in fig 2.5 points A, B, C, D, and E are collinear that AB=BC=CD=DE. then.. (B) AD=AC+_
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SOLUTION :
Given : AB || CD || EF , AB = 6 cm, CD = x cm, BD = 4 cm, and DE = y cm and EF = 10 cm.
In ∆ECD and ∆EAB
∠CED = ∠AEB [common]
∠ECD = ∠EAB [corresponding angles]
∆ECD ~ ∆EAB ……....(1) [By AA similarity]
∴ EC/EA = CD/AB
[Corresponding parts of similar triangles are proportional]
EC / EA = x/6 ……………(2)
In ∆ACD and ∆AEF
∠CAD = ∠EAF [common]
∠ACD = ∠AEF [corresponding angles]
∆ACD ~ ∆AEF [By AA similarity]
∴ AC/AE = CD/EF
[Corresponding parts of similar ∆ are proportional]
AC /AE = x/10 ……………..(3)
On Adding eq 2 & 3
EC/EA+ AC /AE = x/6 + x/10
(EC + AC) /AE =( 5x + 3x)/30
AE / AE = 8x /30
1 = 8x/30
x = 30/8
x = 3.75 cm
From eq (i),
∆ECD ~ ∆EAB
DC/AB = ED /EB
[Corresponding parts of similar ∆ are proportional]
3.75/6 = y/y+4
6y = 3.75(y+4)
6y = 3.75y + 15
6y - 3.75y = 15
2.25y = 15
y = 15/2.25
y = 6.67 cm
Hence,, x = 3.75 cm and y = 6.67 cm.
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