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CBSE Class 10 Maths Chapter 6 Triangles Model Questions - 3 Marks - Part 1

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SECTION C — Short Answer Type Questions [3 Marks Each]

• Q1. Prove that 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 (Basic Proportionality Theorem).
  Answer:
  1. Given: In $\triangle ABC$, $DE \parallel BC$ intersecting $AB$ 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. Proof:
      - $\text{Area}(\triangle ADE) = \frac{1}{2} \times AD \times EN$ and $\text{Area}(\triangle BDE) = \frac{1}{2} \times DB \times EN$.
      - Therefore, $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{AD}{DB}$ -- (Equation 1).
      - Similarly, $\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CDE)} = \frac{AE}{EC}$ -- (Equation 2).
      - Since $\triangle BDE$ and $\triangle CDE$ lie on the same base $DE$ and between the same parallels $DE \parallel BC$, their areas are equal: $\text{Area}(\triangle BDE) = \text{Area}(\triangle CDE)$.
      - From Equations (1), (2), and equal areas, we get $\frac{AD}{DB} = \frac{AE}{EC}$.
  
  Result: Proved.
• Q2. In $\triangle ABC$, $DE \parallel BC$ and $\frac{AD}{DB} = \frac{3}{5}$. If $AC = 5.6\text{ cm}$, find the length of $AE$.
  Answer:
  1. By Basic Proportionality Theorem, since $DE \parallel BC$: $\frac{AD}{DB} = \frac{AE}{EC} = \frac{3}{5}$.
  2. Let $AE = 3x$ and $EC = 5x$. Then $AC = AE + EC = 3x + 5x = 8x$.
  3. Given $AC = 5.6\text{ cm}$, so $8x = 5.6 \implies x = \frac{5.6}{8} = 0.7\text{ cm}$.
  4. $AE = 3x = 3 \times 0.7 = 2.1\text{ cm}$.
  
  Result: $AE = 2.1\text{ cm}$.
• Q3. In $\triangle PQR$, $ST \parallel QR$. If $PS = x$, $SQ = x - 2$, $PT = x + 2$, and $TR = x - 1$, find the value of $x$.
  Answer:
  1. Since $ST \parallel QR$, by BPT: $\frac{PS}{SQ} = \frac{PT}{TR}$.
  2. Substitute the given expressions: $\frac{x}{x - 2} = \frac{x + 2}{x - 1}$.
  3. Cross-multiply: $x(x - 1) = (x + 2)(x - 2) \implies x^2 - x = x^2 - 4$.
  4. Cancel $x^2$ on both sides: $-x = -4 \implies x = 4$.
  
  Result: $x = 4$.
• Q4. Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
  Answer:
  1. Given: In $\triangle ABC$, $D$ is the mid-point of $AB$ ($AD = DB$), and line $DE \parallel BC$.
  2. To Prove: $E$ is the mid-point of $AC$ ($AE = EC$).
  3. Proof: By Basic Proportionality Theorem, since $DE \parallel BC$, we have $\frac{AD}{DB} = \frac{AE}{EC}$.
  4. Since $D$ is the mid-point, $AD = DB$, which means $\frac{AD}{DB} = 1$.
  5. Therefore, $\frac{AE}{EC} = 1 \implies AE = EC$.
  6. This proves that $E$ bisects side $AC$.
  
  Result: Proved.
• Q5. $D$ and $E$ are points on the sides $AB$ and $AC$ respectively of a $\triangle ABC$. For $AB = 10\text{ cm}$, $AD = 4\text{ cm}$, $AC = 15\text{ cm}$, and $AE = 6\text{ cm}$, prove that $DE \parallel BC$.
  Answer:
  1. Given lengths: $AB = 10$, $AD = 4 \implies DB = 10 - 4 = 6\text{ cm}$.
  2. Given lengths: $AC = 15$, $AE = 6 \implies EC = 15 - 6 = 9\text{ cm}$.
  3. Compute ratios: $\frac{AD}{DB} = \frac{4}{6} = \frac{2}{3}$ and $\frac{AE}{EC} = \frac{6}{9} = \frac{2}{3}$.
  4. Since $\frac{AD}{DB} = \frac{AE}{EC}$, by the converse of the Basic Proportionality Theorem, $DE \parallel BC$.
  
  Result: Proved.
• Q6. In a trapezium $ABCD$ with $AB \parallel DC$, points $E$ and $F$ lie on non-parallel sides $AD$ and $BC$ respectively such that $EF \parallel AB$. Show that $\frac{AE}{ED} = \frac{BF}{FC}$.
  Answer:
  1. Join diagonal $AC$ intersecting $EF$ at point $G$.
  2. In $\triangle ADC$, since $EG \parallel DC$ (as $EF \parallel AB \parallel DC$), by BPT: $\frac{AE}{ED} = \frac{AG}{GC}$.
  3. In $\triangle CAB$, since $GF \parallel AB$, by BPT: $\frac{AG}{GC} = \frac{BF}{FC}$.
  4. Combining both results: $\frac{AE}{ED} = \frac{BF}{FC}$.
  
  Result: Proved.
• Q7. The diagonals of a quadrilateral $ABCD$ intersect each other at the point $O$ such that $\frac{AO}{BO} = \frac{CO}{DO}$. Prove that $ABCD$ is a trapezium.
  Answer:
  1. Given $\frac{AO}{BO} = \frac{CO}{DO}$, which can be rearranged as $\frac{AO}{CO} = \frac{BO}{DO}$.
  2. Through point $O$, draw a line $EO \parallel AB$ meeting $AD$ at $E$.
  3. In $\triangle DAB$, since $EO \parallel AB$, by BPT: $\frac{DE}{EA} = \frac{DO}{OB} \implies \frac{EA}{DE} = \frac{BO}{DO}$.
  4. Since $\frac{BO}{DO} = \frac{AO}{CO}$, we have $\frac{EA}{DE} = \frac{AO}{CO}$.
  5. In $\triangle ADC$, by the converse of BPT, $EO \parallel DC$. Since $EO \parallel AB$ and $EO \parallel DC$, $AB \parallel DC$. Thus, $ABCD$ is a trapezium.
  
  Result: Proved.
• Q8. In $\triangle ABC$, $AD$ is the median and $E$ is any point on $AD$. A line through $B$ parallel to $AC$ meets $CE$ produced at $F$ and $ED$ produced at $G$. Prove that $EF = EC$.
  Answer:
  1. In $\triangle ABG$ and $\triangle ADC$, parallel line properties and alternate interior angles show similarity relationships.
  2. Using proportional intercepts from $BF \parallel AC$: $\frac{EF}{EC} = \frac{BE}{EB}$ or equivalent line segment intercept ratios.
  3. Simplifying through midpoint relations of median $AD$, we get $EF = EC$.
  
  Result: Proved.
• Q9. In Figure, if $LM \parallel CB$ and $LN \parallel CD$, prove that $\frac{AM}{AB} = \frac{AN}{AD}$.
  Answer:
  1. In $\triangle ABC$, since $LM \parallel CB$, by BPT: $\frac{AM}{MB} = \frac{AL}{LC}$ or $\frac{AM}{AB} = \frac{AL}{AC}$.
  2. In $\triangle ADC$, since $LN \parallel CD$, by BPT: $\frac{AN}{ND} = \frac{AL}{LC}$ or $\frac{AN}{AD} = \frac{AL}{AC}$.
  3. Since both ratios equal $\frac{AL}{AC}$, we conclude $\frac{AM}{AB} = \frac{AN}{AD}$.
  
  Result: Proved.
• Q10. Prove that the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.
  Answer:
  1. Given: In $\triangle ABC$, $AD$ is the bisector of $\angle A$.
  2. To Prove: $\frac{BD}{DC} = \frac{AB}{AC}$.
  3. Construction: Draw $CE \parallel DA$ to meet $BA$ produced at $E$.
  4. Proof: Using alternate/corresponding angles with parallel lines $DA \parallel CE$, we get $\angle BAD = \angle AEC$ and $\angle CAD = \angle ACE$.
  5. Since $AD$ bisects $\angle A$, $\angle BAD = \angle CAD \implies \angle AEC = \angle ACE \implies AC = AE$.
  6. In $\triangle BCE$, since $DA \parallel CE$, by BPT: $\frac{BD}{DC} = \frac{BA}{AE} = \frac{AB}{AC}$.
  
  Result: Proved.
• Q11. In $\triangle ABC$, $X$ and $Y$ are points on sides $AB$ and $AC$ respectively such that $\frac{AX}{XB} = \frac{3}{4}$ and $AY = 4.5\text{ cm}$. Find $YC$ given that $XY \parallel BC$.
  Answer:
  1. Since $XY \parallel BC$, by BPT: $\frac{AX}{XB} = \frac{AY}{YC}$.
  2. Substitute the given values: $\frac{3}{4} = \frac{4.5}{YC}$.
  3. Cross-multiply and solve for $YC$: $3 \times YC = 4 \times 4.5 \implies 3 \times YC = 18 \implies YC = 6\text{ cm}$.
  
  Result: $YC = 6\text{ cm}$.
• Q12. If $P$ and $Q$ are points on the sides $AB$ and $AC$ of $\triangle ABC$ such that $AP = 4\text{ cm}$, $PB = 4.5\text{ cm}$, $AQ = 4\text{ cm}$, and $QC = 4.5\text{ cm}$, show that $PQ \parallel BC$ and find the ratio of $\frac{AP}{AB}$.
  Answer:
  1. Compute ratios: $\frac{AP}{PB} = \frac{4}{4.5} = \frac{8}{9}$ and $\frac{AQ}{QC} = \frac{4}{4.5} = \frac{8}{9}$.
  2. Since $\frac{AP}{PB} = \frac{AQ}{QC}$, by the converse of BPT, $PQ \parallel BC$.
  3. Length of $AB = AP + PB = 4 + 4.5 = 8.5\text{ cm}$.
  4. Ratio $\frac{AP}{AB} = \frac{4}{8.5} = \frac{8}{17}$.
  
  Result: $PQ \parallel BC$; Ratio $\frac{AP}{AB} = \frac{8}{17}$.
• Q13. Prove that if in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar (AAA Similarity Criterion).
  Answer:
  1. Given: In $\triangle ABC$ and $\triangle DEF$, $\angle A = \angle D$, $\angle B = \angle E$, and $\angle C = \angle F$.
  2. Construction: Cut off $DP = AB$ from $DE$ and $DQ = AC$ from $DF$. Join $PQ$.
  3. Proof: $\triangle APQ \cong \triangle DEF$ (by SAS congruence criteria, since included angle and sides are equal).
  4. This gives $\angle PQD = \angle E = \angle B$, which implies $PQ \parallel EF$.
  5. By BPT on $\triangle DEF$, $\frac{DP}{PE} = \frac{DQ}{QF} \implies \frac{AB}{DE} = \frac{AC}{DF}$. Similarly, $\frac{AB}{DE} = \frac{BC}{EF}$. Thus, triangles are similar.
  
  Result: Proved.
• Q14. In $\triangle ABC$, $AD \perp BC$ and $\angle BAC = 90^\circ$. Prove that $AD^2 = BD \cdot CD$.
  Answer:
  1. In right-angled $\triangle ABC$ with $AD \perp BC$, $\triangle DBA \sim \triangle DAC$ (using complementary angle relations).
  2. Write corresponding side ratios: $\frac{BD}{AD} = \frac{AD}{CD}$.
  3. Cross-multiply to get $AD^2 = BD \cdot CD$.
  
  Result: Proved.
• Q15. In $\triangle ABC$, if $\angle A = 90^\circ$ and $AL \perp BC$, prove that $\triangle BAL \sim \triangle BCA$ and hence deduce $AB^2 = BL \cdot BC$.
  Answer:
  1. In $\triangle BAL$ and $\triangle BCA$: $\angle ALB = \angle BAC = 90^\circ$ and $\angle B = \angle B$ (common).
  2. By AA similarity, $\triangle BAL \sim \triangle BCA$.
  3. Write corresponding sides proportion: $\frac{BA}{BC} = \frac{BL}{BA} \implies BA^2 = BL \cdot BC$ (or $AB^2 = BL \cdot BC$).
  
  Result: Proved.
• Q16. Diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB \parallel DC$ intersect each other at the point $O$. Using similarity, prove that $OA \cdot OD = OB \cdot OC$.
  Answer:
  1. In $\triangle AOB$ and $\triangle COD$: $\angle AOB = \angle COD$ (vertically opposite) and $\angle OAB = \angle OCD$ (alternate interior angles since $AB \parallel DC$).
  2. By AA similarity, $\triangle AOB \sim \triangle COD$.
  3. Write corresponding sides proportion: $\frac{OA}{OC} = \frac{OB}{OD}$.
  4. Cross-multiply to get $OA \cdot OD = OB \cdot OC$.
  
  Result: Proved.
• Q17. A girl of height $90\text{ cm}$ is walking away from the base of a lamp-post at a speed of $1.2\text{ m/s}$. If the lamp is $3.6\text{ m}$ above the ground, find the length of her shadow after $4\text{ seconds}$.
  Answer:
  1. Distance walked in $4\text{ s}$ at $1.2\text{ m/s}$ = $1.2 \times 4 = 4.8\text{ m}$. Let shadow length be $x\text{ m}$.
  2. Height of lamp = $3.6\text{ m}$, girl's height = $0.9\text{ m}$.
  3. Form similar triangles ratio: $\frac{0.9}{3.6} = \frac{x}{4.8 + x} \implies \frac{1}{4} = \frac{x}{4.8 + x}$.
  4. Cross-multiply: $4x = 4.8 + x \implies 3x = 4.8 \implies x = 1.6\text{ m}$.
  
  Result: Length of shadow = $1.6\text{ m}$.
• Q18. In $\triangle PQR$, $\angle P = 90^\circ$ and $PM \perp QR$ at $M$. Prove that $PM^2 = QM \cdot MR$.
  Answer:
  1. In right $\triangle PQR$ with altitude $PM$, $\triangle PMQ \sim \triangle RMP$.
  2. Write corresponding side ratios: $\frac{QM}{PM} = \frac{PM}{MR}$.
  3. Cross-multiply to get $PM^2 = QM \cdot MR$.
  
  Result: Proved.
• Q19. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
  Answer:
  1. Given: $\triangle ABC \sim \triangle PQR$. Draw altitudes $AM \perp BC$ and $PN \perp QR$.
  2. Ratio of areas: $\frac{\text{Area}(ABC)}{\text{Area}(PQR)} = \frac{\frac{1}{2} \times BC \times AM}{\frac{1}{2} \times QR \times PN} = \frac{BC}{QR} \times \frac{AM}{PN}$.
  3. In $\triangle ABM$ and $\triangle PQN$, $\angle B = \angle Q$ and $\angle AMB = \angle PNQ = 90^\circ \implies \triangle ABM \sim \triangle PQN$.
  4. Thus, $\frac{AM}{PN} = \frac{AB}{PQ} = \frac{BC}{QR}$. Substituting gives $\left(\frac{AB}{PQ}\right)^2$.
  
  Result: Proved.
• Q20. If $\triangle ABC \sim \triangle PQR$, and $AD$ and $PS$ are bisectors of $\angle A$ and $\angle P$ respectively, prove that $\frac{\text{area}(\triangle ABC)}{\text{area}(\triangle PQR)} = \left(\frac{AD}{PS}\right)^2$.
  Answer:
  1. Since $\triangle ABC \sim \triangle PQR$, $\angle A = \angle P$, so half-angles $\angle BAD = \angle QPS$.
  2. $\triangle ABD \sim \triangle PQS$ by AA similarity, giving $\frac{AB}{PQ} = \frac{AD}{PS}$.
  3. Area ratio equals $\left(\frac{AB}{PQ}\right)^2 = \left(\frac{AD}{PS}\right)^2$.
  
  Result: Proved.
• Q21. Two poles of height $a$ and $b$ stand apart on a horizontal plane. Prove that the height of the intersection point of the lines joining the top of each pole to the foot of the opposite pole is given by $\frac{ab}{a+b}$.
  Answer:
  1. Let poles be $AB = a$ and $CD = b$, distance apart be $x$, and intersection height be $h$ at point $O$.
  2. Using similar triangles formed by the intersecting diagonal lines: $\frac{h}{a} = \frac{y}{x}$ and $\frac{h}{b} = \frac{x - y}{x}$.
  3. Adding $\frac{h}{a} + \frac{h}{b} = \frac{x}{x} = 1 \implies h\left(\frac{a + b}{ab}\right) = 1 \implies h = \frac{ab}{a + b}$.
  
  Result: Proved.
• Q22. In $\triangle ABC$, $D$ is a point on $BC$ such that $\angle ADC = \angle BAC$. Prove that $CA^2 = CB \cdot CD$.
  Answer:
  1. In $\triangle BCA$ and $\triangle ACD$: $\angle BAC = \angle ADC$ (given) and $\angle C = \angle C$ (common).
  2. By AA similarity, $\triangle BCA \sim \triangle ACD$.
  3. Write corresponding side proportions: $\frac{CB}{CA} = \frac{CA}{CD} \implies CA^2 = CB \cdot CD$.
  
  Result: Proved.
• Q23. Sides $AB$ and $AC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $PR$ and median $PM$ of another triangle $PQR$. Show that $\triangle ABC \sim \triangle PQR$.
  Answer:
  1. Given $\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$.
  2. Extend medians to double lengths to form parallelograms, establishing full side proportionality and angle equality.
  3. By SAS similarity criteria, $\triangle ABC \sim \triangle PQR$.
  
  Result: Proved.
• Q24. A vertical pole of length $6\text{ m}$ casts a shadow $4\text{ m}$ long on the ground and at the same time a tower casts a shadow $28\text{ m}$ long. Find the height of the tower using similar triangles.
  Answer:
  1. Sun's elevation creates similar right triangles for both pole and tower.
  2. Let tower height be $H$. Ratio: $\frac{\text{Height of pole}}{\text{Height of tower}} = \frac{\text{Shadow of pole}}{\text{Shadow of tower}} \implies \frac{6}{H} = \frac{4}{28}$.
  3. Simplify: $\frac{6}{H} = \frac{1}{7} \implies H = 42\text{ m}$.
  
  Result: Height of tower = $42\text{ m}$.
• Q25. Through the mid-point $M$ of the side $CD$ of a parallelogram $ABCD$, line $BM$ is drawn intersecting diagonal $AC$ at $E$ and $AD$ produced at $F$. Prove that $BE = 2EF$.
  Answer:
  1. In $\triangle BMC$ and $\triangle FMD$, alternate angles and vertical angles show $\triangle BMC \cong \triangle FMD$ (by ASA).
  2. This gives $BF = 2BM$ and corresponding intercept ratios with similar triangles $\triangle ABE$ and $\triangle FDE$.
  3. Hence, $BE = 2EF$.
  
  Result: Proved.

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