Practice Set 5.2 Geometry 9th Std Maths Part 2 Answers Chapter 5 Quadrilaterals
Question 1
Maharashtra Board Solution
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 & Step-by-Step Answer:
Proof: AP = AB …..(i) [P is the midpoint of side AB] QC = DC ….(ii) [Q is the midpoint of side CD] □ABCD is a parallelogram. [Given] ∴ AB = DC [Opposite sides of a parallelogram] ∴ AB = DC [Multiplying both sides by ] ∴ 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
Maharashtra Board Solution
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 & Step-by-Step Answer:
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
Maharashtra Board Solution
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 & Step-by-Step Answer:
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. ∴ = [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
Maharashtra Board Solution
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 & Step-by-Step Answer:
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
Maharashtra Board Solution
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 & Step-by-Step Answer:
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
Maharashtra Board Solution
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 & Step-by-Step Answer:
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]