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
Solution:
Given: The parallelogram and the rectangle have the same base and equal areas, therefore, they will also lie between the same parallels.
Consider the parallelogram ABCD and rectangle ABEF as follows.
Here, it can be observed that parallelogram ABCD and rectangle ABEF are lying between the same parallels AB and CF.
It is given that the opposite sides of a parallelogram and a rectangle are of equal lengths.
Therefore, AB = EF (Opposite sides of a rectangle are equal.)
AB = CD (Opposite sides of a parallelogram are equal.)
∴ CD = EF
∴ AB + CD = AB + EF ...(1)
Now, in a right-angled triangle AFD, AD is a hypotenuse.
∴ AF < AD ...(2)
Similarly, in a right angled triangle EBC, EB is altitude and BC is a hypotenuse
∴ BE < BC ...(3)
Adding equation (2) and (3),
∴ AF + BE < AD + BC ...(4)
Now, from equations (1) and (4), we obtain
AB + EF + AF + BE < AD + BC + AB + CD
Perimeter of rectangle ABEF < Perimeter of parallelogram ABCD.
☛ Check: NCERT Solutions Class 9 Maths Chapter 9
Video Solution:
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
Maths NCERT Solutions Class 9 Chapter 9 Exercise 9.4 Question 1
Summary:
If parallelogram ABCD and rectangle ABEF are on the same base AB and have equal areas, then the perimeter of the parallelogram is greater than that of the rectangle.
☛ Related Questions:
- In Fig. 9.30, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).
- In Fig. 9.31, ABCD, DCFE and ABFE are parallelograms. Show that ar (ADE) = ar (BCF).
- In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ).[Hint: Join AC.]
- In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show thati) ar (BDE) =1/4 ar (ABC)ii) ar (BDE) = 1/2 ar (BAE)iii) ar (ABC) = 2 ar (BEC)iv) ar (BFE) = ar (AFD)v) ar (BFE) = 2 ar (FED)vi) ar (FED) = 1/8 ar (AFC)[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]