In Delta ABC,P, Q and R are the midpoints of | vedclass

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(D) In $\Delta ABC$,since $P, Q,$ and $R$ are the midpoints of sides $BC, CA,$ and $AB$ respectively,the triangle is divided into four congruent trian

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$\frac{3}{4}$

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(D) In $\Delta ABC$,since $P, Q,$ and $R$ are the midpoints of sides $BC, CA,$ and $AB$ respectively,the triangle is divided into four congruent triangles: $\Delta ARQ, \Delta RBP, \Delta QPC,$ and $\Delta PQR$.Therefore,$Area(\Delta ARQ) = Area(\Delta RBP) = Area(\Delta QPC) = Area(\Delta PQR) = \frac{1}{4} Area(\Delta ABC)$.The quadrilateral $BCQR$ is composed of three of these triangles: $\Delta RBP, \Delta QPC,$ and $\Delta PQR$.Thus,$Area(BCQR) = Area(\Delta RBP) + Area(\Delta QPC) + Area(\Delta PQR) = \frac{1}{4} Area(\Delta ABC) + \frac{1}{4} Area(\Delta ABC) + \frac{1}{4} Area(\Delta ABC) = \frac{3}{4} Area(\Delta ABC)$.

$1$. $\square ABCD$ is a rhombus.	$a$. $\overline{AC}$ and $\overline{BD}$ bisect each other.
$2$. $\square ABCD$ is a parallelogram.	$b$. $\overline{AC}$ and $\overline{BD}$ bisect each other at right angles.
$3$. $\square ABCD$ is a rectangle.	$c$. $\overline{AC}$ and $\overline{BD}$ are congruent and bisect each other at right angles.
$4$. $\square ABCD$ is a square.	$d$. $\overline{AC}$ and $\overline{BD}$ are congruent and bisect each other.

In $\Delta ABC$,$P, Q$ and $R$ are the midpoints of $\overline{BC}, \overline{CA}$ and $\overline{AB}$ respectively. Then,$Area(BCQR) = \ldots \ldots \ldots \times Area(ABC)$.

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