Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
DifficultIn the figure,$ABCD$ is a parallelogram and $EFCD$ is a rectangle. Also,$AL \perp DC$. Prove that:
$(i)$ $\text{ar}(ABCD) = \text{ar}(EFCD)$
$(ii)$ $\text{ar}(ABCD) = DC \times AL$
$(i)$ $\text{ar}(ABCD) = \text{ar}(EFCD)$
$(ii)$ $\text{ar}(ABCD) = DC \times AL$
(N/A) $(i)$ Since a rectangle is also a parallelogram,and both $ABCD$ and $EFCD$ are on the same base $DC$ and between the same parallels $DC$ and $EF$,therefore,$\text{ar}(ABCD) = \text{ar}(EFCD)$.
$(ii)$ From the result above,$\text{ar}(ABCD) = \text{ar}(EFCD)$.
Since $EFCD$ is a rectangle,its area is $\text{length} \times \text{breadth} = DC \times FC$.
Thus,$\text{ar}(ABCD) = DC \times FC$ $(1)$.
Since $AL \perp DC$ and $EF \parallel DC$,$AL$ is the height of the parallelogram. In the rectangle $AFCL$ (or by considering the parallel lines),we have $AL = FC$ $(2)$.
Substituting $(2)$ in $(1)$,we get $\text{ar}(ABCD) = DC \times AL$.
$(ii)$ From the result above,$\text{ar}(ABCD) = \text{ar}(EFCD)$.
Since $EFCD$ is a rectangle,its area is $\text{length} \times \text{breadth} = DC \times FC$.
Thus,$\text{ar}(ABCD) = DC \times FC$ $(1)$.
Since $AL \perp DC$ and $EF \parallel DC$,$AL$ is the height of the parallelogram. In the rectangle $AFCL$ (or by considering the parallel lines),we have $AL = FC$ $(2)$.
Substituting $(2)$ in $(1)$,we get $\text{ar}(ABCD) = DC \times AL$.
Explore More
Similar Questions
$D$ and $E$ are points on sides $AB$ and $AC$ respectively of $\Delta ABC$ such that $\operatorname{ar}(DBC) = \operatorname{ar}(EBC)$. Prove that $DE \parallel BC$.
Medium
View SolutionDiagonals $AC$ and $BD$ of a quadrilateral $ABCD$ intersect each other at $P$. Show that $ar(APB) \times ar(CPD) = ar(APD) \times ar(BPC)$.
Medium
View Solution$P$ 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 $\operatorname{ar}(RQC) = \frac{3}{8} \operatorname{ar}(ABC)$.
Medium
View SolutionIn the figure,$P$ is a point in the interior of a parallelogram $ABCD$. Show that:
$(i)$ $\operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$
$(ii)$ $\operatorname{ar}(APD) + \operatorname{ar}(PBC) = \operatorname{ar}(APB) + \operatorname{ar}(PCD)$
[Hint: Through $P$,draw a line parallel to $AB$.]
$(i)$ $\operatorname{ar}(APB) + \operatorname{ar}(PCD) = \frac{1}{2} \operatorname{ar}(ABCD)$
$(ii)$ $\operatorname{ar}(APD) + \operatorname{ar}(PBC) = \operatorname{ar}(APB) + \operatorname{ar}(PCD)$
[Hint: Through $P$,draw a line parallel to $AB$.]
Difficult
View SolutionShow that a median of a triangle divides it into two triangles of equal areas.
Medium
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