Question

The length of triangle base is 36. The line which is parallel to the base divides the triangle into two equal area parts. Find a length of the segment between triangle legs.

Answer 1

Answer:

$18\sqrt{2}$

Step-by-step explanation:

Consider triangle ABC with the base AC=36. Let the line DE be parallel to the line AC. If DE||AC, then triangles ABC and DBE are similar. If k is the scale factor of these triangles, then

$\dfrac{DE}{AC}=\dfrac{BG}{BF}=k.$

Thus,

$DE=k\cdot AC=36k,\\ \\BG=k\cdot BF.$

The area of the triangle ABC is

$A_{ABC}=\dfrac{1}{2}AC\cdot BF=\dfrac{1}{2}\cdot 36\cdot BF=18BF.$

The area of the triangle DBE is

$A_{DBE}=\dfrac{1}{2}DE\cdot BG=\dfrac{1}{2}\cdot 36k\cdot kBF=18k^2BF.$

Since line DE divides triangle ABC in two equal area parts, we have that

$A_{DBE}=\dfrac{1}{2}A_{ABC},\\ \\18k^2BF=\dfrac{1}{2}\cdot 18BF,\\ \\k^2=\dfrac{1}{2},\\ \\k=\dfrac{1}{\sqrt{2}}$

and

$DE=\dfrac{1}{\sqrt{2}}\cdot 36=18\sqrt{2}.$