In the given fig., find the value of x in terms of a, b and c .
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Two Triangles are said to be similar if their i) corresponding angles are equal and
ii) corresponding sides are proportional.(the ratio between the lengths of corresponding sides are equal)
•Similarity of triangles should be expressed symbolically using correct correspondence of their vertices
AA similarity: if two angles of one triangle are respectively equal to two angles of another triangle then the two Triangles are similar.
SOLUTION :
GIVEN:
∠KNP =∠ KML = 50°
LM = a , MN = b , NK = c
In ∆KPN and ∆KLM,
∠ KNP = ∠KML = 50° [ given]
∠ K = ∠ K [Common]
∆KPN ~ ∆KLM [by AA similarity criterion of triangles]
KN/ KM = NP/ML [corresponding sides of similar triangles are proportional]
c /(b+c) = x/a [KM = MN + NK]
x(b+c) = c×a
x = ac/ (b+c)
Hence, the value of x = ac/(b+c)
HOPE THIS WILL HELP YOU...
ii) corresponding sides are proportional.(the ratio between the lengths of corresponding sides are equal)
•Similarity of triangles should be expressed symbolically using correct correspondence of their vertices
AA similarity: if two angles of one triangle are respectively equal to two angles of another triangle then the two Triangles are similar.
SOLUTION :
GIVEN:
∠KNP =∠ KML = 50°
LM = a , MN = b , NK = c
In ∆KPN and ∆KLM,
∠ KNP = ∠KML = 50° [ given]
∠ K = ∠ K [Common]
∆KPN ~ ∆KLM [by AA similarity criterion of triangles]
KN/ KM = NP/ML [corresponding sides of similar triangles are proportional]
c /(b+c) = x/a [KM = MN + NK]
x(b+c) = c×a
x = ac/ (b+c)
Hence, the value of x = ac/(b+c)
HOPE THIS WILL HELP YOU...
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