# Ex.9.3 Q16 Areas of Parallelograms and Triangles Solution - NCERT Maths Class 9

## Question

In the given figure,

\(ar (DRC)\) \(= ar (DPC)\) and \(ar (BDP) = ar (ARC)\).

Show that both the quadrilaterals \(ABCD\) and \(DCPR\) are trapeziums.

## Text Solution

**What is known?**

\(ar (DRC) = ar (DPC)\) and \(ar (BDP) = ar (ARC).\)

**What is unknown?**

How we can show that quadrilaterals \(ABCD\) and \(DCPR\) are trapeziums.

**Reasoning:**

We know that if Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.

**Steps:**

It is given that

\(\text{Area} (\Delta DRC )= \text{Area} (\Delta DPC )\)

As \(\Delta DRC\) and \(\Delta DPC\) lie on the same base \(DC\) and have equal areas, therefore, they must lie between the same parallel lines.

According to Theorem \(9.3\) : **Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.**

\[\therefore {{DC || RP}}\]

Therefore, \(DCPR\) is a trapezium.

It is also given that

\(\begin{align}\text { Area }(\Delta {BDP})=\text { Area }(\Delta {ARC})\end{align}\)

Area \((\Delta \, BDP)\) - Area \((\Delta\, DPC)\) = Area \((\triangle \,ARC)\) -Area \((\Delta \, DRC)\)

\(\begin{align} {\therefore \text { Area }(\Delta {BDC})}&={\text { Area }(\Delta {ADC})}\end{align}\)

Since \(\begin{align} (\Delta BDC )\, and \,(\Delta ADC)\end{align}\) are on the same base \(CD\) and have equal areas, they must lie between the same parallel lines.

According to Theorem \(9.3\) : **Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.**

\[\begin{align}\therefore \quad A B \| C D\end{align}\]

Therefore, \(ABCD\) is a trapezium.