# Ex.9.3 Q10 Areas of Parallelograms and Triangles Solution - NCERT Maths Class 9

## Question

Diagonals \(AC\) and \(BD\) of a trapezium \(ABCD\) with \(AB\; || \;DC\) intersect each other at \(O\). Prove that \(ar (AOD) = ar (BOC)\).

## Text Solution

**What is known?**

Diagonals \(AC\) and \(BD\) of a trapezium \(ABCD\) with \(AB || DC\) intersect each other at \(O\).

**What is unknown?**

How we can prove that \(ar (AOD) = ar (BOC).\)

**Reasoning:**

We can use theorem for triangles \(DAC\) and \(DBC\), 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 \(DOC\) from both sides to get the required result.

**Steps:**

It can be observed that \(\begin{align} (\Delta DAC)\, \rm{and} \,(\Delta DBC)\end{align}\) lie on the same base \(DC\) and between the same parallels \(AB\) and \(CD\).

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

\[\begin{align} {ar}\left( \Delta { DAC} \right)&={ar}\left( \Delta { DBC} \right) \\\left[ \begin{array} {ar}\left( \Delta { DAC} \right)- \\ {ar }\left( \Delta { DOC} \right) \\\end{array} \right]&\!\!=\!\!\left[ \begin{array} {ar }\left( \Delta { DBC} \right)- \\ {ar }\left( \Delta { DOC} \right) \\\end{array} \right] \\\end{align}\]

Subtracting area \((\Delta{ DOC})\) from both sides

\[\begin{align} {ar }\left( \Delta{ AOD} \right)&={ ar }\left( \Delta{ BOC} \right) \\\end{align}\]