Question

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 1

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,

$x^2+x^2=169\\2x^2=169\\x=\frac{13}{\sqrt{2} }$

Then the area of isosceles right triangle will be: $A=\frac{1}{2} x^2$

Therefore the area is: $A=\frac{169}{4}$

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.

$S=\frac{a}{1-r}$

Substitute $a=\frac{169}{4} \ and\ r=\frac{1}{4}$ in above formula.

$S=\frac{\frac{169}{4}}{1-\frac{1}{4}}$

$S=\frac{\frac{169}{4}}{\frac{3}{4}}$

$S=\frac{169}{3}$

Hence, the Sum of the areas of theses triangles is 169/3.