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

Answer:

9.6 mi

Step-by-step 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

$\dfrac{CA}{EB}=\dfrac{CM}{EM}\\ \\\dfrac{16}{EB}=\dfrac{12}{EM}\\ \\EM=\dfrac{3}{4}EB$

By the Pythagorean theorem,

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