The hypotenuese of an isosceles right triangle is 13 inches. The midpoints of its sides are connected to form an inscribed triangle, and this process is repeated, creating a third triangle inscribed in the previous one. Find the sum of the areas of these triangles if this process is continued infinitely.

Answer:The Sum of the areas of theses triangles is 169/3.Step-by-step explanation:Consider the provided information.The hypotenuse of an isosceles right triangle is 13 inches.Therefore, [tex]x^2+x^2=169\\2x^2=169\\x=\frac{13}{\sqrt{2} }[/tex]Then the area of isosceles right triangle will be: [tex]A=\frac{1}{2} x^2[/tex]Therefore the area is: [tex]A=\frac{169}{4}[/tex]It is given that sum of the area of these triangles if this process is continued infinitely.We can find the sum of the area using infinite geometric series formula.[tex]S=\frac{a}{1-r}[/tex]Substitute [tex]a=\frac{169}{4} \ and\ r=\frac{1}{4}[/tex] in above formula.[tex]S=\frac{\frac{169}{4}}{1-\frac{1}{4}}[/tex][tex]S=\frac{\frac{169}{4}}{\frac{3}{4}}[/tex][tex]S=\frac{169}{3}[/tex]Hence, the Sum of the areas of theses triangles is 169/3.