Question

the centre of gravity of the ABC equilateral crib is G. If AG= 4 root 3 cm, AB=?​

Answer 1

Answer:

G is the centroid in the equilateral △ ABC. If AB = 10 cm then length of AG is :

[A]\frac{20\sqrt{3}}{3}cm

[B]10\sqrt{3}cm

[C]\frac{10\sqrt{3}}{3}cm

[D]5\sqrt{3}cm

\mathbf{\frac{10\sqrt{3}}{3}cm}

AB = 10cm

BD = 5 cm

∠ADB = 90°

∴ AD = \sqrt{AB^{2} - BD^{2}}

=> \sqrt{10^{2}-5^{2}} = \sqrt{100-25}

=> \sqrt{75} = 5\sqrt{3}cm

AG = \frac{2}{3} AD= \frac{2}{3}\times 5\sqrt{3}

= \frac{10\sqrt{3}}{3}cm

Hence option [C] is the right answer.

please mark me as brainliest answer..