C-14-Q054mediumsingle_mcq

In $\triangle ABC$ , medians $AD$ , $BE$ meet at $G$ . A line through $G$ parallel to $DE$ meets $AC$ at $F$ . Then $AC = ?$

a $3 EF$
b $4 EF$
c $6 EF$
d $2 EF$

ব্যাখ্যা

$DE$ is the midsegment joining midpoints, so $DE = \tfrac12 AC$ . The centroid $G$ divides median $AD$ in $2:1$ , and the line through $G$ parallel to $DE$ gives $EF = \tfrac16 AC$ , so $AC = 6\,EF$ .

এই অধ্যায়ের আরও প্রশ্ন

C-14-Q055In , on , on . Then
C-14-Q056Congruence is a special case of —
C-14-Q057Two congruent figures are —
C-14-Q058Two similar figures are —
C-14-Q059Polygons with equal numbers of sides whose angles are sequentially equal are —
C-14-Q060Two polygons are similar when matching angles are equal AND matching sides are —
C-14-Q061A rectangle and a square are equiangular but generally —
C-14-Q062Two equiangular triangles are always —
C-14-Q063Two similar triangles are always —
C-14-Q064If two equiangular triangles have one matching pair of sides equal, they are —

In $\triangle ABC$ , medians $AD$ , $BE$ meet at $G$ . A line through $G$ parallel to $DE$ meets $AC$ at $F$ . Then $AC = ?$

3 EF

4 EF

6 EF

2 EF

In △ABC\triangle ABC△ABC, medians ADADAD, BEBEBE meet at GGG. A line through GGG parallel to DEDEDE meets ACACAC at FFF. Then AC=?AC = ?AC=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In △ABC\triangle ABC△ABC, medians ADADAD, BEBEBE meet at GGG. A line through GGG parallel to DEDEDE meets ACACAC at FFF. Then AC=?AC = ?AC=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In $\triangle ABC$ , medians $AD$ , $BE$ meet at $G$ . A line through $G$ parallel to $DE$ meets $AC$ at $F$ . Then $AC = ?$

In $\triangle ABC$ , medians $AD$ , $BE$ meet at $G$ . A line through $G$ parallel to $DE$ meets $AC$ at $F$ . Then $AC = ?$