Question

5. AB || CD || EF and AC = CE. Find x if
(i) BD = 3x + 2 and DF = 4x - 3
(ii) BD = 5x – 4 and BF = 4x + 16​

Answer 1

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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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Answer 2

this is very simple solution