## Maharashtra State Board Class 9 Maths Solutions Chapter 5 Quadrilaterals Practice Set 5.2

Question 1.

In the adjoining figure, □ABCD is a parallelogram, P and Q are midpoints of sides AB and DC respectively, then prove □APCQ is a parallelogram.

Given: □ABCD is a parallelogram. P and Q are the midpoints of sides AB and DC respectively.

To prove: □APCQ is a parallelogram.

Solution:

Proof:

AP = \(\frac { 1 }{ 2 }\) AB …..(i) [P is the midpoint of side AB]

QC = \(\frac { 1 }{ 2 }\) DC ….(ii) [Q is the midpoint of side CD]

□ABCD is a parallelogram. [Given]

∴ AB = DC [Opposite sides of a parallelogram]

∴ \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) DC [Multiplying both sides by \(\frac { 1 }{ 2 }\)]

∴ AP = QC ….(iii) [From (i) and (ii)]

Also, AB || DC [Opposite angles of a parallelogram]

i.e. AP || QC ….(iv) [A – P – B, D – Q – C]

From (iii) and (iv),

□APCQ is a parallelogram. [A quadrilateral is a parallelogram if its opposite sides is parallel and congruent]

Question 2.

Using opposite angles test for parallelogram, prove that every rectangle is a parallelogram.

Given:

□ABCD is a rectangle.

To prove: Rectangle ABCD is a parallelogram.

Solution:

Proof:

□ABCD is a rectangle.

∴ ∠A ≅ ∠C = 90° [Given]

∠B ≅ ∠D = 90° [Angles of a rectangle]

∴ Rectangle ABCD is a parallelogram. [A quadrilateral is a parallelogram, if pairs of its opposite angles are congruent]

Question 3.

In the adjoining figure, G is the point of concurrence of medians of ADEF. Take point H on ray DG such that D-G-H and DG = GH, then prove that □GEHF is a parallelogram.

Given: Point G (centroid) is the point of concurrence of the medians of ADEF.

DG = GH

To prove: □GEHF is a parallelogram.

Solution:

Proof:

Let ray DH intersect seg EF at point I such that E-I-F.

∴ seg DI is the median of ∆DEF.

∴ El = FI ……(i)

Point G is the centroid of ∆DEF.

∴ \(\frac { DG }{ GI }\) = \(\frac { 2 }{ 1 }\) [Centroid divides each median in the ratio 2:1]

∴ DG = 2(GI)

∴ GH = 2(GI) [DG = GH]

∴ GI + HI = 2(GI) [G-I-H]

∴ HI = 2(GI) – GI

∴ HI = GI ….(ii)

From (i) and (ii),

□GEHF is a parallelogram [A quadrilateral is a parallelogram, if its diagonals bisect each other]

Question 4.

Prove that quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle.

Given: □ABCD is a parallelogram.

Rays AS, BQ, CQ and DS bisect ∠A, ∠B, ∠C and ∠D respectively.

To prove: □PQRS is a rectangle.

Solution:

Proof:

∠BAS = ∠DAS = x° …(i) [ray AS bisects ∠A]

∠ABQ = ∠CBQ =y° ….(ii) [ray BQ bisects ∠B]

∠BCQ = ∠DCQ = u° …..(iii) [ray CQ bisects ∠C]

∠ADS = ∠CDS = v° ….(iv) [ray DS bisects ∠D]

□ABCD is a parallelogram. [Given]

∴ ∠A + ∠B = 180° [Adjacent angles of a parallelogram are supplementary]

∴ ∠BAS + ∠DAS + ∠ABQ + ∠CBQ = 180° [Angle addition property]

∴ x°+x°+ v° + v° = 180 [From (i) and (ii)]

∴ 2x° + 2v° =180

∴ x + y = 90° ……(v) [Dividing both sides by 2]

Also, ∠A + ∠D= 180° [Adjacent angles of a parallelogram are supplementary]

∴ ∠BAS + ∠DAS + ADS + ∠CDS = 180° [Angle addition property]

∴ x° + x° + v° + v° = 180°

∴ 2x° + 2v° = 180°

∴ x° + v° = 90° …..(vi) [Dividing both sides by 2]

In ∆ARB,

∠RAB + ∠RBA + ∠ARB = 180° [Sum of the measures of the angles of a triangle is 180°]

∴ x° + y° + ∠SRQ = 180° [A – S – R, B – Q – R]

∴ 90° + ∠SRQ = 180° [From (v)]

∴ ∠SRQ = 180°- 90° = 90° …..(vi)

Similarly, we can prove

∠SPQ = 90° …(viii)

In ∆ASD,

∠ASD + ∠SAD + ∠SDA = 180° [Sum of the measures of angles a triangle is 180°]

∴ ∠ASD + x° + v° = 180° [From (vi)]

∴ ∠ASD + 90° = 180°

∴∠ASD = 180°- 90° = 90°

∴ ∠PSR = ∠ASD [Vertically opposite angles]

∴ ∠PSR = 90° …..(ix)

Similarly we can prove

∠PQR = 90° ..(x)

∴ In □PQRS,

∠SRQ = ∠SPQ = ∠PSR = ∠PQR = 90° [From (vii), (viii), (ix), (x)]

∴ □PQRS is a rectangle. [Each angle is of measure 90°]

Question 5.

In the adjoining figure, if points P, Q, R, S are on the sides of parallelogram such that AP = BQ = CR = DS, then prove that □PQRS is a parallelogram.

Given: □ABCD is a parallelogram.

AP = BQ = CR = DS

To prove: □PQRS is a parallelogram.

Solution:

Proof:

□ABCD is a parallelogram. [Given]

∴ ∠B = ∠D ….(i) [Opposite angles of a parallelogram]

Also, AB = CD [Opposite sides of a parallelogram]

∴ AP + BP = DR + CR [A-P-B, D-R-C]

∴ AP + BP = DR + AP [AP = CR]

∴ BP = DR ….(ii)

In APBQ and ARDS,

seg BP ≅ seg DR [From (ii)]

∠PBQ ≅ ∠RDS [From (i)]

seg BQ ≅ seg DS [Given]

∴ ∆PBQ ≅ ∆RDS [SAS test]

∴ seg PQ ≅ seg RS …..(iii) [c.s.c.t]

Similarly, we can prove that

∆PAS ≅ ∆RCQ

∴ seg PS ≅ seg RQ ….(iv) [c.s.c.t]

From (iii) and (iv),

□PQRS is a parallelogram. [A quadrilateral is a parallelogram, if pairs of its opposite angles are congruent]

**Maharashtra Board Class 9 Maths Chapter 5 Quadrilaterals Practice Set 5.2 Intext Questions and Activities**

Question 1.

Points D and E are the midpoints of side AB and side AC of ∆ABC respectively. Point F is on ray ED such that ED = DF. Prove that □AFBE is a parallelogram. For this example write ‘given’ and ‘to prove’ and complete the proof. (Text book pg. no. 66)

Given: D and E are the midpoints of side AB and side AC respectively.

ED = DF

To prove: □AFBE is a parallelogram.

Solution:

Proof:

seg AB and seg EF are the diagonals of □AFBE.

seg AD ≅ seg DB [Given]

seg DE ≅ seg DF [Given]

∴ Diagonals of □AFBE bisect each other.

∴ □AFBE is a parallelogram. [ By test of parallelogram]