Question

The sides of a right triangle are $9,12$ and $15$ cm long. Find the sum of the squares of the medians.

• A. $327.5$ cm
• B. $332.5$ cm
• C. $337.5$ cm
• D. $322.5$ cm

Answer 1

Answer: (C)

The sides of a right triangle are $9,12$ and $15$ cm

$(M_1{)}^2=\frac{1}{4}\left[2\right(12{)}^2+2\times (15{)}^2-(9{)}^2]$

$=\frac{1}{4}[2\times 144+2\times 225-81]$

$=\frac{1}{4}[288+450-81]$

$=\frac{1}{4}\times 657=164.25$ cm

$(M_2{)}^2=\frac{1}{4}\left[2\right(9{)}^2+2\times (15{)}^2-(12{)}^2]$

$=\frac{1}{4}[2\times 81+2\times 225-144]$

$=\frac{1}{4}[162+450-144]$

$=\frac{1}{4}\times 468=117.00$cm

$(M_3{)}^2=\frac{1}{4}\left[2\right(12{)}^2+2\times (9{)}^2-(15{)}^2]$

$=\frac{1}{4}[2\times 144+2\times 81-225]$

$=\frac{1}{4}[288+162-225]$

$=\frac{1}{4}\times 225=56.25$

Then length of medians $=164.25+117.00+56.25=337.5$ cm