Advertisements
Advertisements
Question
D and E are points on the sides AB and AC respectively of a ΔABC such that DE║BC.
If AB = 13.3cm, AC = 11.9cm and EC = 5.1cm, find AD.
Solution
In Δ ABC, it is given that DE || BC.
Applying Thales’ Theorem, we get :
`(AD)/(DB) = (AE)/(EC)`
Adding 1 to both sides, we get :
`(AD)/(DB) +1= (AE)/(EC) + 1`
⟹ `(AB)/(DB) = (AC)/(EC)`
⟹ `13.3/(DB) = (11.9)/(5.1)`
⟹ `DB = (13.3 × 5.1)/11.9 =5.7 cm`
Therefore, AD=AB-DB=13.5-5.7=7.6 cm
APPEARS IN
RELATED QUESTIONS
In the given figure, PS is the bisector of ∠QPR of ΔPQR. Prove that `(QS)/(SR) = (PQ)/(PR)`
In ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC
If AD = 8cm, AB = 12 cm and AE = 12 cm, find CE.
In a ΔABC, D and E are points on the sides AB and AC respectively. For the following case show that DE || BC
AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and EC = 5.5 cm.
M and N are points on the sides PQ and PR respectively of a ΔPQR. For the following case, state whether MN || QR
PM = 4cm, QM = 4.5 cm, PN = 4 cm and NR = 4.5 cm
D and E are the points on the sides AB and AC respectively of a ΔABC such that: AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm. Prove that BC = 5/2 DE.
D and E are points on the sides AB and AC respectively of a ΔABC such that DE║BC.
If AD = 3.6cm, AB = 10cm and AE = 4.5cm, find EC and AC.
In the adjoining figure, ABC is a triangle in which AB = AC. IF D and E are points on AB and AC respectively such that AD = AE, show that the points B, C, E and D are concyclic.
In ΔABC, D is the midpoint of BC and AE⊥BC. If AC>AB, show that `AB^2= AD^2+1/4 BC^2 −BC.DE `
In the given figure, ∠ACB 90° CD ⊥ AB Prove that `(BC^2)/(AC^2)=(BD)/(AD)`
Prove that If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. In the figure, find EC if `(AD)/(DB) = (AE)/(EC)` using the above theorem.
