Problem (21) : Tinggi Segitiga


soal dikirim via email by inung.emka@yahoo.com

 photo segitigat_zps850e149c.png
Nilai t pada gambar dibawah adalah …

Penyelesaian :

Disini kita akan memanfaatkan sifat segitiga sebangun,

Pandang \triangle ABD :

\frac{AD}{AB} = \frac{EF}{FB}

EF = \frac{AD}{AB} FB … (i)

Pandang \triangle ABC :

\frac{AB}{BC} = \frac{AF}{FE}

FE = \frac{BC}{AB} AF … (ii)

Dari pers (i) dan (ii), diperoleh :

(i) = (ii)

\frac{AD}{AB} FB = \frac{BC}{AB} AF

\frac{6}{AB} FB = \frac{12}{AB} AF

FB = 2 AF … (iii)

Substitusi pers (iii) ke pers (i), diperoleh :

EF = \frac{AD}{AB} FB

= \frac{AD}{AF+FB} FB

= \frac{6}{\frac{1}{2}FB+FB} FB

= \frac{6}{\frac{3}{2}FB} FB

= \frac{6.2}{3}

= 4

Jadi panjang t = EF = 4cm

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