Question

The distances of the point of concurrency of the medians of a triangle from its vertices are respectively 1.2cm, 1.4cm, 1.6cm. Find the lengths of its medians.

Answer 1

Lengths of medians

AR = 3.6 cm

BQ = 4.2 cm

CP = 4.8 cm

In triangle centeroid (i.e. point of concurrency) divides median in 2:1

For AR

AO:OR = 2:1

AO/1.2 = 2/1

AO = 2.4cm

AR = AO+OR

AR = 3.6 cm __________(1)

For BQ

BO:OQ = 2:1

BO/1.4 = 2/1

BO = 2.8cm

BQ = AO+OR

BQ = 4.2 cm ___________(2)

For CP

CO:OP = 2:1

CO/1.6 = 2/1

CO = 3.2cm

CP = CO+OP

CP = 4.8 cm ___________(3)

Answer 2

The length of medians AR = 3.6 cm  ,

The length of medians  BQ = 4.2 cm ,

The length of medians  CP = 4.8 cm

Step-by-step explanation:

Given as :

For Triangle  A B C ,

The distances of the point of concurrency of the medians of a triangle from its vertices :

Let The center of the Triangle = O

The distance OR = 1.2 cm

The distance OQ = 1.4 cm

The distance OP  = 1.6 cm

Let The length of medians = AR  , BQ , CP

According to question

In triangle centroid divides median in 2:1

So, For median AR

AO :  OR = 2 : 1

Or, $\dfrac{AO}{OR}$  = $\dfrac{2}{1}$

∴    AO = $\dfrac{2}{1}$ × 1.2 cm

So, The measure of AO = 2.4

The length of AR = AO + OR = 2.4 cm + 1.2 cm = 3.6 cm

Similarly

So, For median BQ

BO :  OQ = 2 : 1

Or, $\dfrac{BO}{OQ}$  = $\dfrac{2}{1}$

∴    BO = $\dfrac{2}{1}$ × 1.4 cm

So, The measure of BO = 2.8 cm

The length of BQ = BO + OQ = 2.8 cm + 1.4 cm = 4.2 cm

Again

So, For median CP

CO :  OP = 2 : 1

Or, $\dfrac{CO}{OP}$  = $\dfrac{2}{1}$

∴    CO = $\dfrac{2}{1}$ × 1.6 cm

So, The measure of CO = 3.2 cm

The length of CP = CO + OP = 3.2 cm + 1.6 cm = 4.8 cm

Hence, The length of medians AR = 3.6 cm  , BQ = 4.2 cm , CP = 4.8 cm  Answer