Question

Given the diagram below, if P is the incenter of AJRL,

Answer 1

Answer:

13. PM=31 14. KN=34 15. OL=36.6 KP=46

Step-by-step explanation:

13. PM≅PN≅PO 14.KM≅KN

15.$x^{2}+31^{2}=48^{2}$

$x^{2} +961=2304$

subtract 961 from both sides

$\sqrt{x^2} =\sqrt{1343}\\x=36.6$

16. $31^{2} +34^{2}=x^{2}$

$961+1156=x^{2}$

add 961 and 1156 together

$\sqrt{2117^2} }=\sqrt{x^{2} }$$x=46.0$