Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
Difficult$P$ 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 $\operatorname{ar} (PRQ) = \frac{1}{2} \operatorname{ar} (ARC)$.
(N/A) Given: In $\Delta ABC$,$P$ is the mid-point of $AB$,$Q$ is the mid-point of $BC$,and $R$ is the mid-point of $AP$.
To prove: $\operatorname{ar} (PRQ) = \frac{1}{2} \operatorname{ar} (ARC)$.
Proof:
$1$. In $\Delta ABQ$,$P$ is the mid-point of $AB$. Since the median divides a triangle into two triangles of equal area,$PQ$ is a median of $\Delta ABQ$.
Therefore,$\operatorname{ar} (APQ) = \frac{1}{2} \operatorname{ar} (ABQ)$.
$2$. In $\Delta APQ$,$R$ is the mid-point of $AP$. Thus,$QR$ is a median of $\Delta APQ$.
Therefore,$\operatorname{ar} (PRQ) = \frac{1}{2} \operatorname{ar} (APQ) = \frac{1}{2} \times (\frac{1}{2} \operatorname{ar} (ABQ)) = \frac{1}{4} \operatorname{ar} (ABQ)$.
$3$. Since $AQ$ is a median of $\Delta ABC$,$\operatorname{ar} (ABQ) = \frac{1}{2} \operatorname{ar} (ABC)$.
Substituting this,$\operatorname{ar} (PRQ) = \frac{1}{4} \times (\frac{1}{2} \operatorname{ar} (ABC)) = \frac{1}{8} \operatorname{ar} (ABC)$.
$4$. Now,consider $\Delta ARC$. Since $R$ is the mid-point of $AP$,$CR$ is a median of $\Delta APC$.
Therefore,$\operatorname{ar} (ARC) = \frac{1}{2} \operatorname{ar} (APC)$.
$5$. Since $CP$ is a median of $\Delta ABC$,$\operatorname{ar} (APC) = \frac{1}{2} \operatorname{ar} (ABC)$.
Therefore,$\operatorname{ar} (ARC) = \frac{1}{2} \times (\frac{1}{2} \operatorname{ar} (ABC)) = \frac{1}{4} \operatorname{ar} (ABC)$.
$6$. Comparing the results: $\operatorname{ar} (PRQ) = \frac{1}{8} \operatorname{ar} (ABC)$ and $\operatorname{ar} (ARC) = \frac{1}{4} \operatorname{ar} (ABC)$.
Clearly,$\frac{1}{8} \operatorname{ar} (ABC) = \frac{1}{2} \times (\frac{1}{4} \operatorname{ar} (ABC))$.
Thus,$\operatorname{ar} (PRQ) = \frac{1}{2} \operatorname{ar} (ARC)$.
To prove: $\operatorname{ar} (PRQ) = \frac{1}{2} \operatorname{ar} (ARC)$.
Proof:
$1$. In $\Delta ABQ$,$P$ is the mid-point of $AB$. Since the median divides a triangle into two triangles of equal area,$PQ$ is a median of $\Delta ABQ$.
Therefore,$\operatorname{ar} (APQ) = \frac{1}{2} \operatorname{ar} (ABQ)$.
$2$. In $\Delta APQ$,$R$ is the mid-point of $AP$. Thus,$QR$ is a median of $\Delta APQ$.
Therefore,$\operatorname{ar} (PRQ) = \frac{1}{2} \operatorname{ar} (APQ) = \frac{1}{2} \times (\frac{1}{2} \operatorname{ar} (ABQ)) = \frac{1}{4} \operatorname{ar} (ABQ)$.
$3$. Since $AQ$ is a median of $\Delta ABC$,$\operatorname{ar} (ABQ) = \frac{1}{2} \operatorname{ar} (ABC)$.
Substituting this,$\operatorname{ar} (PRQ) = \frac{1}{4} \times (\frac{1}{2} \operatorname{ar} (ABC)) = \frac{1}{8} \operatorname{ar} (ABC)$.
$4$. Now,consider $\Delta ARC$. Since $R$ is the mid-point of $AP$,$CR$ is a median of $\Delta APC$.
Therefore,$\operatorname{ar} (ARC) = \frac{1}{2} \operatorname{ar} (APC)$.
$5$. Since $CP$ is a median of $\Delta ABC$,$\operatorname{ar} (APC) = \frac{1}{2} \operatorname{ar} (ABC)$.
Therefore,$\operatorname{ar} (ARC) = \frac{1}{2} \times (\frac{1}{2} \operatorname{ar} (ABC)) = \frac{1}{4} \operatorname{ar} (ABC)$.
$6$. Comparing the results: $\operatorname{ar} (PRQ) = \frac{1}{8} \operatorname{ar} (ABC)$ and $\operatorname{ar} (ARC) = \frac{1}{4} \operatorname{ar} (ABC)$.
Clearly,$\frac{1}{8} \operatorname{ar} (ABC) = \frac{1}{2} \times (\frac{1}{4} \operatorname{ar} (ABC))$.
Thus,$\operatorname{ar} (PRQ) = \frac{1}{2} \operatorname{ar} (ARC)$.
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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:
$(i)$ $\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$
$(iii)$ $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$
$(v)$ $\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$
$(vi)$ $\operatorname{ar}(FED) = \frac{1}{8} \operatorname{ar}(AFC)$
$(i)$ $\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$
$(iii)$ $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$
$(v)$ $\operatorname{ar}(BFE) = 2 \operatorname{ar}(FED)$
$(vi)$ $\operatorname{ar}(FED) = \frac{1}{8} \operatorname{ar}(AFC)$
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View Solution$A$ villager,Itwaari,has a plot of land in the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given an equal amount of land in lieu of his land,adjoining his plot,so as to form a triangular plot. Explain how this proposal will be implemented.
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View SolutionWhich of the following figures lie on the same base and between the same parallels? In such a case,write the common base and the two parallels.
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