In Delta XYZ,the midpoints of the sides are A | vedclass

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(A) $1$. By the Midpoint Theorem,the triangle formed by joining the midpoints of the sides of a triangle has an area equal to $1/4$ of the area of the

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Area of $\Delta XYZ = 96$,Area of $\Delta PQR = 6$

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(A) $1$. By the Midpoint Theorem,the triangle formed by joining the midpoints of the sides of a triangle has an area equal to $1/4$ of the area of the original triangle.$2$. For $\Delta XYZ$,$\Delta ABC$ is formed by joining the midpoints of its sides. Therefore,$	ext{Area}(\Delta ABC) = (1/4) 	imes 	ext{Area}(\Delta XYZ)$.$3$. Given $	ext{Area}(\Delta ABC) = 24$,we have $24 = (1/4) 	imes 	ext{Area}(\Delta XYZ)$,which implies $	ext{Area}(\Delta XYZ) = 24 	imes 4 = 96$.$4$. Similarly,for $\Delta ABC$,$\Delta PQR$ is formed by joining the midpoints of its sides. Therefore,$	ext{Area}(\Delta PQR) = (1/4) 	imes 	ext{Area}(\Delta ABC)$.$5$. Substituting the given value,$	ext{Area}(\Delta PQR) = (1/4) 	imes 24 = 6$.$6$. Thus,the area of $\Delta XYZ$ is $96$ and the area of $\Delta PQR$ is $6$.

In $\Delta XYZ$,the midpoints of the sides are $A, B$ and $C$. In $\Delta ABC$,the midpoints of the sides are $P, Q$ and $R$. If the area of $\Delta ABC$ is $24$,find the area of $\Delta XYZ$ and $\Delta PQR$.

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