In figure below, ar $ (\mathrm{DRC})= $ ar $ (\mathrm{DPC}) $ and ar $ (\mathrm{BDP})=\operatorname{ar}(\mathrm{ARC}) $. Show that both the quadrilaterals $ \mathrm{ABCD} $ and $ \mathrm{DCPR} $ are trapeziums. "
Given:
$ar(\mathrm{DRC})=ar(\mathrm{DPC})$ and $ar(\mathrm{BDP})=\operatorname{ar}(\mathrm{ARC})$
To do:
We have to show that both the quadrilaterals \( \mathrm{ABCD} \) and \( \mathrm{DCPR} \) are trapeziums.
Solution:
$ar (\triangle DPC) = ar(\triangle DRC)$.....……(i)
$ar(\triangle BDP) = ar(\triangle ARC)$……...(ii)
Subtracting (i) from (ii), we get,
$ar(\triangle BDP) - ar(\triangle DPC) = ar(\triangle ARC) - ar(\triangle DRC)$
$ar(\triangle BDC) = ar(\triangle ADC)$
Triangles $BDC$ and $ADC$ are on the same base $DC$.
Therefore,
$DC \| AB$
This implies,
$ABCD$ is a trapezium.
Similarly,
$ar(\triangle DRC) = ar(\triangle DPC)$
Triangles $DRC$ and $DPC$ have the same base $DC$.
Therefore,
$RP \| DC$
This implies,
$DCPR$ is a trapezium.
Hence proved.
Related Articles In figure below, \( \mathrm{P} \) is a point in the interior of a parallelogram \( \mathrm{ABCD} \). Show that(i) \( \operatorname{ar}(\mathrm{APB})+\operatorname{ar}(\mathrm{PCD})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD}) \)(ii) \( \operatorname{ar}(\mathrm{APD})+\operatorname{ar}(\mathrm{PBC})=\operatorname{ar}(\mathrm{APB})+\operatorname{ar}(\mathrm{PCD}) \)[Hint: Through \( \mathrm{P} \), draw a line parallel to \( \mathrm{AB} \).]"\n
In figure below, $ABCD, DCFE$ and $ABFE$ are parallelograms. Show that \( \operatorname{ar}(\mathrm{ADE})=\operatorname{ar}(\mathrm{BCF}) \)."\n
In figure below, $D$ and \( \mathrm{E} \) are two points on \( \mathrm{BC} \) such that \( \mathrm{BD}=\mathrm{DE}=\mathrm{EC} \). Show that \( \operatorname{ar}(\mathrm{ABD})=\operatorname{ar}(\mathrm{ADE})=\operatorname{ar}(\mathrm{AEC}) \)."\n
In figure, \( \mathrm{ABC} \) and \( \mathrm{BDE} \) are two equilateral triangles such that \( \mathrm{D} \) is the mid-point of \( \mathrm{BC} \). If \( \mathrm{AE} \) intersects \( \mathrm{BC} \) at \( \mathrm{F} \), show that(i) \( \operatorname{ar}(\mathrm{BDE})=\frac{1}{4} \operatorname{ar}(\mathrm{ABC}) \)(ii) \( \operatorname{ar}(\mathrm{BDE})=\frac{1}{2} \operatorname{ar}(\mathrm{BAE}) \)(iii) \( \operatorname{ar}(\mathrm{ABC})=2 \) ar \( (\mathrm{BEC}) \)(iv) \( \operatorname{ar}(\mathrm{BFE})=\operatorname{ar}(\mathrm{AFD}) \)(v) \( \operatorname{ar}(\mathrm{BFE})=2 \operatorname{ar}(\mathrm{FED}) \)(vi) \( \operatorname{ar}(\mathrm{FED})=\frac{1}{8} \operatorname{ar}(\mathrm{AFC}) \)[Hint: Join \( \mathrm{EC} \) and \( \mathrm{AD} \). Show that \( \mathrm{BE} \| \mathrm{AC} \) and \( \mathrm{DE} \| \mathrm{AB} \)
In figure below, $AP \| BQ \| \mathrm{CR}$. Prove that \( \operatorname{ar}(\mathrm{AQC})=\operatorname{ar}(\mathrm{PBR}) . \)"\n
In figure below, $PQRS$ and $ABRS$ are parallelograms and \( \mathrm{X} \) is any point on side \( \mathrm{BR} \). Show that(i) \( \operatorname{ar}(\mathrm{PQRS})=\operatorname{ar}(\mathrm{ABRS}) \)(ii) \( \operatorname{ar}(\mathrm{AXS})=\frac{1}{2} \mathrm{ar}(\mathrm{PQRS}) \)"\n
In figure below, \( \mathrm{ABC} \) is a right triangle right angled at \( \mathrm{A} . \mathrm{BCED}, \mathrm{ACFG} \) and \( \mathrm{ABMN} \) are squares on the sides \( \mathrm{BC}, \mathrm{CA} \) and \( \mathrm{AB} \) respectively. Line segment \( \mathrm{AX} \perp \mathrm{DE} \) meets \( \mathrm{BC} \) at Y. Show that:(i) \( \triangle \mathrm{MBC} \cong \triangle \mathrm{ABD} \)(ii) \( \operatorname{ar}(\mathrm{BYXD})=2 \operatorname{ar}(\mathrm{MBC}) \)(iii) \( \operatorname{ar}(\mathrm{BYXD})=\operatorname{ar}(\mathrm{ABMN}) \)(iv) \( \triangle \mathrm{FCB} \cong \triangle \mathrm{ACE} \)(v) \( \operatorname{ar}(\mathrm{CYXE})=2 \operatorname{ar}(\mathrm{FCB}) \)(vi) \( \operatorname{ar}(\mathrm{CYXE})=\operatorname{ar}(\mathrm{ACFG}) \)(vii) ar \( (\mathrm{BCED})=\operatorname{ar}(\mathrm{ABMN})+\operatorname{ar}(\mathrm{ACFG}) \)"
\( P \) 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 \( (\mathrm{PBQ})=\operatorname{ar}(\mathrm{ARC}) \)
In figure below, ABCDE is a pentagon. A line through \( \mathrm{B} \) parallel to \( \mathrm{AC} \) meets \( \mathrm{DC} \) produced at F. Show that(i) \( \operatorname{ar}(\mathrm{ACB})=\operatorname{ar}(\mathrm{ACF}) \)(ii) \( \operatorname{ar}(\mathrm{AEDF})=\operatorname{ar}(\mathrm{ABCDE}) \)"\n
Diagonals \( \mathrm{AC} \) and \( \mathrm{BD} \) of a quadrilateral \( \mathrm{ABCD} \) intersect each other at \( \mathrm{P} \). Show that ar \( (\mathrm{APB}) \times \operatorname{ar}(\mathrm{CPD})=\operatorname{ar}(\mathrm{APD}) \times \operatorname{ar}(\mathrm{BPC}) \).[Hint: From \( \mathrm{A} \) and \( \mathrm{C} \), draw perpendiculars to \( \mathrm{BD} \).]
In figure below, diagonals \( \mathrm{AC} \) and \( \mathrm{BD} \) of quadrilateral \( \mathrm{ABCD} \) intersect at \( \mathrm{O} \) such that \( \mathrm{OB}=\mathrm{OD} \).If \( \mathrm{AB}=\mathrm{CD} \), then show that:(i) \( \operatorname{ar}(\mathrm{DOC})=\operatorname{ar}(\mathrm{AOB}) \)(ii) \( \operatorname{ar}(\mathrm{DCB})=\operatorname{ar}(\mathrm{ACB}) \)(iii) DA\| \( \mathrm{CB} \) or \( \mathrm{ABCD} \) is a parallelogram.[Hint : From D and B, draw perpendiculars to AC.]"\n
In figure below, $E$ is any point on median \( \mathrm{AD} \) of a \( \triangle \mathrm{ABC} \). Show that ar \( (\mathrm{ABE})=\operatorname{ar}(\mathrm{ACE}) \)."\n
In figure below, \( \mathrm{ABC} \) and \( \mathrm{ABD} \) are two triangles on the same base \( \mathrm{AB} \). If line- segment \( \mathrm{CD} \) is bisected by \( \mathrm{AB} \) at \( \mathrm{O} \), show that \( \operatorname{ar}(\mathrm{ABC})=\operatorname{ar}(\mathrm{ABD}) \)."\n
\( \mathrm{D}, \mathrm{E} \) and \( \mathrm{F} \) are respectively the mid-points of the sides \( \mathrm{BC}, \mathrm{CA} \) and \( \mathrm{AB} \) of a \( \triangle \mathrm{ABC} \). Show that(i) BDEF is a parallelogram.(ii) \( \operatorname{ar}(\mathrm{DEF})=\frac{1}{4} \operatorname{ar}(\mathrm{ABC}) \)(iii) \( \operatorname{ar}(\mathrm{BDEF})=\frac{1}{2} \operatorname{ar}(\mathrm{ABC}) \)
Diagonals \( \mathrm{AC} \) and \( \mathrm{BD} \) of a trapezium \( \mathrm{ABCD} \) with \( \mathrm{AB} \| \mathrm{DC} \) intersect each other at \( \mathrm{O} \). Prove that ar \( (\mathrm{AOD})=\operatorname{ar}(\mathrm{BOC}) \).
Kickstart Your Career
Get certified by completing the course
Get Started