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 $ABEF$.
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