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


{\tt{\fbox{\red{Trigonometry}}}}

In the figure given below,

AB ll EF ll CD. If AB = 22.5 cm,

EP = 7.5 cm, PC =15 cm and

DC = 27 cm. Calculate:

(i) EF (ii) AC



Please Solve this, it would be extremely helpful for me. {\tt{\fbox{\red{Trigonometry-example-1

1 Answer

5 votes

Answer:

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 }


(DC)/(EF) =(PC)/(EP)\\\\\\(27)/(EF)=(15)/(7.5)\\\\

Cross multiply

EF * 15 = 27 * 7.5


EF =(27*7.5)/(15)\\\\

EF = 27 * 0.5

EF = 13.5 cm

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


(AB)/(EF)=(AC)/(EC)\\\\(22.5)/(13.5)=(AC)/(22.5)


AC= (22.5*22.5)/(13.5)\\\\AC=37.5 cm

AC = 37.5 cm

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