Question

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

PO = 31, KM = 34, and PL = 48, find each measure.
13. PM
14. KN

15. OL
16. KP

Answer 1

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

Explanation:

13. PM≅PN≅PO 14.KM≅KN

15.
`x^(2)+31^(2)=48^(2)`

`x^(2) +961=2304\\`

subtract 961 from both sides

`√(x^2) =√(1343)\\x=36.6`

16.
`31^(2) +34^(2)=x^(2)`

`961+1156=x^(2)`

add 961 and 1156 together

`√(2117^2) }=\sqrt{x^(2) }`
`x=46.0`