Question

In the given figure AB || CD || EF. If AB = 6 cm, CD = x cm, EF = 4 cm, BD = 5 cm and DE = y cm. Find x and y.

Answer 1

In the given diagram ∆AEF and ∆ACD

∠AEF = ∠ACD = 90°

∠A is common

By AA – Similarity.

∴ ∆AEF ~ ∆ACD

`"AE"/"AC" = "AF"/"AD" = "EF"/"CD"`

`"AE"/"AC" = "EF"/"CD"`

`"AE"/"AC" = 4/x`

AC = `("AE" xx x)/4`  ...(1)

In ∆EAB and ∆ECD,

∠EAB = ∠ECD = 90°

∠E is common

∆ECD ~ ∆EAB

`"EC"/"EA" = "ED"/"EB" = "CD"/"AB"`

`"EC"/"EA" = x/6`

EC = `("EA" xx x)/6`  ...(2)

In ∆AEB, CD || AB

By Basic Proportionality Theorem

`"AB"/"CD" = "EB"/"ED"`

`6/x = (5 + y)/y`

x = `(6y)/(y + 5)`  ...(EC = x)  ...(3)

Add (1) and (2) we get

AC + EC = `("AE" xx x)/4 + (x xx "EA")/6`

AE = `"AE"(x/4 + x/6)`

AE = `"AE"((3x + 2x)/12)`

AE = `"AE" xx ((5x)/12)`

∴ 1 = `(5x)/(12)`

⇒ 5x = 12

x = `(12)/(5)`

= 2.4 cm

Substitute the value of x = 2.4 in (3)

2.4 = `(6y)/(y + 5)`

6y = 2.4y + 12

6y – 2.4y = 12

⇒ 3.6 y = 12

y = `12/3.6 = 120/36 = 10/3` = 3.3 cm

The value of x = `12/5` or 2.4 cm and y = `10/3` or 3.3 cm