---
title: "Triangles: CBSE Class 10 Maths previous year questions"
url: https://www.swavid.com/cbse/class-10/maths/pyq/triangles
---

# Triangles: CBSE Class 10 Maths previous year questions

30 questions from CBSE Class 10 board papers, newest first, each with full working.

## CBSE Class 10 Maths Standard Question Paper 2026 (Set 30/1/1) with Solutions

### Question 7

*1 mark · MCQ*

If $\Delta \text{ABC} \sim \Delta \text{DEF}$ are similar such that $2\text{AB} = \text{DE}$ and $\text{BC} = 8\text{ cm}$, then $\text{EF}$ is equal to :

- 4\text{ cm}
- 8\text{ cm}
- 12\text{ cm}
- 16\text{ cm}

**Solution**

1. Given $\Delta \text{ABC} \sim \Delta \text{DEF}$, the ratio of their corresponding sides is equal: $\frac{\text{AB}}{\text{DE}} = \frac{\text{BC}}{\text{EF}}$.
2. From $2\text{AB} = \text{DE}$, we get $\frac{\text{AB}}{\text{DE}} = \frac{1}{2}$. Substituting $\text{BC} = 8\text{ cm}$, $\frac{1}{2} = \frac{8}{\text{EF}}$, which gives $\text{EF} = 16\text{ cm}$.

**Answer:** (d) $16\text{ cm}$

> Common mistake: Taking the ratio of sides incorrectly as $\frac{\text{AB}}{\text{DE}} = 2$ instead of $\frac{1}{2}$.

### Question 22

*2 marks · Very short answer*

(A) In $\Delta \text{ABC}$, $\text{DE} \parallel \text{BC}$. If $\text{AD} = x$, $\text{DB} = x - 2$, $\text{AE} = x + 2$ and $\text{EC} = x - 1$, then find the value of $x$.

**Solution**

1. Since $DE \parallel BC$ in $\triangle ABC$, by the Basic Proportionality Theorem, $\frac{AD}{DB} = \frac{AE}{EC}$.
2. Substitute the given lengths into the relation: $\frac{x}{x - 2} = \frac{x + 2}{x - 1}$.
3. Cross-multiply to get $x(x - 1) = (x + 2)(x - 2)$, which simplifies to $x^2 - x = x^2 - 4$.
4. Solving for $x$ gives $x = 4$.

**Answer:** $4$

> Common mistake: Incorrect cross-multiplication or algebraic expansion of terms.

### Question 22 (OR)

*2 marks · Very short answer*

(B) In the figure given above, $\Delta \text{ABC} \sim \Delta \text{XYZ}$, then find the values of $x$ and $y$.

**Solution**

1. Since $\triangle ABC \sim \triangle XYZ$, the corresponding sides are proportional: $\frac{AB}{XY} = \frac{AC}{XZ} = \frac{BC}{YZ}$.
2. Substitute the given values $\frac{4}{x} = \frac{y}{6} = \frac{6}{7.2}$.
3. From $\frac{4}{x} = \frac{6}{7.2}$, we get $x = \frac{4 \times 7.2}{6} = 4.8$.
4. From $\frac{y}{6} = \frac{6}{7.2}$, we get $y = \frac{6 \times 6}{7.2} = 5$.

**Answer:** $x = 4.8$, $y = 5$

> Common mistake: Matching the incorrect corresponding sides of the similar triangles.

### Question 34

*5 marks · Proof*

(A) State and prove Basic Proportionality Theorem.

**Solution**

1. Given: A triangle $\Delta ABC$ in which a line parallel to side $BC$ intersects other two sides $AB$ and $AC$ at $D$ and $E$ respectively.
2. To prove: $\frac{AD}{DB} = \frac{AE}{EC}$.
3. Construction: Join $BE$ and $CD$. Draw $DM \perp AC$ and $EN \perp AB$.
4. Proof: Area of $\Delta ADE$ = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AD \times EN$.
5. Area of $\Delta BDE$ = $\frac{1}{2} \times DB \times EN$.
6. Dividing the two areas: $\frac{\text{ar}(\Delta ADE)}{\text{ar}(\Delta BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB}$ (Equation 1).
7. Similarly, $\frac{\text{ar}(\Delta ADE)}{\text{ar}(\Delta DEC)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC}$ (Equation 2).
8. Since $\Delta BDE$ and $\Delta DEC$ are on the same base $DE$ and between the same parallel lines $DE$ and $BC$, their areas are equal: $\text{ar}(\Delta BDE) = \text{ar}(\Delta DEC)$.
9. From Equations 1, 2, and the equal areas, we get $\frac{AD}{DB} = \frac{AE}{EC}$.
10. Hence proved.

**Answer:** Hence proved.

> Common mistake: Not mentioning the reason why $\text{ar}(\Delta BDE) = \text{ar}(\Delta DEC)$.

### Question 34 (OR)

*5 marks · Proof*

(B) In the given figure, $\text{CM}$ and $\text{RN}$ are respectively the medians of $\Delta \text{ABC}$ and $\Delta \text{PQR}$. If $\Delta \text{ABC} \sim \Delta \text{PQR}$, then prove that :
(i) $\Delta \text{AMC} \sim \Delta \text{PNR}$
(ii) $\Delta \text{CMB} \sim \Delta \text{RNQ}$

**Part (i)**

1. Given $\Delta ABC \sim \Delta PQR$, therefore $\frac{AB}{PQ} = \frac{BC}{QR}$ and $\angle B = \angle Q$.
2. Since $CM$ and $RN$ are medians, $BM = \frac{1}{2}BC$ and $QN = \frac{1}{2}QR$.
3. Thus, $\frac{AB}{PQ} = \frac{2BM}{2QN} = \frac{BM}{QN}$, which can also be written as $\frac{AB}{PQ} = \frac{AM}{PN}$ since $AM$ and $PN$ are corresponding halves of proportional sides.
4. In $\Delta AMC$ and $\Delta PNR$, $\frac{AC}{PR} = \frac{AM}{PN}$ and $\angle A = \angle P$.
5. Therefore, by SAS similarity criterion, $\Delta AMC \sim \Delta PNR$.

Answer (i): Hence proved.

**Part (ii)**

1. From $\Delta ABC \sim \Delta PQR$, we have $\angle B = \angle Q$.
2. Also, we know that $\frac{BC}{QR} = \frac{BM}{QN}$.
3. In $\Delta CMB$ and $\Delta RNQ$, $\frac{BC}{QR} = \frac{BM}{QN}$ translates to proportional sides including the median segments, and $\angle B = \angle Q$.
4. Therefore, by SAS similarity criterion, $\Delta CMB \sim \Delta RNQ$.

Answer (ii): Hence proved.

**Answer:** Hence proved.

> Common mistake: Using incorrect corresponding sides when applying the SAS similarity criterion.

## CBSE Class 10 Maths Basic Question Paper 2025 (Set 430/1/1) with Solutions

### Question 7

*1 mark · MCQ*

Which of the following is not the criterion for similarity of triangles ?

- AAA
- SSS
- SAS
- RHS

**Solution**

1. The standard criteria for similarity of triangles in NCERT are AAA (or AA), SSS, and SAS similarity criteria.
2. RHS is a criterion for congruence of right-angled triangles, not a general similarity criterion (though right triangles can be similar by AA).
3. Therefore, RHS is not listed as a standard criterion for triangle similarity.

**Answer:** (D) RHS

> Common mistake: Confusing congruence criteria like RHS with similarity criteria.

### Question 8

*1 mark · MCQ*

From the figures given below, which of the following is true about the measure of $\angle P$?

- $\angle P=60^{\circ}$
- $\angle P=80^{\circ}$
- $\angle P=40^{\circ}$
- The measure of $\angle P$ cannot be determined

**Solution**

1. Check the ratio of the corresponding sides of the two triangles: $\frac{PQ}{BC} = \frac{12}{6} = 2$, $\frac{PR}{AB} = \frac{6\sqrt{3}}{3.8}$ which does not match directly, let us check another pairing.
2. Check $\frac{PQ}{AC} = \frac{12}{3\sqrt{3}} = \frac{4}{\sqrt{3}}$, $\frac{QR}{AB} = \frac{7.6}{3.8} = 2$, $\frac{PR}{BC} = \frac{6\sqrt{3}}{6} = \sqrt{3}$. Let us check SSS ratio: $\frac{AB}{QR} = \frac{3.8}{7.6} = \frac{1}{2}$, $\frac{BC}{PQ} = \frac{6}{12} = \frac{1}{2}$, $\frac{AC}{PR} = \frac{3\sqrt{3}}{6\sqrt{3}} = \frac{1}{2}$.
3. By SSS similarity criterion, $\triangle ABC \sim \triangle RQP$.
4. Therefore, $\angle P = \angle C = 80^{\circ}$ corresponding to the order of vertices in similarity.

**Answer:** (B) $\angle P=80^{\circ}$

> Common mistake: Matching vertices incorrectly without checking side ratios properly.

### Question 22 (a)

*2 marks · Proof*

In the given figure, if PQ || RS, then prove that $\Delta POQ\sim\Delta SOR.$

**Solution**

1. Given: $PQ \parallel RS$.
2. To prove: $\Delta POQ \sim \Delta SOR$.
3. Proof: $\angle PQO = \angle SRO$ (alternate interior angles since $PQ \parallel RS$).
4. $\angle POQ = \angle SOR$ (vertically opposite angles).
5. Therefore, $\Delta POQ \sim \Delta SOR$ by AA similarity criterion. Hence proved.

**Answer:** $\Delta POQ \sim \Delta SOR$ proved using AA similarity.

> Common mistake: Writing wrong correspondence of vertices in similarity statements.

### Question 22 (OR)

*2 marks · Very short answer*

In the given figure, $\Delta OSR \sim \Delta OQP$, $\angle ROQ=125^{\circ}$ and $\angle ORS=70^{\circ}$. Find the measures of $\angle OSR$ and $\angle OQP$.

**Solution**

1. Line $RXQ$ is a straight line, so $\angle ROQ + \angle SOQ = 180^{\circ}$.
2. Substitute $\angle ROQ = 125^{\circ}$ to get $\angle SOQ = 180^{\circ} - 125^{\circ} = 55^{\circ}$.
3. In $\Delta OSR$, the sum of angles is $180^{\circ}$, so $\angle OSR = 180^{\circ} - (125^{\circ} + 70^{\circ})$ is incorrect; instead use exterior angle property or triangle sum: $\angle OSR = 180^{\circ} - 70^{\circ} - \angle SOR$. Since $\Delta OSR \sim \Delta OQP$, $\angle OSR = \angle OPQ$ and $\angle ORS = \angle OQP$.
4. Thus, $\angle OQP = \angle ORS = 70^{\circ}$ and in $\Delta OSR$, $\angle OSR = 180^{\circ} - 125^{\circ} - 70^{\circ}$ using $\angle ROS = 55^{\circ}$, giving $\angle OSR = 180^{\circ} - (55^{\circ} + 70^{\circ}) = 55^{\circ}$.

**Answer:** $\angle OSR = 55^{\circ}$ and $\angle OQP = 70^{\circ}$

> Common mistake: Confusing corresponding angles of similar triangles.

### Question 33

*5 marks · Long answer*

State "Basic Proportionality Theorem" and use it to prove the following:
In a quadrilateral ABCD, diagonals AC and BD intersect each other at O such that $\frac{AO}{BO}=\frac{CO}{DO}$ as shown in the given figure. Prove that ABCD is a trapezium.

**Solution**

1. Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
2. Diagram: Draw a trapezium $ABCD$ with diagonals $AC$ and $BD$ intersecting at $O$. Through $O$, draw a line segment $OE$ parallel to $AB$ meeting $AD$ at $E$.
3. In $\triangle DAB$, $EO \parallel AB$. By Basic Proportionality Theorem, $\frac{AE}{ED} = \frac{BO}{OD}$.
4. We are given that $\frac{AO}{BO} = \frac{CO}{DO}$, which can be rewritten as $\frac{AO}{CO} = \frac{BO}{OD}$.
5. From the first and second relations, we get $\frac{AE}{ED} = \frac{AO}{CO}$.
6. In $\triangle ADC$, a line $EO$ divides sides $AD$ and $AC$ in the same ratio, so by the converse of Basic Proportionality Theorem, $EO \parallel DC$.
7. Since $EO \parallel AB$ and $EO \parallel DC$, we have $AB \parallel DC$.
8. Therefore, quadrilateral $ABCD$ is a trapezium with $AB \parallel DC$. Hence proved.

**Answer:** Hence proved that ABCD is a trapezium.

> Common mistake: Not stating the converse of the Basic Proportionality Theorem while proving $EO \parallel DC$.

## CBSE Class 10 Maths Standard Question Paper 2025 (Set 30/1/1) with Solutions

### Question 8

*1 mark · MCQ*

In triangles $ABC$ and $DEF$, $\angle B = \angle E$, $\angle F = \angle C$ and $AB = 3 DE$. Then, the two triangles are :

- congruent but not similar
- congruent as well as similar
- neither congruent nor similar
- similar but not congruent

**Solution**

1. By AA similarity criterion, $\triangle ABC \sim \triangle DEF$ because corresponding angles are equal.
2. Since $AB = 3 DE$, the ratio of sides is $3:1$, so the triangles are similar but not congruent.

**Answer:** (d) similar but not congruent

> Common mistake: Assuming equal angles imply congruence without checking the side lengths.

### Question 24

*2 marks · Very short answer*

If $\Delta ABC \sim \Delta PQR$ in which $AB = 6$ cm, $BC = 4$ cm, $AC = 8$ cm and $PR = 6$ cm, then find the length of $(PQ + QR)$.

**Solution**

1. Since $\Delta ABC \sim \Delta PQR$, the corresponding sides are proportional: $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$.
2. Substitute the given values: $\frac{6}{PQ} = \frac{4}{QR} = \frac{8}{6}$.
3. From $\frac{6}{PQ} = \frac{8}{6}$, we get $PQ = \frac{36}{8} = \frac{9}{2} = 4.5$ cm. From $\frac{4}{QR} = \frac{8}{6}$, we get $QR = \frac{24}{8} = 3$ cm.
4. Thus, $PQ + QR = 4.5 + 3 = 7.5$ cm.

**Answer:** $7.5$ cm

> Common mistake: Writing the ratio of sides in incorrect correspondence order.

### Question 24 (OR)

*2 marks · Very short answer*

In the given figure, $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$, show that $\Delta PQS \sim \Delta TQR$.

**Solution**

1. Given $\angle 1 = \angle 2$, in $\Delta PQR$, this implies $PQ = PR$ because sides opposite to equal angles are equal.
2. Substitute $PR = PQ$ in the given relation $\frac{QR}{QS} = \frac{QT}{PR}$ to get $\frac{QR}{QS} = \frac{QT}{PQ}$.
3. Consider $\Delta PQS$ and $\Delta TQR$: we have $\frac{QP}{QT} = \frac{QS}{QR}$ and the included angle $\angle Q$ is common.
4. Therefore, by SAS similarity criterion, $\Delta PQS \sim \Delta TQR$. Hence proved.

**Answer:** Hence proved

> Common mistake: Failing to substitute $PR = PQ$ before applying the SAS similarity test.

### Question 33

*5 marks · Proof*

The diagonal BD of a parallelogram ABCD intersects the line segment AE at the point F, where E is any point on the side BC. Prove that $DF \times EF = FB \times FA$.

**Solution**

1. Given: A parallelogram ABCD where diagonal BD intersects line segment AE at F, and E is on BC.
2. To prove: $DF \times EF = FB \times FA$.
3. Consider $\triangle AFD$ and $\triangle EFB$.
4. $\angle AFD = \angle EFB$ (vertically opposite angles).
5. Since ABCD is a parallelogram, $AD \parallel BC$.
6. Therefore, $\angle ADF = \angle EBF$ (alternate interior angles for transversal BD).
7. By AA similarity criterion, $\triangle AFD \sim \triangle EFB$.
8. From the properties of similar triangles, the ratios of corresponding sides are equal: $\frac{AF}{EF} = \frac{DF}{FB}$.
9. Cross-multiplying gives $DF \times EF = FB \times FA$.
10. Hence proved.

**Answer:** Hence proved.

> Common mistake: Taking incorrect pairs of corresponding sides from similar triangles.

### Question 33 (OR)

*5 marks · Proof*

In $\Delta ABC$, if $AD \perp BC$ and $AD^2 = BD \times DC$, then prove that $\angle BAC = 90°$.

**Solution**

1. Given: In $\triangle ABC$, $AD \perp BC$ and $AD^2 = BD \times DC$.
2. To prove: $\angle BAC = 90^\circ$.
3. From right-angled triangle $\triangle ABD$, using Pythagoras theorem, $AB^2 = AD^2 + BD^2$.
4. From right-angled triangle $\triangle ACD$, using Pythagoras theorem, $AC^2 = AD^2 + DC^2$.
5. Adding both equations, $AB^2 + AC^2 = 2AD^2 + BD^2 + DC^2$.
6. Substitute $AD^2 = BD \times DC$ into the equation: $AB^2 + AC^2 = 2(BD \times DC) + BD^2 + DC^2$.
7. The right side becomes $(BD + DC)^2$, which is equal to $BC^2$ since $BD + DC = BC$.
8. Therefore, $AB^2 + AC^2 = BC^2$.
9. By the converse of Pythagoras theorem, $\angle BAC = 90^\circ$.
10. Hence proved.

**Answer:** Hence proved.

> Common mistake: Not recognizing that $2BD \cdot DC + BD^2 + DC^2$ forms the complete square $(BD + DC)^2$.

## CBSE Class 10 Maths Basic Question Paper 2024 (Set 430/1/3) with Solutions

### Question 21

*2 marks · Proof*

D is a point on the side BC of $\triangle ABC$ such that $\angle ADC = \angle BAC$. Show that $AC^2 = BC \times DC$.

**Solution**

1. Given: In $\triangle ABC$, $D$ is a point on $BC$ such that $\angle ADC = \angle BAC$.
2. To prove: $AC^2 = BC \times DC$.
3. In $\triangle ADC$ and $\triangle BAC$, $\angle ADC = \angle BAC$ (Given).
4. Also, $\angle C = \angle C$ (Common angle).
5. Therefore, $\triangle ADC \sim \triangle BAC$ by AA similarity criterion.
6. Since the triangles are similar, the ratios of their corresponding sides are equal: $\frac{AC}{BC} = \frac{DC}{AC}$.
7. Cross-multiplying gives $AC^2 = BC \times DC$. Hence proved.

**Answer:** Hence proved.

> Common mistake: Writing the corresponding vertices incorrectly in similarity statements, leading to wrong side ratios.

### Question 32

*5 marks · Proof*

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that other two sides are divided in the same ratio.

**Solution**

1. Given: A triangle ABC in which a line parallel to side BC intersects AB at D and AC at E.
2. To prove: $\frac{AD}{DB} = \frac{AE}{EC}$
3. Construction: Join BE and CD. Draw $DM \perp AC$ and $EN \perp AB$.
4. Area of $\triangle ADE$, $\text{ar}(\triangle ADE) = \frac{1}{2} \times AD \times EN$
5. Area of $\triangle BDE$, $\text{ar}(\triangle BDE) = \frac{1}{2} \times DB \times EN$
6. Dividing the two areas, $\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \quad \text{--(i)}$
7. Similarly, $\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DEC)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad \text{--(ii)}$
8. Since $\triangle BDE$ and $\triangle DEC$ lie on the same base DE and between the same parallels BC and DE, $\text{ar}(\triangle BDE) = \text{ar}(\triangle DEC)$
9. From (i), (ii) and the above equality, we get $\frac{AD}{DB} = \frac{AE}{EC}$.
10. Hence proved.

**Answer:** $\frac{AD}{DB} = \frac{AE}{EC}$

> Common mistake: Forgetting to state the condition that triangles on the same base and between same parallels are equal in area.

### Question 32 (OR)

*5 marks · Proof*

Sides AB and BC and median AD of a $\triangle ABC$ are respectively proportional to sides PQ and PR and median PM of $\triangle PQR$. Show that $\triangle ABC \sim \triangle PQR$.

**Solution**

1. Given: $\triangle ABC$ and $\triangle PQR$ with medians AD and PM such that $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$.
2. To prove: $\triangle ABC \sim \triangle PQR$.
3. Produce AD to a point E such that $AD = DE$ and join CE. Also produce PM to L such that $PM = ML$ and join RL.
4. In $\triangle ABD$ and $\triangle ECD$, $BD = DC$ (since AD is a median), $\angle ADB = \angle EDC$ (vertically opposite angles), and $AD = ED$ (by construction).
5. So, $\triangle ABD \cong \triangle ECD$ by SAS congruence, which gives $AB = EC$ (by CPCT).
6. Similarly, for $\triangle PQM$ and $\triangle RLM$, we get $PQ = RL$.
7. Since $\frac{AB}{PQ} = \frac{AD}{PM} = \frac{2AD}{2PM} = \frac{AE}{PL}$ and $AB = EC$, $PQ = RL$, we have $\frac{EC}{RL} = \frac{AC}{PR} = \frac{AE}{PL}$, which gives $\triangle AEC \sim \triangle PRL$.
8. Therefore, $\angle 1 = \angle 2$. Similarly, $\angle 3 = \angle 4$, giving $\angle A = \angle P$.
9. Since $\frac{AB}{PQ} = \frac{AC}{PR}$ and $\angle A = \angle P$, $\triangle ABC \sim \triangle PQR$ by SAS similarity.
10. Hence proved.

**Answer:** $\triangle ABC \sim \triangle PQR$

> Common mistake: Not extending the median to double its length to construct a parallelogram-based similarity.

## CBSE Class 10 Maths Standard Question Paper 2024 (Set 30/1/1) with Solutions

### Question 24

*2 marks · Proof*

In the given figure, ABCD is a quadrilateral. Diagonal BD bisects $\angle B$ and $\angle D$ both. Prove that : (i) $\Delta ABD \sim \Delta ACBD$ (ii) $AB = BC$

**Part (i)**

1. Given that diagonal BD bisects $\angle B$ and $\angle D$ of quadrilateral ABCD.
2. In $\triangle ABD$ and $\triangle CBD$, $\angle ABD = \label{CBD}$ since BD bisects $\angle B$.
3. Also, $\angle ADB = \angle CDB$ since BD bisects $\angle D$.
4. By AA similarity criterion, $\triangle ABD \sim \triangle CBD$.

Answer (i): \triangle ABD \sim \triangle CBD

**Part (ii)**

1. From the similarity of $\triangle ABD$ and $\triangle CBD$, the corresponding sides are proportional.
2. Therefore, $\frac{AB}{CB} = \frac{AD}{CD}$.

Answer (ii): AB = BC

**Answer:** Hence proved.

> Common mistake: Confusing the order of vertices in similar triangles.

### Question 34

*5 marks · Proof*

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.

**Solution**

1. Given: A triangle ABC in which a line parallel to side BC intersects AB at D and AC at E.
2. To prove: $\frac{\text{AD}}{\text{DB}} = \frac{\text{AE}}{\text{EC}}$
3. Construction: Join BE and CD. Draw $DM \perp AC$ and $EN \perp AB$.
4. Proof: Area of triangle ADE is $\text{ar}(\text{ADE}) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \text{AD} \times EN$.
5. Similarly, $\text{ar}(\text{BDE}) = \frac{1}{2} \times \text{DB} \times EN$.
6. Dividing the two areas: $\frac{\text{ar}(\text{ADE})}{\text{ar}(\text{BDE})} = \frac{\frac{1}{2} \times \text{AD} \times EN}{\frac{1}{2} \times \text{DB} \times EN} = \frac{\text{AD}}{\text{DB}}$ (Equation 1).
7. Similarly, considering base AE on AC with height DM: $\frac{\text{ar}(\text{ADE})}{\text{ar}(\text{DEC})} = \frac{\frac{1}{2} \times \text{AE} \times DM}{\frac{1}{2} \times \text{EC} \times DM} = \frac{\text{AE}}{\text{EC}}$ (Equation 2).
8. Since triangle BDE and triangle DEC are on the same base DE and between the same parallels BC and DE, their areas are equal: $\text{ar}(\text{BDE}) = \text{ar}(\text{DEC})$ (Equation 3).
9. From Equations 1, 2, and 3, the left-hand sides are equal, so the right-hand sides are equal: $\frac{\text{AD}}{\text{DB}} = \frac{\text{AE}}{\text{EC}}$.
10. Hence proved.

**Answer:** Hence proved.

> Common mistake: Omitting reasons for equal areas of triangles on the same base and between same parallels.

### Question 34 (OR)

*5 marks · Proof*

In the given figure PA, QB and RC are each perpendicular to AC. If $\text{AP} = x, \text{BQ} = y$ and $\text{CR} = z$, then prove that $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$

**Solution**

1. Given: PA, QB and RC are perpendicular to AC. $\text{AP} = x$, $\text{BQ} = y$, $\text{CR} = z$. Let $\text{AQ} = a$ and $\text{QC} = b$.
2. To prove: $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$
3. Proof: Since $PA \perp AC$ and $QB \perp AC$, $PA \parallel QB$.
4. In triangle PAC, since $BQ \parallel AP$, triangle CBQ is similar to triangle CAP by AA similarity.
5. Therefore, $\frac{BQ}{AP} = \frac{CQ}{CA}$, which gives $\frac{y}{x} = \frac{b}{a+b}$ (Equation 1).
6. Similarly, since $QB \perp AC$ and $RC \perp AC$, $QB \parallel RC$.
7. In triangle RAC, since $BQ \parallel RC$, triangle ABQ is similar to triangle ACR by AA similarity.
8. Therefore, $\frac{BQ}{RC} = \frac{AQ}{AC}$, which gives $\frac{y}{z} = \frac{a}{a+b}$ (Equation 2).
9. Adding Equation 1 and Equation 2: $\frac{y}{x} + \frac{y}{z} = \frac{b}{a+b} + \frac{a}{a+b} = \frac{a+b}{a+b} = 1$.
10. Dividing both sides by $y$, we get $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$.
11. Hence proved.

**Answer:** Hence proved.

> Common mistake: Incorrectly setting up the ratios of corresponding sides of similar triangles.

## CBSE Class 10 Maths Basic Question Paper 2023 (Set 430/1/1) with Solutions

### Question 8

*1 mark · MCQ*

The sides of two similar triangles are in the ratio $4 : 7$. The ratio of their perimeters is

- $4 : 7$
- $12 : 21$
- $16 : 49$
- $7 : 4$

**Solution**

1. The ratio of the perimeters of two similar triangles is equal to the ratio of their corresponding sides.
2. Since the ratio of their sides is $4 : 7$, the ratio of their perimeters is also $4 : 7$.

**Answer:** (a) $4 : 7$

> Common mistake: Squaring the ratio of the sides, which applies to the ratio of areas, not perimeters.

### Question 9

*1 mark · MCQ*

In the given figure, $AB \parallel CD$. If $AB = 5 \text{ cm}$, $CD = 2 \text{ cm}$ and $OB = 3 \text{ cm}$, then the length of $OC$ is

- $\frac{15}{2} \text{ cm}$
- $\frac{10}{3} \text{ cm}$
- $\frac{6}{5} \text{ cm}$
- $\frac{3}{5} \text{ cm}$

**Solution**

1. Since $AB \parallel CD$, triangle $AOB$ is similar to triangle $cod$ by AA similarity criterion.
2. Therefore, the ratio of corresponding sides is equal: $\frac{OC}{OB} = \frac{CD}{AB}$.
3. Substituting the given values, $\frac{OC}{3} = \frac{2}{5}$, which gives $OC = \frac{6}{5} \text{ cm}$.
4. Thus, the correct option is (c).

**Answer:** (c) $\frac{6}{5} \text{ cm}$

> Common mistake: Taking the incorrect ratio of sides of similar triangles.

### Question 15

*1 mark · MCQ*

In the above figure, the criterion of similarity by which $\Delta ABC \sim \Delta PQR$ is :

- SSA (Side - Side - Angle) Similarity
- ASA (Angle - Side - Angle) Similarity
- SAS (Side - Angle - Side) Similarity
- AA (Angle - Angle) Similarity

**Solution**

1. In $\Delta ABC$ and $\Delta PQR$, we have $\frac{AB}{PQ} = \frac{2.2}{4.4} = \frac{1}{2}$ and $\frac{BC}{QR} = \frac{3.5}{7} = \frac{1}{2}$.
2. Since the included angles are equal ($\angle B = \angle Q = 50^\circ$), by SAS similarity criterion, $\Delta ABC \sim \Delta PQR$.

**Answer:** (c) SAS (Side - Angle - Side) Similarity

> Common mistake: Confusing SAS similarity with SSA, which is not a valid similarity criterion.

### Question 25

*2 marks · Proof*

In the given figure, $ABC$ and $AMP$ are two right triangles, right angled at $B$ and $M$, respectively. Prove that $\Delta ABC \sim \Delta AMP$.

**Solution**

1. In $\triangle ABC$ and $\triangle AMP$, $\angle ABC = \angle AMP = 90^\circ$ (given).
2. $\angle A = \angle A$ (common angle).
3. Therefore, $\triangle ABC \sim \triangle AMP$ by AA similarity criterion. Hence proved.

**Answer:** Hence proved that $\triangle ABC \sim \triangle AMP$.

> Common mistake: Writing the corresponding vertices incorrectly in the similarity statement.

### Question 31

*3 marks · Proof*

(a) In the given figure, $DE \parallel AC$ and $DF \parallel AE$
Prove that $\frac{BF}{FE} = \frac{BE}{EC}$

**Solution**

1. Given: In $\triangle ABC$, $DE \parallel AC$ and $DF \parallel AE$.
2. To prove: $\frac{BF}{FE} = \frac{BE}{EC}$.
3. In $\triangle ABE$, since $DF \parallel AE$, by Basic Proportionality Theorem, $\frac{BD}{DA} = \frac{BF}{FE}$ (1)
4. In $\triangle ABC$, since $DE \parallel AC$, by Basic Proportionality Theorem, $\frac{BD}{DA} = \frac{BE}{EC}$ (2)
5. From equations (1) and (2), we get $\frac{BF}{FE} = \frac{BE}{EC}$.
6. Hence proved.

**Answer:** Hence proved that $\frac{BF}{FE} = \frac{BE}{EC}$.

> Common mistake: Applying Basic Proportionality Theorem on wrong triangles or taking incorrect ratios of segments.

### Question 31 (OR)

*3 marks · Proof*

(b) The diagonals of a quadrilateral $ABCD$ intersect each other at the point $O$ such that $\frac{AO}{BO} = \frac{CO}{OD}$. Show that quadrilateral $ABCD$ is a trapezium.

**Solution**

1. Given: Quadrilateral $ABCD$ with diagonals intersecting at $O$ such that $\frac{AO}{BO} = \frac{CO}{OD}$.
2. To show: Quadrilateral $ABCD$ is a trapezium.
3. Construction: Through $O$, draw a line $EOF$ parallel to $AB$ meeting $AD$ at $E$ and $BC$ at $F$.
4. In $\triangle DAB$, $EO \parallel AB$, so by Basic Proportionality Theorem, $\frac{AE}{ED} = \frac{BO}{OD}$ (1)
5. Given that $\frac{AO}{BO} = \frac{CO}{OD}$, which can be rewritten as $\frac{AO}{CO} = \frac{BO}{OD}$ (2)
6. From (1) and (2), $\frac{AE}{ED} = \frac{AO}{CO}$.
7. In $\triangle ADC$, line $EO$ divides the sides $AD$ and $AC$ in the same ratio, so by the converse of Basic Proportionality Theorem, $EO \parallel DC$.
8. Since $EO \parallel AB$ and $EO \parallel DC$, we have $AB \parallel DC$.
9. Therefore, quadrilateral $ABCD$ is a trapezium.
10. Hence proved.

**Answer:** Hence proved that quadrilateral $ABCD$ is a trapezium.

> Common mistake: Forgetting to use construction or applying the converse of Thales theorem incorrectly.

## CBSE Class 10 Maths Standard Question Paper 2023 (Set 30/1/1) with Solutions

### Question 22

*2 marks · Short answer*

In the given figure, $XZ$ is parallel to $BC$. $AZ = 3\text{ cm}$, $ZC = 2\text{ cm}$, $BM = 3\text{ cm}$ and $MC = 5\text{ cm}$. Find the length of $XY$.

**Solution**

1. In $\triangle ABC$, $XZ \parallel BC$, so by Basic Proportionality Theorem, $\frac{AZ}{ZC} = \frac{AX}{XB}$.
2. Given $AZ = 3\text{ cm}$ and $ZC = 2\text{ cm}$, so $\frac{AX}{XB} = \frac{3}{2}$.
3. In $\triangle ABM$, since $XY \parallel BM$, by Basic Proportionality Theorem, $\frac{AX}{XB} = \frac{AY}{YM}$.
4. Using $\frac{AX}{XB} = \frac{3}{2}$ and $BM = 3\text{ cm}$, $YM = \frac{2}{5} \times BM = \frac{2}{5} \times 3 = 1.2\text{ cm}$.

**Answer:** $1.2\text{ cm}$

> Common mistake: Confusing the segments of the transversal lines or incorrectly applying the Basic Proportionality Theorem to sub-triangles.

### Question 32

*5 marks · Proof*

If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, prove that the other two sides are divided in the same ratio.

**Solution**

1. Given: A triangle $ABC$ in which a line parallel to side $BC$ intersects other two sides $AB$ and $AC$ at $D$ and $E$ respectively.
2. To prove: $\frac{AD}{DB} = \frac{AE}{EC}$
3. Construction: Join $BE$ and $CD$. Draw $DM \perp AC$ and $EN \perp AB$.
4. Consider the area of triangle $ADE$: $\text{ar}(\triangle ADE) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AD \times EN$.
5. Consider the area of triangle $BDE$: $\text{ar}(\triangle BDE) = \frac{1}{2} \times DB \times EN$.
6. Dividing $\text{ar}(\triangle ADE)$ by $\text{ar}(\triangle BDE)$, we get $\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB}$ (Equation 1).
7. Similarly, considering $\triangle ADE$ with base $AE$ and $\triangle DEC$ with base $EC$, we get $\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DEC)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC}$ (Equation 2).
8. Since $\triangle BDE$ and $\triangle DEC$ are on the same base $DE$ and between the same parallels $BC$ and $DE$, their areas are equal: $\text{ar}(\triangle BDE) = \text{ar}(\triangle DEC)$.
9. From Equations 1, 2, and the equal areas, the left-hand sides are equal, which implies $\frac{AD}{DB} = \frac{AE}{EC}$.
10. Hence proved.

**Answer:** Hence proved that the other two sides are divided in the same ratio.

> Common mistake: Forgetting to state the reason why triangles on the same base and between the same parallels have equal areas.

## CBSE Class 10 Maths Basic Question Paper 2022 (Set 430/1/1) with Solutions

### Question 8 (OR)

*3 marks · Short answer*

Draw a line segment $PQ = 7.5\text{ cm}$. Divide it in the ratio $3 : 1$.

**Solution**

1. Draw a line segment $PQ = 7.5\text{ cm}$.
2. Draw a ray $PX$ making an acute angle with $PQ$.
3. Locate $3 + 1 = 4$ points ($A_1, A_2, A_3, A_4$) on ray $PX$ at equal intervals.
4. Join $A_4$ to $Q$, and draw a line through $A_3$ parallel to $A_4Q$ intersecting $PQ$ at point $R$ to divide it in the ratio $3 : 1$.

**Answer:** Line segment divided in the ratio $3 : 1$.

> Common mistake: Making incorrect number of divisions on the ray or drawing parallel lines inaccurately.

## Related pages

- [Triangles: NCERT solutions](https://www.swavid.com/maths/class/10/chapter/triangles/ncert-solutions)
- [All CBSE Class 10 Maths papers](https://www.swavid.com/cbse/class-10/maths/previous-year-papers)

Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student