PRACTICE QUESTIONS ON BASIC PROPORTIONALITY THEOREM - GMS - Learning Simply
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# PRACTICE QUESTIONS ON BASIC PROPORTIONALITY THEOREM

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# 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/D3/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/D3/and AC = 15 cm find AE.

Solution :

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 :

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

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

24x2 - 8x - 21x + 7  =  20x2 - 12x - 15x + 9

24x2-20x- 29x + 27x + 7 - 9  =  0

4x2- 2x - 2  =  0

2x2- 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

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 :

12  =  8 + DB

DB  =  12 - 8

DB  =  4

 AC  =  18AC  =  AE + EC18  =  12 + ECEC  =  18 - 12EC  =  6 cm

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.

Solution :

 AD  =  1.4AB  =  AD + DB5.6  =  1.4 + DBDB  =  5.6 - 1.4DB  =  4 .2 cm AC  =  7.2AC  =  AE + EC7.2  =  1.8 + ECEC  =  7.2-1.8EC  =  5.4 cm

1.4/4.2  =  1.8/5.4

0.33  =  0.33

Hence DE ||BC.

At the helm of GMS Learning is Principal Balkishan Agrawal, a dedicated and experienced educationist. Under his able guidance, our school has flourished academically and has achieved remarkable milestones in various fields. Principal Agrawal’s visio…