Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
Easy$(1)$ If a planar region formed by a figure $T$ is made up of two non-overlapping planar regions formed by figures $P$ and $Q$,then $\operatorname{ar}(T) = \dots$
$(2)$ Area of a parallelogram $= \dots$
$(2)$ Area of a parallelogram $= \dots$
(N/A) $(1)$ The area of a figure $T$ is the sum of the areas of the two non-overlapping regions $P$ and $Q$. Therefore,$\operatorname{ar}(T) = \operatorname{ar}(P) + \operatorname{ar}(Q)$.
$(2)$ The area of a parallelogram is given by the product of its base and the corresponding altitude (height) to that base. Therefore,$\text{Area} = \text{base} \times \text{corresponding altitude}$.
$(2)$ The area of a parallelogram is given by the product of its base and the corresponding altitude (height) to that base. Therefore,$\text{Area} = \text{base} \times \text{corresponding altitude}$.
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Similar Questions
In the figure,$l, m,$ and $n$ are straight lines such that $l \parallel m$ and $n$ intersects $l$ at $P$ and $m$ at $Q$. $ABCD$ is a quadrilateral such that its vertex $A$ is on $l$. The vertices $C$ and $D$ are on $m$ and $AD \parallel n$. Show that $\operatorname{ar}(ABCQ) = \operatorname{ar}(ABCDP).$
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 Solution$(1)$ If two figures are congruent,they must have $\ldots \ldots$ areas.
$(2)$ Area of figure $A$ is denoted as $\ldots \ldots$ symbolically.
$(2)$ Area of figure $A$ is denoted as $\ldots \ldots$ symbolically.
Easy
View SolutionIn the figure,if parallelogram $ABCD$ and rectangle $ABEM$ are of equal area,then:
Easy
View SolutionIn the figure,$CD \parallel AE$ and $CY \parallel BA$. Prove that $\operatorname{ar}(\triangle CBX) = \operatorname{ar}(\triangle AXY)$.
Medium
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