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
State 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$(1)$ Area of a rhombus $= \frac{1}{2} \times \ldots \ldots \ldots$
$(2)$ Area of a triangle $= \ldots \ldots \ldots$
$(2)$ Area of a triangle $= \ldots \ldots \ldots$
Easy
View SolutionIf $ar(PQRS) = 80 \, cm^2$ for a parallelogram $PQRS$,then $ar(PSR) = \dots \dots \dots cm^2$.
Medium
View Solution$ABCD$ is a quadrilateral whose diagonal $AC$ divides it into two parts of equal area. Then $ABCD$:
Easy
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
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