Answer:
Area [tex]=62.5\sqrt{6}[/tex] square units
[tex]AB=5\sqrt{15}[/tex] units
[tex]BC=5\sqrt{10}[/tex] units
Step-by-step explanation:
In a right triangle the altitude drawn to the hypotenuse is the geometric mean of the segments at which this altitude divides the hypotenuse.
So,
[tex]BD^2=15\cdot 10\\ \\BD^2=150\\ \\BD=\sqrt{150}=5\sqrt{6}\ units[/tex]
a. The area of the triangle ABC is
[tex]A_{ABC}=\dfrac{1}{2}\cdot BD\cdot AC=\dfrac{1}{2}\cdot 5\sqrt{6}\cdot (15+10)=\dfrac{125\sqrt{6}}{2}=62.5\sqrt{6}\ un^2.[/tex]
b. The legs of the right triangle are geometric means of the segment adjacent to this leg and the hypotenuse, so
[tex]AB^2=AD\cdot AC=15\cdot 25\Rightarrow AB=5\sqrt{15}\ units\\ \\BC^2=CD\cdot AC=10\cdot 25\Rightarrow BC=5\sqrt{10}\ units[/tex]
