# Match the followi

A)

Given that DE||BC,
by B.P.T.,

Let AE = x

Then, from the figure, EC = 5.6-x

5x = 3(5.6-x)

5x = 16.8-3x

8x = 16.8

x = 2.1cm

Therefore, (A)-(s)

B)
As Δ ABC
ΔDEF,

3EF = 2BC

3EF = 2 × 6

EF = 4cm

Therefore,(B)-(q)

C)

QR =
QR =
= 6cm

Therefore (C)-(p)

D)

(BPT)

3(2x + 4) = (2x-1)(9x-21)

6x + 12 = 18x2-42x-9x + 21

18x2-57x + 9 = 0

18x2-54x-3x + 9 = 0

18x(x-3)-3(x-3) = 0

(18x-3)(x-3) = 0

So, x = 3 or x =

But for x = , 2x-1<0 which is not possible.
Therefore, (D)-(r)

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