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

These NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Ex 9.3 Questions and Answers are prepared by our highly skilled subject experts.

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

Question 1.
In fig. 9.23, E is any point on median AD of a ∆ABC Show that ar(∆ABE) = ar(∆ACE).

Solution:
Given: In ∆ABC, AD is median. Any point E lie of median AD.
To prove that: ar(∆ABE) = ar(∆ACE)
Proof: In ∆ABC, AD is median.
∴ ar(∆ABD) = ar(∆ACD) ……(i)
(A median of a triangle divides it into two triangles of equal areas)
Now, In ∆EBC, ED is median.
∴ ar(∆EBD) = ar(∆CED) ……(ii)
(A median of a triangle divides it into two triangles of equal areas)
Subtract equation (ii) from equation (i)
ar(∆ABD) – ar(∆EBD) = ar(∆ACB) – ar(∆CED)
∴ ar(∆ABE) = ar(∆ACE)

Question 2.
In a triangle ABC, E is the mid point of median AD. Show that ar(BED) = $$\frac {1}{4}$$ ar(ABC).
Solution:
Given: ABC is a triangle in which AD is median and E is the mid point of median AD.

To prove that: ar(BED) = $$\frac {1}{4}$$ ar(ABC)
Construction: Join BE.
Proof: In ∆ABD, E is the mid point of AD. Therefore, BF is median.
We know that a median of a triangle divide triangles of equal areas.
∴ ar(∆BFD) = ar(∆ABE)
or, ar(∆BED) = $$\frac {1}{4}$$ ar(∆ABD) …..(i)
Now, In ∆ABC,
ar (∆ABD) = ar (∆ACD)
(because AD is median of ∆ABC)
or, ar (∆ABD) = $$\frac {1}{4}$$ ar (∆ABC) …….(ii)
Put Value of ar (∆ABD) in equation (i) we have
ar (∆BED) = $$\frac{1}{2} \times \frac{1}{2}$$ ar (∆ABC)
or, ar (∆BED) = $$\frac {1}{4}$$ ar (∆ABC)

Question 3.
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
Solution:
Given: ABCD is a parallelogram in which diagonal AC and BD bisects each other at O.

To prove that: ar (∆AOB) = ar (∆BOC) = ar (∆COD) = ar (∆DOA)
Proof: In ∆ABC, O is mid point of side AC.
Therefore, BO is median of ∆ABC.
∴ ar (∆AOB) = ar (∆BOC) ……(i)
(Median divide a A in two equal parts)
Now, In ∆ADC, DO is median
∴ ar (∆AOD) = ar (∆COD) …….(ii)
(Median divide a ∆ in two equal parts)
Again in ∆ADB, AO is a median.
∴ ar (∆AOD) = ar (∆AOB) ………(iii)
From equation (i), (ii) and (iii)
ar (∆AOB) = ar (∆BOC) = ar (∆COD) = ar (∆DOA)
Therefore, the diagonals of a parallelogram divide it into four triangles of equal area.

Question 4.
In fig. 9.24, 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:
In ∆ACD, O is mid point of CD.
∴ AO is median
We know that a median of a triangle divides it into triangles of equal areas.
∴ ar (∆AOQ) = ar (∆AOD) …..(i)
Similarly, In ∆BCD, BO is median
∴ ar (∆BOC) = ar (∆BOD)
ar (∆AOC) + ar (∆BOC) = ar (∆AOD) + ar (∆BOD)
So, 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) = $$\frac {1}{4}$$ ar (ABC)
(iii) ar (BDEF) = $$\frac {1}{2}$$ ar (ABC)
Solution:
Given: ABC is a triangle in which D, E and F are the mid points of side BC, AC and AB respectively.
To prove that:
(i) BDEF is a parallelogram
(ii) ar(DEF) = $$\frac {1}{4}$$ ar (ABC)
(iii) ar(BDEF) = $$\frac {1}{2}$$ ar (ABC)

Proof: (i) In ∆ABC,
E and F are the mid point of side AC and AB respectively.
We know that the line joining the mid points of two sides of a triangle is parallel to third side and half of it.
∴ EF || BC …….(i)
and EF = $$\frac {1}{2}$$ BC
or, EF = BD …(ii) (∵ D is mid point of BC)
From (i) and (ii)
BDEF is a parallelogram.

(ii) We have,
BDEF is a parallelogram.
∴ ar (∆BDF) = ar (∆DEF) ……(i)
(Diagonals of a || gm divide it in two equal parts)
Similarly, CDFE is parallelogram.
∴ ar (∆CDE) = ar (∆DEF) …..(ii)
and ar (∆AEF) = ar (∆DEF) ……(iii)
From equation (i), (ii) and (iii)
ar (∆BDF) = ar (∆CDE) = ar (∆AEF) = ar (∆DEF)
Now,
ar (∆BDF) + ar (∆CDE) + ar (∆AFE) + ar (∆DEF) = ar (∆ABC)
or, ar (∆DEF) + ar (∆DEF) + ar (∆DEF) + ar (∆DEF) = ar (∆ABC)
[∵ ar (∆BDF) + ar (∆CDE) = ar (∆AFF)]
or, 4 ar (∆DEF) = ar (∆ABC)
or, ar (∆DEF) = $$\frac {1}{4}$$ ar (∆ABC)

(iii) ar (|| gm BDEF) = 2 ar (∆DEF)
or, ar (|| gm BDEF) = 2 . $$\frac {1}{4}$$ ar (∆ABC)
[∵ ar (∆DEF) = $$\frac {1}{4}$$ ar (∆ABC)]
So, ar (|| gm BEDF) = $$\frac {1}{4}$$ ar (∆ABC)

Question 6.
In fig. 9.25, diagonals AC and BD of quadrilateral ABCD intersect at O, 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:
Given: ABCD is a quadrilateral in which diagonals AC and BD intersect each other at O.
To prove that:
(i) ar(∆DOC) = ar(∆AOB)
(ii) ar (∆DCB) = ar(∆ACB)
(iii) DA || CB or ABCD is a parallelogram.
Construction: Draw DN ⊥ AC and BM ⊥ AC.

Proof: (i) In ∆DON and ∆BOM
∠DNO = ∠BMO (Each 90°)
∠DNO = ∠BOM (Vertically opposite angle)
DO = BO (Given)
By A-A-S congruency condition
∆DON ≅ ∆BOM
So, ar (∆DON) = ar (∆BOM) ……(i)
(Congruent ∆s have equal area)
DN = BM (By C.P.C.T)
Now, In ∆DNC and ∆BMA
∠DNC = ∠BMA (Each 90°)
CD = BA (Given)
DN = BM (Prove above)
By R.H.S Congruency Condition
∆DNC = ∆BMA
So, ar (∆DNC) = ar (∆BMA) …….(ii)
(Congruent ∆s have equal area)
ar (∆DON) + ar (∆DNC) = ar (∆BOM) + ar (∆BMA)
ar (∆DOC) = ar (∆AQB)

(ii) We have,
ar (∆DOC) = ar (∆AOB) (Prove above)
∴ ar (∆DOC) + ar (∆OCB) = ar (∆AOB) + ar (∆QCB) [add ar (∆OCB) both side]
or, ar (∆DCB) = ar (∆ACB)

(iii) We have
ar (∆DCB) = ar (∆ACB)
We know that if two mangles on same base and equal in area, then it must be lie between same parallels.
Therefore, DA || CB …..(A)
∠DCO = ∠BAO (∆DNC ≅ ∆BMA)
which is the pair of alternate interior angles.
CD || BA …….(B)
From (A) and (B),
ABCD is a parallelogram.

Question 7.
D and B are points on sides AB and AC respectively of ∆ABC such that ar (DBC) = ar (EBC).
Prove that DE || BC.

Solution:
Since ∆s BCE and ABCD are equal in area and have a same base BC.
Therefore, altitude from E of ∆BCE = altitude from D of ∆BCD.
or, ∆s BCE and BCD are between the same parallel lines.
DE || BC.

Question 8.
XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively. Show that ar (ABE) = ar (ACF)
Solution:
Given: In ∆ABC, XY || BC, BE || AC and CF || AB.
where E and F lie on XY.
To prove that: ar (∆ABE) = ar (∆ACF)

Proof: We have
XY || BC
or EY || BC ……(i)
and BE || AC
or BE || CY …….(ii)
From, equation (i) and (ii)
EBCY is a parallelogram.
Again, parallelogram EBCY and ∆AEB lie on same base EB and between same parallels BE and AC.
We know that if a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.
So, ar(∆ABE) = $$\frac {1}{2}$$ ar(|| gm FBCY) …….(A)
Now, XY || BC
or, XF || BC ……(iii)
and, CF || AB
or, CF || BX ……(iv)
From equation (iii) and (iv)
BCFX is a parallelogram.
Again, parallelogram BCFX and ∆ACF lie on sane base CF and between same parallels CF and AB.
∴ ar(∆ACF) = $$\frac {1}{2}$$ ar (|| gm BCFX) ……(B)
Now, parallelogram EBCY and parallelogram BCFX lie on same base BC and between same parallels BC and EF.
We know that, parallelograms on the same base and lie between the same parallels having equal area.
∴ ar(|| gm EBCY) = ar (|| gm BCFX) ……(C)
Therefore from equation (A), (B) and (C)
we have ar(∆ABE) = ar(∆ACF)

Question 9.
The side AB of a parallelogram ABCD is produced to any points. A line through A and parallel to CD meet CB produced at Q arid then parallelogram PBQR is completed (see Fig 9.26). Show that ar(ABCD) = ar(PBQR).

Solution:
Given: ABCD is a parallelogram in which P is any point on AB produced and AQ || CP.
To prove that: ar(|| gm ABCD) = ar(|| gm BPRQ)
Construction: Join AC and PQ.
Proof: Since AC and PQ are diagonals of parallelograms ABCD and BPQR respectively.

∴ ar(∆ABC) = $$\frac {1}{2}$$ ar(|| gm ABCD) ……(i)
and ar(∆PBQ) = $$\frac {1}{2}$$ ar(|| gm BPRQ) ……(ii)
Now, ∆s ACQ and AQP are on die same base AQ and between same parallels AQ and CP.
∴ ar(∆ACQ) = ar(∆AQP)
or, ar(∆ACQ) – (∆ABQ) = ar(∆AQP) – ar(∆ABQ)
[Subtracting ar(∆ABQ) from both sides]
or, ar(∆ABC) = ar(∆BPQ)
or, $$\frac {1}{2}$$ ar(|| gm ABCD) = $$\frac {1}{2}$$ ar (|| gm BPRQ) [From equ. (i) and (ii)]
or, ar(|| gm ABCD) = ar (|| gm BPRQ)

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:
Given: ABCD is a trapezium, in which AB || DC, and diagonal AC and BD intersect each other at O.
To prove that: ar(∆AOD) = ar(∆BOC)
Proof: Since ∆DAC and ∆DBC lie on same base BC and between same parallels AB and DC.
∴ ar(∆DAC) = ar(∆DBC)
[∆s on the same base and between same parallels are equal in area]
or, ar (∆DAC) – ar (∆DOC) = ar (∆DBC) – ar (∆DOC) [Subtract ar (∆DOC) both side]
ar (∆AOD) = ar (∆BOC)

Question 11.
In fig. 9.27, ABCDE is a pantagon. A line through B parallel to AC meet DC produced at F.
Show that
(i) ar (ACB) = ar (ACF)
(ii) ar (AEDF) = ar (ABCDE)

Solution:
Given; ABCDE is a pentagon, and AC || BF.
To prove that:
(i) ar (ACB) = ar (ACF)
(a) ar (AEDE) = ar (ABCDE)
Proof:
(i) Since ∆s ACB and ACF lie on the same base AC and between same parallels AC and BF.
∴ ar(∆ACB) = ar(∆ACF)
[∆s on same base and betwen same paralles are equal in area]

(ii) We have,
ar (∆ACB) = ar(∆ACF) [Prove above]
ar(∆ACB) + ar(ACDE) = ar(∆ACF) + ar(ACDE) [Add ar (ACDE) both side]
∴ ar (AEDF) = ar (ABCDE)

Question 12.
A villager Itwaari has a plot oi land of the shape of a quadrilateral. Tht Gram Panchayat of the village dicided b take over some portion of his plot from ore of the comers to construct a Health Centre. Itwaari agrees to the above proposal wih the condition that he should be given eqial amount of land in lieu of his land adjoiniag his plot so as to form a triangular pbt. Explain how this proposal will be implemented.

Solution:
Let a villager Itwaari has a plot of land of the shape of a quadrilateral ABCD.
The Gram Panchayat decided to take comer portion of the land ∆ACD. Itwaari agreed. So the remaining part of the land belong to Itwaari is
ar (ABCD) – ar (∆ACD) = ar (∆ABC)
Now, Itwaari demands the adjoining plot. So, if DE is drawn parallel to AC and join A to E. In this way we add ar (∆ACE) in the remaining plot of Itwaari. In this way proposal will be implemented.
As, ar (∆ABC) + ar (∆ACD) = ar (∆ABC) + ar (∆ACE)
Here ar(∆ACD) = ar(∆ACE)
(Triangles made between two parallels and same base AC)
ar(ABCD) = ar(∆ABE)

Question 13.
ABCD is a trapezium with AB || DC. A line parallel to AC intersect AB at X and BC at Y. Prove that ar (ADX) = ar (ALY).
Solution:
Given: ABCD is a trapezium, in which AB || DC and AC || XY. .
To prove that: ar(∆ADX) = ar(∆ACY)

Construction: Join CX.
Proof: Since ∆s ADX and ACX lie on same base AX and between same parallels AB and DC.
[∆s on same base and between same parallels are equal in area]
Now, ∆ACY and ∆ACX lie on same base AC and betwen same parallels AC and XY.
ar (∆ACY) = ar (∆ACX) …….(ii)
From equation (i) and (ii) we get

Question 14.
In fig 9.28, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

Solution:
Given: AP || BQ || CR
To prove that: ar (∆AQC) = ar (∆PBR)
Proof: Since ∆s ABQ and PBQ lie on same base and betwen same parallels AP and BQ.
∴ ar (∆ABQ) = ar (∆PBQ) ……(i)
[∆s on same base and between same parallels are equal in area]
Again, ∆s BQC and BQR lie on same base BQ and between same parallels BQ and CR.
ar (∆BQC) = ar (∆BQR) ……(ii)
ar(∆ABQ) + ar(∆BQC) = ar(∆PBQ) + ar(∆BQR)
or, ar (∆AQC) = ar (∆PBR).

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:
Given: ABCD is a quadrilateral in which diagonal AC ab BD intersect each other at O, and ar (∆AOD) = ar (∆BOC).
To prove that: ABCD is a trapezium.
Proof: We have
ar(∆AOD) = ar(∆BOC) (Given)
or, ar(∆AOD) + ar(∆DOC) = ar(∆BOC) + ar(∆DOC) [Add ar (∆DOC) both side]
We know that if two triangles on same base and have equal area, then it must lie in between same parallels.
Here, ∆s ADC and BDC lie on same base DC, and
∴ DC must be parallel to AB.
∴ ABCD is a trapezium.

Question 16.
In fig.9.29, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Solution:
We have,
ar(∆BDP) = ar(∆ARC) (Given)
ar(∆BDP) – ar(∆DPC) = ar(∆ARC) – ar(∆DRC)
ar (∆DRC) = ar (∆DBC) (Given)
or, ar (∆BDC) = ar (∆ADC)
Now, ∆s BDC and ADC lie on same base and having equal area therefore they must lie betwen two parallel lines.
Then, AB || DC
Therefore, ABCD is a trapezium.
Again, ar(∆DRC) = ar(∆DPC) (Given)
Here, ∆s DRC and DPC lie on same base DC and having equal area.
∴ DC || RP
Therefore, DCPR is a trapezium.

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