Midsegment of a Triangle - Examples, Exercises and Solutions

In order to calculate the perimeter of triangle $\triangle ADE$we need to find the lengths of its sides,

Let's now refer to the given information that $DE$is a median in $\triangle ABC$and therefore a median in a triangle equals half the length of the side it does not intersect, additionally we'll remember the definition of a median in a triangle as a line segment that extends from the midpoint of one side to the midpoint of another side, we'll write the property mentioned (a) and the fact derived from the given definition (b+c):

a.

$DE=\frac{1}{2}BC$b.

$AD=\frac{1}{2}AB$c.

$AE=\frac{1}{2}AC\\$Additionally, the given data in the drawing are:

d.

$BC=8$e.

$AB=6$f.

$AC=10$Therefore, we will substitute d', e', and f' respectively in a', b', and c', and we get:

g.

$DE=\frac{1}{2}BC=\frac{1}{2}\cdot8=4$h.

$AD=\frac{1}{2}AB=\frac{1}{2}\cdot6=3$i.

$AE=\frac{1}{2}AC=\frac{1}{2}\cdot10=5$

Therefore the perimeter of $\triangle ADE$ is:

j.

$P_{ADE}=DE+AD+AE=4+3+5=12$Therefore the correct answer is answer d.

We are tasked with finding the area of trapezoid DECB in the right triangle ABC where DE is the midsegment parallel to BC. Given $BC = 5$ cm, $AC = 13$ cm, let us calculate the area step-by-step.

• First, use the Pythagorean theorem to find the length of $AB$. Since triangle ABC is a right triangle at $B$, we have:

$AB = \sqrt{AC^2 - BC^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ cm.

• The midsegment DE of triangle ABC, being parallel to the base BC, is half the length of BC:

$DE = \frac{1}{2} \times BC = \frac{1}{2} \times 5 = 2.5$ cm.

• Now, to find the area of trapezoid DECB, which has bases DE and BC, the height is the same as $AB$ (the vertical side of triangle ABC):

$A = \frac{1}{2} \times (b_1 + b_2) \times h = \frac{1}{2} \times (2.5 + 5) \times 12$.

Calculate the expression:

• $A = \frac{1}{2} \times 7.5 \times 12 = \frac{1}{2} \times 90 = 45$ cm².
• Since DECB is half of the total triangle, the area is half of 45.

So, the area of trapezoid DECB is $45 \div 2 = 22.5$ cm².

The solution to the problem is $\boxed{22.5}$.