Question

find the value of x.
(image of a triangle with a segment parallel to the base, creating a smaller similar triangle. the height of the larger triangle is 28, the height of the smaller triangle is 2x. the sides of the larger triangle are marked as equal, and the base segments are marked as equal.)
x = \square

Explanation:

Step1: Identify Midline Theorem

The triangle has a midline (the segment with length \(2x\)) parallel to the base, so by the Midline Theorem, the midline length is half the length of the base. Here, the left side is 28, and the midline relates to the height? Wait, no—wait, the triangle is isoceles? Wait, the markings: the two sides of the big triangle are equal (tick marks), and the base is bisected (tick marks on the base segments). So the segment \(2x\) is a midline, so it should be half of 28? Wait, no, midline in a triangle: the segment connecting midpoints of two sides is parallel to the third side and half its length. Wait, here, the vertical side is 28, and the segment \(2x\) is parallel to the left vertical side? Wait, no, the big triangle is a right triangle? Wait, the left side is vertical (length 28), the base is horizontal, and there's a smaller right triangle inside, with vertical side \(2x\). The two sides of the big triangle (the hypotenuse) have tick marks, so it's an isoceles triangle? Wait, no, the hypotenuse has two tick marks, so it's equal, and the base is bisected (two tick marks on the base segments). So the segment \(2x\) is the midline, so by the Midline Theorem, \(2x=\frac{28}{2}\)? Wait, no, midline length is half the length of the third side. Wait, the left side is 28, the segment \(2x\) is parallel to the left side? No, wait, the big triangle: left side is 28 (vertical), base is horizontal, hypotenuse with tick marks. The smaller triangle: vertical side \(2x\), base is half the big base (since base is bisected). So by the Basic Proportionality Theorem (Thales' theorem) or midline theorem, the line segment \(2x\) is parallel to the left side (28) and half its length? Wait, no, if the base is bisected, then the segment \(2x\) connects the midpoints of the base and the hypotenuse, so it should be parallel to the left side (the other leg) and half its length. So \(2x=\frac{28}{2}\)? Wait, no, \(2x\) is the length, so \(2x = 14\)? No, wait, maybe I got it reversed. Wait, the left side is 28, the segment \(2x\) is the midline, so midline length is half the length of the third side. Wait, the third side here is the left leg (28), so midline (the vertical segment of the smaller triangle) is half of 28? So \(2x=\frac{28}{2}\)? No, that would be \(2x = 14\), then \(x = 7\). Wait, let's re-examine. The big triangle: right triangle? Wait, the left side is vertical, base is horizontal, so right angle at the bottom left. The smaller triangle: right angle at the bottom, with vertical side \(2x\), horizontal side equal to half the big base (since base is bisected). So by the Midline Theorem, the segment connecting midpoints of the base and hypotenuse is parallel to the left leg (28) and half its length. So \(2x=\frac{28}{2}\)? Wait, no, \(2x\) is the length, so \(2x = 14\)? Then \(x = 7\). Wait, that makes sense. So step 1: recognize the midline theorem (segment connecting midpoints of two sides is half the third side). Step 2: set up the equation \(2x=\frac{28}{2}\). Step 3: solve for x.

Step2: Solve for \(x\)

From the midline theorem, \(2x=\frac{28}{2}\)
Simplify \(\frac{28}{2}=14\), so \(2x = 14\)
Divide both sides by 2: \(x=\frac{14}{2}=7\)

Answer:

\(x = 7\)