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aalyn [17]
3 years ago
6

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

Mathematics
1 answer:
sp2606 [1]3 years ago
6 0

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

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