NCERT solutions for class 9 Maths Chapter 9

Looking for NCERT solutions for chapter 9 of class 9 mathematics? In this post we are sharing the solutions to all the exercises of Chapter 9.

NCERT solutions for class 9 Maths Chapter 9 Exercise 9.1





(i)Common base is DC and // lines are AB and DC

(iii) Common base is QR and // lines are PS and QR

(v) Common base is AD and // lines are AD and BQ



NCERT solutions for class 9 Maths Chapter 9 Exercise 9.2



Given: AB = 16 cm, AE = 8 cm and CF = 10 cm

AB = DC = 16 CM (Opposite sides of a //gm are equal)

Area of //gm ABCD = base x height

Since there are two bases and two heights, we can say

Area of //gm ABCD = DC x AE …. (i)

Area of //gm ABCD = AD x CF … (ii)

By (i) and (ii)

AD x CF = DC x AE

AD x 10 = 16 x 8

AD = (16 X 8) / 10

AD = 12.8 cm










Area (//gm ABCD) = base x height

Base could be either DC or AB and height could be either BC or AD

Ar (triangle EHF) = ½ x OE x HF

But OE = ½ x BC and HF= DC

Therefore Ar (triangle EHF) = ½ x ½ x BC x DC => ¼ x ar (ABCD) …(i)




Area of (triangle GHF) = ½ x OG x HF

But OG = ½ x BC and HF= DC

Therefore Area (triangle EHF) = ½ x ½ x BC x DC => ¼ x ar (ABCD) … (ii)

Adding (i) and (ii)




Area (triangle EHF) + Area of (triangle EHF) = ¼ x ar (ABCD) + ¼ x ar (ABCD)

Area of (//gm EFGH) = ½ X (//gm ABCD)








Refer example number 2 in NCERT, there it is proved that a triangle and a //gm on the same base and between same // lines, then

½ x ar (//gm) = ar (triangle)




Ar (APB) = ½ x Ar (ABCD) … (i)

Similarly, Ar (BQC) = ½ x Ar (ABCD) … (ii)

By equating (i) and (ii) we get

Ar (APB) = Ar (BQC)














Construct a line // to DC and AB

When a triangle and a //gm on the same base and between same // lines, then

½ x ar (//gm) = ar (triangle)




Ar (PCD) = ½ x Ar (QRCD) … (i)

Similarly, Ar (APB) = ½ x Ar (QRAB) … (ii)

Adding (i) and (ii) we get

Ar (PCD) + Ar (APB) = ½ x Ar (QRCD) + ½ x Ar (QRAB)

= ½ x [Ar (QRCD) + Ar (QRAB)]

= ½ Ar (ABCD) … (v)






Similarly construct a line RT //to AD and BC




Ar (APD) = ½ x Ar (ARTD) … (iii)

Similarly, Ar (BPC) = ½ x Ar (RBCT) … (iv)

Adding (iii) and (iv) we get

Ar (APD) + Ar (BPC) = ½ x Ar (ARTD) + ½ x Ar (RBCT)

= ½ x [Ar (ARTD) + Ar (RBCT)]

= ½ Ar (ABCD) … (vi)




By equating (v) and (vi)

Ar (PCD) + Ar (APB) = Ar (APD) + Ar (BPC)








PQRS and ABRS are //gms with same base RS

AR (PQRS) = AR (ABRS) (Parallelograms having the same base and between the same // lines have equal area) … (i)

Triangle ASX and //gm ABRS have same base AS and are between the same // lines AS and BR. When a triangle and a //gm on the same base and between same // lines, then

½ x ar (//gm) = ar (triangle)




Ar (ASX) = ½ x Ar (ABRS)

By using (i) we get

Ar (ASX) = ½ x Ar (PQRS)








The field is divided into three parts and all are triangles.

//gm PQRS and triangle APQ have the same base and are between the same parallels, therefore

Area (APQ) = ½ x Area (PQRS)

That means that other two triangles will have the same area as the triangle APQ

So she can show wheat in the triangle APQ and pulses in other two triangles PSA and AQR

NCERT solutions for class 9 Maths Chapter 9 Exercise 9.3



Median divides a triangle into two equal parts therefore

Ar (ABD) = Ar (ADC) … (1)

Ar (BED) = Ar (EDC) … (2)

Subtracting (2) from (1)

Ar (ABD) - Ar (BED) = Ar (ADC) - Ar (EDC)

Ar (ABE) = Ar (ACE)








Median divides a triangle into two equal parts therefore

Ar (ABD) = Ar (ADC) … (1)

Ar (BED) = Ar (EDC) … (2)




Ar (ADC) = ½ x Ar (ABC) … (3)

But E is midpoint of AD, that is, EB is also a median

Therefore Ar (BED) = Ar (AEB)


Ar (ADC) = 2 Ar (BED) …(4)

Using (4) in (3)

2 Ar (BED) = ½ x Ar (ABC)

Ar (BED) = ¼ x Ar (ABC)








Triangle ADC and BDC are on the same base DC and between the same parallels

Therefore, Ar (ADC) = Ar (BDC)

By subtracting Ar (ODC) from both side we get

Ar (AOD) = Ar (BOC) … (i)

Similarly in case of triangles DBC and ABC

Ar (DBC) = Ar (ABC)

By subtracting Ar (BOC) from both side we get

Ar (DOC) = Ar (BOC) … (ii)




Also, diagonals of a //gm are equal and they bisect each other therefore in triangle BDC, OC is the median

Ar (DOC) = Ar (AOB) … (iii)




By (i), (ii) and (iii), we get

Ar (AOD) = Ar (BOC)

Ar (AOD) = Ar (BOC) = Ar (DOC) = Ar (AOC)

Hence proved







Line segment CD is bisected by AB, therefore OB and OA are the medians of triangles BCD and ACD

OB divides the triangle BCD into two equal parts

Ar (BOD) = Ar (BOC) … (i)

Similarly OA divides the triangle ACD into two equal parts

Ar (AOD) = Ar (AOC) … (ii)




Adding (i) and (ii)

Ar (BOD) + Ar (AOD) = Ar (BOC) + Ar (AOC)

Ar (ABD) = Ar (ACD)

Hence proved













If a line is formed by joining the mid-points of the two sides of a triangle, then the line is //to the third side and is half of it


FE// BC and FE = ½ x BC


Or, FE// BD and FE = BD

If FE// BD and are equal that means the distance between them should also be equal and the line segments thus formed will also be // to each other

This proves that BDEF is a parallelogram


//gms BDEF and CDFE both share same base FE with triangle DEF

Therefore ar (DEF) = ½ x ar (BDEF) and ar (DEF) = ½ x ar (CDFE) ... (D)

Or, we can say that ar (DEF) = ar (BFD) and ar (DEF) = ar (DEC) … (A)




Similarly //gms AEDF and CDFE both share same base FD with triangle DEF

Therefore ar (DEF) = ½ x ar (AEDF) and ar (DEF) = ½ x ar (CDFE)

Or, we can say that ar (DEF) = ar (AFE) and ar (DEF) = ar (DEC) … (B)




From (A) and (B), we get




ar (DEF) = ar (BFD) = ar (DEC) = ar (AFE)




Ar (ABC) = ar (DEF) + ar (BFD) + ar (DEC) + ar (AFE)




Ar (ABC) = 4 ar (DEF)

¼ x Ar (ABC) = ar (DEF)





¼ x Ar (ABC) = ar (DEF) … (E)

From (D) we know that

Ar (DEF) = ½ x ar (BDEF)

Using in (E)

¼ x Ar (ABC) = ½ x ar (BDEF)

2 x ¼ x Ar (ABC) = ar (BDEF)

½ x Ar (ABC) = ar (BDEF)







In triangle DOB and AOB

∠DOC = ∠AOB (Vertically opposite angles)

DC = AB (Given)

OD = OB (Given)

Triangles DOB and AOB are congruent as per SSA congruency

OC = OA (by CPCT) … (1)

Therefore ar (DOC) = ar (AOB)


In triangle DOA and COB

∠DOA= ∠COB (Vertically opposite angles)

OA = OC (Proved above)

OD = OB (Given)

Triangles DOA and COB are congruent as per SSA congruency

AD = CB (by CPCT) … (2)

Therefore ar (DOC) = ar (AOB)


DC = AB (Given)

AD = CB (Proved above)





Ar (DBC) = Ar(EBC) (Given)

Also, both the triangles have the same base.

When two triangles have the same base and equal area, the must also be between the same parallels, which in this case are DE and BC

Therefore, DE//BC



XY//BC

Therefore EY//BC and XF //BC (as E and F are the points on the line XY which is // to BC)

Also, BE//AC and CF//AB (Given)

This proves that BEYC and BXFC are parallelograms

//gms BEYC and BXFC have equal base BC and are between the same parallels,


Ar(BEYC) = ar(BXFC)

Subtracting ar (XYCB), we get


Ar(EXB) = ar(FYC) … (1)






Triangles XCF and CAF have same base CF and they lie between the same parallels, CF and AB


ar (XFC) = ar (CAF) … (2)

Similarly, Triangles EAB and EYB have same base EB and they lie between the same parallels, EB and AC


ar (EAB) = ar (EYB) … (3)

Triangles XYB and XYC have same base XY and they lie between the same parallels, XY and BC


ar(XYB) = ar (XYC) … (4)




On adding (1) and (4), we get


ar(EYB) = ar(XFC) …(5)

By (2), (3) and (5)




ar ( ACF) = ar(ABE) (Hence proved)





AQ//CP (Given)

Triangles AQC and AQP have the same base AQ and are between the same parallels CP and AQ

Therefore ar (AQC) = ar (AQP)

Subtracting ar (ABQ), we get


ar (ABC) = ar (QBP) … (1)

But AC and AP are the diagonals of the //gms, ABCD and QBPR

And diagonals divide the //gms into triangles of equal area


ar (ABCD) = ½ ar(ABC) and ar(BPRQ) = ½ ar(QBP) … (2)

By (1) and (2)

ar (ABCD) = ar (BPRQ)








AB//CD (Given)

Triangles ABD and ABC have the same base and lie between the same parallels AB and DC


Ar(ABD)= Ar(ABC)

Subtracting ar (AOB) on both sides

We get


ar(AOD)= ar(BOC) (Hence proved)







AC//BF (Given)

Triangles ACB and ACF have the same base and they lie between the same parallels, AC and BF

Therefore, ar (ACB) = ar (ACF) (Hence proved)





ar (ACB) = ar (ACF) (Proved above)

Adding ar (AEDC) on both sides

We get,

ar (BAEDC) = ar (AEDCF) (Hence proved)
















ABCD is a trapezium with AB//BC, since X is a point of AB, therefore AX//DC

XY//AC (Given)

Triangles ADX and ACX have the same base AX and are between the same parallels AX and DC


ar (ADX) = ar (ACX) … (1)

Similarly, Triangles ACX and ACY have the same base AC and lie between the same parallels XY and AC





ar (ACY) = ar (ACX) … (2)

By (1) and (2)

We get

ar (ADX) = ar (ACY)









AP//BQ//CR

Triangles ABQ and PBQ have the same base BQ and are between the same parallels BQ and AP


ar (ABQ) = ar (PBQ) … (1)




Triangles BQC and BQR have the same base BQ and are between the same parallels BQ and CR


ar (BQC) = ar (BQR) … (2)

Add (1) and (2)

We get

ar (ABQ) + ar (BQC) = ar (PBQ) + ar (BQR)

ar (AQC) = AR (PBR) (Hence proved)





Ar (AOD) = ar (BOC) (Given)

Add ar (DOC) on both sides

We get


Ar (ADC) = ar (BDC)

Triangles ADC and BDC have equal area and they have the same base DC, therefore, they must also be between same parallels. That is, AB must be parallel to DC, therefore ABCD is a trapezium.





Ar (DRC) = ar (DCP) (Given)

Ar (DCP) = Ar (DRC) … (1)

Triangles DRC and DCR have equal area and they have the same base DC, therefore, they must also be between same parallels. That is, DC must be parallel to RP … (4)





Also, Ar (BDP) = ar (ARC) … (2)

Subtracting (1) from (2)

We get

, Ar (DBC) = ar (DAC)

Triangles DBC and DAC have equal area and they have the same base DC, therefore, they must also be between same parallels. That is, DC must be parallel to AB … (5)




By (4) and (5), it is proved that DCRP and ABCD are trapeziums