Question

Consider $\Delta ABC$ the medians AD & CF intersect at right angles at G. If $BC=3cm$ and $AB=4cm$, then the length of $AC$ is (in cm):

• A. $\sqrt{12}$
• B. $3.5$
• C. $\sqrt{5}$
• D. $\sqrt{7}$

Answer 1

The correct option is C $\sqrt{5}$

From the figure,

$△AGC,△AGF,△DGC$ are right angled at $G$

Using Pythagoras theorem,

$A{C}^2=G{C}^2+A{G}^2$

$x^2=4n^2+4m^2$ -----------------(i)

$A{F}^2=F{G}^2+A{G}^2$

$4=n^2+4m^2$ -----------------(ii)

$D{C}^2=G{C}^2+D{G}^2$

$2.25=4n^2+m^2$ -----------------(iii)

Solving (ii) and (iii)

$n^2=\frac{1}{3},m^2=\frac{11}{12}$

Substitute in (i)

$x^2=4n^2+4m^2=\frac{4\left(1\right)}{3}+\frac{4\left(11\right)}{12}=\frac{4}{3}+\frac{11}{3}=\frac{15}{3}$

$x^2=5$

$\therefore x=AC=\sqrt{5}$