# Ex 9.4, 1 (Optional) - Chapter 9 Class 9 - Areas of Parallelograms and Triangles (Deleted)

Last updated at April 16, 2019 by Teachoo

Last updated at April 16, 2019 by Teachoo

Transcript

Ex 9.4, 1 (Optional) Parallelogram 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. Our figure would look like Finding perimeter of ABCD and ABEF Perimeter of ABCD ABCD is a parallelogram So, AB = CD BC = AD (Opposite sides of parallelogram are equal) Perimeter of ABCD = AB + BC + CD + DA = AB + DA + AB + DA = 2AB + 2DA = 2(AB + DA) Perimeter of ABEF ABEF is a rectangle with Length = l = AB Breadth = b = AF Perimeter of ABEF = 2 (l + b) = 2 (AB + AF) Now, As AB is common in both perimeters, we have to prove that DA > AF Now, in right triangle ∆ADF Hypotenuse is the largest side in a right angled triangle. ∴ AD > AF Hence, 2(AB + DA) > 2 (AB + AF) Perimeter of ABCD > Perimeter of ABER ∴ Perimeter of Parallelogram is greater than Perimeter of rectangle.

Ex 9.4 (Optional)

Ex 9.4, 1 (Optional)
Deleted for CBSE Board 2022 Exams
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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.