Advertisements
Advertisements
Question
D and E are points on the sides AB and AC respectively of a ΔABC such that DE║BC.
If `(AD)/(DB) = 4/7` and AC = 6.6cm, find AE.
Solution
In Δ ABC, it is given that DE || BC.
Applying Thales’ theorem, we get :
`(AD)/(DB) = (AE)/(EC)`
⟹ `4/7 =(AE)/(EC)`
Adding 1 to both the sides, we get :
`11/7 = (AC)/(EC)`
⟹ EC = `(6.6 ×7)/11 = 4.2 cm`
Therefore
AE = AC – EC = 6.6 – 4.2 = 2.4 cm
APPEARS IN
RELATED QUESTIONS
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.
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 the given figure, O is a point inside a ΔPQR such that ∠PQR such that ∠POR = 90°, OP = 6cm and OR = 8cm. If PQ = 24cm and QR = 26cm, prove that ΔPQR is right-angled.
State and converse of Thale’s theorem.
In the given figure, in ∆ABC, point D on side BC is such that, ∠BAC = ∠ADC. Prove that, CA2 = CB × CD
A line is parallel to one side of triangle which intersects remaining two sides in two distinct points then that line divides sides in same proportion.
Given: In ΔABC line l || side BC and line l intersect side AB in P and side AC in Q.
To prove: `"AP"/"PB" = "AQ"/"QC"`
Construction: Draw CP and BQ
Proof: ΔAPQ and ΔPQB have equal height.
`("A"(Δ"APQ"))/("A"(Δ"PQB")) = (["______"])/"PB"` .....(i)[areas in proportion of base]
`("A"(Δ"APQ"))/("A"(Δ"PQC")) = (["______"])/"QC"` .......(ii)[areas in proportion of base]
ΔPQC and ΔPQB have [______] is common base.
Seg PQ || Seg BC, hence height of ΔAPQ and ΔPQB.
A(ΔPQC) = A(Δ______) ......(iii)
`("A"(Δ"APQ"))/("A"(Δ"PQB")) = ("A"(Δ "______"))/("A"(Δ "______"))` ......[(i), (ii), and (iii)]
`"AP"/"PB" = "AQ"/"QC"` .......[(i) and (ii)]
ΔABC ~ ΔDEF. If AB = 4 cm, BC = 3.5 cm, CA = 2.5 cm and DF = 7.5 cm, then the perimeter of ΔDEF is ______.
In figure, line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and ∠AEF = ∠AFE. Prove that `(BD)/(CD) = (BF)/(CE)`.
In the given figure, ABC is a triangle in which DE||BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, then find the value of x.
In the given figure ∠CEF = ∠CFE. F is the midpoint of DC. Prove that `(AB)/(BD) = (AE)/(FD)`
