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

Question 1. Choose the correct alternative answer for the following questions.

(i) If two sides of a triangle are 5 cm and 1.5 cm, the length of its third side cannot be _____.
(A) 3.7 cm
(B) 4.1 cm
(C) 3.8 cm
(D) 3.4 cm
Answer:
Sum of the lengths of two sides of a triangle \( > \) length of the third side
Here, 1.5 cm + 3.4 cm = 4.9 cm \( < \) 5 cm
\( \therefore \) Third side \( \neq \) 3.4 cm
Answer: (D) 3.4 cm
In simple words: The sum of any two sides of a triangle must be greater than the length of the third side. If the third side were 3.4 cm, the sum of 1.5 cm and 3.4 cm (4.9 cm) would be less than 5 cm, which violates the triangle inequality theorem.

🎯 Exam Tip: Remember the triangle inequality theorem: the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

(ii) In \( \triangle PQR \), if \( \angle R > \angle Q \), then _____.
(A) QR \( > \) PR
(B) PQ \( > \) PR
(C) 3.8 cm
(D) 3.4 cm
Answer:
Answer: (B) PQ > PR
In simple words: In any triangle, the side opposite to the greater angle is longer. Since \( \angle R \) is greater than \( \angle Q \), the side opposite \( \angle R \) (PQ) must be greater than the side opposite \( \angle Q \) (PR).

🎯 Exam Tip: The relationship between angles and their opposite sides in a triangle is crucial for inequality problems. A larger angle implies a longer opposite side.

(iii) In \( \triangle TPQ \), if \( \angle T = 65^\circ \), \( \angle P = 95^\circ \), Which of the following is a true statement?
(A) PQ \( < \) TP
(B) PQ \( < \) TQ
(C) TQ \( < \) TP \( < \) PQ
(D) PQ \( < \) TP \( < \) TQ
Answer:
\( \angle Q = 180^\circ - (95^\circ + 65^\circ) = 20^\circ \)
\( \therefore \angle Q < \angle T < \angle P \)
\( \therefore \) PT \( < \) PQ \( < \) TQ
Answer: (D) PQ < TP < TQ
In simple words: First, calculate the third angle \( \angle Q \). Then, arrange the angles in ascending order to find the corresponding ascending order of the lengths of the sides opposite those angles.

🎯 Exam Tip: Always find all three angles of a triangle first if two are given. Then, apply the property that the side opposite the largest angle is the longest, and the side opposite the smallest angle is the shortest.

Question 2. \( \triangle ABC \) is isosceles in which AB = AC. Seg BD and seg CE are medians. Show that BD = CE.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC को दर्शाता है जिसमें शीर्ष A पर, भुजाओं AB और AC को दर्शाया गया है। बिंदु E भुजा AB पर है और बिंदु D भुजा AC पर है, जिससे BD और CE त्रिभुज के माध्यिकाएँ हैं। त्रिभुज ABC में आधार भुजा BC है।
Given: In isosceles \( \triangle ABC \), AB = AC. seg BD and seg CE are the medians of \( \triangle ABC \).
To prove: BD = CE
Proof: AE = \( \frac{1}{2} \) AB....(i) [E is the midpoint of side AB]
AD = \( \frac{1}{2} \) AC ....(ii) [D is the midpoint of side AC]
Also, AB = AC ......(iii) [Given]
\( \therefore \) AE = AD ....(iv) [From (i), (ii) and (iii)]
In \( \triangle ADB \) and \( \triangle AEC \),
ℹ️ चित्र व्याख्या (Diagram Explanation): यह दो त्रिभुज ADB और AEC को दर्शाता है, जो एक बड़े त्रिभुज ABC के भीतर बने हैं। इसमें AB और AC भुजाएँ तथा माध्यिकाएँ BD और CE हैं। D, AC का मध्यबिंदु है और E, AB का मध्यबिंदु है।
seg AB = seg AC \( \angle BAD = \angle CAE \)
seg AD = seg AE
\( \therefore \triangle ADB \cong \triangle AEC \)
\( \therefore \) seg BD \( \cong \) seg CE
\( \therefore \) BD = CE
In simple words: Since AB = AC and BD, CE are medians, E and D are midpoints. This makes AE = AD. By SAS congruence, \( \triangle ADB \) and \( \triangle AEC \) are congruent, which means their corresponding parts, the medians BD and CE, are equal.

🎯 Exam Tip: When proving congruence of medians in an isosceles triangle, look for SAS (Side-Angle-Side) congruence criteria using the given equal sides and the common angle.

Question 3. In \( \triangle PQR \), if PQ > PR and bisectors of \( \angle Q \) and \( \angle R \) intersect at S. Show that SQ > SR.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज PQR को दर्शाता है। शीर्ष P ऊपर है, और आधार भुजा QR है। \( \angle Q \) और \( \angle R \) के कोण समद्विभाजक बिंदु S पर प्रतिच्छेद करते हैं, जो त्रिभुज के अंदर है।
Given: In \( \triangle PQR \), PQ > PR and bisectors of \( \angle Q \) and \( \angle R \) intersect at S.
To prove: SQ > SR
Solution:
Proof:
\( \angle SQR = \frac{1}{2} \angle PQR \) ....(i) [Ray QS bisects \( \angle PQR \)]
\( \angle SRQ = \frac{1}{2} \angle PRQ \) ....(ii) [Ray RS bisects \( \angle PRQ \)]
In \( \triangle PQR \),
PQ > PR [Given]
\( \therefore \angle R > \angle Q \) [Angle opposite to greater side is greater.]
\( \therefore \frac{1}{2}(\angle R) > \frac{1}{2}(\angle Q) \) [Multiplying both sides by \( \frac{1}{2} \)]
\( \therefore \angle SRQ > \angle SQR \) ....(iii) [From (i) and (ii)]
In \( \triangle SQR \),
\( \angle SRQ > \angle SQR \) [From (iii)]
\( \therefore \) SQ > SR [Side opposite to greater angle is greater]
In simple words: Since PQ > PR, the angle opposite PQ (\( \angle R \)) is greater than the angle opposite PR (\( \angle Q \)). As SQ and SR are bisectors, half of \( \angle R \) is greater than half of \( \angle Q \), meaning \( \angle SRQ > \angle SQR \). In \( \triangle SQR \), the side opposite the larger angle \( \angle SRQ \) (which is SQ) will be greater than the side opposite the smaller angle \( \angle SQR \) (which is SR).

🎯 Exam Tip: This proof combines the properties of angles opposite to sides and angle bisectors. Clearly state which triangle you are considering at each step (e.g., \( \triangle PQR \) vs. \( \triangle SQR \)).

Question 4. In the adjoining figure, points D and E are on side BC of \( \triangle ABC \), such that BD = CE and AD \( \cong \) AE. Show that \( \triangle ABD \cong \triangle ACE \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC को दर्शाता है जिसमें शीर्ष A ऊपर है। भुजा BC पर दो बिंदु D और E इस प्रकार स्थित हैं कि B, D, E, C एक रेखा में हैं। रेखाखंड AD और AE खींचे गए हैं, और वे लंबाई में बराबर दिखाए गए हैं।
Given: Points D and E are on side BC of \( \triangle ABC \), such that BD = CE and AD = AE.
To prove: \( \triangle ABD \cong \triangle ACE \)
Proof:
In \( \triangle ADE \),
seg AD = seg AE [Given]
\( \therefore \angle AED = \angle ADE \) ...(i) [Isosceles triangle theorem]
Now, \( \angle ADE + \angle ADB = 180^\circ \) ...(ii) [Angles in a linear pair]
\( \therefore \angle AED + \angle AEC = 180^\circ \) ....(iii) [Angles in a linear pair]
\( \therefore \angle ADE + \angle ADB = \angle AED + \angle AEC \) [From (ii) and (iii)]
\( \therefore \angle ADE + \angle ADB = \angle ADE + \angle AEC \) [From (i)]
\( \therefore \angle ADB = \angle AEC \) ....(iv) [Eliminating \( \angle ADE \) from both sides]
In \( \triangle ABD \) and \( \triangle ACE \),
seg BD = seg CE [Given]
\( \angle ADB = \angle AEC \) [From (iv)]
seg AD = seg AE [Given]
\( \therefore \triangle ABD \cong \triangle ACE \) [SAS test]
In simple words: First, establish that \( \angle ADE = \angle AED \) due to \( \triangle ADE \) being isosceles. Then, use the linear pair property for angles on the line BC to show that \( \angle ADB = \angle AEC \). Finally, use the given equal segments (BD = CE, AD = AE) and the proved equal angles (\( \angle ADB = \angle AEC \)) to prove the congruence of \( \triangle ABD \) and \( \triangle ACE \) by the SAS criterion.

🎯 Exam Tip: This problem requires careful step-by-step reasoning, particularly in using linear pairs and the isosceles triangle theorem to deduce equal angles for congruence proof.

Question 5. In the adjoining figure, point S is any point on side QR of \( \triangle PQR \). Prove that: PQ + QR + RP > 2PS
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज PQR को दर्शाता है, जिसमें शीर्ष P ऊपर है और आधार भुजा QR है। बिंदु S भुजा QR पर स्थित है। रेखाखंड PS खींचा गया है, जो शीर्ष P को आधार भुजा QR पर स्थित बिंदु S से जोड़ता है।
Proof:
In \( \triangle PQS \),
PQ + QS \( > \) PS .....(i) [Sum of any two sides of a triangle is greater than the third side]
Similarly, in \( \triangle PSR \),
PR + SR \( > \) PS ...(ii) [Sum of any two sides of a triangle is greater than the third side]
\( \therefore \) PQ + QS + PR + SR \( > \) PS + PS
\( \therefore \) PQ + QS + SR + PR \( > \) 2PS
\( \therefore \) PQ + QR + PR \( > \) 2PS [Q-S-R]
In simple words: Apply the triangle inequality theorem to \( \triangle PQS \) and \( \triangle PSR \) separately. Sum the resulting inequalities and replace (QS + SR) with QR, as S lies on QR, to prove the final inequality.

🎯 Exam Tip: This problem is a classic application of the triangle inequality. Break down the larger triangle into smaller triangles formed by the segment from the vertex to the opposite side.

Question 6. In the adjoining figure, bisector of \( \angle BAC \) intersects side BC at point D. Prove that AB > BD.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC को दर्शाता है जिसमें शीर्ष A ऊपर है। \( \angle BAC \) का समद्विभाजक AD भुजा BC को बिंदु D पर प्रतिच्छेद करता है।
Given: Bisector of \( \angle BAC \) intersects side BC at point D.
To prove: AB > BD
Solution:
Proof:
\( \angle BAD = \angle DAC \) ....(i) [Seg AD bisects \( \angle BAC \)]
\( \angle ADB \) is the exterior angle of \( \triangle ADC \).
\( \therefore \angle ADB > \angle DAC \) ....(ii) [Property of exterior angle]
\( \therefore \angle ADB > \angle BAD \) ....(iii) [From (i) and (ii)]
In \( \triangle ABD \),
\( \angle ADB > \angle BAD \) [From (iii)]
\( \therefore \) AB > BD [Side opposite to greater angle is greater]
In simple words: By the exterior angle property, \( \angle ADB \) is greater than \( \angle DAC \). Since AD bisects \( \angle BAC \), \( \angle DAC \) is equal to \( \angle BAD \). Therefore, \( \angle ADB \) is greater than \( \angle BAD \). In \( \triangle ABD \), the side opposite \( \angle ADB \) (AB) must be greater than the side opposite \( \angle BAD \) (BD).

🎯 Exam Tip: The exterior angle theorem is key here. Remember that an exterior angle of a triangle is greater than either of its interior opposite angles.

Question 7. In the adjoining figure, seg PT is the bisector of \( \angle QPR \). A line through R intersects ray QP at point S. Prove that PS = PR.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज QPR को दर्शाता है। शीर्ष P ऊपर है। PT \( \angle QPR \) का कोण समद्विभाजक है। R से होकर जाने वाली एक रेखा किरण QP को बिंदु S पर प्रतिच्छेद करती है।
Given: Seg PT is the bisector of \( \angle QPR \).
To prove: PS = PR
Construction: Draw seg SR \( \parallel \) seg PT.
Solution:
Proof:
seg PT is the bisector of \( \angle QPR \). [Given]
\( \therefore \angle QPT = \angle RPT \) ....(i)
seg PT \( \parallel \) seg SR [Construction]
and seg QS is their transversal.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज को दर्शाता है जहाँ P एक शीर्ष है, और QR आधार है। PT एक रेखाखंड है। एक समानांतर रेखा SR खींची गई है जो PT के समानांतर है और किरण QP को S पर प्रतिच्छेद करती है।
\( \therefore \angle QPT = \angle PSR \) ...(ii) [Corresponding angles]
seg PT \( \parallel \) seg SR [Construction]
and seg PR is their transversal.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज को दर्शाता है जहाँ P एक शीर्ष है, और R एक बिंदु है। PT एक रेखाखंड है। एक समानांतर रेखा SR खींची गई है जो PT के समानांतर है।
\( \therefore \angle RPT = \angle PRS \) .....(iii) [Alternate angles]
\( \therefore \angle PRS = \angle PSR \) ...(iv) [From (i), (ii) and (iii)]
In \( \triangle PSR \),
\( \angle PRS = \angle PSR \) [From (iv)]
\( \therefore \) PS = PR [Converse of isosceles triangle theorem]
In simple words: By construction, draw SR parallel to PT. Use the properties of parallel lines with transversals to show that \( \angle QPT = \angle PSR \) (corresponding angles) and \( \angle RPT = \angle PRS \) (alternate angles). Since PT bisects \( \angle QPR \), \( \angle QPT = \angle RPT \). Combining these, we get \( \angle PRS = \angle PSR \), which means \( \triangle PSR \) is isosceles with PS = PR.

🎯 Exam Tip: Auxiliary constructions (like drawing a parallel line) are often necessary for proofs involving angle bisectors and side equality. Look for opportunities to create alternate interior or corresponding angles.

Question 8. In the adjoining figure, seg AD \( \perp \) seg BC. Seg AE is the bisector of \( \angle CAB \) and B - E - C. Prove that \( \angle DAE = \frac{1}{2} (\angle C - \angle B) \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC को दर्शाता है जिसमें शीर्ष A ऊपर है। भुजा BC पर दो बिंदु D और E इस प्रकार हैं कि AD \( \perp \) BC और AE \( \angle CAB \) का कोण समद्विभाजक है।
Given: seg AD \( \perp \) seg BC
seg AE is the bisector of \( \angle CAB \).
To prove: \( \angle DAE = \frac{1}{2} (\angle C - \angle B) \) [\( \because\therefore\because \) AD \( \perp \) BC]
\( \therefore \angle DAE = 180^\circ - 90^\circ - \angle AED \)
\( \therefore \angle DAE = 90^\circ - \angle AED \) ....(ii)
Proof:
\( \therefore \angle CAE = \frac{1}{2} \angle A \) ....(i) [seg AE is the bisector of \( \angle CAB \)]
In \( \triangle DAE \),
\( \angle DAE + \angle ADE + \angle AED = 180^\circ \) [Sum of the measures of the angles of a triangle is \( 180^\circ \)]
\( \therefore \angle DAE + 90^\circ + \angle AED = 180^\circ \) [\( \because \) AD \( \perp \) BC]
\( \therefore \angle DAE = 180^\circ - 90^\circ - \angle AED \)
\( \therefore \angle DAE = 90^\circ - \angle AED \) ....(ii)
In \( \triangle ACE \),
\( \therefore \angle ACE + \angle CAE + \angle AEC = 180^\circ \) [Sum of the measures of the angles of a triangle i \( 180^\circ \)]
\( \angle C + \frac{1}{2} \angle A + \angle AED = 180^\circ \) [From (i) and C-D-E]
\( \therefore \angle AED = 180^\circ - \angle C - \frac{1}{2} \angle A \) ......(iii)
\( \therefore \angle DAE = 90^\circ - 180^\circ - \angle C+ \frac{1}{2} \angle A \) [Substituting (iii) in (ii)]
\( \therefore \angle DAE = \angle C + \frac{1}{2} \angle A - 90^\circ \) .....(iv)
In \( \triangle ABC \),
\( \angle A + \angle B + \angle C = 180^\circ \)
\( \therefore \frac{1}{2} \angle A + \frac{1}{2} \angle B + \frac{1}{2} \angle C = 90^\circ \) [Dividing both sides by 2]
\( \therefore \frac{1}{2} \angle A = 90^\circ - \frac{1}{2} \angle C - \frac{1}{2} \angle B \) \( \quad \) (v)
\( \therefore \angle DAE \)
\( = \angle C+ (90^\circ - \frac{1}{2} \angle C - \frac{1}{2} \angle B)-90^\circ \) [Substituting (v) in (iv)]
\( \therefore \angle DAE = \angle C - \frac{1}{2} \angle C - \frac{1}{2} \angle B \)
\( = \frac{1}{2} \angle C - \frac{1}{2} \angle B \)
\( \therefore \angle DAE = \frac{1}{2} (\angle C - \angle B) \)
In simple words: Express \( \angle DAE \) in terms of \( \angle AED \) using \( \triangle DAE \). Then, find \( \angle AED \) in terms of \( \angle C \) and \( \angle A \) from \( \triangle ACE \). Substitute this into the first equation. Finally, use the angle sum property of \( \triangle ABC \) to relate \( \angle A \) to \( \angle B \) and \( \angle C \), simplifying the expression to prove \( \angle DAE = \frac{1}{2} (\angle C - \angle B) \).

🎯 Exam Tip: This proof involves algebraic manipulation of angles within different triangles. Ensure each step is justified by a geometric theorem (e.g., angle sum property, bisector definition, perpendicular lines).

Maharashtra Board Class 9 Maths Chapter 3 Triangles Problem Set 3 Intext Questions And Activities

Question 1. Draw a triangle of any measure on a thick paper. Take a point T on ray QR as shown in the figure given below. Cut two pieces of thick paper which will exactly fit the comers of \( \angle P \) and \( \angle Q \). See that the same two jpieces fit exactly at the comer of \( \angle PRT \) as shown in the figure. (Textbook pg. no. 24)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज PQR को दर्शाता है। QR एक रेखा है और T बिंदु R के दाईं ओर किरण पर है। \( \angle PRT \) एक बाह्य कोण है। त्रिभुज के भीतर \( \angle P \) और \( \angle Q \) को दर्शाया गया है।
In simple words: This activity demonstrates the Exterior Angle Theorem. When you cut out the angles \( \angle P \) and \( \angle Q \) from a triangle and place them adjacent to each other, they will perfectly fit into the exterior angle \( \angle PRT \).

🎯 Exam Tip: Practical activities like this help visualize abstract geometric concepts. The key takeaway is that the measure of an exterior angle of a triangle is equal to the sum of its interior opposite angles.

Question 2. Check the congruence of triangles.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चार अलग-अलग त्रिभुजों (A1B1C1, A2B2C2, A3B3C3, A4B4C4) को दर्शाता है, जो सभी एक मूल त्रिभुज ABC की प्रतियाँ हैं। पहले त्रिभुज को सीधे स्लाइड करके, दूसरे को घुमाकर और तीसरे को पलटकर बनाया गया है, जिससे उनकी सर्वांगसमता प्रदर्शित होती है।
Draw \( \triangle ABC \) of any measure on a card-sheet and cut it out. Place it on a card-sheet. Make a copy of it by drawing its border. Name it as \( \triangle A_1B_1C_1 \). Now slide the \( \triangle ABC \) which is the cut out of a triangle to some distance and make one more copy of it. Name it \( \triangle A_2B_2C_2 \). Then rotate the cut out of triangle ABC a little, as shown in the figure, and make another copy of it. Name the copy as \( \triangle A_3B_3C_3 \). Then flip the triangle ABC, place it on another card-sheet and make a new copy of it. Name this copy as \( \triangle A_4B_4C_4 \). Have you noticed that each of \( \triangle A_1B_1C_1 \), \( \triangle A_2B_2C_2 \), \( \triangle A_3B_3C_3 \) and \( \triangle A_4B_4C_4 \) is congruent with \( \triangle ABC \) ? Because each of them fits exactly with \( \triangle ABC \). Let us verify for \( \triangle A_3B_3C_3 \). If we place \( \angle A \) upon \( \angle A_3 \), \( \angle B \) upon \( \angle B_3 \) and \( \angle C \) upon \( \angle C_3 \), then only they will fit each other and we can say that \( \triangle ABC \cong \triangle A_3B_3C_3 \). We also have AB = A3B3, BC = B3C3, CA = C3A3. Note that, while examining the congruence of two triangles, we have to write their angles and sides in a specific order, that is with a specific one-to-one correspondence. If \( \triangle ABC \cong \triangle PQR \), then we get the following six equations:
(i) \( \angle A = \angle P, \angle B = \angle Q, \angle C = \angle R \)
(ii) and AB = PQ, BC = QR, CA = RP
This means, with a one-to-one correspondence between the angles and the sides of two triangles, we get hree pairs of congruent angles and three pairs of congruent sides. (Textbook pg. no. 29)
In simple words: This activity demonstrates that triangles can be congruent regardless of their position, orientation, or reflection, as long as their corresponding sides and angles are equal. Congruence implies that they can perfectly overlap each other.

🎯 Exam Tip: Congruent figures have identical shape and size. The order of vertices in a congruence statement (e.g., \( \triangle ABC \cong \triangle PQR \)) is critical as it indicates the correct one-to-one correspondence of vertices, sides, and angles.

Question 3. Every student in the group should draw a right angled triangle, one of the angles measuring \( 30^\circ \). The choice of lengths of sides should be their own. Each one should measure the length of the hypotenuse and the length of the side opposite to \( 30^\circ \) angle.
One of the students in the group should fill in the following table.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक समकोण त्रिभुज ABC को दर्शाता है, जिसमें \( \angle B = 90^\circ \)। \( \angle C \) को \( 30^\circ \) और \( \angle A \) को \( 60^\circ \) के रूप में चिह्नित किया गया है।

Did you notice any property of sides of right angled triangle with one of the angles measuring \( 30^\circ \)? (Textbook pg. no. 34)
Answer: We observe that the length of the side opposite to \( 30^\circ \) is half the length of the hypotenuse.
In simple words: This activity reveals the 30-60-90 triangle property: the side opposite the 30-degree angle is always half the length of the hypotenuse.

🎯 Exam Tip: The 30-60-90 triangle property is a fundamental shortcut in geometry. Memorize it: side opposite \( 30^\circ \) = x, side opposite \( 60^\circ \) = \( x\sqrt{3} \), hypotenuse = 2x.

Question 4. The measures of angles of a set square in your compass box are \( 30^\circ, 60^\circ \) and \( 90^\circ \). Verify the property of the sides of the set square. (Textbook pg. no. 34)
[Students should attempt the above activity on their own.]
In simple words: This activity asks you to physically confirm the 30-60-90 triangle rule using a set square. You will find that the side opposite the \( 30^\circ \) angle is half the hypotenuse.

🎯 Exam Tip: Hands-on verification reinforces theoretical knowledge. Connecting classroom concepts to real-world tools like set squares deepens understanding.

Question 5. Draw a triangle ABC. Draw medians AD, BE and CF of the triangle. Let their point of concurrence be G, which is called the centroid of the triangle. Compare the lengths of AG and GD with a divider. Verify that the length of AG is twice the length of GD. Similarly, verify that the length of BG is twice the length of GE and the length of CG is twice the length of GF. Name the property of medians of a triangle observed here. (Textbook pg. no. 37)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज ABC को दर्शाता है जिसमें AD, BE और CF उसकी माध्यिकाएँ हैं। ये तीनों माध्यिकाएँ त्रिभुज के अंदर बिंदु G पर प्रतिच्छेद करती हैं, जिसे केंद्रक कहा जाता है।
Answer: The point of concurrence of medians of the triangle divides each median in the ratio 2: 1.
In simple words: The centroid of a triangle is the point where its three medians intersect. This point divides each median in a 2:1 ratio, with the longer segment being from the vertex to the centroid.

🎯 Exam Tip: The centroid is an important point of concurrence. Remember that it divides each median into two segments in a 2:1 ratio, with the part closer to the vertex being twice the part closer to the midpoint of the side.

Question 6. Draw a triangle ABC on a cardboard. Draw its medians and denote their point of concurrence as G. Cut out the triangle. Now take a pencil. Try to balance the triangle on the flat tip of the pencil. The triangle is balanced only when the point G is on the flat tip of the pencil. This activity shows an important property of the centroid (point of concurrence of the medians) of the triangle. Point it out. (Textbook pg. no. 37)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक त्रिभुज को दर्शाता है जो एक बिंदु पर संतुलित है, जो त्रिभुज के केंद्रक (माध्यिकाओं के संगम बिंदु) का प्रतिनिधित्व करता है।
Answer: The centroid of the triangle is the triangle's centre of gravity. Hence, the triangle in the experiment remains balanced.
In simple words: The centroid is the center of mass or gravity of a triangle. This means if you balance the triangle on a pinpoint, it will stably rest at its centroid.

🎯 Exam Tip: The centroid's property as the center of gravity is a key conceptual understanding. It's often used in physics applications related to equilibrium.

Question 7. Take a photograph on a mobile or a computer. Recall what you do to reduce it or to enlarge it. Also recall what you do to see a part of the photograph in detail. (Textbook pg. no, 45)
In simple words: When you reduce or enlarge a photograph, you are performing a scaling operation, maintaining the proportions. When you zoom in to see detail, you are essentially looking at a magnified, similar portion of the image.

🎯 Exam Tip: This question relates to the concept of similarity. Reducing or enlarging an image creates a similar image, where angles are preserved, and sides are proportional.

Question 8. On a card-sheet, draw a triangle of sides 4 cm, 3 cm and 2 cm. Cut it out. Make 13 more copies of the triangle and cut them out from the card sheet. Note that all these triangular pieces are congruent. Arrange them as shown in the following figure and make three triangles out of them.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक बड़े त्रिभुज के अंदर छोटे त्रिभुजों की एक व्यवस्था को दर्शाता है। शीर्ष A पर एक छोटा त्रिभुज (ऊपर 2 और 4 भुजाओं वाला), फिर आधार पर एक दूसरा त्रिभुज (3 और 4 भुजाओं वाला) और अंत में दोनों के बीच एक तीसरा त्रिभुज (2 और 3 भुजाओं वाला) दिख रहा है।
Number of triangle: 1
ℹ️ चित्र व्याख्या (Diagram Explanation): यह तीन छोटे त्रिभुजों से बने एक बड़े त्रिभुज को दर्शाता है। शीर्ष पर एक छोटा त्रिभुज (भुजाएँ 4, 4, 2), बाईं ओर एक और त्रिभुज (भुजाएँ 4, 3, 3) और दाईं ओर एक तीसरा त्रिभुज (भुजाएँ 2, 3, 2) है।
Number of triangles: 4
ℹ️ चित्र व्याख्या (Diagram Explanation): यह नौ छोटे त्रिभुजों से बने एक बड़े त्रिभुज को दर्शाता है, जिसमें एक बड़े समबाहु त्रिभुज के भीतर एक ग्रिड पैटर्न में छोटे त्रिभुजों की पुनरावृत्ति है।
Number of triangles: 9
\( \triangle ABC \) and \( \triangle DEF \) are similar in the correspondence ABC \( \leftrightarrow \) DEF.
\( \angle A \simeq \angle D, \angle B \simeq \angle E, \angle C \simeq \angle F \)
and \( \frac{AB}{DE} = \frac{4}{8} = \frac{1}{2} ; \frac{BC}{EF} = \frac{3}{6} = \frac{1}{2} ; \frac{AC}{DF} = \frac{2}{4} = \frac{1}{2} \) ...the corresponding sides are in proportion.
Similarly, consider \( \triangle DEF \) and \( \triangle PQR \). Are their angles congruent and sides proportional in the correspondence DEF \( \leftrightarrow \) PQR? (Textbook pg. no. 45)
Answer: Yes.
\( \angle D \simeq \angle P, \angle E \simeq \angle Q, \angle F \simeq \angle R \)
and \( \frac{DE}{PQ} = \frac{8}{12} = \frac{2}{3} ; \frac{EF}{QR} = \frac{6}{9} = \frac{2}{3} ; \frac{DF}{PR} = \frac{4}{6} = \frac{2}{3} \)
In simple words: This activity demonstrates how smaller congruent triangles can form larger triangles, illustrating the concept of similarity. When triangles are similar, their corresponding angles are congruent, and their corresponding sides are in proportion.

🎯 Exam Tip: Similarity is defined by two conditions: congruent corresponding angles and proportional corresponding sides. Both conditions must be met for triangles to be similar.

Question 9. Draw a triangle \( \triangle A_1B_1C_1 \) on a card-sheet and cut it out. Measure \( \angle A_1, \angle B_1, \angle C_1 \)
Draw two more triangles \( \triangle A_2B_2C_2 \) and \( \triangle A_3B_3C_3 \) such that
\( \angle A_1 = \angle A_2 = \angle A_3, \angle B_1 = \angle B_2 = \angle B_3, \angle C_1 = \angle C_2 = \angle C_3 \)
and B1C1 \( > \) B2C2 \( > \) B3C3. Now cut these two triangles also. Measure the lengths of the three triangles. Arrange the triangles in two ways as shown in the figure.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह तीन समरूप त्रिभुजों को दर्शाता है जिनकी भुजाओं की लंबाई भिन्न है लेकिन कोण समान हैं। एक त्रिभुज को दूसरे के ऊपर इस तरह रखा गया है कि उनके शीर्ष A1, A2, A3 संरेखित हैं, और आधार भुजाएँ समानांतर हैं।
ℹ️ चित्र व्याख्या (Diagram Explanation): यह तीन समरूप त्रिभुजों को दर्शाता है जिनके शीर्ष A1, A2, A3 संरेखित हैं। यह बड़े से छोटे त्रिभुजों की एक श्रृंखला दिखाता है, सभी का आकार समान है लेकिन आकार भिन्न है।
Check the ratios \( \frac{A_1B_1}{A_2B_2}, \frac{B_1C_1}{B_2C_2}, \frac{A_1C_1}{A_2C_2} \). You will notice that the ratios are equal.
Similarly, see whether the ratios \( \frac{A_1C_1}{A_3C_3}, \frac{B_1C_1}{B_3C_3}, \frac{A_1B_1}{A_3B_3} \) are equal. What do you observe? (Texthook pg. no. 46)
Answer: From the activity we observe that, when corresponding angles of two triangles are equal, the ratios of their corresponding sides are also equal i.e., their corresponding sides are in the same proportion.
In simple words: This activity demonstrates the AA (Angle-Angle) similarity criterion. If the corresponding angles of two triangles are equal, then their corresponding sides are proportional, and the triangles are similar.

🎯 Exam Tip: The AA similarity criterion is one of the most frequently used similarity tests. If two angles of one triangle are congruent to two angles of another triangle, the triangles are similar.

Question 10. Prepare a map of road surrounding your school or home, upto a distance of about 500 metre. How will you measure the distance between two spots on a road?
While walking, count how many steps cover a distance of about two metre. Suppose, your three steps cover a distance of 2 metre. Considering this proportion 90 steps means 60 metre. In this way you can judge the distances between different spots on roads and also the lengths of roads. You have to judge the measures of angles also where two roads meet each other. Choosing a proper scale for lengths of roads, prepare a map. Try to show shops, buildings, bus stops, rickshaw stand etc. in the map. (Textbook pg. no. 48)
In simple words: This activity requires you to create a scaled map of an area around your school or home. You'll need to use techniques like pacing to estimate distances and visualize angles to represent real-world locations accurately on paper using a suitable scale.

🎯 Exam Tip: Map-making combines scale factors and proportional reasoning, which are direct applications of similarity in real-world contexts. Understanding how to represent large distances on a smaller scale is a practical skill.