NCERT Solutions Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.3

NCERT Solutions Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.3

NCERT Solutions Class 9th Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.3

?Chapter – 9?

✍Areas of Parallelograms and Triangles✍

?Exercise 9.3?

Question 1. In figure, E is any point on median AD of a ∆ABC. Show that ar (ABE) = ar (ACE).

Solution: We have a ∆ABC such that AD is a median.
∴ ar(∆ABD) = ar(∆ACD) …(1)
[∵ A median divides the triangle into two triangles of equal areas]
Similarly, in ∆BEC, we have
ar(∆BED) = ar(∆DEC) …(2)
Subtracting (2) from (1), we have
ar(∆ABD) – ar(∆BED) = ar(∆ACD) – ar(∆DEC)
⇒ ar(∆ABE) = ar(∆ACE).

Question 2. In a triangle ABC, E is the mid-point of median AD. Show that ax (BED) = 1/4ar(ABC).
Solution: We have a ∆ABC and its median AD.
Let us join B and E.

Since, a median divides the triangle into two triangles of equal area.
ar (∆ABD) = 1/2ar(ΔABC) …….(1)
Now, in ∆ABD, BE is a median.
[ ∵ E is the mid-point of AD]
∴ ar(∆BED) = 1/2ar(ΔABC) …(2)
From (1) and (2), we have
ar(∆BED) = 1/2 [1/2ar(ΔABC) ]
⇒ ar(∆BED) = 1/4ar(ΔABC)

Question 3. Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Solution: We have a parallelogram ABCD (say)
such that its diagonals intersect at O.
∵Diagonals of a parallelogram bisect each other.
∴ AO = OC and BO = OD
Let us draw CE ⊥ BD.
Now, ar(∆BOC) = 1/2BO x CE and
ar(∆DOC) = 1/2OD x CE

Since, BO = OD
∴ ar(∆BOC) = ar(∆DOC) …(1)
Similarly, ar(∆AOD) = ar(∆DOC) …(2)
and ar(∆AOB) = ar(∆BOC) …(3)
From (1), (2) and (3), we have
ar(∆AOB) = ar(∆BOC) = ar(∆COD) = ar(∆DOA)
Thus, the diagonals of a parallelogram divide it into four triangles of equal area.

Question 4. In figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that ar(ABC) = ar(ABD)

Solution: we have ∆ABC and ∆ABD are on the same base AB.
∵ CD is bisected at O. [Given]
∴ CO = OD
Now, in ∆ACD, AO is a median
∴ ar(∆OAC) = ar(∆OAD) …(1)
Again, in ∆BCD, BO is a median
∴ ar(∆OBC) = ar(∆ODB) …(2)
Adding (1) and (2), we have
ar(∆OAQ + ar(∆OBQ) = ar(∆OAD) + ar(∆ODB)
⇒ ar(∆ABC) = ar(∆ABD)

Question 5. D,E and F are respectively the mid-points of the sides BC, CA and AB of a ∆ABC. Show that
(i) BDEF is a parallelogram.
(ii) ar(DEF) = 14ar(ABC)
(iii) ar(BDEF) = 14ar(ABC)

Solution: We have ∆ABC such
that D,E and Fare the mid-points of BC, CA and AB respectively.

(i) In ∆ABC, E and F are the mid-points of AC and B D C AB respectively.
∴ EF || BC [Mid-point theorem]
⇒ EF || BD
Also, EF = 1/2(BC)
⇒ EF = BD [D is the mid – point of BC]
Since BDEF is a quadrilateral whose one pair of opposite sides is parallel and of equal lengths.
∴ BDEF is a parallelogram.

(ii) We have proved that BDEF is a parallelogram.
Similarly, DCEF is a parallelogram and DEAF is also a parallelogram.
Now, parallelogram BDEF and parallelogram DCEF are on the same base EF and between the same parallels BC and EF.
∴ ar(||^gm BDEF) = ar(||^gm DCEF)
⇒ 1/2ar(||^gm BDEF) = 1/2ar(||^gm DCEF)
⇒ ar(∆BDF) = ar(∆CDE) …(1)
[Diagonal of a parallelogram divides it into two triangles of equal area]
Similarly, ar(∆CDE) = ar(∆DEF) …(2)
and ar(∆AEF) = ar(∆DEF) …(3)
From (1), (2) and (3), we have
ar(∆AEF) = ar(∆FBD) = ar(∆DEF) = ar(∆CDE)
Thus, ar(∆ABC) = ar(∆AEF) + ar(∆FBD) + ar(∆DEF) + ar(∆CDE) = 4 ar(∆DEF)
⇒ ar(∆DEF) = 14ar(∆ABC)

(iii) We have, ar (||^gm BDEF) = ar(∆BDF) + ar(∆DEF)
= ar(∆DEF) + ar(∆DEF) [∵ ar(∆DEF) = ar(∆BDF)]
2ar(∆DEF) = 2[14ar(∆ABC)]
= 1/2ar(∆ABC)
Thus, ar (||^gm BDEF) = 12ar(∆ABC)

Question 6. In figure, diagonals AC and BD of quadrilateral ABCD intersect at 0 such that OB = OD. If AB = CD, then show that
(i) ar(DOC) = ar(AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram

Solution: We have a quadrilateral ABCD whose diagonals AC and BD intersect at O.
We also have that OB = OD, 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 A AS congruency]
⇒ 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]
⇒ ar(∆DEC) = ar(∆BFA) …(2)
and ∠1 = ∠2 …(3) [By C.P.C.T.]
Adding (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, ∆DCS 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 (3)]
which are alternate interior angles.
So, AB || CD
Hence, ABCD is a parallelogram.

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 same area.

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

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 paralleloam.

Now, the parallelogram BCYE and ∆ABE are on the same base 8E and between the same parallels BE and AC.
∴ ar(∆ABE) = 1/2ar(∥gmBCYE) …..(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) = 1/2ar(∥gmBCFX) …(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 (1), (2) and (3), we get
ar(∆BE) = ar(∆ACF)

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 ax (ABCD) = ar(PBQR).
[Hint Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]

Solution: Let us join AC and PQ.
ABCD is a parallelogram [Given]
and AC is its diagonal, we know that diagonal of a parallelogram divides it into two triangles of equal areas.
∴ ar(∆ABC) = 1/2ar(∥gm ABCD) …(1)
Also, PBQR is a parallelogram [Given]
and QP is its diagonal.
∴ ar(∆BPQ) = 1/2ar(∥gm PBQR) …(2)
Since, ∆ACQ and AAPQ are on the same base AQ and between A 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 (1), (2) and (3), we get
1/2ar(∥gm ABCD) = 1/2ar(∥gm PBQR)
⇒ ar( ||^gm ABCD) = ar(||^gm PBQR)

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: ABCD have a trapezium ABCD having AB || CD 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)

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 between the same parallels and on the same base are equal in area.
∆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)

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(ADAF) = ar(ADCF)
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(ACEF) = ar(AADE) [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)

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 IX]
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 (1) and (2), we have
ar(∆ADX) = ar(∆ACY)

Question 14. In figure, AP || BQ || CR. Prove that ar(AQC) = ax(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 (1) and (2), we have
ar(∆BCQ) + ar(∆ABQ) = ar(∆BQR) + ar(∆PBQ)
⇒ ar(∆AQC) = ar(∆PBR)

Question 15. Diagonals AC and BD of a quadrilateral ABCD intersect at 0 in such a way that ax(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.

Question 16. In figure ax(DRC) = ar(DPC) and ai(BDP) = ar(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Solution: tfclfiftWe 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.r 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 (2) from (1), we get
ar(∆BDP) – ar(∆DPC) = ar(∆ARQ – ar(∆DRQ
⇒ ar(∆BDC) = ar(∆ADC)
And they are on the same base DC.
∴ ABDC and AADC 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.