In Delta ABC, the midpoints of the sides are | vedclass

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(A) The midpoint theorem states that the triangle formed by joining the midpoints of a triangle has an area equal to $1/4$ of the area of the original

Vedclass · Accepted Answer

Area of $\Delta PQR = 80,$ Area of $\Delta ABC = 320$

Vedclass · Answer

(A) The midpoint theorem states that the triangle formed by joining the midpoints of a triangle has an area equal to $1/4$ of the area of the original triangle.Let $	ext{Area}(\Delta PQR) = A_1$ and $	ext{Area}(\Delta ABC) = A_2$.Since $P, Q, R$ are midpoints of $\Delta ABC,$ $	ext{Area}(\Delta PQR) = \frac{1}{4} 	ext{Area}(\Delta ABC).$Similarly,since $X, Y, Z$ are midpoints of $\Delta PQR,$ $	ext{Area}(\Delta XYZ) = \frac{1}{4} 	ext{Area}(\Delta PQR).$Given $	ext{Area}(\Delta XYZ) = 20.$Therefore,$20 = \frac{1}{4} 	ext{Area}(\Delta PQR) \implies 	ext{Area}(\Delta PQR) = 20 	imes 4 = 80.$Now,$	ext{Area}(\Delta PQR) = \frac{1}{4} 	ext{Area}(\Delta ABC) \implies 80 = \frac{1}{4} 	ext{Area}(\Delta ABC) \implies 	ext{Area}(\Delta ABC) = 80 	imes 4 = 320.$Thus,the area of $\Delta PQR$ is $80$ and the area of $\Delta ABC$ is $320.$

In $\Delta ABC,$ the midpoints of the sides are $P, Q, R.$ In $\Delta PQR,$ the midpoints of the sides are $X, Y, Z.$ If the area of $\Delta XYZ$ is $20,$ find the area of $\Delta PQR$ and $\Delta ABC.$

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