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:

Given:

FPST is a trapezoid,

FP=ST, m∠F=45°,

PF=8, PS=10

To Find:

The Length of the mid-segment MN.

Solution:

This trapezoid FPST is isosceles, because FP=ST

Also

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

Draw the height PH. Triangle FPH is right triangle with two angles of measure 45°. This means that

FH = HP --------------(1)

By the Pythagorean theorem,

$FH^2+PH^2=PF^2,$

From (1)

$\\ \\2FH^2=8^2$  

$\\ \\2FH^2=64$

$FH^2=\frac{64}{2}$

$\\FH^2=32$

$\\FH^2= 2\times 2\times 2 \times2 \times 2$

$\\FH = \sqrt{2\times 2\times 2 \times2 \times 2}$

$\\FH=4\sqrt{2}.$

Since trapezoid FPST is isosceles, the base FT has the length

$FT=2FH+PS$

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

Then the length of the mid-segment is

$MN=\dfrac{FT+PS}{2}$

$MN=\dfrac{8\sqrt{2}+10+10}{2}$

$MN=\dfrac{8\sqrt{2}+20}{2}$

$MN=4\sqrt{2}+10.$