Find the missing lengths.
LK=16 and LI=20, find LH and OH.
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QUESTION 1
From Pythagoras Theorem,
[tex]|KI|^2+|KL|^2=|LI|^2[/tex]
[tex]|KI|^2+16^2=20^2[/tex]
[tex]|KI|^2+256=400[/tex]
[tex]|KI|^2=400-256[/tex]
[tex]|KI|^2=144[/tex]
[tex]|KI|=\sqrt{144}[/tex]
[tex]|KI|=12[/tex]
Let [tex]HI=x[/tex], then [tex]LH=20-x[/tex]
From ΔHKI,
[tex]x^2+|HK|^2=12^2[/tex]
[tex]\Rightarrow |HK|^2=144-x^2..(1)[/tex]
From ΔHKL,
[tex]|HK|^2+(20-x)^2=16^2[/tex]
[tex]\Rightarrow |HK|^2=256-(20-x)^2[/tex]
[tex]\Rightarrow |HK|^2=-x^2+40x-144...(2)[/tex]
Solving equation (1) and (2) gives
[tex]144-x^2=-x^2+40x-144[/tex]
[tex]40x=288[/tex]
[tex]x=\frac{36}{5}[/tex]
[tex]|LH|=20-\frac{36}{5} =\frac{64}{5}=12.8 units[/tex]
QUESTION 2
ΔLKI is similar to ΔOHL
[tex]\frac{|OH|}{|KI|}=\frac{|LI|}{|LH|}[/tex]
[tex]\frac{|OH|}{12}=\frac{\frac{64}{5} }{20}[/tex]
[tex]\frac{|OH|}{12}=\frac{16}{25}[/tex]
[tex]|OH|=\frac{16}{25}\times 12[/tex]
[tex]|OH|=\frac{192}{25}=7.68units[/tex]