Basic Proportionality Theorem PYQs

Overview

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

(d)8 cm

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

(d)5 : 8

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

(d)6 cm

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 2021

00:00

State and prove the Basic Proportionality Theorem.

Statement: A line drawn parallel to one side of a triangle divides the other two sides in the same ratio.

Proof: In ΔABC, let DE ∥ BC intersect AB at D and AC at E.

Then, by AA similarity, ΔADE ~ ΔABC.

Hence, $\frac{AD}{AB} = \frac{AE}{AC}$

This implies $\frac{AD}{DB} = \frac{AE}{EC}$

Thus, the theorem is proved.

Final Answer: Basic Proportionality Theorem proved.

Qbpt5 2020

00:00

State the converse of the Basic Proportionality Theorem.

Step 1: If a line divides any two sides of a triangle in the same ratio,

Step 2: Then the line is parallel to the third side.

Final Answer: Converse of BPT stated correctly.

Qbpt6 2019

00:00

In ΔABC, a line DE is drawn parallel to BC intersecting AB and AC in D and E respectively. If AD : DB = 2 : 3, find AE : EC.

(a)2 : 3

(b)3 : 2

(d)3 : 5

Step 1: Since DE ∥ BC, by Basic Proportionality Theorem,

$\frac{AD}{DB} = \frac{AE}{EC}$

Step 2: Given AD : DB = 2 : 3

Step 3: Therefore AE : EC = 2 : 3

Final Answer: 2 : 3 (Option a)

Qbpt7 2018

00:00

In ΔABC, DE ∥ BC. If AD = 3 cm, DB = 5 cm and AC = 16 cm, find AE.

(a)6 cm

(b)8 cm

(d)12 cm

Step 1: By BPT,

$\frac{AD}{DB} = \frac{AE}{EC}$

Step 2: Given AD = 3 cm, DB = 5 cm, so AD/AB = 3/(3+5) = 3/8.

Step 3: Also, AE/AC = 3/8.

$AE = \frac{3}{8} \times 16 = 6$

Final Answer: 6 cm (Option a)

Qbpt8 2017

00:00

In the given figure, DE ∥ BC. If AD = 4 cm, DB = x cm, AE = 6 cm and EC = 9 cm, find x.

(a)4 cm

(b)5 cm

(d)8 cm

Step 1: By Basic Proportionality Theorem,

$\frac{AD}{DB} = \frac{AE}{EC}$

$\frac{4}{x} = \frac{6}{9}$

$x = \frac{4 \times 9}{6} = 6$

Final Answer: 6 cm (Option c)

Qbpt9 2016

00:00

Prove that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Step 1: Let in ΔABC, a line DE intersect AB and AC in D and E respectively such that AD/DB = AE/EC.

Step 2: Consider a line through D parallel to BC meeting AC at E'.

Step 3: By BPT, AD/DB = AE'/E'C.

Step 4: Given AD/DB = AE/EC, so E and E' coincide.

Step 5: Hence DE ∥ BC. Converse proved.

Final Answer: Converse of Basic Proportionality Theorem proved.

Qbpt10 2025

00:00

In ΔABC, a line DE is drawn parallel to BC intersecting AB and AC at D and E respectively. Which of the following is true?

(a)AD / DB = AB / AC

(b)AE / EC = AB / BC

(c)AD / DB = AE / EC

(d)DE / BC = AB / AC

Step 1: By Basic Proportionality Theorem, a line parallel to one side of a triangle divides the other two sides in the same ratio.

Step 2: Therefore,

$\frac{AD}{DB} = \frac{AE}{EC}$

Final Answer: AD / DB = AE / EC (Option c)

Qbpt11 2024

00:00

In ΔABC, DE ∥ BC. If AD = 2 cm, DB = 3 cm and AC = 10 cm, find AE.

(a)4 cm

(b)5 cm

(d)8 cm

Step 1: By BPT,

$\frac{AD}{AB} = \frac{AE}{AC}$

Step 2: Here AB = AD + DB = 2 + 3 = 5 cm

$\frac{2}{5} = \frac{AE}{10}$

$AE = \frac{2}{5} \times 10 = 4$

Final Answer: 4 cm (Option a)

Qbpt12 2023

00:00

Assertion (A): If a line is drawn parallel to one side of a triangle, then it divides the other two sides in the same ratio. Reason (R): If two triangles are similar, then their corresponding sides are proportional.

(a)Both A and R are true and R is the correct explanation of A.

(b)Both A and R are true but R is not the correct explanation of A.

(c)A is true but R is false.

(d)A is false but R is true.

Step 1: The assertion states the Basic Proportionality Theorem, which is true.

Step 2: The reason is also true and is the basis of the proof using similarity of triangles.

Step 3: Therefore, both A and R are true and R explains A.

Final Answer: Both A and R are true and R is the correct explanation of A. (Option a)

Qbpt13 2022

00:00

In a trapezium, the line joining the mid-points of the non-parallel sides is parallel to the parallel sides. Show that it divides the diagonals in the same ratio.

Step 1: Draw diagonals and join mid-points of non-parallel sides.

Step 2: By using similarity of triangles formed and the property of parallel lines, corresponding segments of diagonals are in the same ratio.

Step 3: Hence, the line divides both diagonals proportionally.

Final Answer: The diagonals are divided in the same ratio.

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

(d)8 cm

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

(d)5 : 8

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

(d)6 cm

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 2021

00:00

State and prove the Basic Proportionality Theorem.

Statement: A line drawn parallel to one side of a triangle divides the other two sides in the same ratio.

Proof: In ΔABC, let DE ∥ BC intersect AB at D and AC at E.

Then, by AA similarity, ΔADE ~ ΔABC.

Hence, $\frac{AD}{AB} = \frac{AE}{AC}$

This implies $\frac{AD}{DB} = \frac{AE}{EC}$

Thus, the theorem is proved.

Final Answer: Basic Proportionality Theorem proved.

Qbpt5 2020

00:00

State the converse of the Basic Proportionality Theorem.

Step 1: If a line divides any two sides of a triangle in the same ratio,

Step 2: Then the line is parallel to the third side.

Final Answer: Converse of BPT stated correctly.

Qbpt6 2019

00:00

In ΔABC, a line DE is drawn parallel to BC intersecting AB and AC in D and E respectively. If AD : DB = 2 : 3, find AE : EC.

(a)2 : 3

(b)3 : 2

(d)3 : 5

Step 1: Since DE ∥ BC, by Basic Proportionality Theorem,

$\frac{AD}{DB} = \frac{AE}{EC}$

Step 2: Given AD : DB = 2 : 3

Step 3: Therefore AE : EC = 2 : 3

Final Answer: 2 : 3 (Option a)

Qbpt7 2018

00:00

In ΔABC, DE ∥ BC. If AD = 3 cm, DB = 5 cm and AC = 16 cm, find AE.

(a)6 cm

(b)8 cm

(d)12 cm

Step 1: By BPT,

$\frac{AD}{DB} = \frac{AE}{EC}$

Step 2: Given AD = 3 cm, DB = 5 cm, so AD/AB = 3/(3+5) = 3/8.

Step 3: Also, AE/AC = 3/8.

$AE = \frac{3}{8} \times 16 = 6$

Final Answer: 6 cm (Option a)

Qbpt8 2017

00:00

In the given figure, DE ∥ BC. If AD = 4 cm, DB = x cm, AE = 6 cm and EC = 9 cm, find x.

(a)4 cm

(b)5 cm

(d)8 cm

Step 1: By Basic Proportionality Theorem,

$\frac{AD}{DB} = \frac{AE}{EC}$

$\frac{4}{x} = \frac{6}{9}$

$x = \frac{4 \times 9}{6} = 6$

Final Answer: 6 cm (Option c)

Qbpt9 2016

00:00

Prove that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Step 1: Let in ΔABC, a line DE intersect AB and AC in D and E respectively such that AD/DB = AE/EC.

Step 2: Consider a line through D parallel to BC meeting AC at E'.

Step 3: By BPT, AD/DB = AE'/E'C.

Step 4: Given AD/DB = AE/EC, so E and E' coincide.

Step 5: Hence DE ∥ BC. Converse proved.

Final Answer: Converse of Basic Proportionality Theorem proved.

Qbpt10 2025

00:00

In ΔABC, a line DE is drawn parallel to BC intersecting AB and AC at D and E respectively. Which of the following is true?

(a)AD / DB = AB / AC

(b)AE / EC = AB / BC

(c)AD / DB = AE / EC

(d)DE / BC = AB / AC

Step 1: By Basic Proportionality Theorem, a line parallel to one side of a triangle divides the other two sides in the same ratio.

Step 2: Therefore,

$\frac{AD}{DB} = \frac{AE}{EC}$

Final Answer: AD / DB = AE / EC (Option c)

Qbpt11 2024

00:00

In ΔABC, DE ∥ BC. If AD = 2 cm, DB = 3 cm and AC = 10 cm, find AE.

(a)4 cm

(b)5 cm

(d)8 cm

Step 1: By BPT,

$\frac{AD}{AB} = \frac{AE}{AC}$

Step 2: Here AB = AD + DB = 2 + 3 = 5 cm

$\frac{2}{5} = \frac{AE}{10}$

$AE = \frac{2}{5} \times 10 = 4$

Final Answer: 4 cm (Option a)

Qbpt12 2023

00:00

(a)Both A and R are true and R is the correct explanation of A.

(b)Both A and R are true but R is not the correct explanation of A.

(c)A is true but R is false.

(d)A is false but R is true.

Step 1: The assertion states the Basic Proportionality Theorem, which is true.

Step 2: The reason is also true and is the basis of the proof using similarity of triangles.

Step 3: Therefore, both A and R are true and R explains A.

Final Answer: Both A and R are true and R is the correct explanation of A. (Option a)

Qbpt13 2022

00:00

In a trapezium, the line joining the mid-points of the non-parallel sides is parallel to the parallel sides. Show that it divides the diagonals in the same ratio.

Step 1: Draw diagonals and join mid-points of non-parallel sides.

Step 2: By using similarity of triangles formed and the property of parallel lines, corresponding segments of diagonals are in the same ratio.

Step 3: Hence, the line divides both diagonals proportionally.

Final Answer: The diagonals are divided in the same ratio.