Question 777654
There are two many triangles, with different measurements that comply with that.
Here are 3 examples:
{{{drawing(300,300,-11,14,-1,24,
green(line(-5,19,3.8,19)),green(rectangle(-5,19,-4.5,19.5)),
green(line(13,14,2.1,14)),green(rectangle(13,14,12.3,14.7)),
green(line(0,0,0,13.7)),green(rectangle(0,0,0.7,0.7)),
triangle(-10,0,10,0,0,13.7),
triangle(2.1,14,13,9,13,19),
triangle(3.8,19,-5,17,-5,21),
locate(4,1.2,10),locate(4,6.5,17),
locate(-4.9,20.5,2),locate(-1,21.2,9),
locate(7.8,17,12),locate(12.3,17,5)
)}}}
Either you are missing some information, or you are expected to say that there are infinte solutions. If you had some other information, like the height, or one angle, you could start by
{{{x}}}= half the base
{{{x+7}}}= length of the congruent sides,
and then use that with any extra information
 
IF it were a {{{highlight(right)}}} isosceles triangle, {{{drawing(200,200,-1,11,-1,11,
triangle(0,0,10,0,0,10),
rectangle(0,0,0.5,0.5),
locate(4,0,x+7),locate(0.1,5,x+7),locate(5,6,2x)
)}}}
then the base is the hypotenuse, with length {{{2x}}} and the congruent sides are the legs.
Then, Pythagoras says
{{{(x+7)^2+(x+7)^2=(2x)^2}}}
{{{2(x+7)^2=4x^2}}}
{{{2(x^2+14x+49)=4x^2}}}
{{{2x^2+28x+96=4x^2}}}
{{{0=2x^2-28x-96}}}
Flipping, and dividing both sides by 2
{{{0=2x^2-28x-96}}} --> {{{2x^2-28x-96=0}}} --> {{{x^2-14x-49=0}}}
The solutions to that equation are
{{{x=7 +- 7sqrt(2)}}}
The solution with the negative sign is a negative number, which could not be the length of half the base, 
so {{{x=7 + 7sqrt(2)}}} and {{{x+7= 14+7sqrt(2)=about23.9}}}{{{(rounding)}}}