Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 4 Set 4.3 Constructions of Triangles Solutions

Question 1. Construct ΔPQR, in which ∠Q = 70°, ∠R = 80° and PQ + QR + PR = 9.5 cm.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज PQR की निर्माण प्रक्रिया को दर्शाता है। इसमें एक रेखाखंड TS है जिसकी लंबाई 9.5 cm है। बिंदु T से 35° का कोण और बिंदु S से 40° का कोण बनाते हुए दो किरणें खींची गई हैं जो बिंदु P पर मिलती हैं। इसके बाद PT और PS के लंब समद्विभाजक खींचे गए हैं जो TS को क्रमशः Q और R पर काटते हैं, जिससे आवश्यक त्रिभुज PQR बनता है। i. As shown in the figure, take point T and S on line QR, such that QT = PQ and RS = PR ....(i) QT + QR + RS = TS [T-Q-R, Q-R-S]
\( \implies \) PQ + QR + PR = TS .....(ii) [From (i)] Also, PQ + QR + PR = 9.5 cm ....(iii) [Given]
\( \implies \) TS = 9.5 cm ii. In ΔPQT PQ = QT [From (i)]
\( \implies \) ∠QPT = ∠QTP = x° ....(iv) [Isosceles triangle theorem] In ΔPQT, ∠PQR is the exterior angle.
\( \implies \) ∠QPT + ∠QTP = ∠PQR [Remote interior angles theorem]
\( \implies \) x + x = 70° [From (iv)]
\( \implies \) 2x = 70°
\( \implies \) x = 35°
\( \implies \) ∠PTQ = 35°
\( \implies \) ∠T = 35° Similarly, ∠S = 40° iii. Now, in ΔPTS ∠T = 35°, ∠S = 40° and TS = 9.5 cm Hence, ΔPTS can be drawn. iv. Since, PQ = TQ,
\( \implies \) Point Q lies on perpendicular bisector of seg PT. Also, RP = RS
\( \implies \) Point R lies on perpendicular bisector of seg PS. Points Q and R can be located by drawing the perpendicular bisector of PT and PS respectively.
\( \implies \) ΔPQR can be drawn. Steps of construction: i. Draw seg TS of length 9.5 cm. ii. From point T draw ray making angle of 35°. iii. From point S draw ray making angle of 40°. iv. Name the point of intersection of two rays as P. v. Draw the perpendicular bisector of seg PT and seg PS intersecting seg TS in Q and R respectively. vi. Join PQ and PR. Hence, ΔPQR is the required triangle.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र त्रिभुज PQR के निर्माण के चरणों को दर्शाता है। इसमें एक रेखाखंड TS 9.5 cm का है, जिसके सिरों T और S पर क्रमशः 35° और 40° के कोण बनाते हुए किरणें खींची गई हैं जो बिंदु P पर प्रतिच्छेद करती हैं। PT और PS के लंब समद्विभाजक रेखाखंड TS को क्रमशः Q और R बिंदुओं पर काटते हैं, जिससे अंतिम त्रिभुज PQR प्राप्त होता है। In simple words: To construct this triangle, first draw a base line equal to the perimeter. Then, construct angles at the ends of this base, which are half of the given base angles of the required triangle. The intersection point forms the third vertex. Finally, find the other two vertices by drawing perpendicular bisectors from this intersection point to the base.

🎯 Exam Tip: Pay close attention to halving the base angles when constructing the initial larger triangle. Accuracy in drawing perpendicular bisectors is crucial for correctly locating the vertices of the final triangle.

Question 2. Construct ΔXYZ, in which ∠Y = 58°, ∠X = 46° and perimeter of triangle is 10.5 cm.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र त्रिभुज XYZ की रचना को दर्शाता है, जहाँ एक रेखाखंड WV की लंबाई 10.5 cm है। बिंदु W पर 29° का कोण और बिंदु V पर 23° का कोण बनाया गया है, जिससे किरणें Z पर प्रतिच्छेद करती हैं। WZ और VZ के लंब समद्विभाजक WV को क्रमशः Y और X पर काटते हैं, जिससे वांछित त्रिभुज XYZ बनता है। i. As shown in the figure, take point W and V on line YX, such that YW = ZY and XV = ZX ......(i) YW + YX + XV = WV [W-Y-X, Y-X-V]
\( \implies \) ZY + YX + ZX = WV ......(ii) [From (i)] Also, ZY + YX + ZX = 10.5 cm .....(iii) [Given]
\( \implies \) WV = 10.5 cm [From (ii) and (iii)] ii. In ΔZWY ZY = YW [From (i)]
\( \implies \) ∠YZW = ∠YWZ = x° .....(iv) [Isosceles triangle theorem] In ΔZYW, ∠ZYX is the exterior angle.
\( \implies \) ∠YZW + ∠YWZ = ∠ZYX [Remote interior angles theorem]
\( \implies \) x + x = 58° [From (iv)]
\( \implies \) 2x = 58°
\( \implies \) x = 29°
\( \implies \) ∠ZWY = 29°
\( \implies \) ∠W = 29°
\( \implies \) Similarly, ∠V = 23° iii. Now, in ΔZWV ∠W = 29°, ∠V = 23° and WV = 10.5 cm Hence, ΔZWV can be drawn. iv. Since, ZY = YW
\( \implies \) Point Y lies on perpendicular bisector of seg ZW. Also, ZX = XV
\( \implies \) Point X lies on perpendicular bisector of seg ZV.
\( \implies \) Points Y and X can be located by drawing the perpendicular bisector of ZW and ZV respectively.
\( \implies \) ΔXYZ can be drawn. Steps of construction: i. Draw seg WV of length 10.5 cm. ii. From point W draw ray making angle of 29°. iii. From point V draw ray making angle of 23°. iv. Name the point of intersection of two rays as Z. v. Draw the perpendicular bisector of seg WZ and seg VZ intersecting seg WV in Y and X respectively. vi. Join XY and XX. Hence, ΔXYX is the required triangle
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र त्रिभुज XYZ की अंतिम रचना को दर्शाता है। इसमें एक रेखाखंड WV 10.5 cm का है, जिसके ऊपर Y और X बिंदु स्थित हैं। Z बिंदु, W और V से खींची गई किरणों के प्रतिच्छेद से प्राप्त हुआ है। WZ और VZ के लंब समद्विभाजक क्रमशः Y और X बिंदुओं को निर्धारित करते हैं। अंत में, XY और XZ को जोड़कर आवश्यक त्रिभुज XYZ प्राप्त होता है। In simple words: This construction is similar to the first one. Begin by drawing a line segment equal to the triangle's perimeter. Then, at each end of this segment, construct angles that are half of the given angles of the final triangle. The intersection of these angle rays gives one vertex. The remaining two vertices are found by drawing perpendicular bisectors of the segments formed by the first vertex and the ends of the perimeter line.

🎯 Exam Tip: Remember to calculate half of the given angles (58°/2 = 29° and 46°/2 = 23°) correctly for the initial construction. Labeling the points accurately (W, V, Z, Y, X) helps maintain clarity throughout the construction process.

Question 3. Construct ΔLMN, in which ∠M = 60°, ∠N = 80° and LM + MN + NL = 11 cm.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र त्रिभुज LMN की रचना प्रक्रिया को दर्शाता है। इसमें एक रेखाखंड ST की लंबाई 11 cm है। बिंदु S से 30° का कोण और बिंदु T से 40° का कोण बनाते हुए दो किरणें खींची गई हैं जो बिंदु L पर प्रतिच्छेद करती हैं। इसके बाद LS और LT के लंब समद्विभाजक खींचे गए हैं जो ST को क्रमशः M और N पर काटते हैं, जिससे आवश्यक त्रिभुज LMN बनता है। i. As shown in the figure, take point S and T on line MN, such that MS = LM and NT = LN .....(i) MS + MN + NT = ST [S-M-N, M-N-T]
\( \implies \) LM + MN + LN = ST .....(ii) Also, LM + MN + LN = 11 cm ....(iii)
\( \implies \) ST = 11 cm [From (ii) and (iii)] ii. In ΔLSM LM = MS
\( \implies \) ∠MLS = ∠MSL = x° .....(iv) [isosceles triangle theorem] In ΔLMS, ∠LMN is the exterior angle.
\( \implies \) ∠MLS + ∠MSL = ∠LMN [Remote interior angles theorem]
\( \implies \) x + x = 60° [From (iv)]
\( \implies \) 2x = 60°
\( \implies \) x = 30°
\( \implies \) ∠LSM = 30°
\( \implies \) ∠S = 30° Similarly, ∠T = 40° iii. Now, in ΔLST ∠S = 30°, ∠T = 40° and ST = 11 cm Hence, ΔLST can be drawn. iv. Since, LM = MS
\( \implies \) Point M lies on perpendicular bisector of seg LS. Also LN = NT
\( \implies \) Point N lies on perpendicular bisector of seg LT.
\( \implies \) Points M and N can be located by drawing the perpendicular bisector of LS and LT respectively.
\( \implies \) ΔLMN can be drawn. Steps of construction: i. Draw seg ST of length 11 cm. ii. From point S draw ray making angle of 30°. iii. From point T draw ray making angle of 40°. iv. Name the point of intersection of two rays as L. v. Draw the perpendicular bisector of seg LS and seg LT intersecting seg ST in M and N respectively. vi. Join LM and LN. Hence, ΔLMN is the required triangle.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र त्रिभुज LMN की अंतिम रचना को दर्शाता है। इसमें एक रेखाखंड ST 11 cm का है, जिसके ऊपर M और N बिंदु स्थित हैं। L बिंदु, S और T से खींची गई किरणों के प्रतिच्छेद से प्राप्त हुआ है। LS और LT के लंब समद्विभाजक क्रमशः M और N बिंदुओं को निर्धारित करते हैं। अंत में, LM और LN को जोड़कर आवश्यक त्रिभुज LMN प्राप्त होता है। In simple words: To construct ΔLMN, start by drawing a segment equal to the perimeter (11 cm). At its endpoints, construct angles that are half of the given angles ∠M and ∠N (30° and 40° respectively). The intersection of these rays gives point L. Then, draw perpendicular bisectors of the segments connecting L to the endpoints of the perimeter line; these bisectors will intersect the perimeter line at M and N, completing ΔLMN.

🎯 Exam Tip: Always double-check the calculation of half angles (60°/2 = 30°, 80°/2 = 40°) before starting the construction. Accurate drawing of the initial angles and the subsequent perpendicular bisectors is key to obtaining the correct final triangle.