⇒ $x^2 = AD^2 - AB^2$.
3. $AC^2 = AB^2 + BC^2 = AB^2 + 4x^2 = AB^2 + 4(AD^2 - AB^2) = 4AD^2 - 3AB^2$.

Final Answer: $AC^2 = 4\,AD^2 - 3\,AB^2$. Proved.