Question 1158998
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The given triangle is isosceles with congruent lateral sides of 15 inches long
and the base of 18 inches.


The altitude of this triangle drawn to the 18-inches side, divide this triangle in two right-angled triangles.


This altitude is one leg of these triangles, while the other legs are 18/2 = 9 inches long.


From the Pythagorean theorem, the length of the altitude is  {{{sqrt(15^2-9^2)}}} = 12 inches.


Then the area of the given triangle is  {{{(1/2)*18*12)}}} = 9*12 = 108 sq.inches.


Let the radius of the inscribed circle be r.


Now use the formula  

    area = {{{(1/2)*P*r}}}

for the triangle, where P is the perimeter P of the triangle   P = 15+15+12 = 42 inches.


The area of the triangle is 108 sq.inches.


Hence,  108 = {{{(1/2)*42*r}}},  which gives

    r = {{{(2*108)/42}}} = {{{(2*6*18)/42}}} = {{{(2*18)/7}}} = {{{36/7}}} inches.   <U>ANSWER</U>
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