Question

Right triangle ABC has legs AC and BC of lengths 16mi and 24mi accordingly. Find the distance of point B from the line that contains the median AM.

Answer 1

9.6 mi

Explanation:

in right triangle ABC,

AC = 16 mi

BC = 24 mi

If AM is the median, then CM = MB = 12 mi

Consider triangles CAM and EBM. In these triangles,

• angles CMA and EMB are congruent as vertical angles;
• angles ACM and MEB are congruent as right angles.

So, triangles CAM and EBM are similar by AA postulate.

Similar triangles have proportional corresponding sides, so

`(CA)/(EB)=(CM)/(EM)\\ \\(16)/(EB)=(12)/(EM)\\ \\EM=(3)/(4)EB`

By the Pythagorean theorem,

`MB^2=EB^2+EM^2\\ \\12^2=EB^2+\left((3)/(4)EB\right)^2\\ \\144=EB^2+(9)/(16)EB^2\\ \\EB^2\left(1+(9)/(16)\right)=144\\ \\EB^2\cdot (25)/(16)=144\\ \\EB^2=144\cdot (16)/(25)\\ \\EB=12\cdot (4)/(5)\\ \\EB=(48)/(5)\\ \\EB=9.6\ mi`