Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
Easy$(1)$ The part of the plane enclosed by a simple closed figure is called a $\ldots \ldots \ldots$
$(2)$ $\ldots \ldots \ldots$ of the planar region corresponding to a closed figure is called its area.
$(2)$ $\ldots \ldots \ldots$ of the planar region corresponding to a closed figure is called its area.
(A) $(1)$ The part of the plane enclosed by a simple closed figure is called a $\text{planar region}$.
$(2)$ The $\text{magnitude}$ or $\text{measure}$ of the planar region corresponding to a closed figure is called its area.
$(2)$ The $\text{magnitude}$ or $\text{measure}$ of the planar region corresponding to a closed figure is called its area.
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Similar Questions
$ABCD$ is a parallelogram and $BC$ is produced to a point $Q$ such that $AD = CQ$. If $AQ$ intersects $DC$ at $P$,show that $\operatorname{ar}(BPC) = \operatorname{ar}(DPQ)$.
Medium
View Solution$(1)$ 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 $= \dots$
Easy
View SolutionIn $\triangle ABC,$ if $L$ and $M$ are the points on $AB$ and $AC,$ respectively such that $LM \parallel BC.$ Prove that $\operatorname{ar}(\triangle LOB) = \operatorname{ar}(\triangle MOC).$
Medium
View SolutionThe area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals $12 \, cm$ and $16 \, cm$ is (in $cm^2$)
Easy
View SolutionIf $E, F, G$ and $H$ are respectively the midpoints of the sides of a parallelogram $ABCD$,show that $ar(EFGH) = \frac{1}{2} ar(ABCD)$.
Difficult
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