Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
Easy$(1)$ Area of a rhombus $= \frac{1}{2} \times \ldots \ldots \ldots$
$(2)$ Area of a triangle $= \ldots \ldots \ldots$
$(2)$ Area of a triangle $= \ldots \ldots \ldots$
(N/A) $(1)$ The area of a rhombus is given by $\frac{1}{2} \times \text{product of its diagonals}$.
$(2)$ The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{corresponding altitude}$.
$(2)$ The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{corresponding altitude}$.
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Similar Questions
$(1)$ 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.
Easy
View SolutionIn square $ABCD$,$AC = 16 \text{ cm}$,then $\operatorname{ar}(ABCD) = \dots \text{ cm}^2$.
Easy
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 $QS$.
Medium
View SolutionIn trapezium $PQRS, PQ || RS$ and diagonals $PR$ and $QS$ intersect at point $M$. Prove that,$ar(PQS) = ar(QPR)$.
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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