Class 9MathematicsAreas of Parallelograms and TrianglesTextbook - Areas of Parallelograms and Triangles
DifficultParallelogram $ABCD$ 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.
(N/A) Given: Parallelogram $ABCD$ and rectangle $ABEF$ are on the same base $AB$ and $ar(ABCD) = ar(ABEF)$.
$1$. Since they are on the same base $AB$ and have equal areas,their heights must be equal. Let the height be $h = AF = BE$.
$2$. In a rectangle,the opposite sides are equal,so $AB = EF$. In a parallelogram,opposite sides are equal,so $AB = CD$. Thus,$CD = EF$.
$3$. The perimeter of rectangle $ABEF = 2(AB + AF)$.
$4$. The perimeter of parallelogram $ABCD = 2(AB + BC)$.
$5$. In the right-angled triangle $\triangle BEC$,$BC$ is the hypotenuse and $BE$ is one of the sides. Therefore,$BC > BE$.
$6$. Adding $AB$ to both sides of the inequality $BC > BE$,we get $AB + BC > AB + BE$.
$7$. Multiplying by $2$,we get $2(AB + BC) > 2(AB + BE)$.
$8$. Since $AB = EF$,the perimeter of the rectangle is $2(AB + BE) = AB + EF + BE + AF$.
$9$. Thus,the perimeter of the parallelogram $ABCD$ is greater than the perimeter of the rectangle $ABEF$.
$1$. Since they are on the same base $AB$ and have equal areas,their heights must be equal. Let the height be $h = AF = BE$.
$2$. In a rectangle,the opposite sides are equal,so $AB = EF$. In a parallelogram,opposite sides are equal,so $AB = CD$. Thus,$CD = EF$.
$3$. The perimeter of rectangle $ABEF = 2(AB + AF)$.
$4$. The perimeter of parallelogram $ABCD = 2(AB + BC)$.
$5$. In the right-angled triangle $\triangle BEC$,$BC$ is the hypotenuse and $BE$ is one of the sides. Therefore,$BC > BE$.
$6$. Adding $AB$ to both sides of the inequality $BC > BE$,we get $AB + BC > AB + BE$.
$7$. Multiplying by $2$,we get $2(AB + BC) > 2(AB + BE)$.
$8$. Since $AB = EF$,the perimeter of the rectangle is $2(AB + BE) = AB + EF + BE + AF$.
$9$. Thus,the perimeter of the parallelogram $ABCD$ is greater than the perimeter of the rectangle $ABEF$.
Explore More
Similar Questions
In the figure,$ABC$ is a right-angled triangle with the right angle at $A$. $BCED$,$ACFG$,and $ABMN$ are squares constructed on the sides $BC$,$CA$,and $AB$ respectively. The line segment $AX \perp DE$ meets $BC$ at $Y$. Show that: $\Delta FCB \cong \Delta ACE$.
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 SolutionIn 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: $\Delta MBC \cong \Delta ABD$.
Medium
View Solution$ABCD$ is a trapezium with $AB || DC$. $A$ line parallel to $AC$ intersects $AB$ at $X$ and $BC$ at $Y$. Prove that $\operatorname{ar}(ADX) = \operatorname{ar}(ACY)$. [Hint: Join $CX$.]
Difficult
View SolutionIn the figure,$ABC$ and $ABD$ are two triangles on the same base $AB$. If line segment $CD$ is bisected by $AB$ at $O$,show that $\operatorname{ar}(ABC) = \operatorname{ar}(ABD)$.
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