Question

in △lto, m∠lot = 90°, lm is an ∠ - bisector of ∠olt. mt⊥lt, lm∩ot = a, lt = 18cm, m∠otl = 30°. find a_{△loa} and a_{△lat}. answer: a_{△loa}= cm² a_{△lat}=

Explanation:

Step1: Use angle - bisector and right - triangle properties

Since $LM$ is the angle - bisector of $\angle OLT$ and $MT\perp LT$, $MO = MT$ (the distance from a point to the two sides of an angle bisector is equal). In right - triangle $OLT$, $\angle LOT = 90^{\circ}$, $\angle OTL=30^{\circ}$, and $LT = 18$ cm. Then, using the trigonometric relation $\cos\angle OTL=\frac{OT}{LT}$, we have $OT = LT\cos30^{\circ}=18\times\frac{\sqrt{3}}{2}=9\sqrt{3}$ cm and $OL = LT\sin30^{\circ}=9$ cm.

Step2: Prove triangles congruent

Triangles $LOM$ and $LTM$ are congruent by the Hypotenuse - Leg (HL) congruence criterion (right - triangles with common hypotenuse $LM$ and $MO = MT$).

Step3: Use angle - angle similarity

Triangles $LOA$ and $LTM$ are similar. Let's consider the fact that $\angle OLA=\angle TLM$ (angle - bisector) and $\angle LOA=\angle LTM = 90^{\circ}$.
We know that the area of a triangle is given by $A=\frac{1}{2}bh$.
Let's assume some ratio of similarity. Since $\angle OTL = 30^{\circ}$, in right - triangle $OLT$, if we consider the properties of $30 - 60-90$ triangles.
Let's find the area of $\triangle LOT$ first: $A_{\triangle LOT}=\frac{1}{2}\times OL\times OT=\frac{1}{2}\times9\times9\sqrt{3}=\frac{81\sqrt{3}}{2}$ $cm^{2}$.
Let's use the angle - bisector theorem. Let $\frac{OA}{AT}=\frac{OL}{LT}$. Since $OL = 9$ cm and $LT = 18$ cm, $\frac{OA}{AT}=\frac{1}{2}$. Also, $\triangle LOA$ and $\triangle LTA$ have the same height (the perpendicular distance from $L$ to $OT$).
The area of $\triangle LOT$ can be written as $A_{\triangle LOT}=A_{\triangle LOA}+A_{\triangle LTA}$.
Let $A_{\triangle LOA}=x$ and $A_{\triangle LTA}=y$. Then $x + y=\frac{81\sqrt{3}}{2}$, and since $\frac{x}{y}=\frac{OA}{AT}=\frac{1}{2}$ (because triangles $\triangle LOA$ and $\triangle LTA$ have the same height and the ratio of their bases is $\frac{OA}{AT}$), we can substitute $y = 2x$ into $x + y=\frac{81\sqrt{3}}{2}$.
We get $x+2x=\frac{81\sqrt{3}}{2}$, $3x=\frac{81\sqrt{3}}{2}$, $x = \frac{27\sqrt{3}}{2}$ $cm^{2}$ and $y = 2x=27\sqrt{3}$ $cm^{2}$.

Answer:

$A_{\triangle LOA}=\frac{27\sqrt{3}}{2}$ $cm^{2}$, $A_{\triangle LTA}=27\sqrt{3}$ $cm^{2}$