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Basic Proportionality Theorem PYQs

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This page provides comprehensive Class 10 Maths Basic Proportionality Theorem PYQs | Triangles. Basic Proportionality Theorem previous year questions for Class 10 Maths Triangles. Practice CBSE board PYQs with step-by-step solutions on SJMaths.

CBSE Class 10 Previous Year Questions with Step-by-Step Solutions

Qbpt1 2024
00:00
In the given figure, DE ∥ BC. If AD = 2.4 cm, DB = 4 cm and AE = 2 cm, find the length of AC.
(a)5 cm
(b)6 cm
(c)7 cm
(d)8 cm
A B C D E 2.4 4 2 EC
Step 1: Since DE ∥ BC, by Basic Proportionality Theorem,
$$\frac{AD}{DB} = \frac{AE}{EC}$$
Step 2: Substitute values:
$$\frac{2.4}{4} = \frac{2}{EC}$$
Step 3: Solve for EC:
$$EC = \frac{4 \times 2}{2.4} = 3.33\text{ cm}$$
Step 4: AC = AE + EC = 2 + 3.33 = 5.33 cm (approx)
Final Answer: AC ≈ 5.33 cm
Qbpt2 2023
00:00
In ΔABC, DE ∥ BC. Find the ratio AD : DB if AE : EC = 3 : 5.
(a)3 : 5
(b)5 : 3
(c)8 : 5
(d)5 : 8
A B C D E 3 5
Step 1: By Basic Proportionality Theorem,
$$\frac{AD}{DB} = \frac{AE}{EC}$$
Step 2: Given AE : EC = 3 : 5
Step 3: Hence AD : DB = 3 : 5
Final Answer: 3 : 5 (Option a)
Qbpt3 2022
00:00
In the given figure, DE ∥ BC. If AD = x, DB = 6 cm, AE = 4 cm and EC = 8 cm, find x.
(a)2 cm
(b)3 cm
(c)4 cm
(d)6 cm
A B C D E x 6 4 8
Step 1: By BPT,
$$\frac{AD}{DB} = \frac{AE}{EC}$$
$$\frac{x}{6} = \frac{4}{8}$$
$$x = 6 \times \frac{1}{2} = 3$$
Final Answer: 3 cm (Option b)
Qbpt4 2026
00:00
In ΔDEF, AB ∥ EF. The value of x is:
(a)0, 2
(b)2 only
(c)-2
(d)1
D E F A B 2x 3x+1 x 2x - 1/2
Step 1: In $\triangle DEF$, since $AB \parallel EF$, by Basic Proportionality Theorem:
$$\frac{DA}{AE} = \frac{DB}{BF}$$
Step 2: Substitute the algebraic expressions:
$$\frac{2x}{3x+1} = \frac{x}{2x - 1/2}$$
Step 3: Since side length $x \neq 0$, divide both sides by $x$:
$$\frac{2}{3x+1} = \frac{1}{2x - 1/2}$$
Step 4: Cross-multiply:
$$2\left(2x - \frac{1}{2}\right) = 3x + 1 \implies 4x - 1 = 3x + 1 \implies x = 2$$
Final Answer: 2 only (Option b)
Qbpt5 2025
00:00
In the adjoining figure, PQ ∥ XY ∥ BC, AP = 2 cm, PX = 1.5 cm and BX = 4 cm. If QY = 0.75 cm, then AQ + CY =
(a)6 cm
(b)4.5 cm
(c)3 cm
(d)5.25 cm
A B C P Q X Y
Step 1: Since $PQ \parallel XY \parallel BC$, by Basic Proportionality Theorem:
$$\frac{AP}{AQ} = \frac{PX}{QY} = \frac{XB}{YC}$$
Step 2: Use $\frac{AP}{AQ} = \frac{PX}{QY}$:
$$\frac{2}{AQ} = \frac{1.5}{0.75} = 2 \implies AQ = 1\text{ cm}$$
Step 3: Use $\frac{XB}{YC} = 2$:
$$\frac{4}{YC} = 2 \implies YC = 2\text{ cm}$$
Step 4: Calculate $AQ + CY = 1 + 2 = 3\text{ cm}$.
Final Answer: 3 cm (Option c)
Qbpt6 2024C
00:00
In the given figure, if M and N are points on the sides OP and OS respectively of ΔOPS, such that MN ∥ PS, then the length of OP is:
(a)6.8 cm
(b)17 cm
(c)15.3 cm
(d)9.6 cm
O P S M N 8.5 cm 4.8 cm 6 cm
Step 1: In $\triangle OPS$, $MN \parallel PS$. By Basic Proportionality Theorem:
$$\frac{OM}{MP} = \frac{ON}{NS}$$
Step 2: Given $MP = 8.5\text{ cm}$, $ON = 4.8\text{ cm}$, and $NS = 6\text{ cm}$:
$$\frac{OM}{8.5} = \frac{4.8}{6} = 0.8 \implies OM = 6.8\text{ cm}$$
Step 3: Calculate total length $OP = OM + MP$:
$$OP = 6.8 + 8.5 = 15.3\text{ cm}$$
Final Answer: 15.3 cm (Option c)
Qbpt7 2024
00:00
Assertion (A): ABCD is a trapezium with DC ∥ AB. E and F are points on AD and BC respectively, such that EF ∥ AB. Then AE/ED = BF/FC.
Reason (R): Any line parallel to parallel sides of a trapezium divides the non-parallel sides proportionally.
(a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c)Assertion (A) is true but Reason (R) is false.
(d)Assertion (A) is false but Reason (R) is true.
D C A B E F
Step 1: Joining diagonal $AC$ intersecting $EF$ at $G$ divides $\triangle ADC$ and $\triangle ABC$.
Step 2: In $\triangle ADC$, $EG \parallel DC \implies \frac{AE}{ED} = \frac{AG}{GC}$.
Step 3: In $\triangle ABC$, $GF \parallel AB \implies \frac{AG}{GC} = \frac{BF}{FC}$.
Step 4: Thus $\frac{AE}{ED} = \frac{BF}{FC}$. Both (A) and (R) are true, and (R) is the correct explanation.
Final Answer: Option (a)
Qbpt8 2024
00:00
In ΔABC, DE ∥ BC (as shown in the figure). If AD = 2 cm, BD = 3 cm, BC = 7.5 cm, then the length of DE (in cm) is:
(a)2.5
(b)3
(c)5
(d)6
A B C D E 2 3 BC=7.5 DE
Step 1: Since $DE \parallel BC$, $\triangle ADE \sim \triangle ABC$.
Step 2: $\frac{AD}{AB} = \frac{DE}{BC}$.
Step 3: Here $AB = AD + BD = 2 + 3 = 5\text{ cm}$:
$$\frac{2}{5} = \frac{DE}{7.5} \implies DE = 3\text{ cm}$$
Final Answer: 3 (Option b)
Qbpt9 2024
00:00
In the given figure, in ΔABC, DE ∥ BC. If AD = 2.4 cm, DB = 4 cm and AE = 2 cm, then the length of AC is:
(a)10/3 cm
(b)3/10 cm
(c)16/3 cm
(d)1.2 cm
A B C D E 2.4 4 2
Step 1: By BPT: $\frac{AD}{DB} = \frac{AE}{EC} \implies \frac{2.4}{4} = \frac{2}{EC} \implies EC = \frac{10}{3}\text{ cm}$.
Step 2: $AC = AE + EC = 2 + \frac{10}{3} = \frac{16}{3}\text{ cm}$.
Final Answer: 16/3 cm (Option c)
Qbpt10 2023C
00:00
In the given figure, DE ∥ BC and all measurements are given in centimetres. The length of AE is:
(a)2 cm
(b)2.25 cm
(c)2.5 cm
(d)2.75 cm
A B C D E 3 4 AE 3
Step 1: By BPT: $\frac{AD}{DB} = \frac{AE}{EC} \implies \frac{3}{4} = \frac{AE}{3} \implies AE = 2.25\text{ cm}$.
Final Answer: 2.25 cm (Option b)
Qbpt11 2023
00:00
In the given figure, DE ∥ BC. If AD = 2 units, DB = 3 units and AE = 3 units, EC = x units, then the value of x is:
(a)2
(b)3
(c)5
(d)9/2
A B C D E 2 3 3 x
Step 1: By BPT: $\frac{AD}{DB} = \frac{AE}{EC} \implies \frac{2}{3} = \frac{3}{x} \implies x = \frac{9}{2}$.
Final Answer: 9/2 (Option d)
Qbpt12 2023
00:00
In ΔABC, PQ ∥ BC. If PB = 6 cm, AP = 4 cm, AQ = 8 cm, find the length of AC.
(a)12 cm
(b)20 cm
(c)6 cm
(d)14 cm
A B C P Q 4 6 8
Step 1: By BPT: $\frac{AP}{PB} = \frac{AQ}{QC} \implies \frac{4}{6} = \frac{8}{QC} \implies QC = 12\text{ cm}$.
Step 2: $AC = AQ + QC = 8 + 12 = 20\text{ cm}$.
Final Answer: 20 cm (Option b)
Qbpt13 2023
00:00
In the given figure, PQ ∥ AC. If BP = 4 cm, AP = 2.4 cm and BQ = 5 cm, then length of BC is:
(a)8 cm
(b)3 cm
(c)0.3 cm
(d)25/3 cm
C A B P Q 2.4 cm 4 cm 5 cm
Step 1: $\triangle BPQ \sim \triangle BAC \implies \frac{BP}{BA} = \frac{BQ}{BC}$.
Step 2: $BA = 4 + 2.4 = 6.4\text{ cm} \implies \frac{4}{6.4} = \frac{5}{BC} \implies BC = 8\text{ cm}$.
Final Answer: 8 cm (Option a)
Qbpt14 Term I, 2021-22
00:00
In the figure given below, what value of x will make PQ ∥ AB?
(a)2
(b)3
(c)4
(d)5
C A B P Q x+3 3x+19 x 3x+4
Step 1: By converse of BPT: $\frac{CP}{PA} = \frac{CQ}{QB} \implies \frac{x+3}{3x+19} = \frac{x}{3x+4}$.
Step 2: $(x+3)(3x+4) = x(3x+19) \implies 3x^2 + 13x + 12 = 3x^2 + 19x \implies 6x = 12 \implies x = 2$.
Final Answer: 2 (Option a)
Qbpt15 2020
00:00
In figure, DE ∥ BC. If AD/DB = 3/2 and AE = 2.7 cm, then EC is equal to:
(a)2.0 cm
(b)1.8 cm
(c)4.0 cm
(d)2.7 cm
A B C D E 2.7
Step 1: By BPT: $\frac{AD}{DB} = \frac{AE}{EC} \implies \frac{3}{2} = \frac{2.7}{EC} \implies EC = 1.8\text{ cm}$.
Final Answer: 1.8 cm (Option b)
Qbpt16 2019C
00:00
In figure, GC ∥ BD and GE ∥ BF. If AC = 3 cm and CD = 7 cm, then find the value of AE/AF.
A B D F C G E 3 cm 7 cm
Step 1: In $\triangle ABD$, $GC \parallel BD \implies \frac{AG}{GB} = \frac{AC}{CD} = \frac{3}{7}$.
Step 2: In $\triangle ABF$, $GE \parallel BF \implies \frac{AE}{EF} = \frac{AG}{GB} = \frac{3}{7}$.
Step 3: $\frac{AE}{AF} = \frac{AE}{AE + EF} = \frac{3}{3 + 7} = \frac{3}{10}$.
Final Answer: 3/10
Qbpt17 Board Term I, 2017
00:00
In ΔABC, X is middle point of AC. If XY ∥ AB, then prove that Y is middle point of BC.
Step 1: By BPT in $\triangle ABC$: $\frac{CX}{XA} = \frac{CY}{YB}$.
Step 2: Since $X$ is midpoint of $AC$, $CX = XA \implies \frac{CX}{XA} = 1$.
Step 3: Thus $\frac{CY}{YB} = 1 \implies CY = YB$. Hence $Y$ is the midpoint of $BC$.
Final Answer: Y is the midpoint of BC. (Proved)
Qbpt18 Board Term I, 2017
00:00
In ΔABC, D and E are point on side AB and AC respectively, such that DE ∥ BC. If AE = 2 cm, AD = 3 cm and BD = 4.5 cm, then find CE.
A B C D E 3 4.5 2
Step 1: By BPT: $\frac{AD}{BD} = \frac{AE}{CE} \implies \frac{3}{4.5} = \frac{2}{CE} \implies CE = 3\text{ cm}$.
Final Answer: 3 cm
Qbpt19 Board Term I, 2017
00:00
In ΔABC, DE ∥ BC, then find the value of x if AD = x, DB = x+1, AE = x+3, EC = x+5.
A B C D E x x+1 x+3 x+5
Step 1: By BPT: $\frac{x}{x+1} = \frac{x+3}{x+5} \implies x(x+5) = (x+1)(x+3)$.
Step 2: $x^2 + 5x = x^2 + 4x + 3 \implies x = 3$.
Final Answer: 3
Qbpt20 Board Term I, 2017
00:00
In given figure, DE ∥ BC. If AD/DB = 3/4 and AC = 14 cm, find EC.
A B C D E
Step 1: By BPT: $\frac{AE}{EC} = \frac{3}{4} \implies \frac{AC}{EC} = \frac{7}{4}$.
Step 2: $\frac{14}{EC} = \frac{7}{4} \implies EC = 8\text{ cm}$.
Final Answer: 8 cm
Qbpt21 2026
00:00
In ΔABC, DE ∥ BC. If AD = x, DB = x - 2, AE = x + 2 and EC = x - 1, then find the value of x.
A B C D E x x-2 x+2 x-1
Step 1: By BPT: $\frac{x}{x-2} = \frac{x+2}{x-1} \implies x(x-1) = (x-2)(x+2)$.
Step 2: $x^2 - x = x^2 - 4 \implies x = 4$.
Final Answer: 4
Qbpt22 2025
00:00
In the adjoining figure, AD/BD = AE/EC and ∠BDE = ∠CED, prove that ABC is an isosceles triangle.
A B C D E
Step 1: Since $\frac{AD}{BD} = \frac{AE}{EC}$, by converse of BPT, $DE \parallel BC$.
Step 2: So $\angle BDE = \angle B$ and $\angle CED = \angle C$ (corresponding angles).
Step 3: Since $\angle BDE = \angle CED$, $\angle B = \angle C \implies AB = AC$. Thus $\triangle ABC$ is isosceles.
Final Answer: ΔABC is isosceles. (Proved)
Qbpt23 2024C
00:00
PQRS is a trapezium with PQ ∥ SR. If M and N are two points on the non-parallel sides PS and QR respectively, such that MN is parallel to PQ, then show that PM/MS = QN/NR.
D C A B E F
Step 1: Join $PR$ intersecting $MN$ at $O$.
Step 2: In $\triangle PRS$, $MO \parallel SR \implies \frac{PM}{MS} = \frac{PO}{OR}$ (by BPT).
Step 3: In $\triangle PQR$, $ON \parallel PQ \implies \frac{PO}{OR} = \frac{QN}{NR}$ (by BPT).
Step 4: Hence $\frac{PM}{MS} = \frac{QN}{NR}$. (Proved)
Final Answer: PM/MS = QN/NR (Proved)
Qbpt24 NCERT, 2020
00:00
In the given figure, DE ∥ AC and DF ∥ AE. Prove that BF/FE = BE/EC.
A B C D E F
Step 1: In $\triangle ABC$, $DE \parallel AC \implies \frac{BD}{DA} = \frac{BE}{EC}$.
Step 2: In $\triangle ABE$, $DF \parallel AE \implies \frac{BD}{DA} = \frac{BF}{FE}$.
Step 3: Therefore, $\frac{BF}{FE} = \frac{BE}{EC}$. (Proved)
Final Answer: BF/FE = BE/EC (Proved)
Qbpt25 2020
00:00
In figure, if PQ ∥ BC and PR ∥ CD, prove that QB/AQ = DR/AR.
A B D C Q R P
Step 1: In $\triangle ABC$, $PQ \parallel BC \implies \frac{AQ}{QB} = \frac{AP}{PC}$.
Step 2: In $\triangle ACD$, $PR \parallel CD \implies \frac{AR}{RD} = \frac{AP}{PC}$.
Step 3: Equating both gives $\frac{AQ}{QB} = \frac{AR}{RD} \implies \frac{QB}{AQ} = \frac{DR}{AR}$. (Proved)
Final Answer: QB/AQ = DR/AR (Proved)
Qbpt26 2026, 2024
00:00
State and prove Basic Proportionality Theorem.
**OR**
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.
A B C D E
Given: In $\triangle ABC$, $DE \parallel BC$ intersects $AB$ at $D$ and $AC$ at $E$.
To Prove: $\frac{AD}{DB} = \frac{AE}{EC}$.
Construction: Join $BE, CD$. Draw $DM \perp AC$ and $EN \perp AB$.
Proof: $\text{ar}(\triangle ADE) = \frac{1}{2} AD \cdot EN$, $\text{ar}(\triangle BDE) = \frac{1}{2} DB \cdot EN \implies \frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \frac{AD}{DB}$.
Similarly, $\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DEC)} = \frac{AE}{EC}$.
Since $\triangle BDE$ and $\triangle DEC$ have same base $DE$ and between $DE \parallel BC$, $\text{ar}(\triangle BDE) = \text{ar}(\triangle DEC)$.
Hence, $\frac{AD}{DB} = \frac{AE}{EC}$. (Proved)
Final Answer: Basic Proportionality Theorem Proved.
Qbpt27 2019C
00:00
ABCD is a trapezium with AB ∥ CD. E and F are points on non parallel sides AD and BC respectively, such that EF ∥ AB. Show that AE/ED = BF/FC.
D C A B E F
Step 1: Join diagonal $AC$ intersecting $EF$ at $G$.
Step 2: In $\triangle ADC$, $EG \parallel DC \implies \frac{AE}{ED} = \frac{AG}{GC}$.
Step 3: In $\triangle ABC$, $GF \parallel AB \implies \frac{AG}{GC} = \frac{BF}{FC}$.
Step 4: Hence $\frac{AE}{ED} = \frac{BF}{FC}$. (Proved)
Final Answer: AE/ED = BF/FC (Proved)
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