Question : $\triangle$ LON and $\triangle$ LMN are two right-angled triangles with common hypotenuse LN such that $\angle$ LON = $90^{\circ}$ and $\angle$ LMN = $90^{\circ}$. LN is the bisector of $\angle$ OLM. If LN = 29 cm and ON = 20 cm, then what is the perimeter (in cm) of $\triangle$ LMN?
Option 1: 67
Option 2: 62
Option 3: 65
Option 4: 70
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Correct Answer: 70
Solution :
Given: $\triangle$LON and $\triangle$LMN are two right-angled triangles with common hypotenuse LN such that $\angle$LON = $90^{\circ}$ and $\angle$LMN = $90^{\circ}$. LN is the bisector of $\angle$OLM also, LN = 29 cm and ON = 20 cm.
In $\triangle$LON,
LO = $\sqrt{29^2-20^2}=\sqrt{441}$ = 21 cm
So, the perimeter of $\triangle$LON = 21 + 20 + 29 = 70 cm
Now, between $\triangle$LON and $\triangle$LMN,
$\angle$LON = $\angle$LMN, (right angle)
$\angle$OLN = $\angle$MLN, (since LN is the bisector of $\angle$OLM)
and LN = LN (general side)
So, $\triangle$LON and $\triangle$LMN are congruent triangles,
Therefore, the perimeter of $\triangle$LMN is also 70 cm.
Hence, the correct answer is 70.
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