Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 4 Set 4.1 Constructions of Triangles Solutions

Question 1. Construct APQR, in which QR = 4.2 cm, m∠Q = 40° and PQ + PR = 8.5 cm.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): एक त्रिभुज PQR की रचना है। आधार QR 4.2 सेमी है। कोण Q 40° है। एक किरण QT खींची गई है जिस पर बिंदु S 8.5 सेमी पर है, ताकि PQ + PR = QS। P, SR के लंब समद्विभाजक और QT के प्रतिच्छेदन पर स्थित है। As shown in the rough figure draw seg QR = 4.2 cm
Draw a ray QT making an angle of 40° with QR
Take a point S on ray QT, such that QS = 8.5 cm
Now, QP + PS = QS [Q-P-S]
.. QP + PS = 8.5 cm .......(i)
Also, PQ + PR = 8.5 cm ......(ii) [Given]
.. QP + PS = PQ + PR [From (i) and (ii)]
.. PS = PR
.. Point P is on the perpendicular bisector of seg SR
.. The point of intersection of ray QT and perpendicular bisector of seg SR is point P.
Steps of construction:
(i) Draw seg QR of length 4.2 cm.
(ii) Djraw ray QT, such that \( \angle RQT = 40^\circ \).
(iii) Mark point S on ray QT such that \( \text{I(QS)} = 8.5 \text{ cm} \).
(iv) Join points R and S.
(v) Draw perpendicular bisector of seg RS intersecting ray QT. Name the point as P.
(vi) Join the points P and R.
Hence, \( \triangle PQR \) is the required triangle.
In simple words: To construct triangle PQR, we first draw the base QR and angle Q. Then, we mark a point S on an extended ray such that QS equals the sum of the other two sides. The third vertex P is found by drawing the perpendicular bisector of RS, which is a key geometric technique.

🎯 Exam Tip: Accuracy in drawing the base, measuring the angle, and constructing the perpendicular bisector is crucial for a correct solution. Ensure all construction lines are visible.

Question 2. Construct \( \triangle XYZ \), in which YZ = 6 cm, XY + XZ = 9 cm, \( \angle XYZ = 50^\circ \).
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): एक त्रिभुज XYZ की रचना है। आधार YZ 6 सेमी है। कोण Y 50° है। एक किरण YT खींची गई है जिस पर बिंदु W 9 सेमी पर है, ताकि XY + XZ = YW। X, WZ के लंब समद्विभाजक और YT के प्रतिच्छेदन पर स्थित है। As shown in the rough figure draw seg YZ = 6 cm
Draw a ray YT making an angle of 50° with YZ
Take a point W on ray YT, such that YW = 9 cm
Now, YX + XW = YW [Y-X-W]
.:. YX + XW = 9 cm ....(i)
Also, XY + XZ = 9 cm ....(ii) [Given]
.:. YX + XW = XY + XZ [From (i) and (ii) ]
.:. XW = XZ
.. Point X is on the perpendicular bisector of seg WZ
.. The point of intersection of ray YT and perpendicular bisector of seg WZ is j point X.
Steps of construction:
(i) Draw seg YZ of length 6 cm.
(ii) Draw ray YT, such that \( \angle ZYT = 50^\circ \).
(iii) Mark point W on ray YT such that \( \text{I(YW)} = 9 \text{ cm} \).
(iv) Join points W and Z.
(v) Draw perpendicular bisector of seg WZ intersecting ray YT. Name the point as X.
(vi) Join the points X and Z.
Hence, \( \triangle XYZ \) is the required triangle.
In simple words: This construction follows a similar pattern to Question 1. We establish the base YZ and angle Y, then locate a point W on an extended ray equal to the sum of the other two sides. The final vertex X is determined by the intersection of the ray and the perpendicular bisector of WZ.

🎯 Exam Tip: Ensure precise angle measurement and segment marking. Errors in these initial steps will propagate through the entire construction. Practice using a protractor and ruler accurately.

Question 3. Construct \( \triangle ABC \), in which BC = 6.2 cm, \( \angle ACB = 50^\circ \), AB + AC = 9.8 cm.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): एक त्रिभुज ABC की रचना है। आधार BC 6.2 सेमी है। कोण C 50° है। एक किरण CT खींची गई है जिस पर बिंदु D 9.8 सेमी पर है, ताकि AB + AC = CD। A, DB के लंब समद्विभाजक और CT के प्रतिच्छेदन पर स्थित है। As shown in the rough figure draw seg CB = 6.2 cm
Draw a ray CT making an angle of 50° with CB
Take a point D on ray CT, such that
CD = 9.8 cm
Now, CA + AD = CD [C-A-D]
.. CA + AD = 9.8 cm .......(i)
Also, AB + AC = 9.8 cm ......(ii) [Given]
.. CA + AD = AB + AC [From (i) and (ii)]
.. AD = AB
.. Point A is on the perpendicular bisector of seg DB
.. The point of intersection of ray CT and perpendicular bisector of seg DB is point A.
Steps of construction:
(i) Draw seg BC of length 6.2 cm.
(ii) Draw ray CT, such that \( \angle BCT = 50^\circ \).
(iii) Mark point D on ray CT such that \( \text{I(CD)} = 9.8 \text{ cm} \).
(iv) Join points D and B.
(v) Draw perpendicular bisector of seg DB intersecting ray CT. Name the point as A.
(vi) Join the points A and B.
Hence, \( \triangle ABC \) is the required triangle.
In simple words: To construct triangle ABC, begin by drawing the base BC and angle C. Extend a ray from C, and mark a point D such that CD equals the given sum of AB and AC. The vertex A is then located at the intersection of this ray and the perpendicular bisector of DB.

🎯 Exam Tip: Clearly label all points and segments according to the construction steps. A neat and organized drawing aids in demonstrating your understanding and securing marks.

Question 4. Construct \( \triangle ABC \), in which BC = 3.2 cm, \( \angle ACB = 45^\circ \) Solution:and perimeter of AABC is 10 cm.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): एक त्रिभुज ABC की रचना है। आधार BC 3.2 सेमी है। कोण C 45° है। परिमाप 10 सेमी है। एक किरण CT खींची गई है जिस पर बिंदु D 6.8 सेमी पर है (जो AB + AC है)। A, DB के लंब समद्विभाजक और CT के प्रतिच्छेदन पर स्थित है। Perimeter of \( \triangle ABC \) = AB + BC + AC
.. 10 = AB + 3.2 + AC
.. AB + AC = 10 - 3.2
.. AB + AC = 6.8 cm
Now, In \( \triangle ABC \)
BC = 3.2 cm, \( \angle ACB = 45^\circ \) and AB + AC = 6.8 cm ....(i)
As shown in the rough figure draw j seg BC = 3.2 cm
Draw a ray CT making an angle of 45° with CB
Take a point D on ray CT, such that
CD = 6.8 cm
Now, CA + AD = CD [C-A-D]
.. CA + AD = 6.8 cm ...(ii)
Also, AB + AC = 6.8 cm ....(iii) [From (i)]
.. CA + AD = AB + AC [From (ii) and (iii)]
.. AD = AB
.. Point A is on the perpendicular bisector of seg DB
.. The point of intersection of ray CT and perpendicular bisector of seg DB is point A.
Steps of construction:
(i) Draw seg BC of length 3.2 cm.
(ii) Draw ray CT, such that \( \angle BCT = 45^\circ \).
(iii) Mark point D on ray CT such I(CD) = 6.8 cm. that
(iv) Join points D and B.
(V) Draw perpendicular bisector of seg DB intersecting ray CT. Name the point as A.
(vi) Join the points A and B.
Hence, \( \triangle ABC \) is the required triangle.
In simple words: This problem requires calculating the sum of two sides (AB + AC) from the given perimeter before starting the construction. Once this sum is known, the method of drawing the base, the angle, and then finding the third vertex using a perpendicular bisector remains the same as previous problems.

🎯 Exam Tip: The crucial first step is to correctly deduce the sum of the two unknown sides from the perimeter. Errors in this calculation will lead to an incorrect construction. Double-check your arithmetic.