# PRACTICE QUESTIONS ON BASIC PROPORTIONALITY THEOREM

**Question 1 :**

In ΔABC, D and E are points on the sides AB and AC respectively such that DE ∥ BC (i) If AD/DB = 3/4 and AC = 15 cm find AE.

(ii) If AD = 8x −7 , DB = 5x −3 , AE = 4x −3 and EC = 3x −1 find the value of x.

**Question 2 :**

ABCD is a trapezium in which AB || DC and P,Q are points on AD and BC respectively, such that PQ || DC if PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD

**Question 3 :**

In ΔABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE ||BC

(i) AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm.

(ii) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm.

Solutions :

**Question 1 :**

In ΔABC, D and E are points on the sides AB and AC respectively such that DE ∥ BC (i) If AD/DB = 3/4 and AC = 15 cm find AE.

**Solution :**

AD/DB = AE/EC

AC = 15 (Given)

Let AE = x and EC = 15 - x

3/4 = x/(15-x)

3(15 - x) = 4x

45 - 3x = 4x

4x + 3x = 45

7x = 45

x = 45/7

x = 6.43

(ii) If AD = 8x −7 , DB = 5x −3 , AE = 4x −3 and EC = 3x −1 find the value of x.

**Solution :**

AD/DB = AE/EC

(8x - 7)/(5x - 3) = (4x - 3)/(3x - 1)

(8x - 7) (3x - 1) = (4x - 3)(5x - 3)

24x^{2} - 8x - 21x + 7 = 20x^{2} - 12x - 15x + 9

24x^{2}-20x^{2 }- 29x + 27x + 7 - 9 = 0

4x^{2}- 2x - 2 = 0

2x^{2}- x - 1 = 0

(2x + 1) (x - 1) = 0

x = -1/2 and x = 1

Hence the value of x is 1.

**Question 2 :**

ABCD is a trapezium in which AB || DC and P,Q are points on AD and BC respectively, such that PQ || DC if PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD

**Solution :**

Let us draw the a rough picture based on the given information.

Let us join DB.

In triangle DAB,

PD/PA = DT/TB ----(1)

In triangle DBC,

BT/TD = BQ/QC

By taking reciprocal on both sides,

TD/BT = QC/BQ ----(2)

(1) = (2)

PD/PA = QC/BQ

18/PA = 15/35

PA = 18(35)/15

PA = 42

AD = 42 + 18

AD = 60 cm

**Question 3 :**

In ΔABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE ||BC

(i) AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm.

**Solution :**

AD = 8 AB = AD + DB 12 = 8 + DB DB = 12 - 8 DB = 4
AD/DB = AE/EC 8/4 = 12/6 2 = 2 Hence DE ||BC. (ii) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm.
AD/DB = AE/EC 1.4/4.2 = 1.8/5.4 0.33 = 0.33 Hence DE ||BC. |