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In figure 1, G\!C \parallel B\!D and  G\!E \parallel BF. If AC = 3 \; cm and C\!D = 7 \; cm, then find the value of  \frac{AE}{AF} .

 

 

 

 
 
 
 
 

Answers (1)

Given that GC \parallel BD and GE \parallel BF. Therefore by basic proportionality theorem which states that if a line is drawn parallel to one side of a triangle the other two sides in distinct points, then the other two sides are divided in same ratio.

\frac{AC}{AD}=\frac{AG}{BF} \;\;\;\;\;\; -(1)

Similarly GE \parallel BF

\frac{AE}{AF}=\frac{AG}{BF} \;\;\;\;\;\; -(2)

From eqn (1) and (2),

\frac{AC}{AD}=\frac{AE}{AF}

\frac{3}{7} = \frac{AE}{AF}

Hence the value of  \frac{AE}{AF} = \frac{3}{7}.

 

Posted by

Ravindra Pindel

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