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 ΔABC. Show that
ar (ΔABE) = ar (ΔACE).
Solution.
To prove: ar (ΔABE) = ar (ΔACE)
Proof: Since AD is the median of ΔABC.
∴ ar (ΔABD) = ar (ΔACD) …..…(i)
[Since a median of a triangle divides it into two triangles of equal areas]
Also, ED is the median of ΔEBC.
∴ ar (ΔBED) = ar (ΔCED) …..… (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 (ΔABD) – ar (ΔBED) = ar (ΔACD) – ar (ΔCED)
⇒ ar (ΔABE) = ar (ΔACE)
Hence proved.
Ex 9.3 Class 9 Maths Question 2.
In a ΔABC, E is the mid-point of
median AD. Show that
ar (ΔBED) = ¼ ar (ΔABC).
Solution.
To prove: ar (ΔBED) = ¼ ar (ΔABC)
Proof: We know that the median a triangle divides it into two triangles of equal areas.
∴ ar (ΔABD) = ar (ΔADC)
⇒ ar (ΔABD) = ½ ar (ΔABC) …..…(i)
In ΔABD, BE is the median because E is the mid-point of AD.
∴ ar (ΔBED) = ar (ΔBAE)
⇒ ar (ΔBED) = ½ ar (ΔABD)
⇒ ar (ΔBED) = ½ . ½ ar (ΔABC) [From eq. (i)]
⇒ ar (ΔBED) = ¼ ar (ΔABC)
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.
To prove: Diagonals AC and BD divide parallelogram ABCD into four triangles of equal areas.
i.e., ar (ΔOAB) = ar (ΔOBC) = ar (ΔOCD) = ar (ΔOAD)
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 ΔABC, BO is a median.
∴ ar (ΔOAB) = ar (ΔOBC) ….… (i)
In ΔABD, AO is a median.
∴ ar (ΔOAB) = ar (ΔOAD) ….… (ii)
Similarly, in ΔACD, DO is the median.
∴ ar (ΔOAD) = ar (ΔOCD)
…..…(iii)
From eqs. (i), (ii) and (iii), we get
ar
(ΔOAB) = ar (ΔOBC) = ar (ΔOCD) = ar (ΔOAD)
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 (ΔABC) = ar( ΔABD).
Solution.
To prove: ar (ΔABC) = ar (ΔABD)
Proof: Since the line segment CD is bisected by AB at O.
∴ OC = OD
In ΔACD, we have OC = OD
So, AO is the median of ΔACD.
Since we know that the median divides a triangle into two triangles of equal areas.
∴ ar (ΔAOC) = ar (ΔAOD) …..… (i)
Similarly, in ΔBCD, BO is the median.
∴ ar (ΔBOC) = ar (ΔBOD) …..… (ii)
On adding eqs. (i) and (ii), we get
ar (ΔAOC) + ar (ΔBOC) = ar (ΔAOD) + ar (ΔBOD)
⇒ ar (ΔABC) = ar (ΔABD)
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 ΔABC. Show that
(i)
BDEF is a parallelogram.
(ii)
ar (ΔDEF) = ¼ ar (ΔABC).
(iii)
ar (||gm BDEF) = ½ ar (ΔABC).
Solution.
are respectively D, E and F.
To prove:
(i) BDEF is a parallelogram.
(ii) ar (ΔDEF) = ¼ ar (ΔABC)
(iii) ar (||gm BDEF) = ½ ar (ΔABC)
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 (ΔBDF)
= ar (ΔDEF) ……….(i)
Since
FDCE is also parallelogram.
Therefore, ar (ΔDEF)
= ar (ΔDEC) ……….(ii)
Again,
AFDE is also parallelogram.
ar
(ΔAFE)
= ar (ΔDEF) ……….(iii)
From
eqs. (i), (ii) and (iii), we get
ar
(ΔDEF)
= ar (ΔBDF)
= ar (ΔDEC)
= ar (ΔAFE) ……….(iv)
Now,
ar (ΔABC)
= ar (ΔDEF)
+ ar (ΔBDF)
+ ar (ΔDEC)
+ ar (ΔAFE)
……….(v)
⇒ ar (ΔABC) = ar (ΔDEF) + ar (ΔDEF) + ar (ΔDEF) + ar (ΔDEF) [Using
(iv) and (v)]
⇒ ar (ΔABC) = 4 × ar
(ΔDEF)
⇒ ar (ΔDEF) = ¼ ar
(ΔABC)
(iii) ar (||gm
BDEF) = ar (ΔBDF)
+ ar (ΔDEF)
= ar (ΔDEF)
+ ar (ΔDEF) [Using (iv)]
⇒ ar (||gm
BDEF) = 2 ar (ΔDEF)
⇒ ar (||gm
BDEF) = 2 × ¼ ar (ΔABC)
⇒ ar (||gm
BDEF) = ½ ar (ΔABC)
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 (ΔDOC) = ar (ΔAOB)
(ii)
ar (ΔDCB) = ar (ΔACB)
(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.
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.
∴ 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).]
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 (∆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]
Adding ar (quad. AEDC) to both sides, we get
⇒ 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.
∴ ar (∆DAF) = ar (∆DCF)
Subtracting ar (∆DEF) from both sides, we get
ar (∆DAF) – ar (∆DEF) = ar (∆DCF) – ar (∆DEF)
⇒ ar (∆ADE) = ar (∆CEF)
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]
Adding ar (quad. ABCE) to both sides, we get
ar (∆CEF) + ar (quad. ABCE) = ar (∆ADE) + ar (quad. ABCE)
⇒ 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
ar (∆ADX) = ar (∆ACY)
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]
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.
∴ Quadrilateral DCPR is a trapezium.
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)
⇒ ar (∆BDC) = ar (∆ADC)
And they are on the same base DC.
∴ ∆BDC and ∆ADC must lie between the
same parallels.
So, AB || DC, i.e., a pair of opposite sides of
quadrilateral ABCD is parallel.
∴ Quadrilateral ABCD is a trapezium.
NCERT Solutions for Maths Class 10