Question 982458
The area of a triangle is {{{(1/2)(base)(altitude)}}} .
The sides of an isosceles triangle come in two lengths:
the length of the two congruent legs, and
the length of the other side, which we usually consider the base.
However, you can take any side of a triangle as the base,
and the perpendicular to that base from the other vertex is the altitude,
so, in that case,
{{{20cm}}} is the altitude to a base measuring {{{B}}}{{{cm}}} , and
{{{30cm}}} is the altitude to a base measuring {{{b}}}{{{cm}}} .
Calculating the area both ways, we get the same result, so
{{{(1/2)*B*20=(1/2)*b*30}}}--->{{{2B=3b}}}--->{{{B=(3/2)b=1.5b}}} .
Could the legs of the isosceles triangle measure {{{B}}}{{{cm}}} ?
If so, the triangle looks like this
{{{drawing(300,420,-7.5,7.5,-1.7,19.3,
triangle(0,0,0,18.97,-6.325,0),
triangle(0,0,0,18.97,6.325,0),
rectangle(0,0,-0.5,0.5),locate(0.1,10,30),
locate(-3.16,9.46,B),locate(-3.6,0,"b / 2"),
arrow(-3.7,-0.4,-6.325,-0.4),arrow(-2.5,-0.4,0,-0.4),
red(arc(-6.325,0,5,5,-70.53,0)),
locate(-4.8,1.5,red(theta))
)}}} and {{{cos(red(theta))=(0.5b)/(1.5b)=0.5/1.5=1/3}}} ---> {{{red(theta)=70.53^o}}} (rounded).
Could the legs of the isosceles triangle measure {{{b}}}{{{cm}}} ?
{{{drawing(350,200,-3.5,3.5,-1,3,
triangle(0,0,0,2.646,-3,0),
triangle(0,0,0,2.646,3,0),
rectangle(0,0,-0.5,0.5),locate(0.1,1.5,20),
locate(-1.5,1.323,b),locate(-1.8,0,"B / 2"),
arrow(-1.8,-0.2,-3,-0.2),arrow(-1.2,-0.2,0,-0.2),
green(arc(-3,0,2.5,2.5,-41.41,0)),
locate(-2.2,0.7,green(alpha))
)}}} We would have an angle {{{alpha}}} such that
{{{cos(alpha)=(0.5B)/b=0.5*(1.5b)/b=0.5*1.5=0.75}}}--->{{{alpha=41.41^o}}} (rounded).