Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
EasyIn trapezium $ABCD$,$AB || CD$ and diagonals $AC$ and $BD$ intersect at point $O$. Prove that $ar(AOD) = ar(BOC)$.
(N/A) Given: In trapezium $ABCD$,$AB || CD$ and diagonals $AC$ and $BD$ intersect at point $O$.
To prove: $ar(AOD) = ar(BOC)$.
Proof:
$1$. Triangles $ADC$ and $BDC$ lie on the same base $CD$ and between the same parallels $AB || CD$.
$2$. Therefore,$ar(ADC) = ar(BDC)$ (Triangles on the same base and between the same parallels are equal in area).
$3$. Subtract $ar(DOC)$ from both sides:
$ar(ADC) - ar(DOC) = ar(BDC) - ar(DOC)$.
$4$. From the figure,$ar(ADC) - ar(DOC) = ar(AOD)$ and $ar(BDC) - ar(DOC) = ar(BOC)$.
$5$. Thus,$ar(AOD) = ar(BOC)$.
Hence proved.
To prove: $ar(AOD) = ar(BOC)$.
Proof:
$1$. Triangles $ADC$ and $BDC$ lie on the same base $CD$ and between the same parallels $AB || CD$.
$2$. Therefore,$ar(ADC) = ar(BDC)$ (Triangles on the same base and between the same parallels are equal in area).
$3$. Subtract $ar(DOC)$ from both sides:
$ar(ADC) - ar(DOC) = ar(BDC) - ar(DOC)$.
$4$. From the figure,$ar(ADC) - ar(DOC) = ar(AOD)$ and $ar(BDC) - ar(DOC) = ar(BOC)$.
$5$. Thus,$ar(AOD) = ar(BOC)$.
Hence proved.
Explore More
Similar Questions
In $\triangle ABC$,$D$ is the mid-point of $AB$ and $P$ is any point on $BC$. If $CQ \parallel PD$ meets $AB$ in $Q$,then prove that $\operatorname{ar}(\triangle BPQ) = \frac{1}{2} \operatorname{ar}(\triangle ABC)$.
Medium
View SolutionIn the given figure,$ABCD$ is a trapezium with $AB \parallel DC$. $E$ is a point on extended $BC$. Prove that $ar(BDE) = ar(ACED)$.
Medium
View SolutionIn $\Delta ABC$,point $D$ lies on side $BC$. $E$ is the midpoint of $AD$. Prove that,$ar(\Delta EBC) = \frac{1}{2} ar(\Delta ABC)$.
Medium
View SolutionThe figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides $8 \, cm$ and $6 \, cm$ is:
Easy
View SolutionIn the figure,$ABCD$ is a parallelogram. Points $P$ and $Q$ on $BC$ trisect $BC$ into three equal parts. Prove that $\operatorname{ar}(APQ) = \operatorname{ar}(DPQ) = \frac{1}{6} \operatorname{ar}(ABCD)$.
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo