Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
MediumShow that a median of a triangle divides it into two triangles of equal areas.
(N/A) Let $\Delta ABC$ be a triangle and let $AD$ be one of its medians,where $D$ is the midpoint of $BC$.
We wish to show that $\operatorname{ar}(ABD) = \operatorname{ar}(ACD)$.
Since the formula for the area of a triangle involves an altitude,let us draw $AN \perp BC$.
Now,$\operatorname{ar}(ABD) = \frac{1}{2} \times \text{base} \times \text{altitude} = \frac{1}{2} \times BD \times AN$.
Since $AD$ is a median,$BD = CD$.
Substituting $BD = CD$ in the expression for $\operatorname{ar}(ABD)$,we get:
$\operatorname{ar}(ABD) = \frac{1}{2} \times CD \times AN$.
Since $\frac{1}{2} \times CD \times AN$ is the formula for the area of $\Delta ACD$,we have:
$\operatorname{ar}(ABD) = \operatorname{ar}(ACD)$.
Thus,a median of a triangle divides it into two triangles of equal areas.
We wish to show that $\operatorname{ar}(ABD) = \operatorname{ar}(ACD)$.
Since the formula for the area of a triangle involves an altitude,let us draw $AN \perp BC$.
Now,$\operatorname{ar}(ABD) = \frac{1}{2} \times \text{base} \times \text{altitude} = \frac{1}{2} \times BD \times AN$.
Since $AD$ is a median,$BD = CD$.
Substituting $BD = CD$ in the expression for $\operatorname{ar}(ABD)$,we get:
$\operatorname{ar}(ABD) = \frac{1}{2} \times CD \times AN$.
Since $\frac{1}{2} \times CD \times AN$ is the formula for the area of $\Delta ACD$,we have:
$\operatorname{ar}(ABD) = \operatorname{ar}(ACD)$.
Thus,a median of a triangle divides it into two triangles of equal areas.
Explore More
Similar Questions
In the figure,$ABCD$ is a quadrilateral and $BE \parallel AC$. $BE$ meets $DC$ produced at $E$. Show that the area of $\Delta ADE$ is equal to the area of the quadrilateral $ABCD$.
Medium
View SolutionIn the figure,$D$ and $E$ are two points on $BC$ such that $BD = DE = EC$. Show that $ar(ABD) = ar(ADE) = ar(AEC)$.
Can you now answer the question that you have left in the 'Introduction' of this chapter,whether the field of Budhia has been actually divided into three parts of equal area?
Can you now answer the question that you have left in the 'Introduction' of this chapter,whether the field of Budhia has been actually divided into three parts of equal area?
Medium
View SolutionIn the figure,$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC$. If $AE$ intersects $BC$ at $F$,show that:
$(i)$ $\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$
$(iii)$ $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$
$(v)$ $\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$
$(vi)$ $\operatorname{ar}(FED) = \frac{1}{8} \operatorname{ar}(AFC)$
$(i)$ $\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$
$(iii)$ $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$
$(v)$ $\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$
$(vi)$ $\operatorname{ar}(FED) = \frac{1}{8} \operatorname{ar}(AFC)$
Medium
View Solution$P$ and $Q$ are respectively the mid-points of sides $AB$ and $BC$ of a triangle $ABC$ and $R$ is the mid-point of $AP$. Show that $\operatorname{ar}(RQC) = \frac{3}{8} \operatorname{ar}(ABC)$.
Medium
View Solution$D, E$ and $F$ are respectively the mid-points of the sides $BC, CA$ and $AB$ of a $\Delta ABC$. Show that $\operatorname{ar}( BDEF ) = \frac{1}{2} \operatorname{ar}( ABC )$
Medium
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