# CBSE Important Questions Class 9 Maths Chapter 9

**Important Questions Class 9 Mathematics Chapter 9 – ****Areas of Parallelograms and Triangles**

Chapter 9 of Mathematics Class 9, Areas of Parallelograms and Triangles, is covered in the unit Mensuration. Chapter 9 discusses topics like triangles on the same base and between the same parallels and areas of parallelograms and triangles. These topics are fairly difficult for students to understand in Class 9, and thus mastery calls for adequate revision.

You may acquire a better understanding of the subject and the concepts by looking at the important questions Class 9 Mathematics Chapter 9. As learning resources, Extramarks offers students NCERT Solutions, Exemplar solutions, and test questions so they can practice a range of math problems and work through their challenges. To understand the kind of problems from Mathematics Chapter 9, Areas of Parallelograms and Triangles, and their weightage, students are also urged to complete sample papers and the previous years’ question papers.

Use the important questions Class 9 Mathematics Chapter 9 to learn how to calculate the areas of parallelograms and triangles. Students will master the key theorems and formulas of parallelograms and triangles in the chapter more quickly and effectively by completing the important questions Class 9 Mathematics Chapter 9.

Students can completely rely on Chapter 9 Class 9 Mathematics important questions in order to facilitate learning and in understanding the concepts of areas of parallelograms and triangles based on various scenarios. While practising the exercise problems from the NCERT book, students can also use these solved questions as a reference source.

**Important Questions Class 9 Mathematics Chapter 9- With Solutions**

We have offered thorough, detailed solutions to assist students in solving their questions and gaining a better understanding of the chapters. The important questions Class 9 Mathematics Chapter 9 are all thoroughly explained in the NCERT Solutions that are provided here. The CBSE Board’s most recent instructions for the exam format and syllabus are followed while preparing this NCERT Solutions.

A few Mathematics Class 9 Chapter 9 important questions are provided here, along with their answers:

**Question 1: The median of a triangle divides it into two**

**(A) triangles of equal area**

**(B) isosceles triangles**

**(C) right triangles**

**(D) congruent triangles**

**Solution 1:**

(A) triangles of equal area

**Explanation:**

The median of the triangle divides it into two triangles of equal area.

**Question 2: In which of the following figures (refer image given below) do you find two polygons lying on the same base and in between the same parallels?**

**Solution 2: (D)**

**Explanation:**

In figure (D), both the parallelograms, PQRA and BQRS, are on the same base QR and between the same parallels QR and PS.

**Question 3: In the figure given below, the area of the parallelogram**

**ABCD is:**

**(A) AD × DL**

**(B) BC × BN**

**(C) DC × DL**

**(D) AB × BM**

**Solution 3:**

(C) DC × DL

**Explanation:**

Area of the parallelogram = Base × Corresponding altitude

= AB × DL … (equation 1)

As the opposite sides of the parallelogram are equal,

We get,

AB = DC

Substituting this in eq(1), we get,

Area of parallelogram = AB × DL

= DC × DL

**Question 4: In the figure given below, if the parallelogram ABCD and the rectangle ABEF are of an equal area, then:**

**(A) Perimeter of ABCD = ½ (Perimeter of ABEM) **

**(B) Perimeter of ABCD < Perimeter of ABEM**

**(C) Perimeter of ABCD > Perimeter of ABEM**

**(D) Perimeter of ABCD = Perimeter of ABEM**

**Solution 4: (C) **Perimeter of ABCD > Perimeter of ABEM

**Explanation:**

In the rectangle ABEM,

AB = EM …(equation 1) [sides of the rectangle]

In the parallelogram ABCD,

CD = AB …(equation 2)

Adding equations (1) and (2),

we have

AB + CD = EM + AB …eq.(i)

We know,

The perpendicular distance between the two parallel sides of the parallelogram is always less than the length of the other parallel sides.

BE < BC as well as AM < AD

[since, in a right-angled triangle, the hypotenuse is always greater than the other side]

By adding both the above inequalities, we have

SE + AM <BC + AD or BC + AD> BE + AM

By adding AB + CD on both sides, we have

AB + CD + BC + AD> AB + CD + BE + AM

⇒ AB+BC + CD + AD> AB+BE + EM+ AM [Thus, CD = AB = EM]

Thus,

We have,

The perimeter of the parallelogram ABCD > perimeter of the rectangle ABEM.

**Write true/false and justify your answer:**

**Question 5: ABCD is the parallelogram, and X is a midpoint of AB. If area (AXCD) = 24 cm****2****, then area (ABC) = 24 cm****2****.**

**Solution 5: False**

**Explanation:**

In ΔABC, X is the midpoint of AB

Considering the given question,

Hence area (ΔAXC) = area (ΔBXC) = ½ area(ΔABC)

Thus,

area (ΔAXC) = area (ΔBXC) = 12 cm2

Area of parallelogram ABCD = 2 x area (ΔABC)

= 2 x 24 = 48 cm2

But, as per the question,

area of parallelogram ABCD = area (AXCD) + area (ΔBXC)

= 24 + 12 = 36 cm2

Thus, it contradicts here.

So, if area (AXCD) = 24 cm2, then area (ABC) ≠ 24 cm2

**Question 6: PQRS is the rectangle that is inscribed in the quadrant of the circle of radius 13 cm. A is a point on PQ. If PS = 5 cm, then area (PAS) = 30 cm****2****.**

**Solution 6: True. **

Since A is a point on PQ, then the area (PAS) ≠ 30 cm2

But, the statement can only be true if PA is equal to PS.

Justification:

As per the question,

PQRS is the rectangle which is inscribed in a quadrant of a circle of radius 13 cm.

Given,

A is any point on PQ

PA<PQ

area (△PQR) = ½ ×PQ×QR = ½ ×12×5 = 30cm2

PS=5 cm

Suppose PA<PQ,

area(△PAS) < area(△PQR)

area(△PAS) < 30 cm2

Suppose PA=PQ,

area(△PAS) = ½ ×PQ×PS

= ½ ×12×5

= 30cm2

**Question 7: In figure given below, ABCD is the parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.**

**Solution 7:**

Given,

AB = CD = 16 cm (Since opposite sides of the parallelogram)

CF = 10 cm and AE = 8 cm

We know,

Area of the parallelogram = Base × Altitude

= CD×AE = AD×CF

⇒ 16×8 = AD×10

⇒ AD = 128/10 cm

⇒ AD = 12.8 cm

**Question 8: If E, F, G and H are, respectively, the mid-points of sides of the parallelogram ABCD, show that area (EFGH) = 1/2 area (ABCD).**

**Solution 8:**

Given,

E, F, G and H are the mid-points of sides of the parallelogram ABCD, respectively.

To Prove,

area (EFGH) = ½ area (ABCD)

Construction,

Here, H and F are joined.

Proof:

AD || BC and AD = BC (Since Opposite sides of the parallelogram)

⇒ ½ AD = ½ BC

And,

AH || BF and DH || CF

⇒ AH = BF and DH = CF (Since H and F are the midpoints)

∴, ABFH and HFCD are parallelograms.

Here,

We know ΔEFH and parallelogram ABFH both lie on the same common base FH and in between the same parallel lines AB and HF.

Therefore, the area of EFH = ½ area of ABFH — equation (i)

Also, the area of GHF = ½ area of HFCD — equation(ii)

Adding equation (i) and (ii),

area of ΔEFH + area of ΔGHF = ½ area of ABFH + ½ area of HFCD

⇒ area of EFGH = area of ABFH

Therefore, area (EFGH) = ½ area (ABCD)

**Question 9: P and Q are the two points lying on sides DC and AD, respectively, of the parallelogram ABCD.**

**Show that area(APB) = area (BQC).**

**Solution 9:**

ΔAPB and the parallelogram ABCD lie on the same base AB and in between the same parallel AB and DC.

area(ΔAPB) = ½ area(parallelogram ABCD) — (i)

Similarly,

area(ΔBQC) = ½ area(parallelogram ABCD) — (ii)

From (i) and (ii), we have

area(ΔAPB) = area(ΔBQC)

**Question 10: In figure given below, P is the point in the interior of a parallelogram ABCD. Show that**

**(i) area(APB) + area(PCD) = ½ area(ABCD)**

**(ii) area(APD) + area(PBC) = area(APB) + area(PCD)**

**[Hint: Through point P, draw a line parallel to AB.]**

**Solution 10:**

(i) A line GH is drawn parallel to AB which passes through P.

In the parallelogram,

AB || GH (by construction) —equation (i)

∴ AD || BC ⇒ AG || BH — equation (ii)

From equations (i) and (ii),

ABHG is the parallelogram.

Now,

ΔAPB and the parallelogram ABHG are lying on the same base AB and in between the same parallel lines AB and GH.

∴ area(ΔAPB) = ½ area(ABHG) — equation (iii)

also,

ΔPCD and the parallelogram CDGH are lying on the same base CD and in between the same parallel lines CD and GH.

∴ area(ΔPCD) = ½ area(CDGH) — (iv)

Adding equations (iii) and (iv),

area(ΔAPB) + area(ΔPCD) = ½ [area(ABHG)+area(CDGH)]

⇒ area(APB)+ area(PCD) = ½ area(ABCD)

(ii) A line EF is drawn parallel to AD and passes through P.

In the parallelogram,

AD || EF (by construction) — (i)

Therefore.

AB || CD ⇒ AE || DF — (ii)

From equations (i) and (ii),

AEDF is a parallelogram.

Now,

ΔAPD and parallelogram AEFD lying on the same base AD and in-between the same parallel lines AD and EF.

∴area(ΔAPD) = ½ area(AEFD) — (iii)

Also, ΔPBC and parallelogram BCFE lie on the same base BC and in between the same parallel lines BC and EF.

∴area(ΔPBC) = ½ area(BCFE) — (iv)

Adding equations (iii) and (iv),

area(ΔAPD)+ area(ΔPBC) = ½ {area(AEFD)+area(BCFE)}

⇒area(APD)+area(PBC) = area(APB)+area(PCD)

**Question 11: In the figure given below, PQRS and ABRS are the parallelograms, and X is the point on the side BR. Show that**

**(i) area (PQRS) = area (ABRS)**

**(ii) area(AXS) = ½ area (PQRS)**

**Solution 11:**

(i) Parallelogram PQRS and ABRS lie on the same base SR and in between the same parallel lines SR and PB.

∴ area (PQRS) = area (ABRS) — equation (i)

(ii) ΔAXS and the parallelogram ABRS are lying on the same base AS and in between the same parallel lines AS and BR.

∴ area(ΔAXS) = ½ area(ABRS) — equation (ii)

From (i) and (ii), we find that

area(ΔAXS) = ½ area (PQRS)

**Question 12: In the figure given below, E is the point on a median AD of the ΔABC. Show that area (ABE) = area(ACE).**

**Solution 12:**

Given,

AD is the median of the ΔABC. Thus, it will divide ΔABC into two triangles of equal area.

∴area(ABD) = area(ACD) — equation(i)

and,

ED is the median of ΔABC.

∴area(EBD) = area(ECD) — equation(ii)

Subtracting equation (ii) from (i),

area (ABD) – area (EBD) = area(ACD) – area(ECD)

⇒area(ABE) = area(ACE)

**Question 13: In the triangle ABC, E is any mid-point of the median AD. Show that area (BED) = ¼ area(ABC).**

**Solution 13:**

area(BED) = (1/2)×BD×DE

As E is the midpoint of AD,

AE = DE

and AD is the median on side BC of triangle ABC,

BD = DC,

DE = (1/2) AD — equation (i)

BD = (1/2)BC — equation (ii)

From equations (i) and (ii), we have,

area (BED) = (1/2)×(1/2)BC × (1/2)AD

⇒ area(BED) = (1/2)×(1/2)area(ABC)

⇒ area(BED) = ¼ area(ABC)

**Question 14: Show that the diagonals of the parallelogram divide it into four triangles of an equal area.**

**Solution 14:**

O is the midpoint of AC and BD. (diagonals bisect each other)

In ΔABC, BO is the median.

∴area(AOB) = area(BOC) — equation(i)

also,

In ΔBCD, CO is the median.

∴area(BOC) = area(COD) — equation(ii)

In ΔACD, OD is the median.

∴area(AOD) = area(COD) — equation(iii)

In ΔABD, AO is the median.

∴area(AOD) = area(AOB) — equation(iv)

From equations (i), (ii), (iii) and (iv), we have,

area(BOC) = area(COD) = area(AOD) = area(AOB)

Thus, we have the diagonals of a parallelogram divided into four triangles of equal area.

**Question 15: In the figure given below, ABC and ABD are the two triangles on the same base AB. If the line-segment CD is bisected by AB at point O, show that: area(ABC) = area(ABD).**

**Solution 15:**

In ΔABC, AO is the median. (CD is bisected by AB at O)

∴area(AOC) = area(AOD) — (i)

also,

ΔBCD, BO is the median. (CD is bisected by AB at O)

∴area(BOC) = area(BOD) — (ii)

Adding equation (i) and (ii),

We have,

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

⇒area(ABC) = area(ABD)

**Question 16: D, E and F are, respectively, the mid-points of the sides BC, CA and AB respectively of a ΔABC.**

**Show that**

**(i) BDEF is the parallelogram.**

**(ii) area(DEF) = ¼ area(ABC)**

**(iii) area (BDEF) = ½ area (ABC)**

**Solution 16:**

(i) In ΔABC,

EF || BC and EF = ½ BC (by the mid-point theorem)

and,

BD = ½ BC (D is the midpoint)

Thus, BD = EF

and,

BF and DE are parallel and equal to each other.

Thus, the pair of opposite sides are equal in length and parallel to each other.

∴ BDEF is a parallelogram.

(ii) Proceeding from the result of (i),

BDEF, DCEF, and AFDE are parallelograms.

A diagonal of the parallelogram divides it into two triangles of equal area.

∴area(ΔBFD) = area(ΔDEF) (For parallelogram BDEF) — equation(i)

also,

area(ΔAFE) = area(ΔDEF) (For parallelogram DCEF) — equation(ii)

area(ΔCDE) = area(ΔDEF) (For parallelogram AFDE) — equation(iii)

From (i), (ii) and (iii)

area(ΔBFD) = area(ΔAFE) = area(ΔCDE) = area(ΔDEF)

⇒ area(ΔBFD) +area(ΔAFE) +area(ΔCDE) +area(ΔDEF) = area(ΔABC)

⇒ 4 area(ΔDEF) = area(ΔABC)

⇒ area(DEF) = ¼ area(ABC)

(iii) Area (parallelogram BDEF) = area(ΔDEF) +area(ΔBDE)

⇒ area(parallelogram BDEF) = area(ΔDEF) +area(ΔDEF)

⇒ area(parallelogram BDEF) = 2× area(ΔDEF)

⇒ area(parallelogram BDEF) = 2× ¼ area(ΔABC)

⇒ area(parallelogram BDEF) = ½ area(ΔABC)

**Question 17: In the figure given below, the diagonals AC and BD of the quadrilateral ABCD intersect at O such that OB = OD.**

**If AB = CD, show that:**

**(i) area (DOC) = area (AOB)**

**(ii) area (DCB) = area (ACB)**

**(iii) DA || CB or ABCD is a parallelogram.**

**[Hint: From D and B, draw the perpendiculars to AC.]**

**Solution 17:**

Given,

OB = OD and AB = CD

Construction,

DE ⊥ AC and BF ⊥ AC are drawn.

Proof:

(i) In ΔDOE and ΔBOF,

∠DEO = ∠BFO (Since perpendiculars)

∠DOE = ∠BOF (Since vertically opposite angles)

OD = OB (Given in the question)

∴, ΔDOE ≅ ΔBOF by AAS congruence rule.

∴, DE = BF (By CPCT rule) — (i)

and, area(ΔDOE) = area(ΔBOF) (Congruent triangles) — (ii)

Now,

In ΔDEC and ΔBFA,

∠DEC = ∠BFA (Since perpendiculars)

CD = AB (Given in the question)

DE = BF (From i)

∴ ΔDEC ≅ ΔBFA by RHS congruence condition.

∴ area(ΔDEC) = area(ΔBFA) (Congruent triangles) — (iii)

Adding (ii) and (iii),

area(ΔDOE) + area(ΔDEC) = area(ΔBOF) + area(ΔBFA)

⇒ area (DOC) = area (AOB)

(ii) area(ΔDOC) = area(ΔAOB)

Adding area(ΔOCB) in LHS and RHS, we get,

⇒area(ΔDOC) + area(ΔOCB) = area(ΔAOB) + area(ΔOCB)

⇒ area(ΔDCB) = area(ΔACB)

(iii) When the two triangles have the same base and equal areas, the triangles will be in between the same parallel lines

area(ΔDCB) = area(ΔACB)

DA || BC — (iv)

For the quadrilateral ABCD, one pair of opposite sides are equal (AB = CD), and the other pair of opposite sides are parallel.

Thus ABCD is a parallelogram.

**Question 18: D and E are points on sides AB and AC, respectively, of ΔABC such that area(DBC) = area(EBC). Prove that DE || BC.**

**Solution 18:**

ΔDBC and ΔEBC are on the same base BC and have equal areas.

So, they will lie between the same parallel lines.

Thus, DE || BC.

**Question 19: XY is a line parallel to the side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that**

**area(ΔABE) = area(ΔACF)**

**Solution 19:**

Given,

XY || BC, BE || AC and CF || AB

To show,

area(ΔABE) = area(ΔACF)

Proof:

BCYE is a parallelogram as ΔABE and parallelogram BCYE are on the same base BE and in between the same parallel lines BE and AC.

∴area(ABE) = ½ area(BCYE) … equation (1)

Now,

CF || AB and XY || BC

⇒ CF || AB and XF || BC

⇒ BCFX is a parallelogram.

As ΔACF and parallelogram BCFX are on the same base CF and in between the same parallel AB and FC.

So, area (ΔACF)= ½ area (BCFX) … equation (2)

But,

parallelogram BCFX and BCYE are on the same base BC and in between the same parallels BC and EF.

∴area (BCFX) = area(BCYE) … equation (3)

From equations (1), (2) and (3), we get

area (ΔABE) = area(ΔACF)

⇒ area(BEYC) = area(BXFC)

Since the parallelograms are on the same base BC and in-between the same parallels EF and BC–(iii)

And,

△AEB and parallelogram BEYC is on the same base BE, and in-between the same parallels BE and AC.

⇒ area(△AEB) = ½ area(BEYC) — (iv)

Similarly,

△ACF and parallelogram BXFC on the same base CF and between the same parallels CF and AB.

⇒ area(△ ACF) = ½ area(BXFC) — (v)

From (iii), (iv) and (v),

area(△ABE) = area(△ACF)

**Question 20: The side AB of a parallelogram ABCD is produced to a point P. A line is drawn through A and parallel to CP meets CB produced at Q, and then parallelogram PBQR is completed (see the figure given below). Show that**

**area(ABCD) = area(PBQR).**

**[Hint: Join AC and PQ. Now compare area(ACQ) and area(APQ).]**

**Solution 20:**

AC and PQ are joined.

area(△ACQ) = area(△APQ) (On the same base AQ and between the same parallel lines AQ and CP)

⇒ area(△ACQ)-area(△ABQ) = area(△APQ)-area(△ABQ)

⇒area(△ABC) = area(△QBP) — (i)

AC and QP are the diagonals ABCD and PBQR.

∴area(ABC) = ½ area(ABCD) — (ii)

area(QBP) = ½ area(PBQR) — (iii)

From (ii) and (ii),

½ area(ABCD) = ½area(PBQR)

⇒ area(ABCD) = area(PBQR)

**Question 21: In the figure given below, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F.**

**Show that**

**(i) area(△ACB) = area(△ACF)**

**(ii) area(AEDF) = area(ABCDE)**

**Solution 21:**

(i). Both the △ACB and △ACF lie on the same base AC and in between the same parallels AC and BF.

∴ area(△ACB) = area(△ ACF)

(ii). area (△ACB) = area (△ACF)

⇒ area (△ACB) + area (ACDE) = area (△ACF)+area (ACDE)

⇒ area (ABCDE) = area (AEDF)

**Question 22: ABCD is a trapezium with AB || DC. A line that is parallel to AC intersects AB at X and BC at Y. Prove that area (△ADX) = area (△ACY). [Hint: Join CX.]**

**Solution 22:**

Given,

ABCD is a trapezium with AB || DC.

XY || AC

Construction,

Join CX

To Prove,

area(ADX) = area(ACY)

Proof:

area(△ADX) = area(△AXC) — (i) (Since they are on the same base AX and in-between the same parallels AB and CD)

also,

area(△ AXC)=area(△ ACY) — (ii) (Since they are on the same base AC and in-between the same parallels XY and AC.)

(i) and (ii),

area(△ADX) = area(△ACY)

**Question 23: In the figure given below, AP || BQ || CR. Prove that area(△AQC) = area(△PBR).**

**Solution 23:**

Given,

AP || BQ || CR

To Prove,

area(AQC) = area(PBR)

Proof:

area(△AQB) = area(△PBQ) — (i) (Since they are on the same base BQ and between the same parallels AP and BQ.)

also,

area(△BQC) = area(△BQR) — (ii) (Since they are on the same base BQ and between the same parallels BQ and CR.)

Adding (i) and (ii),

area(△AQB)+area(△BQC) = area(△PBQ)+area(△BQR)

⇒area(△ AQC) = area(△ PBR)

**Question 24: Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that area(△AOD) = area(△BOC). Prove that ABCD is a trapezium.**

**Solution 24:**

Given,

area(△AOD) = area(△BOC)

To prove,

ABCD is a trapezium.

Proof:

area (△AOD) = area (△BOC)

⇒ area(△AOD) + area(△AOB) = area(△BOC)+area(△AOB)

⇒ area(△ADB) = area(△ACB)

Areas of △ADB and △ACB are equal. ∴, they must be lying between the same parallel lines.

∴ AB ∥ CD

∴ABCD is a trapezium.

**Question 25: In the figure given below, area(DRC) = area(DPC) and area(BDP) = area(ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.**

**Solution 25:**

Given,

area(△DRC) = area(△DPC)

area(△BDP) = area(△ARC)

To Prove,

ABCD and DCPR are trapeziums.

Proof:

area(△BDP) = area(△ARC)

⇒ area(△BDP) – area(△DPC) = area(△DRC)

⇒ area(△BDC) = area(△ADC)

∴ area(△BDC) and area(△ADC) are lying in between the same parallel lines.

∴ AB ∥ CD

ABCD is a trapezium.

Similarly,

area(△DRC) = area(△DPC).

∴ area(△DRC) and area(△DPC) are lying in between the same parallel lines.

∴ DC ∥ PR

∴DCPR is a trapezium.

**Question 26: A parallelogram ABCD and a rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than that of the rectangle.**

**Solution 26:**

Given,

Parallelogram ABCD and the rectangle ABEF have the same base AB and equal areas.

To prove,

The perimeter of parallelogram ABCD is greater than the perimeter of rectangle ABEF.

Proof,

We know the opposite sides of a parallelogram and a rectangle are equal.

AB = DC [As ABCD is a paralleogram]

and, AB = EF [As ABEF is the rectangle]

DC = EF … equation (i)

Adding AB on both sides, we get,

⇒AB + DC = AB + EF … equation (ii)

We know the perpendicular segment is the shortest of all the segments and can be drawn to a given line from any point not lying on it.

BE < BC and AF < AD

⇒ BC > BE and AD > AF

⇒ BC+AD > BE+AF … (iii)

Adding equations (ii) and (iii), we get

AB+DC+BC+AD > AB+EF+BE+AF

⇒ AB+BC+CD+DA > AB+ BE+EF+FA

⇒ the perimeter of parallelogram ABCD > the perimeter of the rectangle ABEF.

The perimeter of the parallelogram is greater than that of the rectangle.

Hence proved.

**Question 27: In the figure given below, ABCD, DCFE, and ABFE are parallelograms. Show that area (ADE) = area (BCF).**

**Solution 27:**

Given,

ABCD, DCFE and ABFE are parallelograms

To prove,

area (△ADE) = area (△BCF)

Proof,

In △ADE and △BCF,

AD = BC [Since opposite sides of parallelogram ABCD]

DE = CF [Since opposite sides of parallelogram DCFE]

AE = BF [Since opposite sides of parallelogram ABFE]

△ADE ≅ △BCF [Using SSS Congruence theorem]

area(△ADE) = area(△BCF) [ By CPCT]

**Question 28: In the figure given below, ABCD is a parallelogram and BC is produced to a point Q in such a way that AD = CQ. If AQ intersects DC at P, show that area (BPC) = area (DPQ).**

**[Hint: Join AC.]**

**Solution 28:**

Given:

ABCD is a parallelogram

AD = CQ

To prove:

area (△BPC) = area (△DPQ)

Proof:

In △ADP and △QCP,

∠APD = ∠QPC [Vertically Opposite Angles]

∠ADP = ∠QCP [Alternate Angles]

AD = CQ [given]

△ABO ≅ △ACD [AAS congruency]

DP = CP [CPCT]

In △CDQ, QP is median. [Since, DP = CP]

Since the median of a triangle divides it into two parts of equal areas,

area(△DPQ) = area(△QPC) —(i)

In △PBQ, PC is the median. [Since, AD = CQ and AD = BC ⇒ BC = QC]

As the median of the triangle divides it into two parts of equal areas.

area(△QPC) = area(△BPC) —(ii)

From the equation (i) and (ii), we get

area(△BPC) = area(△DPQ)

**Question 29: In the figure given below, ABC and BDE are two equilateral triangles in such a way that D is the midpoint of BC. If AE intersects BC at F, show that:**

**(i) area (BDE) =1/4 area (ABC)**

**(ii) area (BDE) = ½ area (BAE)**

**(iii) area (ABC) = 2 area (BEC)**

**(iv) area (BFE) = area (AFD)**

**(v) area (BFE) = 2 area (FED)**

**(vi) area (FED) = 1/8 area (AFC)**

**Solution 29:**

(i) We assume that G and H are the midpoints of the sides AB and AC, respectively.

We join the midpoints with line-segment GH. Here, GH is parallel to the third side.

BC will be half of the length of BC by the mid-point theorem.

∴ GH =1/2 BC and GH || BD

∴ GH = BD = DC and GH || BD (As D is the midpoint of BC)

In a similar manner,

GD = HC = HA

HD = AG = BG

ΔABC is divided into four equal equilateral triangles ΔBGD, ΔAGH, ΔDHC and ΔGHD

We can say that,

ΔBGD = ¼ ΔABC

Considering, ΔBDG and ΔBDE

BD = BD (Common base)

Since both triangles are equilateral triangles, we can say that,

BG = BE

DG = DE

ΔBDG ≅ΔBDE [By SSS congruency rule]

area (ΔBDG) = area (ΔBDE)

area (ΔBDE) = ¼ area (ΔABC)

Hence proved.

(ii) area(ΔBDE) = area(ΔAED) (Since common base DE and DE||AB)

area(ΔBDE)−area(ΔFED) = area(ΔAED)−area (ΔFED)

area(ΔBEF) = area(ΔAFD) …(i)

Now,

area(ΔABD) = area(ΔABF)+area(ΔAFD)

area(ΔABD) = area(ΔABF)+area(ΔBEF) [From equation (i)]

area(ΔABD) = area(ΔABE) …(ii)

AD is the median of ΔABC.

area(ΔABD) = ½ area (ΔABC)

= (4/2) area (ΔBDE)

= 2 area (ΔBDE)…(iii)

From (ii) and (iii), we obtain

2 area (ΔBDE) = area (ΔABE)

area (BDE) = ½ area (BAE)

Hence proved.

(iii) area(ΔABE) = area(ΔBEC) [Common base BE and BE || AC]

area(ΔABF) + area(ΔBEF) = area(ΔBEC)

From eqn (i), we get,

area(ΔABF) + area(ΔAFD) = area(ΔBEC)

area(ΔABD) = area(ΔBEC)

½ area(ΔABC) = area(ΔBEC)

area(ΔABC) = 2 area(ΔBEC)

Hence proved.

(iv) Both ΔBDE and ΔAED lie on the same base (DE) and in between the parallel lines DE and AB.

∴area (ΔBDE) = area (ΔAED)

Subtracting area(ΔFED) from L.H.S and R.H.S,

We get,

∴area (ΔBDE)−area (ΔFED) = area (ΔAED)−area (ΔFED)

∴area (ΔBFE) = area(ΔAFD)

Hence proved.

(v) We assume that h is the height of vertex E, corresponding to the side BD in ΔBDE.

Also, we assume that H is the height of vertex A, corresponding to the side BC in ΔABC.

While solving Question (i),

We saw that,

area (ΔBDE) = ¼ area (ΔABC)

While solving Question (iv),

We saw that,

area (ΔBFE) = area (ΔAFD).

∴area (ΔBFE) = area (ΔAFD)

= 2 areas (ΔFED)

Hence, ar (ΔBFE) = 2 ar (ΔFED)

Hence proved.

(vi) area (ΔAFC) = area (ΔAFD) + area(ΔADC)

= 2 area (ΔFED) + (1/2) area(ΔABC) [using (v)

= 2 area(ΔFED) + ½ [4area(ΔBDE)] [Using result of Question (i)]

= 2 area (ΔFED) +2 area(ΔBDE)

Since ΔBDE and ΔAED are on the same base and in between the same parallels

= 2 area(ΔFED) +2 area (ΔAED)

= 2 area (ΔFED) +2 [area (ΔAFD) +area (ΔFED)]

= 2 area (ΔFED) +2 area (ΔAFD) +2 area(ΔFED) [From question (viii)]

= 4 area (ΔFED) +4 area (ΔFED)

⇒area(ΔAFC) = 8 area (ΔFED)

⇒area (ΔFED) = (1/8) area (ΔAFC)

Hence proved.

**Question 30: Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that**

**area(APB)×area (CPD) = area (APD)×area (BPC).**

**[Hint: From A and C, draw perpendiculars to BD.]**

**Solution 30:**

Given:

The diagonal AC and BD of a quadrilateral ABCD intersect each other at point E.

Construction:

From A, draw AM perpendicular to BD

From C, draw CN perpendicular to BD

To Prove,

area(ΔAED) area(ΔBEC) = area (ΔABE) ×area(ΔCDE)

Proof,

area(ΔABE) = ½ ×BE×AM………….. (i)

area(ΔAED) = ½ ×DE×AM………….. (ii)

Dividing eq. (ii) by (i), we get,

area(AED)/area(ABE) = DE/BE…….. (iii)

Similarly,

area(CDE)/area(BEC) = DE/BE ……. (iv)

From eq. (iii) and (iv) , we get

area(AED)/area(ABE) = area(CDE)/area(BEC)

area(ΔAED)×area(ΔBEC) = area(ΔABE)×area (ΔCDE)

Hence proved.

**Question 31: P and Q are, respectively, the mid-points of sides AB and BC of a triangle ABC, and R is the mid-point of AP, showing that:**

**(i) area (PRQ) = ½ area (ARC)**

**(ii) area (RQC) = (3/8) area (ABC)**

**(iii) area (PBQ) = area (ARC)**

**Solution 31:**

(i) We know that the median divides the triangle into two triangles of equal area,

PC is the median of ABC.

area (ΔBPC) = area (ΔAPC) ……….(i)

RC is the median of APC.

area (ΔARC) = ½ area (ΔAPC) ……….(ii)

PQ is the median of BPC.

area (ΔPQC) = ½ area (ΔBPC) ……….(iii)

From eq. (i) and (iii), we get,

area (ΔPQC) = ½ area (ΔAPC) ……….(iv)

From eq. (ii) and (iv), we get,

area (ΔPQC) = area (ΔARC) ……….(v)

P and Q are the mid-points of AB and BC, respectively [given in the question]

PQ||AC

and PA = ½ AC

Since the triangles between the same parallel are equal in area, we get

area (ΔAPQ) = area (ΔPQC) ……….(vi)

From eq. (v) and (vi), we obtain,

area (ΔAPQ) = area (ΔARC) ……….(vii)

R is the mid-point of AP.

RQ is the median of APQ.

area (ΔPRQ) = ½ area (ΔAPQ) ……….(viii)

From (vii) and (viii), we get,

area (ΔPRQ) = ½ area (ΔARC)

Hence proved.

(ii) PQ is the median of ΔBPC

area (ΔPQC) = ½ area (ΔBPC)

= (½) ×(1/2 )area (ΔABC)

= ¼ area (ΔABC) ……….(ix)

Also,

area (ΔPRC) = ½ area (ΔAPC) [From (iv)]

area (ΔPRC) = (1/2)×(1/2)area ( ABC)

= ¼ area(ΔABC) ……….(x)

Add eq. (ix) and (x), we get,

area (ΔPQC) + area (ΔPRC) = (1/4)×(1/4)area (ΔABC)

area (quad. PQCR) = ¼ area (ΔABC) ……….(xi)

Subtracting area (ΔPRQ) from L.H.S and R.H.S,

area (quad. PQCR)–area(ΔPRQ) = ½ area (ΔABC)–area(ΔPRQ)

area (ΔRQC) = ½ area (ΔABC) – ½ area (ΔARC) [From result (i)]

area (ΔARC) = ½ area (ΔABC) –(1/2)×(1/2)area (ΔAPC)

area (ΔRQC) = ½ area (ΔABC) –(1/4)area (ΔAPC)

area (ΔRQC) = ½ area (ΔABC) –(1/4)×(1/2)area (ΔABC) [ As, PC is median of ΔABC]

area (ΔRQC) = ½ area (ΔABC)–(1/8)area (ΔABC)

area (ΔRQC) = [(1/2)-(1/8)]area (ΔABC)

area (ΔRQC) = (3/8)area (ΔABC)

(iii) area (ΔPRQ) = ½ area (ΔARC) [From result (i)]

2 area (ΔPRQ) = area (ΔARC) ……………..(xii)

area (ΔPRQ) = ½ area (ΔAPQ) [RQ is the median of APQ] ……….(xiii)

But, we know that

area (ΔAPQ) = area (ΔPQC) [From the reason mentioned in eq. (vi)] ……….(xiv)

From eq. (xiii) and (xiv), we get,

area (ΔPRQ) = ½ area (ΔPQC) ……….(xv)

At the same time,

area (ΔBPQ) = area (ΔPQC) [PQ is the median of ΔBPC] ……….(xvi)

From eq. (xv) and (xvi), we get,

area (ΔPRQ) = ½ area (ΔBPQ) ……….(xvii)

From eq. (xii) and (xvii), we get,

2×(1/2)area(ΔBPQ)= area(ΔARC)

⟹ area (ΔBPQ) = area (ΔARC)

Hence proved.

**Question 32: In the figure given below, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB, respectively. The line segment AX ^ DE meets BC at Y. Show that:**

**(i) ΔMBC ≅ ΔABD**

**(ii) area(BYXD) = 2area(MBC)**

**(iii) area(BYXD) = area(ABMN)**

**(iv) ΔFCB ≅ ΔACE**

**(v) area(CYXE) = 2area(FCB)**

**(vi) area(CYXE) = area(ACFG)**

**(vii) area(BCED) = area(ABMN)+area(ACFG)**

**Solution 32:**

(i) We know each angle of a square is 90°. Hence, ∠ABM = ∠DBC = 90º

∴∠ABM+∠ABC = ∠DBC+∠ABC

∴∠MBC = ∠ABD

In ∆MBC and ∆ABD,

∠MBC = ∠ABD (Proved earlier)

MB = AB (Sides of square ABMN)

BC = BD (Sides of square BCED)

∴ ∆MBC ≅ ∆ABD (SAS congruency)

(ii) We have

∆MBC ≅ ∆ABD

∴area (∆MBC) = area (∆ABD) … (i)

It is given that AX ⊥ DE and BD ⊥ DE (Since adjacent sides of the square BDEC)

∴ BD || AX (Since two lines perpendicular to the same line are parallel to each other)

∆ABD and the parallelogram BYXD are on the same base BD and in between the same parallels BD and AX.

Area (∆YXD) = 2 Area (∆MBC) [From equation (i)] … (ii)

(iii) The ∆MBC and the parallelogram ABMN are lying on the same base MB and between the same parallels MB and NC.

2 area (∆MBC) = area (ABMN)

area (∆YXD) = area(ABMN) [From equation (ii)] … (iii)

(iv) We know each angle of a square is 90°.

∴∠FCA = ∠BCE = 90º

∴∠FCA+∠ACB = ∠BCE+∠ACB

∴∠FCB = ∠ACE

In ∆FCB and ∆ACE,

∠FCB = ∠ACE

FC = AC (Sides of the square ACFG)

CB = CE (Sides of the square BCED)

∆FCB ≅ ∆ACE (SAS congruency rule)

(v) AX ⊥ DE and CE ⊥ DE (Adjacent sides of square BDEC) [given]

Hence,

CE || AX (Since two lines perpendicular to the same line are parallel to each other)

We consider BACE and parallelogram CYXE

BACE and the parallelogram CYXE are on the same base CE and in between the same parallels CE and AX.

∴area (∆YXE) = 2area (∆ACE) … (iv)

We had proved that

∴ ∆FCB ≅ ∆ACE

area (∆FCB) ≅ area (∆ACE) … (v)

From equations (iv) and (v), we get

area (CYXE) = 2 area (∆FCB) … (vi)

(vi) We consider BFCB and parallelogram ACFG

BFCB and parallelogram ACFG lie on the same base CF and in between the same parallels CF and BG.

∴area (ACFG) = 2 area (∆FCB)

∴area (ACFG) = area (CYXE) [From equation (vi)] … (vii)

(vii) From the figure, we can say that

area (BCED) = area (BYXD)+area (CYXE)

∴area (BCED) = area (ABMN)+area (ACFG) [From equations (iii) and (vii)].

**Question 33:** **In the figure given below, PSDA is a parallelogram. The points Q and R are taken on PS in such a way that PQ = QR = RS and PA || QB || RC. Prove that area (PQE) = area (CFD).**

**Solution 33:**

As per the question,

PA║QB║RC║SD & PQ=QR=RS

Following the equal intercept theorem,

We know if three or more parallel lines make equal intercepts on traversal, then they make equal intercepts on any other form of traversal.

Hence, we get,

PE=EF=FD & AB=BC=CD

From ΔPQE & ΔDCF,

We get,

∠PEQ=∠DFC

PE=DF

∠QPE=∠CDF

So, ΔPQE ≅ ΔDCF

Since congruent figures have equal areas,

We get,

area ΔPQE= area ΔDCF

Hence proved.

**Question 34:** **X and Y are the points on the side LN of a triangle LMN in such a way that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See the figure given below). Prove that area (LZY) = area (MZYX)**

**Solution 34:**

According to the question,

LX=XY=YN

XZ II LM

We have,

area(LZX)+(XZY)=area(LZY) — (1)

area(MXZ)+ar(XZY)=area(MZYX) — (2)

Both triangles, LZX and MXZ, are on the same base XZ and between the same parallels, LM and XZ.

area(LZX)=area(MXZ)

Adding equations (1) and (2),

We get,

area(LZY)=area(MZYX)

Hence proved.

**Question 35:** **The area of a parallelogram ABCD is 90 cm****2**** (see the figure given below). Find**

**(i) area (ABEF)**

**(ii) area (ABD)**

**(iii) area (BEF)**

**Solution 35:**

As per the question, area of the parallelogram, ABCD = 90 cm2

(i) We know, parallelograms on the same base and in between the same parallel are equal in area.

Here, the parallelograms ABCD and ABEF are on the same base AB and between the same parallels AB and CF.

Therefore, area (ABEF) = area (ABCD) = 90 cm2

(ii) We know,

If the triangle and the parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram.

Here,

ΔABD and the parallelogram ABCD are on the same base AB and in between the same parallels AB and CD.

Therefore, ar (ΔABD) = ½ ar (ABCD)

= ½ x 90 = 45 cm2

(iii) We know that,

If the triangle and the parallelogram are on the same base and between the same parallels, then the area of a triangle is equal to half the area of the parallelogram.

Here, ΔBEF and the parallelogram ABEF are on the same base EF and in between the same parallels AB and EF.

Therefore, area (ΔBEF) = ½ area (ABEF) = ½ x 90 = 45 cm2

**Question 36: In the △ ABC, D is the midpoint of AB and P is a point on BC. If CQ || PD meets AB in Q (see the figure given below), then prove that area (BPQ) = ½ area (ABC).**

**Solution 36:** As per the question,

In the figure,

In the △ ABC, D is the mid-point of AB.

P is a point on BC.

If CQ || PD meets AB at Q

To prove: ar (BPQ) = ½ area (ABC)

Construction: We join DC

Proof:- Since D is the midpoint of AB. So, in △ ABC, the CD is the median.

area(△ BCD) = ½ area (△ ABC) ….. equation(1)

We know,

△ PDQ and △ PDC are lying on the same base PD and in between the same parallel lines PD QC.

Therefore, area(△ PDQ) = area(△ PDC) ……………. (2)

From (1) and (2)

area(△ BCD) = ½ area(△ ABC)

area(△ BCD = area(△ BPD) + area(△ PDC) = ½ area(△ ABC)

Area of △ PDC = PDQ

So, area(△ BPQ) = area(△ BPD) + area(△ PDQ) = ½ area(△ ABC)

Therefore, area(△BPQ) = ½ area(△ ABC)

Hence proved.

**Question 37:** **A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced such that they meet at F. Prove that area (ADF) = area (ABFC).**

**Solution 37:** As per the question, ABCD is a parallelogram.

E is any point on BC.

AE and DC are produced so that they meet at F.

To prove: area (ΔADF) = area (ABFC).

Proof:

We know, triangles on the same base and in between the same parallels are equal in area.

Here,

We have ΔABC and ΔABF on the same base AB and between the same parallels, AB || CF.

Area (ΔABC) = area (ΔABF) … equation(1)

We also know that,

The diagonal of a parallelogram divides it into two triangles of equal area.

So, area (ΔABC) = area (ΔACD) … equation(2)

Now, area (ΔADF) = area (ΔACD) + area (ΔACF)

∴ Area (ΔADF) = area (ΔABC) + area (ΔACF) … (From equation (2))

⇒ Area (ΔADF) = area (ΔABF) + area (ΔACF) … (From equation (1))

⇒ Area (ΔADF) = area (ABFC)

**Question 38:** **The diagonals of the parallelogram ABCD intersect at point O. Through O, a line is drawn that intersects AD at P and BC at Q. Show that PQ divides the parallelogram into two equal parts of the same area.**

**Solution 38:** According to the question,

The diagonals of a parallelogram ABCD intersect at O.

Through point O, a line is drawn to intersect AD at P and BC at Q.

To prove: Area (parallelogram PDCQ) = area (parallelogram PQBA).

Proof: AC is the diagonal of parallelogram ABCD

∴ area(ΔABC) = area(ΔACD) = ½ area (||gm ABCD) …(1)

In ΔAOP and ΔCOQ,

AO = CO

Since the diagonals of the parallelogram bisect each other,

We have ∠AOP = ∠COQ

∠OAP = ∠OCQ (Vertically opposite angles)

∴ ΔAOP = ΔCOQ ( Alternate interior angles)

∴ area(ΔAOP) = area(ΔCOQ) (By ASA Congruence Rule condition)

We know, that congruent figures have equal areas.

Thus, area(ΔAOP) + area(parallelogram OPDC) = area(ΔCOQ) + area(parallelogram OPDC)

⇒ area(ΔACD) = area(parallelogram PDCQ)

⇒ ½ area(|| gm ABCD) = area(parallelogram PDCQ)

From equation (1),

We have, area(parallelogram PQBA) = area(parallelogram PDCQ)

⇒ area(parallelogram PDCQ) = area(parallelogram PQBA).

Hence proved.

**Question 39:** **The medians BE and CF of the triangle ABC intersect at G. Prove that area of △ GBC = area of quadrilateral AFGE.**

**Solution 39:**

As per the question,

We have BE & CF are medians.

E is the midpoint of AC.

F is the midpoint of AB.

∴ ΔBCE = ΔBEA … ( i )

ΔBCF = ΔCAF

Construct join EF,

By the midpoint theorem,

We get FE || BC

We know that Δ lying on the same base and in between the same parallels are equal in area.

∴ ΔFBC = ΔBCE

ΔFBC – ΔGBC = ΔBCE – ΔGBC

⇒ ΔFBG = ΔCGE (ΔGBC is common)

⇒ ΔCGE = ΔFBG …( ii )

Subtracting equation (ii) from (i)

We get, ΔBCE – ΔCGE = ΔBEA – ΔFBG

∴ ΔBGC =quadrilateral AFGE.

**Question 40:** **In the below figure, CD || AE and CY || BA. Prove that area (CBX) = area (AXY).**

**Solution 40:** According to the question,

From the figure,

We get CD||AE

and CY || BA

To prove: ar (ΔCBX) = ar (ΔAXY) .

Proof:

We know that,

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

Here,

ΔABY and ΔABC both lie on the same base AB and in between the same parallels CY and BA.

area (ΔABY) = area (ΔABC)

⇒ area (ABX) + area (AXY) = area (ABX) + area (CBX)

Solving and canceling area (ABX),

We get,

⇒ area (AXY) = area (CBX)

**Question 41:** **ABCD is the trapezium where AB || DC, DC = 30 cm and AB = 50 cm. If X and Y are the midpoints of AD and BC, respectively, prove that area (DCYX) = 7/9 area (XYBA).**

**Solution 41:**

As per the question,

We know,

ABCD is the trapezium with AB || DC

Construction: We join DY and produce it to meet AB produced at P.

In ∆BYP and ∆CYD

∠BYP = ∠CYD (vertically opposite angles)

Since alternate opposite angles of DC || AP and BC are the transversal

∠DCY = ∠PBY

Since Y is the midpoint of BC,

BY = CY

Thus ∆BYP ≅ ∆CYD (by ASA cogence criterion)

So, DY = YP and DC = BP

⇒ Y is the midpoint of AD

∴ XY || AP and XY = ½ AP (by midpoint theorem)

⇒ XY = ½ AP

= ½ (AB + BP)

= ½ (AB + DC)

= ½ (50 + 30)

= ½ × 80 cm

= 40 cm

Since X is the midpoint of AD,

And Y is the midpoint of BC.

Hence, trapezium DCYX and ABYX are of the same height, h, cm.

Now

⇒ 9 area(DCXY) = 7 area(XYBA)

⇒ area(DCXY) = 7/9 area(XYBA)

Hence proved.

**Question 42: If E, F, G and H are the mid-points of the sides of a parallelogram ABCD, respectively, show that area (EFGH) = 1/2 area(ABCD).**

**Solution 42:**

Given E, F, G and H are the mid-points of the sides of the parallelogram ABCD, respectively.

To prove: area (EFGH) = ½ area(ABCD)

Construction: H and F are joined.

Proof:

AD || BC and AD = BC (Since opposite sides of a ||gm)

⇒ ½ AD = ½ BC

And,

AH || BF and DH || CF

⇒ AH = BF and DH = CF (Since H and F are the midpoints)

Thus, ABFH and HFCD are parallelograms.

Now,

We know ΔEFH and ||gm ABFH both lie on the same FH, i.e., the common base and in between the same parallel lines AB and HF.

∴ area of EFH = ½ area of ABFH — equation (i)

And,

area of GHF = ½ area of HFCD —equation (ii)

Adding equations (i) and (ii),

area of ΔEFH + area of ΔGHF = ½ area of ABFH + ½ area of HFCD

⇒ area (EFGH) = ½ area(ABCD)

**Question 43: In a triangle ABC, E is the midpoint of the median AD. Show that area(BED) = 1/4 area(ABC).**

**Solution 43:**

area(BED) = ½ × BD × DE

E is the midpoint of AD,

AE = DE

As AD is the median on side BC of triangle ABC,

BD = DC

DE = ½ AD — (i)

BD = ½ BC — (ii)

From equations (i) and (ii), we get,

area(BED) = (1/2) × (½) BC × (1/2)AD

⇒ area(BED) = (½) × (½) area(ABC)

⇒area(BED) = ¼ area(ABC)

**Question 44: D, E and F are the mid-points of the sides BC, CA and AB, respectively, of a ΔABC.**

**Show that**

**(i) BDEF is a parallelogram.**

**(ii) area(DEF) = ¼ area(ABC)**

**(iii) area(BDEF) = ½ area(ABC)**

**Solution 44:**

(i) In ΔABC,

EF || BC and EF = ½ BC (by the midpoint theorem)

and,

BD = ½ BC (D is the midpoint)

Thus, BD = EF

also,

BF and DE are parallel and equal to each other.

Thus, the pair of opposite sides are equal in length and parallel to each other.

∴ BDEF is a parallelogram.

(ii) Proceeding from the result of (i),

BDEF, DCEF, and AFDE are parallelograms.

A diagonal of a parallelogram divides it into two triangles of equal area.

∴ area(ΔBFD) = area(ΔDEF) (For ||gm BDEF) — (i)

also,

area(ΔAFE) = area(ΔDEF) (For ||gm DCEF) — (ii)

area(ΔCDE) = area(ΔDEF) (For ||gm AFDE) — (iii)

From (i), (ii) and (iii)

area(ΔBFD) = area(ΔAFE) = area(ΔCDE) = area(ΔDEF)

⇒ area(ΔBFD) + area(ΔAFE) + area(ΔCDE) + area(ΔDEF) = area(ΔABC)

⇒ 4 area(ΔDEF) = area(ΔABC)

⇒area(DEF) = ¼ area(ABC)

(iii)Area (||gm BDEF) = area(ΔDEF) + area(ΔBDE)

⇒ area(BDEF) = area(ΔDEF) + area(ΔDEF)

⇒ area(BDEF) = 2× area(ΔDEF)

⇒ area(BDEF) = 2× ¼ area(ΔABC)

⇒ area(BDEF) = (1/2)area(ΔABC)

**Question 45:**

**XY is a line parallel to the side BC of the triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that**

**area(ΔABE) = area(ΔACF)**

**Solution 45:**

Given: XY || BC, BE || AC and CF || AB

To show: area(ΔABE) = area(ΔACF)

Proof:

BCYE is a parallelogram as ΔABE and parallelogram BCYE are on the same base BE and in between the same parallel lines BE and AC.

∴ area(ABE) = ½ area(BCYE) … (1)

Now,

CF || AB and XY || BC

⇒ CF || AB and XF || BC

⇒ BCFX is a parallelogram

As ΔACF and ||gm BCFX are on the same base CF and in between the same parallel AB and FC.

∴ area (ΔACF)= ½ area (BCFX) … (2)

But,

parallelogram BCFX and parallelogram BCYE are on the same base BC and in between the same parallels BC and EF.

∴ area (BCFX) = area(BCYE) … (3)

From (1), (2) and (3), we get

area (ΔABE) = area(ΔACF)

⇒ area(BEYC) = area(BXFC)

As the parallelograms are on the same base BC and in-between the same parallels EF and BC…..(4)

Also,

△AEB and ||gm BEYC are on the same base BE and in-between the same parallels BE and AC.

⇒ area(△AEB) = ½ area(BEYC) ……(5)

In a similar manner,

△ACF and parallelogram BXFC on the same base CF and between the same parallels CF and AB.

⇒ area(△ ACF) = ½ area(BXFC) ……..(6)

From (4), (5) and (6),

area(△AEB) = area(△ACF)

**Question 46:**

**The parallelogram ABCD and the rectangle ABEF are on the same base AB and have equal areas. Show that the perimeter of the parallelogram is greater than the perimeter of the rectangle.**

**Solution 46:**

Given,

Parallelogram ABCD and the rectangle ABEF have the same base AB and equal areas.

To prove,

The perimeter of parallelogram ABCD is greater than the perimeter of rectangle ABEF.

Proof,

We know the opposite sides of the ||gm and the rectangle are equal.

AB = DC [As ABCD is the ||gm]

and, AB = EF [As ABEF is the rectangle]

DC = EF … equation(i)

Adding AB on both sides, we get,

⇒ AB + DC = AB + EF … equation(ii)

As we know, the perpendicular segment is the shortest of all the segments that can be drawn to a given line from the point not lying on it.

BE < BC and AF < AD

⇒ BC > BE and AD > AF

⇒ BC + AD > BE + AF …equation (iii)

Adding equations (ii) and (iii), we get

AB + DC + BC + AD > AB + EF + BE + AF

⇒ AB + BC + CD + DA > AB + BE + EF + FA

⇒ perimeter of the ||gm ABCD > perimeter of the rectangle ABEF.

The perimeter of the given parallelogram is greater than that of the rectangle.

**Question 47:**

**The diagonals AC and BD of the quadrilateral ABCD intersect each other at E. Show that**

**area(ΔAED) × area(ΔBEC) = area (ΔABE) × area (ΔCDE).**

**[Hint: From A and C, draw perpendiculars to BD.]**

**Solution 47:**

Given: The diagonal AC and BD of a quadrilateral ABCD intersect each other at E.

Construction:

From A, we draw AM perpendicular to BD

From C, we draw CN perpendicular to BD

To prove: area(ΔAED) area(ΔBEC) = area (ΔABE) area (ΔCDE)

Proof:

area(ΔABE) = ½ ×BE×AM………….. (i)

area(ΔAED) = ½ ×DE×AM………….. (ii)

Dividing eq. (ii) by (i), we get,

area(ΔAED)/area(ΔABE) = [1/2×DE×AM]/ [1/2×BE×AM]

= DE/BE ………..(iii)

Similarly,

area(ΔCDE)/area(ΔBEC) = DE/BE …….(iv)

From eq. (iii) and (iv), we get;

area(ΔAED)/area(ΔABE) = area(ΔCDE)/area(ΔBEC)

Therefore, area(ΔAED) × area(ΔBEC) = area (ΔABE) × area (ΔCDE)

**Benefits of Solving Important Questions Class 9 Mathematics Chapter 9**

The important questions Class 9 Mathematics Chapter 9 address all the exercise questions from the NCERT textbook and were created by our subject experts for the benefit of the students. These Important Questions Class 9 Mathematics Chapter 9 provide questions that students should complete in order to better prepare for their exams. Students should also read other study materials, such as sample papers, previous year papers, important questions, etc.

Since there won’t be enough time to finish the entire book for the exam, they can utilise these materials as a revision tool for a quick study. Students may study well and take the tests without anxiety thanks to these Class 9 Mathematics Chapter 9 important questions.

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To help students become more adept at solving problems, knowledgeable tutors have clearly formulated the solutions. Students can consult the study materials offered by Extramarks to get a clearer understanding of the areas of parallelograms and triangles.

- CBSE revision notes
- CBSE syllabus
- CBSE sample papers
- CBSE past years’ question papers
- CBSE extra questions and solutions

**Q.1 ****Prove that parallelogram on the same base and between the same parallels are equal in area. **

**Marks:**4

**Ans**

__Given:__– Two parallelogram ABCD and EFCD, on the same base DC and between the same parallel line l and m.

__To prove:__– ar (ABCD) = ar (PQCD)

__Proof:-__ In ADE and BCF

DAE = DAE(Corresponding angles )

AED = BFC(Corresponding angles)

AD = BC (Opposite sides of the ||^{gm} ABCD)

So, AED â‰… BFC [By AAS]

Therefore ,

ar(ADE) = ar(BCF) (congruent figure have equal areas)

Now ar(ABCD) = ar(ADE)+ar(EDCB)

= ar(BCF)+ar(EDCB) [ ar(ADE) = ar(BCF)]

= ar(EFCD)

Hence proved.

**Q.2 ****Prove that the bisector of any two consecutive angles of parallelogram intersect at right angle.**

**Marks:**2

**Ans**

Given:- ABCD is a parallelogram. Bisectors of D and C intersect each other at P.

To Prove:- DPC = 90.

Proof:-Since PD bisects angle D and PC bisects angle A, therefore,

$\begin{array}{l}\mathrm{PDC}+\mathrm{PCD}=\frac{1}{2}\mathrm{D}+\frac{1}{2}\mathrm{C}\\ \text{}=\text{}\frac{1}{2}\left(\mathrm{D}+\mathrm{C}\right)\\ \text{}=\text{}\frac{1}{2}\left({180}^{\u02dc}\right)\text{}\left[\begin{array}{l}\mathrm{D}\text{and}\mathrm{C}\text{areinterioranglesonthe}\\ \text{samesideofthetransversal.}\end{array}\right]\\ \text{}=\text{}{90}^{\u02dc}\\ \text{Byusinganglesumpropertyoftriangle,InPDC,weget}{\text{P=90}}^{\text{o}}\text{.}\end{array}$

**Q.3 ****In the following figure, PQRS is a trapezium in which PQ || SR. Prove that ar(QOR) = ar(POS).**

**Marks:**2

**Ans**

Since PQR and PQS are on the same base PQ and between the same parallels PQ and SR.

´ ar(PQR) = ar(PQS)

ar(PQR) ar(POQ) = ar(PQS) ar(POQ)

ar(QOR) = ar(POS)

**Q.4 **

**In the figure given above, PQRS is a parallelogram, and PB is produced to a point A such that PS =AR. If PA intersects SR at B, show that: ar(SBA) = ar(PBR).**

**Marks:**1

**Ans**

Since PSAR

So, ar(PSA) = ar(PSR)(2)

[Triangles on the same base (PS) and between the same parallels PSAR]

ar(PSA) ar(PSB) = ar(PSR) ar(PSB)

ar(SBA) = ar(PBR)

**Q.5 ****In the given figure,**

P is the mid-point of side AB of quadrilateral ABCD such that ar(APC) = ar(BPD), and ar(ACB) = ar(DBC).

Prove that ABCD is a parallelogram.

**Marks:**1

**Ans**

Given that: ar(ACB) = ar(DBC)

And, ACB and DBC lie on the same base BC

ACB and DBC lie between the same parallels,

i.e., ADBC. (1)

Also, it is given that: AP = PB and ar(APC) = ar(BPD)

APC and BPD lie between same parallels,

i.e., ABDC. (2)

From (1) and (2),

Both the pairs of opposite sides of ABCD are parallel,

Hence, ABCD is a parallelogram.

## FAQs (Frequently Asked Questions)

### 1. Is it vital to learn all the questions provided in Important questions Class 9 Mathematics Chapter 9?

Yes, it is vital to attempt all of the questions in Important questions Class 9 Mathematics Chapter 9 in order to attempt all types of questions that may come in the exam. The solutions are offered in understandable language for students to assist them in solving any type of question. These answers were developed by Extramarks professionals with strong mathematical backgrounds.

### 2. What kind of questions can be expected in exams from NCERT Solutions for Class 9 Mathematics Chapter 9?

The chapter- Areas of Parallelogram and Triangles for Class 9 is pretty important from an exam perspective. There probably shall be five questions altogether from the unit: two multiple-choice questions, two short-type questions, and one long-type question.