Let
a and b----------> the bases of an isosceles trapezoid
c------------------> the legs of an isosceles trapezoid
we know that
if one of the legs is 2/3 of the sum of the bases
so
c=(2/3)*(a+b)
see the attached figure to better understand the problem
[the area of an isosceles trapezoid]=(1/2)*[a+b]*h
A=45 cm²
h=5 cm
45=(1/2)*[a+b]*5--------> 90=[a+b]*5-------> 18=[a+b]------> equation 1
[the perimeter of an isosceles trapezoid]=[a+b]+2*[(2/3)*(a+b)]
[the perimeter of an isosceles trapezoid]=(a+b)+(4/3)*(a+b)---> (7/3)*(a+b)
remember equation 1
[a+b]=18
then
[perimeter]=(7/3)*18-----> 42 cm
the answer is
the perimeter is 42 cm
