# Ex.9.4 Q1 Areas of Parallelograms and Triangles Solution - NCERT Maths Class 9

## Question

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.

## Text Solution

**What is known?**

Parallelogram \(ABCD\) and rectangle \(ABEF\) are on the same base \(AB\) and have equal areas.

**What is unknown?**

How we can show that the perimeter of the parallelogram is greater than that of the rectangle.

**Reasoning:**

When we compare the sides of rectangle and parallelogram which are on same base, we can see two sides of parallelogram are opposite to \(90\) degree which shows that these two are longer than rectangle two sides and as we find the perimeter of both quadrilateral, we will get the required result.

**Steps:**

As the parallelogram and the rectangle have the same base and equal area, therefore, these 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 between thesame parallels \(AB\) and \(CF\).

We know that opposite sides of a parallelogram or a rectangle are of equal lengths. Therefore,

\[\begin{align}&{AB}={EF} \text { (For rectangle) } \\ &{AB}={CD}(\text { For parallelogram }) \\ &\therefore {CD}={EF} \\ &\therefore {AB}+{CD}={AB}+{EF} \quad \ldots(1)\end{align}\]

Of all the line segments that can be drawn to a given line from a point not lying on it, the perpendicular line segment is the shortest.

\[ \therefore {AF}<{AD} \]

And similarly, \(BE < BC\)

\[ \therefore {AF}+{BE}<{AD}+{BC}\quad \ldots(2) \]

From Equations (\(1\)) and (\(2\)), we obtain

\[ \begin{align} &{AB}+{EF}+{AF}+{BE} \\ & <{AD}+{BC}+{AB}+{CD} \end{align} \]

Perimeter of rectangle \(ABEF\) \(<\) Perimeter of parallelogram \(ABCD\).