Question 122278
Triangle is ABC.
AB side is {{{sqrt(11)}}}.
AC side is {{{sqrt(11)}}}.
BC side is {{{3}}}. 
Split BC in half at point Z to form two equal line segments BZ=ZC=3/2. 
You can now form two right triangles, ABZ and AZC.
Call the length of AZ side, H. 
Using the Pythagorean theorem, you can find the height, H, because,
{{{H^2+(3/2)^2=(sqrt(11))^2}}}
{{{H^2+(9/4)=11}}}
{{{H^2=11-(9/4)}}}
{{{H^2=44/4-9/4}}}
{{{H^2=36/4}}}
{{{H^2=9}}}
{{{H=3}}}
For the area of the original isoceles triangle,
{{{A=(1/2)bH}}}
{{{A=(1/2)(3/2)(3)}}}
{{{A=9/4}}}
A is 2.25 square units. 
Does that answer match the answer key?