Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 5 Set 5.5 Quadrilaterals Solutions

Question 1. In the adjoining figure, points X, Y, Z are the midpoints of of ∆ABC respectively, cm. Find the lengths of side AB, side BC and side AC AB = 5 cm, AC = 9 cm and BC = 11c.m. Find the lengths of XY, YZ, XZ.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC है जिसके भुजाओं AB, BC और AC पर क्रमशः X, Y, Z बिंदु मध्यबिंदु के रूप में दर्शाए गए हैं। एक रेखाखंड XY, YZ और XZ त्रिभुज के अंदर बन रहे हैं, जो मध्यबिंदुओं को जोड़ते हैं।
Answer: Solution: i. AC = 9 cm [Given] Points X and Y are the midpoints of sides AB and BC respectively. [Given]
Therefore, XY = \( \frac{1}{2} \) AC [Midpoint theorem]
= \( \frac{1}{2} \) x 9 = 4.5 cm ii. AB = 5 cm [Given] Points Y and Z are the midpoints of sides BC and AC respectively. [Given]
Therefore, YZ = \( \frac{1}{2} \) AB [Midpoint theorem]
= \( \frac{1}{2} \) x 5 = 2.5 cm iii. BC = 11 cm [Given] Points X and Z are the midpoints of sides AB and AC respectively. [Given]
Therefore, XZ = \( \frac{1}{2} \) BC [Midpoint theorem]
= \( \frac{1}{2} \) x 11 = 5.5 cm l(XY) = 4.5 cm, l(YZ) = 2.5 cm, l(XZ) = 5.5 cm
In simple words: Using the midpoint theorem, each segment connecting the midpoints of two sides of a triangle is half the length of the third side. We applied this rule to find the lengths of XY, YZ, and XZ based on the given side lengths of triangle ABC.

🎯 Exam Tip: Remember the Midpoint Theorem states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length. This is a fundamental concept for solving such problems.

Question 2. In the adjoining figure, PQRS and MNRL are rectangles. If point M is the midpoint of side PR, then prove that,
i. SL = LR
ii. LN = \( \frac{1}{2} \) SQ.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो आयतों PQRS और MNRL को दर्शाता है। बिंदु M, विकर्ण PR का मध्यबिंदु है। रेखाखंड LN को SQ के साथ तुलना करने के लिए प्रदर्शित किया गया है।
Answer: Given: PQRS and MNRL are rectangles. M is the midpoint of side PR. Solution: Toprove:
i. SL = LR
ii. LN = \( \frac{1}{2} \) (SQ) Proof: i. PQRS and MNRL are rectangles. [Given]
Therefore, \( \angle \)S = \( \angle \)L = 90° [Angles of rectangles]
\( \angle \)S and \( \angle \)L form a pair of corresponding angles on sides SP and LM when SR is their transversal.
Therefore, seg ML || seg PS ...(i) [Corresponding angles test] In \( \triangle \)PRS, Point M is the midpoint of PR and seg ML || seg PS. [Given] [From (i)]
Therefore, Point L is the midpoint of seg SR. ......(ii) [Converse of midpoint theorem]
Therefore, SL = LR ii. Similarly for \( \triangle \)PRQ, we can prove that,
ℹ️ चित्र व्याख्या (Diagram Explanation): यह त्रिभुज PRQ को दर्शाता है जिसके अंदर एक रेखाखंड LN है। बिंदु N को QR के मध्यबिंदु के रूप में दिखाया गया है, जो एक समान संबंध स्थापित करने में मदद करेगा।
Point N is the midpoint of seg QR. ....(iii) In \( \triangle \)RSQ, Points L and N are the midpoints of seg SR and seg QR respectively. [From (ii) and (iii)]
Therefore, LN = \( \frac{1}{2} \) SQ [Midpoint theorem]
In simple words: We used the properties of rectangles (parallel sides, 90-degree angles) and the midpoint theorem along with its converse. First, we proved L is the midpoint of SR by showing ML is parallel to PS. Then, we extended this idea to find N as the midpoint of QR, finally applying the midpoint theorem in triangle RSQ to prove LN is half of SQ.

🎯 Exam Tip: For geometry proofs, always clearly state your 'Given', 'To Prove', and 'Proof' sections. Use standard theorems (like Midpoint Theorem, Converse of Midpoint Theorem, properties of rectangles) and provide justifications for each step to ensure full marks.

Question 3. In the adjoining figure, \( \triangle \)ABC is an equilateral triangle. Points F, D and E are midpoints of side AB, side BC, side AC respectively. Show that \( \triangle \)FED is an equilateral triangle.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समबाहु त्रिभुज ABC है। इसकी भुजाओं AB, BC, और AC के मध्यबिंदु क्रमशः F, D, और E हैं। इन मध्यबिंदुओं को जोड़कर एक नया त्रिभुज FED बनाया गया है, जिसे समबाहु सिद्ध करना है।
Answer: Given: \( \triangle \)ABC is an equilateral triangle. Points F, D and E are midpoints of side AB, side BC, side AC respectively. To prove: \( \triangle \)FED is an equilateral triangle. Solution: Proof: \( \triangle \)ABC is an equilateral triangle. [Given]
Therefore, AB = BC = AC ....(i) [Sides of an equilateral triangle] Points F, D and E are midpoints of side AB, side BC and AC respectively.
Therefore, FD = \( \frac{1}{2} \) AC....(ii) [Midpoint theorem] Points D and E are the midpoints of sides BC and AC respectively.
Therefore, DE = \( \frac{1}{2} \) AB.....(iii) [Midpoint theorem] Points F and E are the midpoints of sides AB and AC respectively.
Therefore, FE = \( \frac{1}{2} \) BC
Therefore, FD = DE = FE [From (i), (ii), (iii) and (iv) ]
Therefore, \( \triangle \)FED is an equilateral triangle.
In simple words: Since triangle ABC is equilateral, all its sides are equal. By applying the midpoint theorem, each side of the inner triangle FED is half the length of a corresponding side of triangle ABC. As all sides of triangle ABC are equal, it follows that all sides of triangle FED are also equal, making it an equilateral triangle.

🎯 Exam Tip: When proving properties of triangles, especially involving midpoints, the Midpoint Theorem is crucial. Clearly state the given conditions and how they lead to the application of the theorem for each side of the inner triangle.

Question 4. In the adjoining figure, seg PD is a median of \( \triangle \)PQR. Point T is the midpoint of seg PD. Produced QT intersects PR at M. Show that \( \frac{PM}{PR} \) = \( \frac{1}{3} \). [Hint: Draw DN || QM]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज PQR है जिसमें PD एक माध्यिका है, जिसका अर्थ है कि D, QR का मध्यबिंदु है। T, PD का मध्यबिंदु है। रेखा QT को बढ़ाया गया है जो PR को M पर काटती है। इसमें एक सहायक रेखा DN को QM के समानांतर खींचा गया है।
Answer: Given: seg PD is a median of \( \triangle \)PQR. Point T is the midpoint of seg PD. To Prove: \( \frac{PM}{PR} \) = \( \frac{1}{3} \) Construction: Draw seg DN ||seg QM such that P-M-N and M-N-R. Proof: In \( \triangle \)PDN, Point T is the midpoint of seg PD and seg TM || seg DN [Given]
Therefore, Point M is the midpoint of seg PN. [Construction and Q-T-M]
Therefore, PM = MN [Converse of midpoint theorem] In \( \triangle \)QMR, Point D is the midpoint of seg QR and seg DN || seg QM [Construction]
ℹ️ चित्र व्याख्या (Diagram Explanation): यह त्रिभुज PQR के अंदर के बिंदुओं Q, M, N, D, R को दर्शाता है। इसमें दिखाया गया है कि N, MR का मध्यबिंदु है, जो PM/PR अनुपात को सिद्ध करने के लिए आवश्यक है।
Therefore, Point N is the midpoint of seg MR. [Converse of midpoint theorem]
Therefore, RN = MN .....(ii)
Therefore, PM = MN = RN .....(iii) [From (i) and (ii)] Now, PR = PM + MN + RN [ P-M-R-Q-T-M]
Therefore, PR = PM + PM + PM [From (iii) ]
Therefore, PR = 3PM
Therefore, \( \frac{PM}{PR} \) = \( \frac{1}{3} \)
In simple words: We used the midpoint theorem and its converse by drawing an auxiliary line DN parallel to QM. This allowed us to establish that M is the midpoint of PN and N is the midpoint of MR, leading to PM = MN = RN. By summing these segments, we found that PR is three times PM, thus proving the desired ratio.

🎯 Exam Tip: Auxiliary constructions are often key in complex geometry proofs. When a ratio involving medians or segments is required, consider drawing parallel lines to apply the midpoint theorem or basic proportionality theorem effectively. Clearly label points and segments from construction.