Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
Medium$D, E$ and $F$ are respectively the mid-points of the sides $BC, CA$ and $AB$ of a $\Delta ABC$. Show that $\operatorname{ar}(DEF) = \frac{1}{4} \operatorname{ar}(ABC)$.
(N/A) Given: $D, E, F$ are mid-points of sides $BC, CA, AB$ respectively in $\Delta ABC$.
To prove: $\operatorname{ar}(DEF) = \frac{1}{4} \operatorname{ar}(ABC)$.
Proof:
$1$. Since $D$ and $F$ are mid-points of $BC$ and $AB$ respectively,by the Mid-point Theorem,$DF \parallel AC$ and $DF = \frac{1}{2} AC = AE$. Thus,$AFDE$ is a parallelogram.
$2$. Similarly,$BDEF$ and $CDFE$ are parallelograms.
$3$. In parallelogram $BDEF$,the diagonal $DF$ divides it into two triangles of equal area: $\operatorname{ar}(\Delta BDF) = \operatorname{ar}(\Delta DEF)$.
$4$. In parallelogram $AFDE$,the diagonal $DE$ divides it into two triangles of equal area: $\operatorname{ar}(\Delta AFE) = \operatorname{ar}(\Delta DEF)$.
$5$. In parallelogram $CDFE$,the diagonal $DE$ divides it into two triangles of equal area: $\operatorname{ar}(\Delta CDE) = \operatorname{ar}(\Delta DEF)$.
$6$. Therefore,$\operatorname{ar}(\Delta ABC) = \operatorname{ar}(\Delta AFE) + \operatorname{ar}(\Delta BDF) + \operatorname{ar}(\Delta CDE) + \operatorname{ar}(\Delta DEF)$.
$7$. Substituting the equal areas: $\operatorname{ar}(\Delta ABC) = \operatorname{ar}(\Delta DEF) + \operatorname{ar}(\Delta DEF) + \operatorname{ar}(\Delta DEF) + \operatorname{ar}(\Delta DEF) = 4 \operatorname{ar}(\Delta DEF)$.
$8$. Hence,$\operatorname{ar}(\Delta DEF) = \frac{1}{4} \operatorname{ar}(\Delta ABC)$.
To prove: $\operatorname{ar}(DEF) = \frac{1}{4} \operatorname{ar}(ABC)$.
Proof:
$1$. Since $D$ and $F$ are mid-points of $BC$ and $AB$ respectively,by the Mid-point Theorem,$DF \parallel AC$ and $DF = \frac{1}{2} AC = AE$. Thus,$AFDE$ is a parallelogram.
$2$. Similarly,$BDEF$ and $CDFE$ are parallelograms.
$3$. In parallelogram $BDEF$,the diagonal $DF$ divides it into two triangles of equal area: $\operatorname{ar}(\Delta BDF) = \operatorname{ar}(\Delta DEF)$.
$4$. In parallelogram $AFDE$,the diagonal $DE$ divides it into two triangles of equal area: $\operatorname{ar}(\Delta AFE) = \operatorname{ar}(\Delta DEF)$.
$5$. In parallelogram $CDFE$,the diagonal $DE$ divides it into two triangles of equal area: $\operatorname{ar}(\Delta CDE) = \operatorname{ar}(\Delta DEF)$.
$6$. Therefore,$\operatorname{ar}(\Delta ABC) = \operatorname{ar}(\Delta AFE) + \operatorname{ar}(\Delta BDF) + \operatorname{ar}(\Delta CDE) + \operatorname{ar}(\Delta DEF)$.
$7$. Substituting the equal areas: $\operatorname{ar}(\Delta ABC) = \operatorname{ar}(\Delta DEF) + \operatorname{ar}(\Delta DEF) + \operatorname{ar}(\Delta DEF) + \operatorname{ar}(\Delta DEF) = 4 \operatorname{ar}(\Delta DEF)$.
$8$. Hence,$\operatorname{ar}(\Delta DEF) = \frac{1}{4} \operatorname{ar}(\Delta ABC)$.
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Similar Questions
In 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 SolutionDiagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB || DC$ intersect each other at $O$. Prove that $ar(AOD) = ar(BOC).$
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 SolutionIn the figure,$AP || BQ || CR$. Prove that $\operatorname{ar}(AQC) = \operatorname{ar}(PBR)$.
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
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