Question

In the given figure, 𝐴𝐵, 𝐶𝐷 and 𝐸𝐹 are parallel lines. Given 𝐴𝐵 = 7.5 𝑐𝑚, 𝐷𝐶 = 𝑦 𝑐𝑚, 𝐸𝐹 = 4.5 𝑐𝑚, 𝐵𝐶 = 𝑥 𝑐𝑚 and 𝐶𝐸 = 3 𝑐𝑚. Calculate the value of 𝑥 and 𝑦.
(Given: ∆FCD ~ ∆FAB, ∆BCD ~ BEF)

Answer 1

.Given, 𝐴𝐵, 𝐶𝐷 and 𝐸𝐹 are parallel lines. Given 𝐴𝐵 = 7.5 𝑐𝑚, 𝐷𝐶 = 𝑦 𝑐𝑚, 𝐸𝐹 = 4.5 𝑐𝑚, 𝐵𝐶 = 𝑥 𝑐𝑚 and 𝐶𝐸 = 3 𝑐𝑚.

To Find: Value of x and y.

We know that a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio.

Now, consider ΔFAB.

CD || AB [Given]

By BPT, we get

⇒ DF/BD = CF/AC...................................................... (i)

Since, ΔFCD ~ ΔFAB,

DF/BF = CD/AB = y/7.5

BF = 7.5 DF/y ............................................................... (ii)

Consider ΔBEF.

CD || EF [Given]

By BPT, we get

⇒ BD/DF = BC/CE = x/3 ............................................ (iii)

Since, ΔBCD ~ ΔBEF,

BD/BF = BC/BE = CD/EF

BD/BF = x/(x+3) = y/4.5 .............................................. (iv)

BF = 4.5 BD/y ................................................................(v)

From (ii) and (v),

7.5 DF/y = 4.5 BD/y

BD/DF = 7.5/4.5 = 5/3 ................................................... (vi)

From (iii) and (vi),

x/3 = 5/3

x = 5 cm

From (iv), we get

x/(x+3) = y/4.5

Substituing the value of x , we get

5/(5+3) = y/4.5

y = (5×4.5)/8 = 2.8125 cm

Hence, x = 5 and y = 2.8125.