Question

The given figure shows an equilateral triangle $ABC$ with each side $15$ cm. Also $DE\parallel BC$, $DF\parallel AC$, and $EG\parallel AB$. If $DE+DF+EG=20$ cm, then the length of $FG$ is

• A. $5$ cm
• B. $10$ cm
• C. $2.5$ cm
• D. $7.5$ cm

Answer 1

The correct option is A $5$ cm

Given, $△ABC$ is an equilateral triangle.$DE\parallel BC$, $DF\parallel AC$
$EG\parallel AB$
In $△ADE$ and $△ABC$
$\angle ADE=\angle ABC=60$ ...(Corresponding angles)
$\angle AED=\angle ACB=60$ ...(Corresponding angles)
Thus, $△ADE$ is an equilateral triangle.
Similarly, $△DBF$ and $△EGC$ are equilateral triangle.
In $△DBF$ and $△EGC$
$\angle DBF=\angle ECG$
$\angle BDF=\angle GEC$
$BD=EC$ ...(since, $AB=AC$ and $AD=AE$, Thus, $AB-AD=AC-AE$)
Thus, $△DBF\cong △EGC$ ...(ASA rule)
or, $BF=GC$ ...(CPCT)
Therefore, $DE+DF=AD+DB=AB=15$
and, $DE+DF+EG=20$
Thus, $EG=5$ cm
Hence, $EG=GC=BF=5$
or $BC=15$
$BF+FG+GC=15$
$FG=5$ cm