Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 5 Set 5.2 Quadrilaterals Solutions

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.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक चतुर्भुज ABCD को दर्शाता है जहाँ P भुजा AB का मध्यबिंदु है और Q भुजा DC का मध्यबिंदु है। चित्र में P और Q को जोड़ने वाली एक रेखा है, जो APCQ को एक नया चतुर्भुज बनाती है।
Answer: 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]
\( \therefore \) AB = DC [Opposite sides of a parallelogram]
\( \therefore \frac{1}{2} \) AB = \( \frac{1}{2} \) DC [Multiplying both sides by \( \frac{1}{2} \)]
\( \therefore \) 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),
\( \implies \) □APCQ is a parallelogram. [A quadrilateral is a parallelogram if its opposite sides is parallel and congruent] In simple words: We prove □APCQ is a parallelogram by showing one pair of opposite sides (AP and QC) are both equal in length and parallel to each other, using the properties of the given parallelogram ABCD and midpoints P and Q.

🎯 Exam Tip: Remember to clearly state the geometric reasons for each step (e.g., "midpoint," "opposite sides of a parallelogram") to earn full marks.

Question 2. Using opposite angles test for parallelogram, prove that every rectangle is a parallelogram.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक आयत ABCD को दर्शाता है जिसके सभी चार कोण 90 डिग्री के हैं। आयत के शीर्षों को A, B, C, D के रूप में चिह्नित किया गया है और भुजाएँ स्पष्ट रूप से दिखाई गई हैं।
Answer: Given: □ABCD is a rectangle. To prove: Rectangle ABCD is a parallelogram. Solution: Proof: □ABCD is a rectangle.
\( \therefore \angle \)A = \( \angle \)C = \( 90^\circ \) [Given]
\( \angle \)B = \( \angle \)D = \( 90^\circ \) [Angles of a rectangle]
\( \implies \) Rectangle ABCD is a parallelogram. [A quadrilateral is a parallelogram, if pairs of its opposite angles are congruent] In simple words: Since all angles in a rectangle are \( 90^\circ \), its opposite angles are equal. A quadrilateral with congruent opposite angles is defined as a parallelogram, so every rectangle is also a parallelogram.

🎯 Exam Tip: When proving a shape is another type of shape, always refer to the specific conditions required for the target shape (here, congruent opposite angles for a parallelogram).

Question 3. In the adjoining figure, G is the point of concurrence of medians of △DEF. Take point H on ray DG such that D-G-H and DG = GH, then prove that □GEHF is a parallelogram.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज DEF को दर्शाता है जिसमें G माध्यिकाओं का संगामी बिंदु (केंद्रक) है। बिंदु H किरण DG पर इस प्रकार स्थित है कि D, G, H संरेखीय हैं और DG = GH है। यह ज्यामितीय संरचना GEHF को एक चतुर्भुज के रूप में दिखाती है।
Answer: Given: Point G (centroid) is the point of concurrence of the medians of △DEF. DG = GH To prove: □GEHF is a parallelogram. Solution: Proof:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज DEF को दर्शाता है जिसमें माध्यिका DI है। बिंदु G केंद्रक है। किरण DH को आगे बढ़ाया गया है जहाँ H एक बिंदु है जैसे कि DG = GH, और I भुजा EF का मध्यबिंदु है। Let ray DH intersect seg EF at point I such that E-I-F.
\( \therefore \) seg DI is the median of △DEF.
\( \therefore \) EI = FI ......(i) Point G is the centroid of △DEF.
\( \implies \frac{DG}{GI} = \frac{2}{1} \) [Centroid divides each median in the ratio 2:1]
\( \therefore \) DG = 2(GI)
\( \therefore \) GH = 2(GI) [DG = GH]
\( \therefore \) GI + HI = 2(GI) [G-I-H]
\( \therefore \) HI = 2(GI) - GI
\( \therefore \) HI = GI ....(ii) From (i) and (ii),
\( \implies \) □GEHF is a parallelogram [A quadrilateral is a parallelogram, if its diagonals bisect each other] In simple words: By using the properties of a centroid (G) dividing the median DI in a 2:1 ratio, and the given condition DG = GH, we establish that the diagonals GF and EH bisect each other, thus proving GEHF is a parallelogram.

🎯 Exam Tip: Understanding the centroid's property of dividing medians in a 2:1 ratio is crucial here. Also, recall that if diagonals bisect each other, the quadrilateral is a parallelogram.

Question 4. Prove that quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समांतर चतुर्भुज ABCD को दर्शाता है। कोण A, B, C, D के कोण समद्विभाजक खींचे गए हैं, जो अंदर एक नया चतुर्भुज PQRS बनाते हैं। कोणों को \( x^\circ, y^\circ, u^\circ, v^\circ \) के रूप में दर्शाया गया है।
Answer: Given: □ABCD is a parallelogram. Rays AS, BQ, CQ and DS bisect \( \angle \)A, \( \angle \)B, \( \angle \)C and \( \angle \)D respectively. To prove: □PQRS is a rectangle. Solution: Proof: \( \angle \)BAS = \( \angle \)DAS = \( x^\circ \) ...(i) [ray AS bisects \( \angle \)A] \( \angle \)ABQ = \( \angle \)CBQ = \( y^\circ \) ....(ii) [ray BQ bisects \( \angle \)B] \( \angle \)BCQ = \( \angle \)DCQ = \( u^\circ \) .....(iii) [ray CQ bisects \( \angle \)C] \( \angle \)ADS = \( \angle \)CDS = \( v^\circ \) ....(iv) [ray DS bisects \( \angle \)D] □ABCD is a parallelogram. [Given]
\( \therefore \angle \)A + \( \angle \)B = \( 180^\circ \) [Adjacent angles of a parallelogram are supplementary]
\( \therefore \angle \)BAS + \( \angle \)DAS + \( \angle \)ABQ + \( \angle \)CBQ = \( 180^\circ \) [Angle addition property]
\( \therefore x^\circ+x^\circ+ y^\circ + y^\circ = 180^\circ \) [From (i) and (ii)]
\( \therefore 2x^\circ + 2y^\circ = 180^\circ \)
\( \therefore x^\circ + y^\circ = 90^\circ \) ......(v) [Dividing both sides by 2] Also, \( \angle \)A + \( \angle \)D= \( 180^\circ \) [Adjacent angles of a parallelogram are supplementary]
\( \therefore \angle \)BAS + \( \angle \)DAS + ADS + \( \angle \)CDS = \( 180^\circ \) [Angle addition property]
\( \therefore x^\circ + x^\circ + v^\circ + v^\circ = 180^\circ \)
\( \therefore 2x^\circ + 2v^\circ = 180^\circ \)
\( \therefore x^\circ + v^\circ = 90^\circ \) .....(vi) [Dividing both sides by 2] In △ARB, \( \angle \)RAB + \( \angle \)RBA + \( \angle \)ARB = \( 180^\circ \) [Sum of the measures of the angles of a triangle is \( 180^\circ \)]
\( \therefore x^\circ + y^\circ + \angle \)SRQ = \( 180^\circ \) [A - S - R, B - Q - R]
\( \therefore 90^\circ + \angle \)SRQ = \( 180^\circ \) [From (v)]
\( \therefore \angle \)SRQ = \( 180^\circ - 90^\circ = 90^\circ \) .....(vii) Similarly, we can prove \( \angle \)SPQ = \( 90^\circ \) ...(viii) In △ASD, \( \angle \)ASD + \( \angle \)SAD + \( \angle \)SDA = \( 180^\circ \) [Sum of the measures of angles a triangle is \( 180^\circ \)]
\( \therefore \angle \)ASD + \( x^\circ + v^\circ = 180^\circ \) [From (vi)]
\( \therefore \angle \)ASD + \( 90^\circ = 180^\circ \)
\( \therefore \angle \)ASD = \( 180^\circ - 90^\circ = 90^\circ \)
\( \therefore \angle \)PSR = \( \angle \)ASD [Vertically opposite angles]
\( \therefore \angle \)PSR = \( 90^\circ \) .....(ix) Similarly we can prove \( \angle \)PQR = \( 90^\circ \) ..(x)
\( \therefore \) In □PQRS,
\( \angle \)SRQ = \( \angle \)SPQ = \( \angle \)PSR = \( \angle \)PQR = \( 90^\circ \) [From (vii), (viii), (ix), (x)]
\( \implies \) □PQRS is a rectangle. [Each angle is of measure \( 90^\circ \)] In simple words: By proving that all interior angles of the quadrilateral PQRS, formed by the angle bisectors, are \( 90^\circ \) (right angles), we establish that PQRS is a rectangle. This relies on the property that adjacent angles of a parallelogram are supplementary and the sum of angles in a triangle is \( 180^\circ \).

🎯 Exam Tip: Focus on using the property of supplementary adjacent angles in a parallelogram and the angle sum property of a triangle to derive the \( 90^\circ \) angles within the inner quadrilateral. Clearly mark corresponding angle pairs.

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.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समांतर चतुर्भुज ABCD को दर्शाता है जिसमें भुजाओं पर चार बिंदु P, Q, R, S हैं। बिंदु P भुजा AB पर, Q भुजा BC पर, R भुजा CD पर, और S भुजा DA पर है। इन बिंदुओं को जोड़ने पर एक आंतरिक चतुर्भुज PQRS बनता है। यह भी दिया गया है कि AP = BQ = CR = DS।
Answer: Given: □ABCD is a parallelogram. AP = BQ = CR = DS To prove: □PQRS is a parallelogram. Solution: Proof: □ABCD is a parallelogram. [Given]
\( \therefore \angle \)B = \( \angle \)D ....(i) [Opposite angles of a parallelogram] Also, AB = CD [Opposite sides of a parallelogram]
\( \therefore \) AP + BP = DR + CR [A-P-B, D-R-C]
\( \therefore \) AP + BP = DR + AP [AP = CR]
\( \therefore \) BP = DR ....(ii) In △PBQ and △RDS, seg BP \( \cong \) seg DR [From (ii)] \( \angle \)PBQ = \( \angle \)RDS [From (i)] seg BQ \( \cong \) seg DS [Given]
\( \therefore \) △PBQ \( \cong \) △RDS [SAS test]
\( \therefore \) seg PQ \( \cong \) seg RS .....(iii) [c.s.c.t] Similarly, we can prove that △PAS \( \cong \) △RCQ
\( \therefore \) seg PS \( \cong \) seg RQ ....(iv) [c.s.c.t] From (iii) and (iv),
\( \implies \) □PQRS is a parallelogram. [A quadrilateral is a parallelogram, if pairs of its opposite angles are congruent] In simple words: By proving that the triangles formed at the corners of the parallelogram (like △PBQ and △RDS) are congruent using the SAS test, we establish that opposite sides of □PQRS are congruent. A quadrilateral with congruent opposite sides is a parallelogram.

🎯 Exam Tip: The key to this proof is using triangle congruence (SAS test here) to show that opposite sides of the inner quadrilateral are equal. Clearly identify the corresponding parts of congruent triangles (c.s.c.t.).

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)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC को दर्शाता है जिसमें D भुजा AB का मध्यबिंदु है और E भुजा AC का मध्यबिंदु है। एक किरण ED को F तक बढ़ाया गया है, जहाँ ED = DF है। यह संरचना AFBE को एक चतुर्भुज के रूप में दिखाती है जिसके विकर्ण AB और EF हैं।
Answer: 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 \( \cong \) seg DF [Given]
\( \therefore \) Diagonals of □AFBE bisect each other.
\( \implies \) □AFBE is a parallelogram. [ By test of parallelogram] In simple words: Since D is the midpoint of AB and it's given that ED = DF, we can conclude that the diagonals AB and EF of quadrilateral AFBE bisect each other. A quadrilateral whose diagonals bisect each other is a parallelogram.

🎯 Exam Tip: The midpoint theorem and the property that a quadrilateral is a parallelogram if its diagonals bisect each other are crucial for this proof. Identify the diagonals correctly.