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
If $E, F, G$ and $H$ are respectively the mid-points of the sides of a parallelogram $ABCD$,show that $\text{ar}(EFGH) = \frac{1}{2} \text{ar}(ABCD)$.
Difficult
View Solution$D, E$ and $F$ are respectively the mid-points of the sides $BC, CA$ and $AB$ of a $\Delta ABC$. Show that $BDEF$ is a parallelogram.
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 Solution$A$ villager,Itwaari,has a plot of land in the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given an equal amount of land in lieu of his land,adjoining his plot,so as to form a triangular plot. Explain how this proposal will be implemented.
Difficult
View SolutionIn the figure,$PQRS$ and $ABRS$ are parallelograms and $X$ is any point on side $BR$. Show that:
$(i)$ $ar(PQRS) = ar(ABRS)$
$(ii)$ $ar(AXS) = 1/2 \, ar(PQRS)$
$(i)$ $ar(PQRS) = ar(ABRS)$
$(ii)$ $ar(AXS) = 1/2 \, ar(PQRS)$
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