Question

Given: FPST is a trapezoid, FP=ST, m∠F=45°, PF=8, PS=10 Find: The midsegment MN.

Answer 1

Answer:

$4\sqrt{2}+10$

Step-by-step explanation:

In trapezoid FPST, FP=ST, then this trapezoid is isosceles, so

$m\angle F=m\angle T=45^{\circ}.$

Draw the height PH. Triangle FPH is right triangle with two angles of measure 45°. This means that FH=HP. By the Pythagorean theorem,

$FH^2+PH^2=PF^2,\\ \\2FH^2=8^2,\\ \\FH^2=32,\\ \\FH=4\sqrt{2}.$

Since trapezoid FPST is isosceles, the base FT hasthe length

$FT=2FH+PS=8\sqrt{2}+10.$

Then the length of the midline is

$MN=\dfrac{FT+PS}{2}=\dfrac{8\sqrt{2}+10+10}{2}=4\sqrt{2}+10.$