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

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**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.