Question

if p is the incenter of δ jkl, po = 31, km = 34, and pl = 48, find each measure.
pm = 24 × kn = 34 √
ol = 18 × kp = 46 √

Explanation:

Step1: Recall the property of the in - center

The in - center of a triangle is equidistant from the sides of the triangle. So, \(PM = PO\) (since \(P\) is the in - center and \(PM\perp JK\), \(PO\perp JL\)). Given \(PO = 31\), so \(PM=31\).

Step2: Recall the property of the angle bisector and congruent right - triangles

Since \(P\) is the in - center and \(KN\) and \(KM\) are segments from the vertex \(K\) to the points of tangency (by the property of the in - center, the lengths of two tangent segments from a single external point to a circle (or in the context of the in - center and the sides of the triangle) are equal). Given \(KM = 34\), so \(KN = 34\).

Step3: Use the Pythagorean theorem in right - triangle \(POL\)

In right - triangle \(POL\), if \(PL = 48\) and \(PO = 31\), then by the Pythagorean theorem \(OL=\sqrt{PL^{2}-PO^{2}}=\sqrt{48^{2}-31^{2}}=\sqrt{(48 + 31)(48-31)}=\sqrt{79\times17}=\sqrt{1343}\approx36.6\). But if we assume there is a mis - understanding and we consider the fact that if we use the property of the in - center and the fact that \(KM = 34\), \(PL = 48\) and \(PO = 31\) (maybe a wrong - figure assumption, but if we consider the formula for the length of the tangent from the in - center to the side, but actually, since \(PM = PO\) (in - center property: distances to sides are equal), \(PM = 31\). For \(OL\), if we assume a wrong approach (maybe a mis - labeled figure where we consider some other relation, but the correct formula is \(OL=\sqrt{PL^{2}-PO^{2}}\). However, if we consider the fact that \(KM = KN\) (tangents from \(K\) to the incircle), \(PM = PN=PO\) (distances from in - center to sides).

Step4: Use the Pythagorean theorem in right - triangle \(KMP\)

In right - triangle \(KMP\), \(KM = 34\), \(PM = 31\). Then \(KP=\sqrt{KM^{2}+PM^{2}}=\sqrt{34^{2}+31^{2}}=\sqrt{1156 + 961}=\sqrt{2117}\approx46\) (by calculation \(34^{2}=1156\), \(31^{2}=961\), \(1156+961 = 2117\), \(\sqrt{2117}\approx46\)).

Answer:

\(PM = 31\), \(KN = 34\), \(OL=\sqrt{48^{2}-31^{2}}\approx36.6\), \(KP=\sqrt{34^{2}+31^{2}}\approx46\)