Q10 of 95 Page 5

In the adjoining figure, DE || BC, find x.

Given: AD = 4, DB =x – 4, AE = x – 3 and EC = 3x – 19

and DE || BC


Basic Proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in the same ratio.


So, by basic proportionality theorem




⇒ 4(3x – 19) = (x – 4)(x – 3)


⇒ 12x – 76 = x2 – 3x – 4x + 12


⇒ 12x – 76 = x2 –7x + 12


⇒ x2 –7x + 12 – 12x + 76 = 0


⇒ x2 –19x + 88 = 0


Solving the Quadratic equation by splitting themiddle term, we get,


⇒ x2 –11x – 8x + 88 = 0


⇒ x(x – 11) – 8(x – 11) = 0


⇒ (x – 8)(x – 11) =0


⇒ x = 8 and 11


More from this chapter

All 95 →
8

If DE has been drawn parallel to side BC of ΔABC cutting AB and AC at points D and E respectively, such that , then find the value of .

 9

In the adjoining figure, P and Q are points on sides AB and AC respectively of mix such that PQ || BC and AP= 8 cm, AB =12 cm, AQ = 3x cm, QC = (x + 2) cm. Find x.

 11

If D and E are points on sides AB and AC respectively of ΔABC and AB = 12 cm, AD = 8 cm, AE = 12 cm, AC =18 cm, then prove that DE || BC.

 12

P and Q are points on sides AB and AC respectively of Δ ABC. For each of the following cases, state whether PQ || BC.

(i) AP= 8 cm, PB = 3 cm, AC = 22 cm and AQ =16 cm.


(ii) AB= 1.28 cm, AC = 2.56 cm, AP= 0.16 cm and AQ = 0.32 cm


(iii) AB = 5 cm, AC =10 cm, AP= 4 cm, AQ = 8 cm.


(iv) AP= 4 cm, PB= 4.5 cm, AQ = 4 cm, QC = 5 cm.