Question

Please Solve this, it would be extremely helpful for me.

Answer 1

Answer:

Step-by-step explanation:

1) ΔCPD & ΔEPF

∠CPD = ∠EPF   { Vertically opposite angles}

∠CDP = ∠PFE {CD║EF, FD is transversal, Alternate interior angles are equal}

ΔCPD ≈ΔEPF  {AA criteria for similarity }

$\frac{DC}{EF} =\frac{PC}{EP}\\\\\\\frac{27}{EF}=\frac{15}{7.5}$

Cross multiply

EF * 15 = 27 * 7.5

$EF =\frac{27*7.5}{15}$

EF = 27 * 0.5

EF = 13.5 cm

ii) EF // AB, so Triangles ACB & ECF are similar triangles

$\frac{AB}{EF}=\frac{AC}{EC}\\\\\frac{22.5}{13.5}=\frac{AC}{22.5}$

$AC= \frac{22.5*22.5}{13.5}\\\\AC=37.5 cm$

AC = 37.5 cm