Question 204722
For the first isosceles triangle:
{{{a+b+c = 23}}} where b is the base and, as in any isosceles triangle, a = c, so we can write:
{{{2a+b = 23}}} or {{{highlight(b = 23-2a)}}}
For the second isosceles triangle:
{{{2a+b+2c = 41}}} where b is the base and, as in any isosceles triangle, 2a = 2c, so we can write:
{{{4a+b = 41}}} or {{{highlight(b = 41-2a)}}}
Now the bases, b, are equal, so we can set these two equations equal:
{{{23-2a = 41-4a}}} Simplify and solve for a. Add 4a to both sides.
{{{23+2a = 41}}} Subtract 23 from both sides.
{{{2a = 18}}} Divide both sides by 2.
{{{a = 9}}} and {{{c = 9}}}
{{{b = 23-2a}}}
{{{b = 23-18}}}
{{{b = 5}}}
First triangle:
a = 9, b = 5, and c = 9
Second triangle:
2a = 18, b = 5, and 2c = 18