Question 641203
The formula for the area of a triangle is
{{{A=(1/2)bh}}}, where {{{b}}} represents the base length,
and {{{h}}} represents the height.
That first sentence suggests that this should be an easy problem, for students knowing just a little algebra and geometry.
If the perimeter is 28 inches and one side other than the base measures 10 inches,
the length of the other two sides would be
{{{b}}} and {{{28-10-b=18-b}}}
If we there was some information about the angles in that triangle,
it could be an easy problem, and some information about the angles may be in the drawing of the triangle.
 
Without any additional information, your best bet is using Heron's formula, based on the semiperimeter, s.
Heron's formula says that the area for a triangle with side lengths a, b, and c, and perimeter 2s is
{{{area=sqrt(s(s-a)(s-b)(s-c))}}}
However, that makes no use of {{{A=(1/2)bh}}}.
 
Maybe the triangle drawing looks sort of like this,
{{{drawing(300,300,-5.5,5.5,-10,1,
rectangle(0,-9.145,0.3,-8.845),
triangle(0,0,-4,-9.165,4,-9.145),
locate(-2.9,-5,10),locate(-0.2,-9.4,b),
line(-2.2,-4.382,-1.8,-4.782),
line(2.2,-4.382,1.8,-4.782),
green(line(0,-9.145,0,0)),locate(0.1,-4.2,h),
locate(-2.2,-7.8,b/2),locate(1.8,-7.8,b/2),
arrow(-0.3,-9.6,-4,-9.6),arrow(0.2,-9.6,4,-9.6)
)}}} , indicating somehow that is it an isosceles triangle 
(maybe with those short cross-lines showing that the slanted sides are congruent,
or with arcs indicating that the base angles are congruent,
or with a 10 next to each slanted side).
In that case, applying Pythagoras to one of the right triangles, we find
{{{h=sqrt(10^2-(b/2)^2)=sqrt(400-b^2)/2}}}
 
If he triangle drawing looks sort of like this,
{{{drawing(300,300,-1,10,-1,10,
triangle(0,0,0,9,9,0),
locate(5,5,10),locate(4.5,0,b),
rectangle(0,0,0.3,0.3)
)}}} , which is a right triangle, then there is a typo in the problem,
because a right triangle with a perimeter of 28 inches can NEVER have a hypotenuse measuring 10 inches.
That's too bad, because in a right triangle, the two perpendicular legs can be taken as base and height, and that would make it a really easy problem..