In Fig. 9.5, if parallelogram ABCD and rectangle ABEF are of equal area, then :
a. Perimeter of ABCD = Perimeter of ABEM
b. Perimeter of ABCD < Perimeter of ABEM
c. Perimeter of ABCD > Perimeter of ABEM
d. Perimeter of ABCD = 1/2 (Perimeter of ABEM)
Solution:
In rectangle ABEM,
AB = EM (sides of a rectangle)
In parallelogram ABCD,
CD = AB
By addition we get
AB + CD = EM + AB
As the perpendicular distance between two parallel sides of a parallelogram is less than the length of the other parallel sides.
BE < BC and AM < AD
In a right angled triangle, hypotenuse is greater than the other side
By adding the above inequalities
BE + AM < BC + AD
Or BC + AD > BE + AM
Let us add AB + CD on both sides
AB + CD + BC + AD > AB + CD + BE + AM
So we get
AB + BC + CD + AD > AB + BE + EM + AM (CD = AB = EM)
Perimeter of parallelogram ABCD > Perimeter of rectangle ABEM
Therefore, the perimeter of the parallelogram ABCD > perimeter of the rectangle ABEM.
✦ Try This: The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 10 cm and 5 cm is :
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 9
NCERT Exemplar Class 9 Maths Exercise 9.1 Problem 5
In Fig. 9.5, if parallelogram ABCD and rectangle ABEF are of equal area, then : a. Perimeter of ABCD = Perimeter of ABEM, b. Perimeter of ABCD < Perimeter of ABEM, c. Perimeter of ABCD > Perimeter of ABEM, d. Perimeter of ABCD = 1/2 (Perimeter of ABEM)
Summary:
In Fig. 9.5, if parallelogram ABCD and rectangle ABEF are of equal area, then perimeter of parallelogram ABCD > perimeter of rectangle ABEM
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