Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
MediumShow that the diagonals of a parallelogram divide it into four triangles of equal area.
(N/A) Let $ABCD$ be a parallelogram such that its diagonals $AC$ and $BD$ intersect at $O$.
Since the diagonals of a parallelogram bisect each other,we have $AO = OC$ and $BO = OD$.
Draw $CE \perp BD$.
Now,$\text{ar}(\Delta BOC) = \frac{1}{2} \times BO \times CE$ and $\text{ar}(\Delta DOC) = \frac{1}{2} \times OD \times CE$.
Since $BO = OD$,it follows that $\text{ar}(\Delta BOC) = \text{ar}(\Delta DOC) \quad ... (1)$.
Similarly,by drawing perpendiculars from $A$ to $BD$,we can show that $\text{ar}(\Delta AOD) = \text{ar}(\Delta AOB) \quad ... (2)$.
Also,considering $\Delta ABD$ and $\Delta CBD$ on the same base $BD$ and between the same parallels $AD \parallel BC$,we know that $\text{ar}(\Delta ABD) = \text{ar}(\Delta CBD)$.
Since $\text{ar}(\Delta ABD) = \text{ar}(\Delta AOB) + \text{ar}(\Delta AOD)$ and $\text{ar}(\Delta CBD) = \text{ar}(\Delta BOC) + \text{ar}(\Delta DOC)$,we have $\text{ar}(\Delta AOB) + \text{ar}(\Delta AOD) = \text{ar}(\Delta BOC) + \text{ar}(\Delta DOC)$.
Using the relations derived,all four triangles have equal area: $\text{ar}(\Delta AOB) = \text{ar}(\Delta BOC) = \text{ar}(\Delta COD) = \text{ar}(\Delta DOA)$.
Thus,the diagonals of a parallelogram divide it into four triangles of equal area.
Since the diagonals of a parallelogram bisect each other,we have $AO = OC$ and $BO = OD$.
Draw $CE \perp BD$.
Now,$\text{ar}(\Delta BOC) = \frac{1}{2} \times BO \times CE$ and $\text{ar}(\Delta DOC) = \frac{1}{2} \times OD \times CE$.
Since $BO = OD$,it follows that $\text{ar}(\Delta BOC) = \text{ar}(\Delta DOC) \quad ... (1)$.
Similarly,by drawing perpendiculars from $A$ to $BD$,we can show that $\text{ar}(\Delta AOD) = \text{ar}(\Delta AOB) \quad ... (2)$.
Also,considering $\Delta ABD$ and $\Delta CBD$ on the same base $BD$ and between the same parallels $AD \parallel BC$,we know that $\text{ar}(\Delta ABD) = \text{ar}(\Delta CBD)$.
Since $\text{ar}(\Delta ABD) = \text{ar}(\Delta AOB) + \text{ar}(\Delta AOD)$ and $\text{ar}(\Delta CBD) = \text{ar}(\Delta BOC) + \text{ar}(\Delta DOC)$,we have $\text{ar}(\Delta AOB) + \text{ar}(\Delta AOD) = \text{ar}(\Delta BOC) + \text{ar}(\Delta DOC)$.
Using the relations derived,all four triangles have equal area: $\text{ar}(\Delta AOB) = \text{ar}(\Delta BOC) = \text{ar}(\Delta COD) = \text{ar}(\Delta DOA)$.
Thus,the diagonals of a parallelogram divide it into four triangles of equal area.
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$D$ and $E$ are points on sides $AB$ and $AC$ respectively of $\Delta ABC$ such that $\operatorname{ar}(DBC) = \operatorname{ar}(EBC)$. Prove that $DE \parallel BC$.
Medium
View SolutionIn the figure,$ABCD$ is a parallelogram and $BC$ is produced to a point $Q$ such that $AD = CQ$. If $AQ$ intersects $DC$ at $P$,show that $\operatorname{ar}(BPC) = \operatorname{ar}(DPQ)$. [Hint: Join $AC$.]
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View SolutionIn the figure,$ABCDE$ is a pentagon. $A$ line through $B$ parallel to $AC$ meets $DC$ produced at $F$. Show that:
$(i)$ $ar(ACB) = ar(ACF)$
$(ii)$ $ar(AEDF) = ar(ABCDE)$
$(i)$ $ar(ACB) = ar(ACF)$
$(ii)$ $ar(AEDF) = ar(ABCDE)$
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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:
$(i)$ $\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$
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$(iii)$ $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$
$(v)$ $\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$
$(vi)$ $\operatorname{ar}(FED) = \frac{1}{8} \operatorname{ar}(AFC)$
$(i)$ $\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$
$(iii)$ $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$
$(v)$ $\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$
$(vi)$ $\operatorname{ar}(FED) = \frac{1}{8} \operatorname{ar}(AFC)$
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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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