Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
EasyState 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 $= (\text{Side})^2$.
$(1)$ 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 $= (\text{Side})^2$.
(A) $(1)$ False. The area of a parallelogram is given by the product of its base and the corresponding altitude,not half of it.
$(2)$ False. The area of a rhombus is half the product of its diagonals,i.e.,$\frac{1}{2} \times d_1 \times d_2$.
$(3)$ True. The area of a square is indeed the square of its side length.
$(2)$ False. The area of a rhombus is half the product of its diagonals,i.e.,$\frac{1}{2} \times d_1 \times d_2$.
$(3)$ True. The area of a square is indeed the square of its side length.
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$XYZW$ is a square. If $XY = 17 \text{ cm}$,then find the area of $XYZW$ in $\text{cm}^2$.
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View SolutionTwo parallelograms are on equal bases and between the same parallels. The ratio of their areas is
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View Solution$PQRS$ is a square. $T$ and $U$ are the mid-points of $PS$ and $QR$ respectively. Find the area of $\Delta OTS$,if $PQ = 8 \, cm$,where $O$ is the point of intersection of $TU$ and $QS$.
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View SolutionIn the given figure,$P$ is a point in the interior of parallelogram $ABCD$. Show that,
$(1) \operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$
$(2) \operatorname{ar}(APD) + \operatorname{ar}(PBC) = \operatorname{ar}(APB) + \operatorname{ar}(PCD)$
$(1) \operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$
$(2) \operatorname{ar}(APD) + \operatorname{ar}(PBC) = \operatorname{ar}(APB) + \operatorname{ar}(PCD)$
Difficult
View SolutionWrite True or False and justify your answer:
In the figure,$ABCD$ and $EFGD$ are two parallelograms and $G$ is the mid-point of $CD.$ Then $\operatorname{ar}(\triangle DPC) = \frac{1}{2} \operatorname{ar}(EFGD).$
In the figure,$ABCD$ and $EFGD$ are two parallelograms and $G$ is the mid-point of $CD.$ Then $\operatorname{ar}(\triangle DPC) = \frac{1}{2} \operatorname{ar}(EFGD).$
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