Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 3 Set 3.4 Triangles Solutions

Question 1. In the adjoining figure, point A is on the bisector of \( \angle XYZ \). If AX = 2 cm, then find AZ.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज XYZ को दर्शाता है, जहाँ YA कोण XYZ का समद्विभाजक है। बिंदु A समद्विभाजक पर स्थित है, और AX भुजा YX पर तथा AZ भुजा YZ पर लंब हैं।
Answer: AX = 2 cm [Given] Point A lies on the bisector of \( \angle XYZ \). [Given] Point A is equidistant from the sides of \( \angle XYZ \). [Every point on the bisector of an angle is equidistant from the sides of the angle] \( \implies \) AZ = AX \( \implies \) AZ = 2 cm In simple words: If a point lies on the bisector of an angle, it is equidistant from the two arms of the angle. Since AX is the distance to one arm and AZ to the other, they must be equal.

🎯 Exam Tip: Remember the angle bisector theorem: any point on the angle bisector is equidistant from the sides of the angle. This is a fundamental concept for proofs.

Question 2. In the adjoining figure, \( \angle RST \) = 56°, seg PT \( \perp \) ray ST, seg PR \( \perp \) ray SR and seg PR \( \cong \) seg PT. Find the measure of \( \angle RSP \). State the reason for your answer.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक त्रिभुज TSR है जिसमें बिंदु P आंतरिक रूप से स्थित है। PR भुजा SR पर लंब है और PT भुजा ST पर लंब है। दिखाया गया है कि भुजाएँ PR और PT की लंबाई समान है।
Answer: seg PT \( \perp \) ray ST, seg PR \( \perp \) ray SR [Given] seg PR = seg PT \( \implies \) Point P lies on the bisector of \( \angle TSR \) [Any point equidistant from the sides of an angle is on the bisector of the angle] \( \implies \) Ray SP is the bisector of \( \angle RST \). \( \angle RST \) = 56° [Given] \( \implies \angle RSP \) = \( \frac{1}{2} \angle RST \) \( \implies \angle RSP \) = \( \frac{1}{2} \) x 56° \( \implies \angle RSP \) = 28° In simple words: Since point P is equidistant from the sides ST and SR, it must lie on the angle bisector of \( \angle RST \). Therefore, ray SP divides \( \angle RST \) into two equal halves.

🎯 Exam Tip: The converse of the angle bisector theorem is crucial here: if a point is equidistant from the sides of an angle, it lies on the angle bisector. Ensure you correctly apply the definition of an angle bisector to find the half-angle.

Question 3. In \( \triangle PQR \), PQ = 10 cm, QR = 12 cm, PR triangle. 8 cm. Find out the greatest and the smallest angle of the triangle.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज PQR को दर्शाता है, जिसकी भुजाओं की लंबाई दी गई है: PQ 10 सेमी है, QR 12 सेमी है, और PR 8 सेमी है।
Answer: In \( \triangle PQR \), PQ = 10 cm, QR = 12 cm, PR = 8 cm [Given] Since, 12 > 10 > 8 \( \implies \) QR > PQ > PR \( \implies \angle QPR > \angle PRQ > \angle PQR \) [Angle opposite to greater side is greater] \( \implies \) In \( \triangle PQR \), \( \angle QPR \) is the greatest angle and \( \angle PQR \) is the smallest angle. In simple words: In any triangle, the angle opposite the longest side is the greatest angle, and the angle opposite the shortest side is the smallest angle. We arrange the side lengths from longest to shortest to find the corresponding angles.

🎯 Exam Tip: This question tests the fundamental property of triangles relating side lengths to opposite angles. Clearly state the side lengths and their order, then deduce the order of the opposite angles.

Question 4. In \( \triangle FAN \), \( \angle F \) = 80°, \( \angle A \) = 40°. Find out the greatest and the smallest side of the triangle. State the reason.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सामान्य त्रिभुज FAN को दर्शाता है, जिसमें कोण और भुजाओं के संबंध का अध्ययन किया गया है।
Answer: In \( \triangle FAN \), \( \angle F + \angle A + \angle N \) = 180° [Sum of the measures of the angles of a triangle is 180°] \( \implies \) 80° + 40° + \( \angle N \) = 180° \( \implies \angle N \) = 180° - 80° - 40° \( \implies \angle N \) = 60° Since, 80° > 60° > 40° \( \implies \angle F > \angle N > \angle A \) \( \implies \) AN > FA > FN [Side opposite to greater angle is greater] \( \implies \) In \( \triangle FAN \), AN is the greatest side and FN is the smallest side. In simple words: First, find the third angle using the angle sum property of a triangle. Then, compare the angles to identify the greatest and smallest. The side opposite the greatest angle is the longest, and the side opposite the smallest angle is the shortest.

🎯 Exam Tip: Always calculate all three angles of the triangle first if not given. The side opposite the largest angle is the longest, and the side opposite the smallest angle is the shortest. This is the converse of the theorem used in Question 3.

Question 5. Prove that an equilateral triangle is equiangular.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक समबाहु त्रिभुज ABC को दर्शाता है, जिसमें सभी भुजाएँ और कोण समान होते हैं।
Answer: Given: \( \triangle ABC \) is an equilateral triangle. To prove: \( \triangle ABC \) is equiangular i.e. \( \angle A = \angle B = \angle C \) Proof: In \( \triangle ABC \), seg AB = seg BC [Sides of an equilateral triangle] \( \implies \angle C = \angle A \) (ii) [Isosceles triangle theorem] In \( \triangle ABC \), seg BC = seg AC [From (i)] \( \implies \angle A = \angle B \) (iii) [Isosceles triangle theorem] From (ii) and (iii), we get: \( \angle A = \angle B = \angle C \) \( \implies \triangle ABC \) is equiangular. In simple words: An equilateral triangle has all three sides equal. By applying the isosceles triangle theorem (angles opposite equal sides are equal) to pairs of sides, we can show that all three angles are also equal, thus it is equiangular.

🎯 Exam Tip: This is a standard proof. Clearly state the given and what needs to be proven. Use the property that an equilateral triangle has equal sides and apply the isosceles triangle theorem twice to establish the equality of all angles.

Question 6. Prove that, if the bisector of \( \angle BAC \) of \( \triangle ABC \) is perpendicular to side BC, then AABC is an isosceles triangle.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक त्रिभुज ABC है जहाँ AD कोण BAC का समद्विभाजक है। साथ ही, AD भुजा BC पर लंबवत है, जिससे यह एक समद्विबाहु त्रिभुज की विशेषता को दर्शाता है।
Answer: Given: Seg AD is the bisector of \( \angle BAC \). seg AD \( \perp \) seg BC To prove: \( \triangle ABC \) is an isosceles triangle. Proof. In \( \triangle ABD \) and \( \triangle ACD \), \( \angle BAD = \angle CAD \) [seg AD is the bisector of \( \angle BAC \)] seg AD = seg AD [Common side] \( \angle ADB = \angle ADC \) [Each angle is of measure 90°] \( \implies \triangle ABD \cong \triangle ACD \) [ASA test] \( \implies \) seg AB = seg AC [c. s. c. t.] \( \implies \triangle ABC \) is an isosceles triangle. In simple words: By using the given information that AD is both an angle bisector and an altitude, we can prove that \( \triangle ABD \) and \( \triangle ACD \) are congruent using the ASA test. This congruence implies that AB = AC, making \( \triangle ABC \) an isosceles triangle.

🎯 Exam Tip: This proof relies on identifying congruent triangles. Look for common sides, given equal angles (angle bisector), and right angles (perpendicular). The ASA congruence criterion is key to showing that the sides AB and AC are equal.

Question 7. In the adjoining figure, if seg PR \( \cong \) seg PQ, show that seg PS > seg PQ.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज PQR को दर्शाता है, जहाँ भुजा QR को S तक बढ़ाया गया है। इसमें भुजा PR और PQ की लंबाई समान दिखाई गई है।
Answer: Proof. In \( \triangle PQR \), seg PR = seg PQ [Given] \( \implies \angle PQR = \angle PRQ \) ....(i) [Isosceles triangle theorem] \( \angle PRQ \) is the exterior angle of \( \triangle PRS \). \( \implies \angle PRQ > \angle PSR \) ....(ii) [Property of exterior angle] From (i) and (ii), \( \implies \angle PQR > \angle PSR \) i.e. \( \angle Q > \angle S \) ....(iii) In \( \triangle PQS \), \( \angle Q > \angle S \) [From (iii)] \( \implies \) PS > PQ [Side opposite to greater angle is greater] \( \implies \) seg PS > seg PQ In simple words: First, use the isosceles triangle property to show that \( \angle PQR = \angle PRQ \). Then, use the exterior angle property to show \( \angle PRQ > \angle S \). Combining these, we find \( \angle Q > \angle S \). In \( \triangle PQS \), the side opposite the larger angle \( \angle Q \) (which is PS) must be greater than the side opposite the smaller angle \( \angle S \) (which is PQ).

🎯 Exam Tip: This problem combines the isosceles triangle theorem with the exterior angle theorem. Carefully track the angles and sides in both triangles involved (\( \triangle PQR \) and \( \triangle PQS \)) to build the logical argument for the inequality.

Question 8. In the adjoining figure, in \( \triangle ABC \), seg AD and seg BE are altitudes and AE = BD. Prove that seg AD = seg BE.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक त्रिभुज ABC है जिसमें AD और BE ऊँचाईयाँ हैं। AD भुजा BC पर है और BE भुजा AC पर है।
Answer: Proof: In \( \triangle ADB \) and \( \triangle BEA \), seg BD = seg AE [Given] \( \angle ADB = \angle BEA \) = 90° [Given, as AD and BE are altitudes] seg AB = seg BA [Common side] \( \implies \triangle ADB \cong \triangle BEA \) [Hypotenuse-side test] \( \implies \) seg AD = seg BE [c. s. c. t.] In simple words: We can prove that \( \triangle ADB \) and \( \triangle BEA \) are congruent using the Hypotenuse-Side (HS) congruence test, as they are right-angled triangles with a common hypotenuse (AB) and one pair of equal corresponding sides (BD = AE). Once congruent, their corresponding parts, AD and BE, must also be equal.

🎯 Exam Tip: When dealing with altitudes in right-angled triangles, consider using the Hypotenuse-Side (RHS) congruence criterion. Clearly identify the hypotenuse, the right angle, and the corresponding equal sides to prove congruence.

Maharashtra Board Class 9 Maths Chapter 3 Triangles Practice Set 3.4 Intext Questions and Activities

Question 1. As shown in the given figure, draw \( \triangle XYZ \) such that side XZ > side XY. Find which of \( \angle Z \) and \( \angle Y \) is greater. (Textbook pg. no. 41)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज XYZ को दर्शाता है, जिसमें भुजा XZ की लंबाई भुजा XY से अधिक है।
Answer: From the given figure, \( \angle Z \) = 25° and \( \angle Y \) = 51° \( \implies \angle Y \) is greater. In simple words: By observing the diagram and the given angles, we directly compare their values. 51° is greater than 25°, so \( \angle Y \) is the greater angle.

🎯 Exam Tip: For observation-based questions with given values, simply compare the numbers. Ensure you are comparing the correct angles as per the question.