Question

In the following figure, ABCD to a trapezium with AB // DC. If AB = 9 cm, DC = 18 cm, CF= 13.5 cm, AP = 6 cm and BE= 15 cm, Calculate : (i) EC (ii) AF (iii) PE

Answer 1

In $\Delta$ APD and $\Delta$ FPD
$\angle$ APB = $\angle$ FPD
$\angle$ BAP = $\angle$ DFP
$\Delta$ APB $\sim$ $\Delta$FPD
(AA creation)
$\frac{AP}{FB}$ = $\frac{AB}{FD}$

$\frac{6}{FB}$ = $\frac{9}{31.5}$

FP = 21 cm
So AF = AP + PF = 6 + 21 = 27 cm

In $\Delta$ AEB and $\Delta$ FEC
$\angle$ AEB = $\angle$FEC
$\angle$ BAE = $\angle$CFE
So
$\Delta$ AEB and ​​​​​​​$\Delta$ FEC are similar
(AA creation)
$\frac{AE}{FE}$ = $\frac{BE}{CE}$​​​​​​​ = $\frac{AB}{FC}$​​​​​​​
$\frac{AE}{FE}$​​​​​​​ = $\frac{9}{13.5}$​​​​​​​

$\frac{AE-EF}{FE}$​​​​​​​ = $\frac{9}{13.5}$​​​​​​​​​​​​​​
$\frac{AE-EF}{FE}$​​​​​​​ - 1 = $\frac{9}{13.5}$​​​​​​​​​​​​​​​​​​​​​

$\frac{27}{EF}$​​​​​​​​​​​​​​​​​​​​​ = $\frac{22.5}{13.5}$​​​​​​​​​​​​​​​​​​​​​

EF = 16.2 cm

Now PE = PF - EF = 21 - 16.2 = 4.8 cm