The height of the trapezoid is [tex]h=11.2[/tex]
Explanation:
AKLM is a trapezoid.
The measurements of the trapezoid are AK=13, LM=14, KL=5, AM=20
We need to find the height of the trapezoid.
Let M' be a point on AM that is 5 units toward point A from M.
Let B be a point on AM such that KB⊥AM. Let x = AB; then BM' = 15 -x.
Using Pythagorean theorem, we have,
[tex]x^{2}+h^{2}=13^{2}[/tex] --------(1)
[tex](15-x)^{2}+h^{2}=14^{2}[/tex]
[tex]225-30x+x^{2} +h^{2}=14^{2}[/tex]-----------(2)
Subtracting the two equations, we have,
[tex]\left(225-30 x+x^{2}+h^{2}\right)-\left(x^{2}+h^{2}\right)=196-169[/tex]
Simplifying, we get,
[tex]225-30x=27[/tex]
Subtracting both sides of the equation by 225, we get,
[tex]30x=198[/tex]
Dividing by 30, we get,
[tex]x=6.6[/tex]
Substituting [tex]x=6.6[/tex] in the equation [tex]x^{2}+h^{2}=13^{2}[/tex], we get,
[tex](6.6)^{2}+h^{2}=169[/tex]
[tex]43.56+h^2=169[/tex]
[tex]h=\sqrt{125.44}[/tex]
[tex]h=11.2[/tex]
Thus, the height of the trapezoid is [tex]h=11.2[/tex]