Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
Medium$P$ 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 $\operatorname{ar}(RQC) = \frac{3}{8} \operatorname{ar}(ABC)$.
(N/A) Given: $P$ is the mid-point of $AB$,$Q$ is the mid-point of $BC$,and $R$ is the mid-point of $AP$.
$1$. Since $P$ is the mid-point of $AB$,$CP$ is a median of $\Delta ABC$. Therefore,$\operatorname{ar}(\Delta PBC) = \frac{1}{2} \operatorname{ar}(\Delta ABC)$.
$2$. In $\Delta PBC$,$PQ$ is a median (since $Q$ is the mid-point of $BC$). Therefore,$\operatorname{ar}(\Delta PBQ) = \operatorname{ar}(\Delta PQC) = \frac{1}{2} \operatorname{ar}(\Delta PBC) = \frac{1}{2} \times \frac{1}{2} \operatorname{ar}(\Delta ABC) = \frac{1}{4} \operatorname{ar}(\Delta ABC)$.
$3$. In $\Delta APQ$,$RQ$ is a median (since $R$ is the mid-point of $AP$). Therefore,$\operatorname{ar}(\Delta RQC)$ is not directly related,but consider $\Delta RQC$ as part of $\Delta AQC$.
Alternatively,$\operatorname{ar}(\Delta RQC) = \operatorname{ar}(\Delta RQP) + \operatorname{ar}(\Delta PQC)$.
Since $R$ is the mid-point of $AP$,$\operatorname{ar}(\Delta RQP) = \frac{1}{2} \operatorname{ar}(\Delta APQ)$.
Since $P$ is the mid-point of $AB$,$\operatorname{ar}(\Delta APQ) = \frac{1}{2} \operatorname{ar}(\Delta ABQ) = \frac{1}{2} \times \frac{1}{2} \operatorname{ar}(\Delta ABC) = \frac{1}{4} \operatorname{ar}(\Delta ABC)$.
So,$\operatorname{ar}(\Delta RQP) = \frac{1}{2} \times \frac{1}{4} \operatorname{ar}(\Delta ABC) = \frac{1}{8} \operatorname{ar}(\Delta ABC)$.
$4$. Thus,$\operatorname{ar}(\Delta RQC) = \operatorname{ar}(\Delta RQP) + \operatorname{ar}(\Delta PQC) = \frac{1}{8} \operatorname{ar}(\Delta ABC) + \frac{1}{4} \operatorname{ar}(\Delta ABC) = \frac{1+2}{8} \operatorname{ar}(\Delta ABC) = \frac{3}{8} \operatorname{ar}(\Delta ABC)$.
$1$. Since $P$ is the mid-point of $AB$,$CP$ is a median of $\Delta ABC$. Therefore,$\operatorname{ar}(\Delta PBC) = \frac{1}{2} \operatorname{ar}(\Delta ABC)$.
$2$. In $\Delta PBC$,$PQ$ is a median (since $Q$ is the mid-point of $BC$). Therefore,$\operatorname{ar}(\Delta PBQ) = \operatorname{ar}(\Delta PQC) = \frac{1}{2} \operatorname{ar}(\Delta PBC) = \frac{1}{2} \times \frac{1}{2} \operatorname{ar}(\Delta ABC) = \frac{1}{4} \operatorname{ar}(\Delta ABC)$.
$3$. In $\Delta APQ$,$RQ$ is a median (since $R$ is the mid-point of $AP$). Therefore,$\operatorname{ar}(\Delta RQC)$ is not directly related,but consider $\Delta RQC$ as part of $\Delta AQC$.
Alternatively,$\operatorname{ar}(\Delta RQC) = \operatorname{ar}(\Delta RQP) + \operatorname{ar}(\Delta PQC)$.
Since $R$ is the mid-point of $AP$,$\operatorname{ar}(\Delta RQP) = \frac{1}{2} \operatorname{ar}(\Delta APQ)$.
Since $P$ is the mid-point of $AB$,$\operatorname{ar}(\Delta APQ) = \frac{1}{2} \operatorname{ar}(\Delta ABQ) = \frac{1}{2} \times \frac{1}{2} \operatorname{ar}(\Delta ABC) = \frac{1}{4} \operatorname{ar}(\Delta ABC)$.
So,$\operatorname{ar}(\Delta RQP) = \frac{1}{2} \times \frac{1}{4} \operatorname{ar}(\Delta ABC) = \frac{1}{8} \operatorname{ar}(\Delta ABC)$.
$4$. Thus,$\operatorname{ar}(\Delta RQC) = \operatorname{ar}(\Delta RQP) + \operatorname{ar}(\Delta PQC) = \frac{1}{8} \operatorname{ar}(\Delta ABC) + \frac{1}{4} \operatorname{ar}(\Delta ABC) = \frac{1+2}{8} \operatorname{ar}(\Delta ABC) = \frac{3}{8} \operatorname{ar}(\Delta ABC)$.
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$(i)$ $\operatorname{ar}(BDE) = \frac{1}{4} \operatorname{ar}(ABC)$
$(ii)$ $\operatorname{ar}(BDE) = \frac{1}{2} \operatorname{ar}(BAE)$
$(iii)$ $\operatorname{ar}(ABC) = 2 \operatorname{ar}(BEC)$
$(iv)$ $\operatorname{ar}(BFE) = \operatorname{ar}(AFD)$
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