Question

1. in △abc, d and e are mid - points of ab and ac. if ab = 12 and ac = 16, what is the length of de?

a. 9
b. 10
c. 8
d. 6

Explanation:

Step1: Apply mid - point theorem

The mid - point theorem states that the line segment joining the mid - points of two sides of a triangle is parallel to the third side and half its length. In \(\triangle ABC\), \(D\) and \(E\) are mid - points of \(AB\) and \(AC\). So \(DE=\frac{1}{2}BC\). First, we need to find \(BC\) using the Pythagorean theorem if \(\triangle ABC\) is a right - triangle (assuming no other information about the angle, we can also use the property of similar triangles). Since \(D\) and \(E\) are mid - points, we can also use the vector or coordinate - free approach. By the mid - point theorem, \(DE=\frac{1}{2}\sqrt{AB^{2}+AC^{2}}\) (if \(\angle A = 90^{\circ}\), but we can also use the general property that \(DE=\frac{1}{2}BC\)). In a non - right triangle, the mid - point theorem still holds. Here, we know that \(DE=\frac{1}{2}\sqrt{AB^{2}+AC^{2}}\) (by the distance formula in a triangle with mid - points). Substitute \(AB = 12\) and \(AC=16\).

Step2: Calculate the value of \(DE\)

\[

$$\begin{align*} DE&=\frac{1}{2}\sqrt{12^{2}+16^{2}}\\ &=\frac{1}{2}\sqrt{144 + 256}\\ &=\frac{1}{2}\sqrt{400}\\ &=\frac{1}{2}\times20\\ & = 10 \end{align*}$$

\]

Answer:

b. 10