# Important Questions for Chapter 9 - Areas of Parallelograms and Triangles

If you are preparing for your final exam or any other test, you can use this free set of important questions for CBSE class 9th Maths PDF for practice.

**Important questions based on NCERT syllabus for Chapter 9 - Areas of Parallelograms and Triangles:**

*Question 1*: ABC and ABD are two triangles on the same base AB. If line- segment CD is bisected by AB at O, show that area(△ABC) = area(△ABD).

*Solution*:

Consider △ACD.

Line-segment CD is bisected by AB at O. Therefore, AO is the median of △ACD

So, Area (△ACO) = Area (△ADO) ... (1)

Considering △BCD, BO is the median.

So, Area (△BCO) = Area (△BDO) ... (2)

Adding equations (1) and (2), we obtain

Area (△ACO) + Area (△BCO) = Area (△ADO) + Area (△BDO)

Area (△ABC) = Area (△ABD)

*Question 2*: In the given figure, AP ∥ BQ ∥ CR. Prove that ar (△AQC) = ar (△PBR).

*Solution*:

Since △ABQ and △PBQ lie on the same base BQ and are between the same parallels AP and BQ,

∴ Area (△ABQ) = Area (△PBQ) ... (1)

Again, △BCQ and △BRQ lie on the same base BQ and are between the same parallels BQ and CR.

∴ Area (△BCQ) = Area (△BRQ) ... (2)

On adding equations (1) and (2), we obtain

Area (△ABQ) + Area (△BCQ) = Area (△PBQ) + Area (△BRQ)

⇒ Area (△AQC) = Area (△PBR)

*Question 3*: Show that a median of a triangle divides it into two triangles of equal areas.

*Solution*:

Let ABC be a triangle and let AD be one of its medians

We need to show that ar (△ABD) = ar (△ACD).

Since the formula for area involves altitude, let us draw AN perpendicular to BC.

Now ar(△ABD) = 1/2 × base × altitude (of triangle △ABD)

=1/2 × BD × AN

=12 × CD × AN (As BD = CD)

=1/2 × base × altitude (of △ACD)

= ar(△ACD)

Hence proved that the median of a triangle divides it into 2 equal halves.