An isosceles trapezoid MNQP has parallel sides MN = 120 m and PQ = 200 m, and non-parallel sides MP = NQ = 50 m. Calculate the lengths of the segments on the base and the height of the trapezoid.

The given figure is an isosceles trapezoid $MNQP$ with parallel sides $MN$ and $PQ$.
Given lengths:
$MN = 120 \text{ m}$
$PQ = 200 \text{ m}$
$MP = 50 \text{ m}$
$NQ = 50 \text{ m}$

Draw perpendiculars from $M$ and $N$ to the base $PQ$, meeting $PQ$ at points $R$ and $S$ respectively. This forms a rectangle $MRSN$ and two right-angled triangles $\triangle MPR$ and $\triangle NQS$.
Since $MRSN$ is a rectangle, $RS = MN = 120 \text{ m}$.
In an isosceles trapezoid, the segments $PR$ and $SQ$ are equal.
The total length of the base $PQ$ is the sum of these segments:

$$PQ = PR + RS + SQ$$ $$200 \text{ m} = PR + 120 \text{ m} + SQ$$

Let $PR = SQ = x$.

$$200 \text{ m} = x + 120 \text{ m} + x$$ $$200 \text{ m} = 2x + 120 \text{ m}$$

Subtract $120 \text{ m}$ from both sides:

$$2x = 200 \text{ m} - 120 \text{ m}$$ $$2x = 80 \text{ m}$$

Divide by 2:

$$x = \frac{80 \text{ m}}{2}$$ $$x = 40 \text{ m}$$

So, $PR = 40 \text{ m}$ and $SQ = 40 \text{ m}$.

Consider the right-angled triangle $\triangle MPR$.
The hypotenuse is $MP = 50 \text{ m}$.
One leg is $PR = 40 \text{ m}$.
The other leg is $MR$, which is the height $h$.
Using the Pythagorean theorem ($a^2 + b^2 = c^2$):

$$MP^2 = MR^2 + PR^2$$ $$(50 \text{ m})^2 = h^2 + (40 \text{ m})^2$$ $$2500 \text{ m}^2 = h^2 + 1600 \text{ m}^2$$

Subtract $1600 \text{ m}^2$ from both sides:

$$h^2 = 2500 \text{ m}^2 - 1600 \text{ m}^2$$ $$h^2 = 900 \text{ m}^2$$

Take the square root of both sides:

$$h = \sqrt{900 \text{ m}^2}$$ $$h = 30 \text{ m}$$

The height of the trapezoid is $\boxed{\text{30 m}}$.

The formula for the area of a trapezoid is $A = \frac{1}{2}(a+b)h$, where $a$ and $b$ are the lengths of the parallel sides and $h$ is the height.
Here, $a = MN = 120 \text{ m}$, $b = PQ = 200 \text{ m}$, and $h = 30 \text{ m}$.

$$A = \frac{1}{2}(MN + PQ)h$$ $$A = \frac{1}{2}(120 \text{ m} + 200 \text{ m})(30 \text{ m})$$ $$A = \frac{1}{2}(320 \text{ m})(30 \text{ m})$$ $$A = (160 \text{ m})(30 \text{ m})$$ $$A = 4800 \text{ m}^2$$

The area of the trapezoid is $\boxed{\text{4800 m}^2}$.

The perimeter is the sum of the lengths of all its sides.

$$P = MN + NQ + QP + PM$$ $$P = 120 \text{ m} + 50 \text{ m} + 200 \text{ m} + 50 \text{ m}$$ $$P = 420 \text{ m}$$

The perimeter of the trapezoid is $\boxed{\text{420 m}}$.