Question

points j, k and l are the midpoints of the sides of \\( \triangle xyz \\). find the following lengths.

a. \\( lk = \\)
b. \\( yl = \\)
c. \\( yk = \\)
d. \\( jk = \\)
e. \\( xk = \\)
f. \\( jl = \\)

Explanation:

Step1: Midpoint and Midsegment Properties

Since \(J,K,L\) are mid - points of the sides of \(\triangle XYZ\), we use the mid - segment theorem. The mid - segment of a triangle is parallel to the third side and half its length. Also, if a point is a mid - point of a side of a triangle, it divides the side into two equal parts.

Step2: Calculate \(LK\)

\(LK\) is a mid - segment parallel to \(XZ\). Given \(XZ = 12\), by the mid - segment theorem \(LK=\frac{1}{2}XZ\). So \(LK = 6\).

Step3: Calculate \(YL\)

Since \(L\) is the mid - point of \(ZY\) and \(ZL = 14\), then \(YL=ZL = 14\).

Step4: Calculate \(YK\)

Since \(K\) is the mid - point of \(XY\) and \(XY = 20\), then \(YK=\frac{1}{2}XY\). So \(YK = 10\).

Step5: Calculate \(JK\)

\(JK\) is a mid - segment parallel to \(ZY\). Given \(ZY=14 + 14=28\), by the mid - segment theorem \(JK=\frac{1}{2}ZY\). So \(JK = 14\).

Step6: Calculate \(XK\)

Since \(K\) is the mid - point of \(XY\) and \(XY = 20\), then \(XK=\frac{1}{2}XY\). So \(XK = 10\).

Step7: Calculate \(JL\)

\(JL\) is a mid - segment parallel to \(XY\). Given \(XY = 20\), by the mid - segment theorem \(JL=\frac{1}{2}XY\). So \(JL = 10\).

Answer:

a. \(LK = 6\)
b. \(YL=14\)
c. \(YK = 10\)
d. \(JK = 14\)
e. \(XK = 10\)
f. \(JL = 10\)