NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.3

# NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.3

## NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.3

NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.3 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Chapter 9 Areas of Parallelograms and Triangles Ex 9.3.

Ex 9.3 Class 9 Maths Question 1.
In the given figure, E is any point on median AD of a Î”ABC. Show that
ar (Î”ABE) = ar (Î”ACE).

Solution.

Given: AD is a median of Î”ABC and E is any point on AD.
To prove: ar (Î”ABE) = ar (Î”ACE)
Proof: Since AD is the median of Î”ABC.
ar (Î”ABD) = ar (Î”ACD)             …..…(i)
[Since a median of a triangle divides it into two triangles of equal areas]
Also, ED is the median of Î”EBC.
ar (Î”BED) = ar (Î”CED)             …..… (ii)
[Since a median of a triangle divides it into two triangles of equal areas]

On subtracting eq. (ii) from eq. (i), we get
ar (Î”ABD) – ar (Î”BED) = ar (Î”ACD) – ar (Î”CED)
ar (Î”ABE) = ar (Î”ACE)
Hence proved.

Ex 9.3 Class 9 Maths Question 2.
In a Î”ABC, E is the mid-point of median AD. Show that
ar (Î”BED) = ¼ ar (Î”ABC).

Solution.

Given: ABC is a triangle and E is the mid-point of the median AD.
To prove: ar (Î”BED) = ¼ ar (Î”ABC)
Proof: We know that the median a triangle divides it into two triangles of equal areas.
ar (Î”ABD) = ½ ar (Î”ABC)                    …..…(i)
In Î”ABD, BE is the median because E is the mid-point of AD.
ar (Î”BED) = ar (Î”BAE)
ar (Î”BED) = ½ ar (Î”ABD)
ar (Î”BED) = ½ . ½ ar (Î”ABC)          [From eq. (i)]
ar (Î”BED) = ¼ ar (Î”ABC)
Hence proved.

Ex 9.3 Class 9 Maths Question 3.
Show that the diagonals of a parallelogram divide it into four triangles of equal area.

Solution.

Given: ABCD is a parallelogram and its diagonals AC and BD intersect each other at O.
To prove: Diagonals AC and BD divide parallelogram ABCD into four triangles of equal areas.

i.e., ar (Î”OAB) = ar (Î”OBC) = ar (Î”OCD) = ar (Î”OAD)

Proof: We know that the diagonals of a parallelogram bisect each other, so we have
OA = OC and OB = OD.
Also, we know that a median of a triangle divides it into two triangles of equal areas.
Now, as in Î”ABC, BO is a median.
ar (Î”OAB) = ar (Î”OBC)              ….… (i)
In Î”ABD, AO is a median.
ar (Î”OAB) = ar (Î”OAD)              ….… (ii)
Similarly, in Î”ACD, DO is the median.
ar (Î”OAD) = ar (Î”OCD)                …..…(iii)
From eqs. (i), (ii) and (iii), we get

ar (Î”OAB) = ar (Î”OBC) = ar (Î”OCD) = ar (Î”OAD)

Thus, the diagonals of a parallelogram divide it into four triangles of equal area.

Hence proved.

Ex 9.3 Class 9 Maths Question 4.
In the given figure, ABC and ABD are two triangles on the same base AB. If the line-segment CD is bisected by AB at O, show that ar (Î”ABC) = ar( Î”ABD).

Solution.

Given: ABC and ABD are two triangles on the same base AB.
To prove: ar (Î”ABC) = ar (Î”ABD)
Proof: Since the line segment CD is bisected by AB at O.
OC = OD
In Î”ACD, we have OC = OD
So, AO is the median of Î”ACD.
Since we know that the median divides a triangle into two triangles of equal areas.
ar (Î”AOC) = ar (Î”AOD)             …..… (i)
Similarly, in Î”BCD, BO is the median.
ar (Î”BOC) = ar (Î”BOD)                 …..… (ii)
On adding eqs. (i) and (ii), we get

ar (Î”AOC) + ar (Î”BOC) = ar (Î”AOD) + ar (Î”BOD)
ar (Î”ABC) = ar (Î”ABD)

Hence proved.

9.3 Class 9 Maths Question 5.
D, E and F are respectively the mid-points of the sides BC, CA and AB of a Î”ABC. Show that
(i) BDEF is a parallelogram.
(ii) ar (Î”DEF) = ¼ ar (Î”ABC).
(iii) ar (||gm BDEF) = ½ ar (Î”ABC).

Solution.

Given: ABC is a triangle in which the mid-points of sides BC, CA and AB
are respectively D, E and F.
To prove:
(i) BDEF is a parallelogram.
(ii) ar (Î”DEF) = ¼ ar (Î”ABC)
(iii) ar (||gm BDEF) = ½ ar (Î”ABC)
Proof:
(i) Since, E and F are the mid-points of AC and AB.
BC || FE and FE = ½ BC = BD              [By mid-point theorem]
Thus, BD || FE and BD = FE
Similarly, BF || DE and BF = DE

Hence, BDEF is a parallelogram.
[
a pair of opposite sides are equal and parallel]

(ii) Similarly, we can prove that both FDCE and AFDE are also parallelograms.
Now, BDEF is a parallelogram, so its diagonal FD divides it into two triangles of equal areas.

Therefore, ar (Î”BDF) = ar (Î”DEF)         ……….(i)

Since FDCE is also parallelogram.

Therefore, ar (Î”DEF) = ar (Î”DEC)         ……….(ii)

Again, AFDE is also parallelogram.

ar (Î”AFE) = ar (Î”DEF)                ……….(iii)

From eqs. (i), (ii) and (iii), we get

ar (Î”DEF) = ar (Î”BDF) = ar (Î”DEC) = ar (Î”AFE)         ……….(iv)

Now, ar (Î”ABC) = ar (Î”DEF) + ar (Î”BDF) + ar (Î”DEC) + ar (Î”AFE)      ……….(v)

ar (Î”ABC) = ar (Î”DEF) + ar (Î”DEF) + ar (Î”DEF) + ar (Î”DEF)        [Using (iv) and (v)]

ar (Î”ABC) = 4 × ar (Î”DEF)

ar (Î”DEF) = ¼ ar (Î”ABC)

(iii) ar (||gm BDEF) = ar (Î”BDF) + ar (Î”DEF)

= ar (Î”DEF) + ar (Î”DEF)              [Using (iv)]

ar (||gm BDEF) = 2 ar (Î”DEF)

ar (||gm BDEF) = 2 × ¼ ar (Î”ABC)

ar (||gm BDEF) = ½ ar (Î”ABC)

Ex 9.3 Class 9 Maths Question 6.
In figure, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that
(i) ar (Î”DOC) = ar (Î”AOB)
(ii) ar (Î”DCB) = ar (Î”ACB)
(iii) DA || CB or ABCD is a parallelogram.

Solution:
We have a quadrilateral ABCD whose diagonals AC and BD intersect at O.
We have also given that OB = OD and AB = CD.

Let us draw DE AC and BF AC.

(i) In ∆DEO and ∆BFO, we have
DO = BO                     [Given]
DOE = BOF           [Vertically opposite angles]
DEO = BFO           [Each 90°]
∆DEO ∆BFO       [By AAS congruency rule]
DE = BF                  [By C.P.C.T.]
and ar (∆DEO) = ar (∆BFO)                    ….…(1)
Now, in ∆DEC and ∆BFA, we have
DEC = BFA             [Each 90°]
DE = BF                        [Proved above]
DC = BA                        [Given]
∆DEC ∆BFA          [By RHS congruency rule]
ar (∆DEC) = ar (∆BFA)                        ….…(2)
and
1 = 2      [By C.P.C.T.]                 ….…(3)
Adding eq. (1) and (2), we have
ar (∆DEO) + ar (∆DEC) = ar (∆BFO) + ar (∆BFA)
ar (∆DOC) = ar (∆AOB)

(ii) Since, ar (∆DOC) = ar (∆AOB)          [Proved above]
Adding ar (∆BOC) on both sides, we have
ar (∆DOC) + ar (∆BOC) = ar (∆AOB) + ar (∆BOC)
ar (∆DCB) = ar (∆ACB)

(iii) Since, ∆DCB and ∆ACB are both on the same base CB and having equal areas.
They lie between the same parallels CB and DA.
CB || DA
Also
1 = 2,                                     [By eq. (3)]
which are alternate interior angles.
So, AB || CD
Hence, ABCD is a parallelogram.

Ex 9.3 Class 9 Maths Question 7.
D and E are points on sides AB and AC respectively of ∆ABC such that ar (DBC) = ar (EBC). Prove that DE || BC.

Solution:
We have ∆ABC and points D and E are such that ar (DBC) = ar (EBC).
Since ∆DBC and ∆EBC are on the same base BC and having the same area.

They must lie between the same parallels DE and BC.
Hence, DE || BC.

Ex 9.3 Class 9 Maths Question 8.
XY is a line parallel to side BC of a ∆ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that ar (ABE) = ar (ACF)

Solution:
We have a ∆ABC such that XY || BC, BE || AC and CF || AB.
Since, XY || BC and BE || CY
BCYE is a parallelogram.

Now, the parallelogram BCYE and ∆ABE are on the same base BE and between the same parallels BE and AC.
ar (∆ABE) = ½ ar (gm BCYE)              ……..(1)
Again, CF || AB                [Given]
XY || BC                            [Given]
CF || BX and XF || BC
BCFX is a parallelogram.
Now, ∆ACF and parallelogram BCFX are on the same base CF and between the same parallels AB and CF.
ar (∆ACF) = ½ ar (gm BCFX)               ………(2)
Also, parallelogram BCFX and parallelogram BCYE are on the same base BC and between the same parallels BC and EF.

ar (||gm BCFX) = ar (||gm BCYE)          ………(3)
From eqs. (1), (2) and (3), we get
ar (∆ABE) = ar (∆ACF)

Hence proved.

Ex 9.3 Class 9 Maths Question 9.
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 A parallelogram PBQR is completed (see figure). Show that ar  (ABCD) = ar (PBQR).
[Hint: Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]

Solution:
Join AC and PQ.
ABCD is a parallelogram.           [Given]
AC is its diagonal.

We know that diagonal of a parallelogram divides it into two triangles of equal areas.
ar (∆ABC) = ½ ar (gm ABCD)                 …..…(1)
Also, PBQR is a parallelogram.     [Given]

QP is its diagonal.
ar (∆BPQ) = ½ ar (gm PBQR)                 …..…(2)
Since, ∆ACQ and ∆APQ are on the same base AQ and between the same parallels AQ and CP.

ar (∆ACQ) = ar (∆APQ)
ar (∆ACQ) – ar (∆ABQ) = ar (∆APQ) – ar (∆ABQ)
[Subtracting ar (∆ABQ) from both sides]
ar (∆ABC) = ar (∆BPQ)                       …..…(3)
From eqs. (1), (2) and (3), we get
½ ar (
gm ABCD) = ½ ar (gm PBQR)
ar (||gm ABCD) = ar (||gm PBQR)

Hence proved.

Ex 9.3 Class 9 Maths Question 10.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar (AOD) = ar (BOC)

Solution:
We have a trapezium ABCD having AB || DC and its diagonals AC and BD intersect each other at O.
Since, triangles on the same base and between the same parallels have equal areas.
∆ABD and ∆ABC are on the same base AB and between the same parallels AB and DC.
ar (∆ABD) = ar (∆ABC)
Subtracting ar (∆AOB) from both sides, we get
ar (∆ABD) – ar (∆AOB) = ar (∆ABC) – ar (∆AOB)
ar (∆AOD) = ar (∆BOC)

Hence proved.

Ex 9.3 Class 9 Maths Question 11.
In figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar (ACB) = ar (ACF)
(ii) ar (AEDF) = ar (ABCDE)

Solution:

We have a pentagon ABCDE in which BF || AC and DC is produced to F.
(i) Since, the triangles on the same base and between the same parallels are equal in areas.
∆ACB and ∆ACF are on the same base AC and between the same parallels AC and BF.
ar (∆ACB) = ar (∆ACF)

(ii) Since, ar (∆ACB) = ar (∆ACF)            [Proved above]
ar (∆ACB) + ar (quad. AEDC) = ar (∆ACF) + ar (quad. AEDC)
ar (ABCDE) = ar (AEDF)

ar (AEDF) = ar (ABCDE)                 Hence proved.

Ex 9.3 Class 9 Maths Question 12.
A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.

Solution:
We have a plot in the form of a quadrilateral ABCD.
Let us draw DF || AC and join AF and CF.

Now, ∆DAF and ∆DCF are on the same base DF and between the same parallels AC and DF.
ar (∆DAF) = ar (∆DCF)

Subtracting ar (∆DEF) from both sides, we get
ar (∆DAF) – ar (∆DEF) = ar (∆DCF) – ar (∆DEF)
The portion of ∆ADE can be taken over by the Gram Panchayat by adding the land (∆CEF) to his (Itwaari) land so as to form a triangular plot, i.e. ∆ABF.
Let us prove that ar (∆ABF) = ar (quad. ABCD).

We have, ar (CEF) = ar (ADE)           [Proved above]
ar (∆ABF) = ar (quad. ABCD)

Ex 9.3 Class 9 Maths Question 13.
ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY). [Hint: Join CX]

Solution:
We have a trapezium ABCD such that AB || DC.
XY || AC meets AB at X and BC at Y. Let us join CX.

∆ADX and ∆ACX are on the same base AX and between the same parallels AX and DC.
ar (∆ADX) = ar (∆ACX)                   …..…(1)
∆ACX and ∆ACY are on the same base AC and between the same parallels AC and XY.
ar (∆ACX) = ar (∆ACY)                   …..…(2)
From eq. (1) and (2), we have

Ex 9.3 Class 9 Maths Question 14.
In figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

Solution:
We have, AP || BQ || CR
∆BCQ and ∆BQR are on the same base BQ and between the same parallels BQ and CR.
ar (∆BCQ) = ar (∆BQR)                    …..…(1)
∆ABQ and ∆PBQ are on the same base BQ and between the same parallels AP and BQ.
ar (∆ABQ) = ar (∆PBQ)                     …..…(2)
Adding eqs. (1) and (2), we get

Ar (∆BCQ) + ar (∆ABQ) = ar (∆BQR) + ar (∆PBQ)
ar (∆AQC) = ar (∆PBR)                  Hence proved.

Ex 9.3 Class 9 Maths Question 15.
Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.

Solution:
We have a quadrilateral ABCD and its diagonals AC and BD intersect at O such that
ar (∆AOD) = ar (∆BOC)         [Given]

Adding ar (∆AOB) to both sides, we have
ar (∆AOD) + ar (∆AOB) = ar (∆BOC) + ar (∆AOB)
ar (∆ABD) = ar (∆ABC)
Also, they are on the same base AB.
Since, the triangles are on the same base and having equal area.
They must lie between the same parallels.
AB || DC
Now, ABCD is a quadrilateral having a pair of opposite sides parallel.
So, ABCD is a trapezium.

Ex 9.3 Class 9 Maths Question 16.
In figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Solution:
We have, ar (∆DRC) = ar (∆DPC)                   [Given]
And they are on the same base DC.
∆DRC and ∆DPC must lie between the same parallels.
So, DC || RP, i.e., a pair of opposite sides of quadrilateral DCPR is parallel.
Again, we have ar (∆BDP) = ar (∆ARC)    [Given]        …..…(1)
Also, ar (∆DPC) = ar (∆DRC)                      [Given]        …..…(2)
Subtracting eq. (2) from (1), we get
ar (∆BDP) – ar (∆DPC) = ar (∆ARC) – ar (∆DRC)