Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
MediumIn parallelogram $ABCD$,diagonals $AC$ and $BD$ intersect at point $O$. Point $P$ lies on line segment $BO$. Prove that,$ar(ADO) = ar(CDO)$.
(N/A) $1$. In a parallelogram,the diagonals bisect each other. Therefore,$O$ is the midpoint of $AC$.
$2$. Consider $\triangle ADC$. Since $O$ is the midpoint of $AC$,$DO$ is the median of $\triangle ADC$.
$3$. $A$ median of a triangle divides it into two triangles of equal area.
$4$. Therefore,$ar(ADO) = ar(CDO)$.
$2$. Consider $\triangle ADC$. Since $O$ is the midpoint of $AC$,$DO$ is the median of $\triangle ADC$.
$3$. $A$ median of a triangle divides it into two triangles of equal area.
$4$. Therefore,$ar(ADO) = ar(CDO)$.
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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 SolutionIn which of the following figures,do you find two polygons on the same base and between the same parallels?
Easy
View SolutionIf the medians of a $\Delta ABC$ intersect at $G$,show that $\operatorname{ar}(AGB) = \operatorname{ar}(AGC) = \operatorname{ar}(BGC) = \frac{1}{3} \operatorname{ar}(ABC)$.
Difficult
View SolutionState whether each of the following statements is true or false:
$(1)$ If $ar(ABC) = 96 \, cm^2$ for the parallelogram $ABCD$,then $ar(ABCD) = 192 \, cm^2$.
$(2)$ Area of a right triangle = Product of the sides forming the right angle.
$(1)$ If $ar(ABC) = 96 \, cm^2$ for the parallelogram $ABCD$,then $ar(ABCD) = 192 \, cm^2$.
$(2)$ Area of a right triangle = Product of the sides forming the right angle.
Medium
View Solution$ABCD$ is a quadrilateral whose diagonal $AC$ divides it into two parts of equal area. Then $ABCD$:
Easy
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