In Delta ABC, M and N are the midpoints of ov | vedclass

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(B) Given that $M$ and $N$ are the midpoints of sides $AB$ and $AC$ of $\Delta ABC$ respectively.According to the Midpoint Theorem, the line segment j

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$22.5$

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(B) Given that $M$ and $N$ are the midpoints of sides $AB$ and $AC$ of $\Delta ABC$ respectively.According to the Midpoint Theorem, the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.Therefore, $MN \parallel BC$ and $MN = \frac{1}{2} BC$.Since $MN \parallel BC$, $\Delta AMN$ is similar to $\Delta ABC$ by the $AA$ similarity criterion ($\angle AMN = \angle ABC$ and $\angle ANM = \angle ACB$ as corresponding angles).The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.Thus, $\frac{	ext{Area}(\Delta AMN)}{	ext{Area}(\Delta ABC)} = \left( \frac{MN}{BC} ight)^2$.Substituting $MN = \frac{1}{2} BC$, we get $\frac{	ext{Area}(\Delta AMN)}{	ext{Area}(\Delta ABC)} = \left( \frac{1}{2} ight)^2 = \frac{1}{4}$.Given $	ext{Area}(\Delta ABC) = 90$, we have $	ext{Area}(\Delta AMN) = \frac{1}{4} 	imes 90 = 22.5$.

Part $I$	Part $II$
$1.$ In $\Delta ABC$ and $\Delta XYZ, \frac{AB}{XY} = \frac{BC}{YZ}$ and $\angle B \cong \angle Y$	$a. SSS$ condition
$2.$ In $\Delta ABC$ and $\Delta XYZ, \angle A \cong \angle X, \angle B \cong \angle Y$ and $\angle C \cong \angle Z$	$b. SAS$ condition
$3.$ In $\Delta ABC$ and $\Delta XYZ, \frac{AB}{XY} = \frac{BC}{YZ} = \frac{CA}{ZX}$	$c. AAA$ condition
-	$d. \text{None of the conditions apply}$

In $\Delta ABC$, $M$ and $N$ are the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively. If the area of $\Delta ABC$ is $90$, find the area of $\Delta AMN$.

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