# $P$ and $Q$ are respectively the mid-points of sides $\mathrm{AB}$ and $\mathrm{BC}$ of a triangle $\mathrm{ABC}$ and $\mathrm{K}$ is the mid-point of $\mathrm{AP}$, show that(i) $\operatorname{ar}(\mathrm{PRQ})=\frac{1}{2} \operatorname{ar}(\mathrm{ARC})$(ii) ar $(\mathrm{RQC})=\frac{3}{8}$ ar $(\mathrm{ABC})$(iii) ar $(\mathrm{PBQ})=\operatorname{ar}(\mathrm{ARC})$

Given:

$P$ and $Q$ are respectively the mid-points of sides $\mathrm{AB}$ and $\mathrm{BC}$ of a triangle $\mathrm{ABC}$ and $\mathrm{K}$ is the mid-point of $\mathrm{AP}$

To do:

We have to show that

(i) $\operatorname{ar}(\mathrm{PRQ})=\frac{1}{2} \operatorname{ar}(\mathrm{ARC})$
(ii) ar $(\mathrm{RQC})=\frac{3}{8}$ ar $(\mathrm{ABC})$
(iii) ar $(\mathrm{PBQ})=\operatorname{ar}(\mathrm{ARC})$

Solution:

We know that,

The median of a triangle divides it into two triangles of equal area.

$PC$ is the median of $\triangle ABC$.

$ar (\triangle BPC) = ar (\triangle APC)$……….(i)

$RC$ is the median of $\triangle APC$.

$ar (\triangle ARC) = \frac{1}{2}ar (\triangle APC)$……….(ii)

$PQ$ is the median of $\triangle BPC$.

This implies,

$ar (\triangle PQC) = \frac{1}{2}ar (\triangle\triangle BPC)$……….(iii)

From (i) and (iii), we get,

$ar (\triangle PQC) = \frac{1}{2} ar (\triangle APC)$……….(iv)

From (ii) and (iv), we get,

$ar (\triangle PQC) = ar (\triangle ARC)$……….(v)

$P$ and $Q$ are the mid-points of $AB$ and $BC$ respectively.

$PQ \| AC$

$PA = \frac{1}{2}AC$

The triangles between the same parallels are equal in area.

This implies,

$ar (\triangle APQ) = ar (\triangle PQC)$……….(vi)

From (v) and (vi), we get,

$ar (\triangle APQ) = ar (\triangle ARC)$……….(vii)

$R$ is the mid-point of $AP$.

$RQ$ is the median of $\triangle APQ$.

This implies,

$ar (\triangle PRQ) = \frac{1}{2}ar (\triangle APQ)$……….(viii)

From (vii) and (viii), we get,

$ar (\triangle PRQ) = \frac{1}{2}ar (\triangle ARC)$

(ii) $PQ$ is the median of $\triangle BPC$

$ar (\triangle PQC) = \frac{1}{2} ar (\triangle BPC)$

$= \frac{1}{2}\times[\frac{1}{2}ar (\triangle ABC)]$

$= \frac{1}{4}ar (\triangle ABC)$……….(ix)

$ar (\triangle PRC) = \frac{1}{2} ar (\triangle APC)$         [From (iv)]

$ar (\triangle PRC) = \frac{1}{2}\times[\frac{1}{2}ar (\triangle ABC)]$

$= \frac{1}{4}ar (\triangle ABC)$……….(x)

Adding (ix) and (x), we get,

$ar (\triangle PQC) + ar (\triangle PRC) =(\frac{1}{4}+\frac{1}{4})ar (\triangle ABC)$

$ar (PQCR) = \frac{1}{2} ar (\triangle ABC)$……….(xi)

Subtracting $ar(\triangle PRQ)$ from both sides,

$ar (PQCR)-ar (\triangle PRQ) = \frac{1}{2}ar (\triangle ABC)-ar (\triangle PRQ)$

$ar (\triangle RQC) = \frac{1}{2}ar (\triangle ABC) - \frac{1}{2} ar (\triangle ARC)$            (Proved)

$ar (\triangle ARC) = \frac{1}{2} ar (\triangle ABC) -\frac{1}{2} \times [\frac{1}{2}ar (\triangle APC)]$

$ar (\triangle RQC) = \frac{1}{2}ar (\triangle ABC)-(\frac{1}{4})ar (\triangle APC)$

$ar (\triangle RQC) = \frac{1}{2}ar (\triangle ABC)-\frac{1}{4} \times [\frac{1}{2}ar (\triangle ABC)]$        (Since $PC$ is the median of $\triangle ABC$)

$ar (\triangle RQC) = \frac{1}{2}ar (\triangle ABC)-\frac{1}{8}ar (\triangle ABC)$

$ar (\triangle RQC) = [(\frac{1}{2}-(\frac{1}{8})]ar (\triangle ABC)$

$ar (\triangle RQC) = \frac{3}{8}ar (\triangle ABC)$

(iii) $ar (\triangle PRQ) = \frac{1}{2}ar (\triangle ARC)$

$2ar (\triangle PRQ) = ar (\triangle ARC)$……………..(xii)

$ar (\triangle PRQ) = \frac{1}{2}ar (\triangle APQ)$          ($RQ$ is the median of APQ$)……….(xiii)$ar (\triangle APQ) = ar (\triangle PQC)$From (xiii) and (xiv), we get,$ar (\triangle PRQ) = \frac{1}{2} ar (\triangle PQC)$……….(xv)$ar (\triangle BPQ) = ar (\triangle PQC)$($PQ$is the median of \triangle BPC$)……….(xvi)

From (xv) and (xvi), we get,

$ar (\triangle PRQ) =\frac{1}{2}ar (\triangle BPQ)$……….(xvii)

From (xii) and (xvii), we get,

$2\times(\frac{1}{2})ar(\triangle BPQ)= ar (\triangle ARC)$

$ar (\triangle BPQ) = ar (\triangle ARC)$

Hence Proved.

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Updated on: 10-Oct-2022

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