Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
EasyWhich of the following figures lie on the same base and between the same parallels? Write the common base and the two parallels for the figure for which the answer is affirmative.
(N/A) In the given figure,we have a parallelogram $ABCD$ and a trapezium $ABST$.
For figures to lie on the same base and between the same parallels,they must share a common side as a base,and their remaining vertices must lie on a line parallel to that base.
Here,the parallelogram $ABCD$ and the trapezium $ABST$ share the common base $AB$.
However,the vertices $D$ and $C$ of the parallelogram lie on the line $DC$,while the vertices $T$ and $S$ of the trapezium lie on the line $TS$.
Since the line $DC$ is not the same as the line $TS$,the two figures do not lie between the same parallels.
Therefore,the answer is negative.
For figures to lie on the same base and between the same parallels,they must share a common side as a base,and their remaining vertices must lie on a line parallel to that base.
Here,the parallelogram $ABCD$ and the trapezium $ABST$ share the common base $AB$.
However,the vertices $D$ and $C$ of the parallelogram lie on the line $DC$,while the vertices $T$ and $S$ of the trapezium lie on the line $TS$.
Since the line $DC$ is not the same as the line $TS$,the two figures do not lie between the same parallels.
Therefore,the answer is negative.
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Similar Questions
In $\Delta PQR$,$PM$ is a median and $N$ is the midpoint of $PM$. If $\text{ar}(PQN) = 36 \text{ cm}^2$,then $\text{ar}(PQR) = \dots \text{ cm}^2$.
Medium
View SolutionState 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 $= (\text{Side})^2$.
Easy
View SolutionIn the figure,$ABCDE$ is any pentagon. $BP$ is drawn parallel to $AC$ and meets $DC$ produced at $P$,and $EQ$ is drawn parallel to $AD$ and meets $CD$ produced at $Q$. Prove that $\operatorname{ar}(ABCDE) = \operatorname{ar}(APQ)$.
Medium
View Solution$A$ point $E$ is taken on the side $BC$ of a parallelogram $ABCD$. $AE$ and $DC$ are produced to meet at $F$. Prove that $\operatorname{ar}(\triangle ADF) = \operatorname{ar}(ABFC)$.
Difficult
View SolutionIn $\Delta PQR$,$PM$ is a median. If $\text{ar}(\Delta PMQ) = 36 \text{ cm}^2$,then find $\text{ar}(\Delta PMR)$ in $\text{cm}^2$.
Medium
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