Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
MediumIn $\triangle ABC$,$AD$ is a median. $E$ is the midpoint of $BD$ and $O$ is the midpoint of $AE$. Prove that $ar(AOB) = \frac{1}{8} ar(ABC)$.
(N/A) $1$. Since $AD$ is the median of $\triangle ABC$,it divides the triangle into two triangles of equal area. Therefore,$ar(ABD) = \frac{1}{2} ar(ABC)$.
$2$. In $\triangle ABD$,$AE$ is a median because $E$ is the midpoint of $BD$. Thus,$ar(ABE) = \frac{1}{2} ar(ABD)$.
$3$. Substituting the value from step $1$: $ar(ABE) = \frac{1}{2} \times (\frac{1}{2} ar(ABC)) = \frac{1}{4} ar(ABC)$.
$4$. In $\triangle ABE$,$BO$ is a median because $O$ is the midpoint of $AE$. Thus,$ar(AOB) = \frac{1}{2} ar(ABE)$.
$5$. Substituting the value from step $3$: $ar(AOB) = \frac{1}{2} \times (\frac{1}{4} ar(ABC)) = \frac{1}{8} ar(ABC)$.
$6$. Hence,it is proved that $ar(AOB) = \frac{1}{8} ar(ABC)$.
$2$. In $\triangle ABD$,$AE$ is a median because $E$ is the midpoint of $BD$. Thus,$ar(ABE) = \frac{1}{2} ar(ABD)$.
$3$. Substituting the value from step $1$: $ar(ABE) = \frac{1}{2} \times (\frac{1}{2} ar(ABC)) = \frac{1}{4} ar(ABC)$.
$4$. In $\triangle ABE$,$BO$ is a median because $O$ is the midpoint of $AE$. Thus,$ar(AOB) = \frac{1}{2} ar(ABE)$.
$5$. Substituting the value from step $3$: $ar(AOB) = \frac{1}{2} \times (\frac{1}{4} ar(ABC)) = \frac{1}{8} ar(ABC)$.
$6$. Hence,it is proved that $ar(AOB) = \frac{1}{8} ar(ABC)$.
Explore More
Similar Questions
In $\Delta ABC$,$\angle B = 90^{\circ}$,$AB = 8\, \text{cm}$ and $BC = 15\, \text{cm}$,then $\text{ar}(\Delta ABC) = \dots \text{cm}^2$.
Medium
View SolutionWrite True or False and justify your answer:
$PQRS$ is a rectangle inscribed in a quadrant of a circle of radius $13 \, cm$. $A$ is any point on $PQ$. If $PS = 5 \, cm$,then $\text{ar}(PAS) = 30 \, cm^2$.
$PQRS$ is a rectangle inscribed in a quadrant of a circle of radius $13 \, cm$. $A$ is any point on $PQ$. If $PS = 5 \, cm$,then $\text{ar}(PAS) = 30 \, cm^2$.
Difficult
View Solution$ABCD$ is a quadrilateral whose diagonal $AC$ divides it into two parts of equal area. Then $ABCD$:
Easy
View SolutionIn the given figure,$P$ is a point in the interior of parallelogram $ABCD$. Show that,
$(1) \operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$
$(2) \operatorname{ar}(APD) + \operatorname{ar}(PBC) = \operatorname{ar}(APB) + \operatorname{ar}(PCD)$
$(1) \operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$
$(2) \operatorname{ar}(APD) + \operatorname{ar}(PBC) = \operatorname{ar}(APB) + \operatorname{ar}(PCD)$
Difficult
View SolutionIn which of the following figures,do you find two polygons on the same base and between the same parallels?
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo