# Ex.9.3 Q9 Areas of Parallelograms and Triangles Solution - NCERT Maths Class 9

## Question

The side \(AB\) of a parallelogram \(ABCD\) is produced to any point \(P\). \(A\) line through \(A\) and parallel to \(CP\) meets \(CB\) produced at \(Q\) and then parallelogram \(PBQR\) is completed (see the following figure). Show that \(ar (ABCD) = ar (PBQR)\).

[Hint: Join \(AC\) and \(PQ\). Now compare area \((ACQ)\) and area \((APQ)\)]

## Text Solution

**What is known?**

The side AB of a parallelogram \(ABCD\) is produced to any point \(p\). A line through A and parallel to \(CP\) meets \(CB\) produced at \(Q\) and then parallelogram \(PBQR\) is completed.

**What is unknown?**

How we can show that \(ar (ABCD) = ar (PBQR).\)

**Reasoning:**

First of all, we can join \(AC\) and \(PQ\). Now we can use theorem for triangles \(ACQ\) and \(AQP\) if two triangles are on same base and between same pair of parallel lines then both will have equal area. Now we can subtract common area of triangle \(ABQ\) from both sides. Now we can compare the result with half of which we have to show.

**Steps:**

Let us join \(AC\) and \(PQ\).

\(\Delta ACQ\) and \(\Delta AQP\) are on the same base \(AQ\) and between the same parallels \(AQ\) and \(CP\).

According to Theorem 9.2: **Two triangles on the same base (or equal bases) and between the same parallels are equal in area.**

\[\begin{align} \text {ar}(\Delta {ACQ})&=\text {ar}(\Delta {APQ})\end{align}\]

\[\,\left[ \begin{array}{l}{\rm{ar}}(\Delta {{ACQ}})\, -\\{\rm{ar}}(\Delta {{ABQ}})\end{array} \right]\!\!=\!\!\left[ \begin{array}{l}{\rm{ar}}(\Delta {{APQ}})\, - \\{{ar}}(\Delta {{ABQ}})\end{array} \right]\]

Subtracting Area \((\Delta {ABQ})\) on both the sides.

\[\begin{align} (\Delta {ABC})&=\text {ar}(\Delta {QBP}) ...(1)\end{align}\]

Since and are diagonals of parallelograms and respectively,

\[\begin{align}\text {ar}(\Delta {ABC})&=\frac{1}{2} \text {ar}({ABCD})...(2) \\ \text {ar}(\Delta {QBP})&=\frac{1}{2} {ar(PBQR) } ...(3)\end{align}\]

From Equations (), (), and (), we obtain

\[\begin{align}\frac{1}{2} \text {ar}({ABCD})&=\frac{1}{2} \text {ar}(PBQR) \\ \text {ar}({ABCD})&=\text {ar}({PBQR})\end{align}\]