In parallelogram $ABCD$,$P$ is the midpoint of $CD$. Then,$\operatorname{ar}(ABCD) : \operatorname{ar}(PBC) = \dots$

  • A
    $1:4$
  • B
    $4:1$
  • C
    $1:2$
  • D
    $4:1$

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Similar Questions

State whether each of the following statements is true or false:
$(1)$ Area of a parallelogram $= \text{base} \times \text{corresponding altitude}$.
$(2)$ Area of a rhombus $= \frac{1}{2} \times \text{Product of its diagonals}$.
$(3)$ Area of a square $= (\text{Side})^2$.

$(1)$ In $\Delta ABC$,$AD$ is an altitude. If $BC = 8 \text{ cm}$ and $AD = 5 \text{ cm}$,then $\text{ar}(\Delta ABC) = \dots \text{ cm}^2$.
$(2)$ $A$ $\dots$ of a triangle divides the triangle into two triangles of equal area.

The perimeter of square $ABCD$ is $16 \, cm$,then $ar(ABCD) = \ldots \ldots \ldots \, cm^2$.

$PQRS$ is a rectangle. If $PQ = 20 \, cm$ and $\operatorname{ar}(PQRS) = 300 \, cm^2$,then $SP = \dots \, cm$.

In the figure,$CD \parallel AE$ and $CY \parallel BA$. Prove that $\operatorname{ar}(\triangle CBX) = \operatorname{ar}(\triangle AXY)$.

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