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) BC

(ii) AF

• A. BC = 22 cm, AF = 25 cm
• B. BC = 20 cm, AF = 25 cm
• C. BC = 22.5 cm, AF = 27 cm
• D. BC = 22.5 cm, AF = 29.7 cm

Answer 1

The correct option is C

BC = 22.5 cm, AF = 27 cm

In the figure,

ABCD is a trapezium

AB||DC

AB=9cm, DC =18 cm, CF =13.5 cm

AP =6 cm and BE =15 cm

To find:

(i) BC

(ii) AF

In $\Delta AEB$ and $\Delta ECF$

$\angle AEB=\angle CEF$ (Vertically opposite angles)

$\angle ABE=\angle ECF$ (Alternate angles)

$\therefore \Delta AEB\sim \Delta ECF$ (AA axiom)

$\Rightarrow \frac{AB}{CF}=\frac{BE}{EC}\Rightarrow \frac{9}{13.5}=\frac{15}{EC}$

$\Rightarrow BC=\frac{15\times 13.5}{9}=22.5cm$

Similarly $\Delta APB$ and $\Delta DPF$

$\angle APB=\angle DPF$ (Vertically opposite angles)

$\angle ABP=\angle PDE$ (alternate angles)

$\therefore \Delta APB\sim \Delta DPE$ (AA axiom)

$\therefore \frac{AB}{DF}=\frac{AP}{PF}$

$\frac{9}{18+13.5}=\frac{6}{PF}\Rightarrow \frac{9}{31.5}=\frac{6}{PF}$

$PF=\frac{6\times 31.5}{9}=21cm$

$\therefore AF=AP+PF$

=6+21 =27 cm