Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
MediumIn the figure,$PQRS$ and $ABRS$ are parallelograms and $X$ is any point on side $BR$. Show that:
$(i)$ $ar(PQRS) = ar(ABRS)$
$(ii)$ $ar(AXS) = 1/2 \, ar(PQRS)$
$(i)$ $ar(PQRS) = ar(ABRS)$
$(ii)$ $ar(AXS) = 1/2 \, ar(PQRS)$
(N/A) We have two parallelograms $PQRS$ and $ABRS$ and a point $X$ on $BR$.
$(i)$ To prove that $ar(PQRS) = ar(ABRS)$:
Since parallelogram $PQRS$ and parallelogram $ABRS$ are on the same base $RS$ and between the same parallels $RS$ and $PB$,their areas are equal.
Therefore,$ar(PQRS) = ar(ABRS)$.
$(ii)$ To prove that $ar(AXS) = 1/2 \, ar(PQRS)$:
Since $\Delta AXS$ and parallelogram $ABRS$ are on the same base $AS$ and between the same parallels $AS$ and $BR$,the area of the triangle is half the area of the parallelogram.
Therefore,$ar(AXS) = 1/2 \, ar(ABRS)$.
Since $ar(PQRS) = ar(ABRS)$ (from part $i$),
Substituting this into the equation,we get:
$ar(AXS) = 1/2 \, ar(PQRS)$.
$(i)$ To prove that $ar(PQRS) = ar(ABRS)$:
Since parallelogram $PQRS$ and parallelogram $ABRS$ are on the same base $RS$ and between the same parallels $RS$ and $PB$,their areas are equal.
Therefore,$ar(PQRS) = ar(ABRS)$.
$(ii)$ To prove that $ar(AXS) = 1/2 \, ar(PQRS)$:
Since $\Delta AXS$ and parallelogram $ABRS$ are on the same base $AS$ and between the same parallels $AS$ and $BR$,the area of the triangle is half the area of the parallelogram.
Therefore,$ar(AXS) = 1/2 \, ar(ABRS)$.
Since $ar(PQRS) = ar(ABRS)$ (from part $i$),
Substituting this into the equation,we get:
$ar(AXS) = 1/2 \, ar(PQRS)$.
Explore More
Similar Questions
In the figure,$ABC$ is a right triangle right-angled at $A$. $BCED$,$ACFG$,and $ABMN$ are squares on the sides $BC$,$CA$,and $AB$ respectively. Line segment $AX \perp DE$ meets $BC$ at $Y$. Show that: $\operatorname{ar}(BYXD) = 2 \operatorname{ar}(MBC)$
Medium
View SolutionParallelogram $ABCD$ and rectangle $ABEF$ are on the same base $AB$ and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.
Difficult
View SolutionIn the figure,$ABCD$ is a quadrilateral and $BE \parallel AC$. $BE$ meets $DC$ produced at $E$. Show that the area of $\Delta ADE$ is equal to the area of the quadrilateral $ABCD$.
Medium
View SolutionShow that the diagonals of a parallelogram divide it into four triangles of equal area.
Medium
View Solution$A$ farmer has a field in the form of a parallelogram $PQRS$. She took any point $A$ on $RS$ and joined it to points $P$ and $Q$. In how many parts is the field divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?
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