Maharashtra Board Class 10 Maths Chapter 1 Similarity Set 1.3 Solutions

Question 1. In the adjoining figure, ∠ABC = 75°, ∠EDC = 75°. State which two triangles are similar and by which test? Also write the similarity of these two triangles by a proper one to one correspondence.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो त्रिभुजों, ∆ABC और ∆EDC को दर्शाता है। बिंदु C दोनों त्रिभुजों का उभयनिष्ठ शीर्ष है। भुजा BC पर बिंदु E स्थित है और भुजा AC पर बिंदु D स्थित है। कोण ABC और कोण EDC दोनों 75 डिग्री माप के हैं।
Answer:
Solution:
In ∆ABC and ∆EDC,
∠ABC = ∠EDC [Each angle is of measure 75°]
∠ACB = ∠ECD [Common angle]
\( \therefore \triangle ABC \sim \triangle EDC\) [AA test of similarity]
One to one correspondence is
ABC → EDC
In simple words: Two triangles, ∆ABC and ∆EDC, are similar because they share a common angle (∠C) and have another pair of corresponding angles (∠B and ∠E) equal to 75°, satisfying the AA similarity test. The correspondence is ABC to EDC.

🎯 Exam Tip: Remember to correctly identify common angles and given equal angles to apply the AA test effectively. Proper one-to-one correspondence is crucial for full marks.

Question 2. Are the triangles in the adjoining figure similar? If yes, by which test?
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो अलग-अलग त्रिभुजों, ∆PQR और ∆LMN को दर्शाता है। त्रिभुज PQR की भुजाओं की माप PQ=6, QR=8, और PR=10 है, जबकि त्रिभुज LMN की भुजाओं की माप LM=3, MN=4, और LN=5 है।
Answer:
Solution:
In ∆PQR and ∆LMN,
\(\frac{PQ}{LM} = \frac{6}{3} = \frac{2}{1}\) (i)
\(\frac{QR}{MN} = \frac{8}{4} = \frac{2}{1}\) (ii)
\(\frac{PR}{LN} = \frac{10}{5} = \frac{2}{1}\) (iii)
\( \therefore \frac{PQ}{LM} = \frac{QR}{MN} = \frac{PR}{LN}\) [From (i), (ii) and (iii)]
\( \therefore \triangle PQR \sim \triangle LMN\) [SSS test of similarity]
In simple words: The triangles ∆PQR and ∆LMN are similar by the SSS (Side-Side-Side) test because the ratios of their corresponding sides are all equal (\(2:1\)).

🎯 Exam Tip: For SSS similarity, calculate the ratios of all three pairs of corresponding sides. If all ratios are equal, the triangles are similar.

Question 3. As shown in the adjoining figure, two poles of height 8 m and 4 m are perpendicular to the ground. If the length of shadow of smaller pole due to sunlight is 6 m, then how long will be the shadow of the bigger pole at the same time?
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो खंभों और उनकी परछाइयों को दर्शाते हुए दो समकोण त्रिभुज दिखाता है। एक छोटा खंभा (PR) 4 मीटर ऊंचा है जिसकी परछाई (QR) 6 मीटर लंबी है, और एक बड़ा खंभा (AC) 8 मीटर ऊंचा है जिसकी परछाई (BC) 'x' मीटर लंबी है।
Answer:
Solution:
Here, AC and PR represents the bigger and smaller poles, and BC and QR represents their shadows respectively.
Now, \( \triangle ACB \sim \triangle PRQ\) [:: Vertical poles and their shadows form similar figures]
\( \therefore \frac{AC}{PR} = \frac{CB}{RQ}\) [Corresponding sides of similar triangles]
\( \frac{8}{4} = \frac{x}{6}\)

\( \implies x = \frac{8 \times 6}{4}\)

\( \implies x = 12\ m\)
\( \therefore \) The shadow of the bigger pole will be 12 metres long at that time.
In simple words: Since the poles and their shadows form similar triangles due to the constant angle of elevation of the sun, we can use the ratio of corresponding sides to find the unknown shadow length. The bigger pole's shadow is 12 meters.

🎯 Exam Tip: Remember that at the same time, the angle of elevation of the sun is constant, leading to similar triangles formed by vertical objects and their shadows. Set up correct ratios for corresponding sides.

Question 4. In ∆ABC, AP ⊥ BC, BQ ⊥ AC, B – P – C, A – Q – C, then prove that ∆CPA ~ ∆CQB. If AP = 7, BQ = 8, BC = 12, then find AC.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज ∆ABC दिखाता है, जिसमें शीर्ष A से भुजा BC पर एक लंब (AP) और शीर्ष B से भुजा AC पर एक लंब (BQ) खींचा गया है। AP और BQ दोनों त्रिभुज के शीर्षलंब हैं, जो क्रमशः भुजाओं BC और AC पर 90 डिग्री का कोण बनाते हैं।
Answer:
Solution:
In \( \triangle CPA\) and \( \triangle CQB\),
\( \angle CPA = \angle CQB\) [Each angle is of measure 90°]
\( \angle ACP = \angle BCQ\) [Common angle]
\( \therefore \triangle CPA \sim \triangle CQB\) [AA test of similarity]
\( \therefore \frac{AC}{BC} = \frac{AP}{BQ}\) [Corresponding sides of similar triangles]
\( \frac{AC}{12} = \frac{7}{8}\)

\( \implies AC = x = \frac{12 \times 7}{8}\)

\( \implies AC = 10.5\ Units\)
In simple words: By proving two angles are equal (one 90° due to perpendiculars, the other common angle ∠C), we establish similarity between ∆CPA and ∆CQB using the AA test. Then, using the property of corresponding sides of similar triangles, we find AC to be 10.5 units.

🎯 Exam Tip: When altitudes are involved, look for right angles and common angles. The AA similarity test is frequently applicable in such scenarios.

Question 5. Given: In trapezium PQRS, side PQ || side SR, AR = 5 AP, AS = 5 AQ, then prove that SR = 5 PQ.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक समलंब चतुर्भुज PQRS को दर्शाता है जिसमें भुजा PQ भुजा SR के समानांतर है। इस समलंब चतुर्भुज के विकर्ण PR और SQ बिंदु A पर प्रतिच्छेद करते हैं, जिससे अंदर दो त्रिभुज बनते हैं।
Answer:
Solution:
side PQ || side SR [Given]
and seg SQ is their transversal.
\( \therefore \angle QSR = \angle SQP\) [Alternate angles]
\( \therefore \angle ASR = \angle AQP\) (i) [Q – A – S]
In \( \triangle ASR\) and \( \triangle AQP\),
\( \angle ASR = \angle AQP\) [From (i)]
\( \angle SAR = \angle QAP\) [Vertically opposite angles]
\( \triangle ASR \sim \triangle AQP\) [AA test of similarity]
\( \therefore \frac{AS}{AQ} = \frac{SR}{PQ}\) (ii) [Corresponding sides of similar triangles]
But, AS = 5 AQ [Given]
\( \therefore \frac{AS}{AQ} = \frac{5}{1}\) (iii)
\( \therefore \frac{SR}{PQ} = \frac{5}{1}\) [From (ii) and (iii)]
\( \therefore SR = 5 PQ\)
In simple words: By using alternate interior angles and vertically opposite angles, we prove that ∆ASR and ∆AQP are similar. Then, using the ratio of corresponding sides and the given relation AS = 5 AQ, we derive that SR = 5 PQ.

🎯 Exam Tip: Parallel lines and transversals are key to identifying alternate angles. Vertically opposite angles are always congruent, a common tool for similarity proofs.

Question 6. Id trapezium ABCD (adjoining figure), side AB || side DC, diagonals AC and BD intersect in point O. If AB = 20, DC = 6, OB = 15, then find OD.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक समलंब चतुर्भुज ABCD को दर्शाता है, जहाँ भुजा AB भुजा DC के समानांतर है। इस चतुर्भुज के विकर्ण AC और BD एक दूसरे को बिंदु O पर काटते हैं, जिससे प्रतिच्छेदन बिंदु पर दो त्रिभुज बनते हैं।
Answer:
Solution:
side AB || side DC [Given]
and seg BD is their transversal.
\( \therefore \angle DBA = \angle BDC\) [Alternate angles]
\( \therefore \angle OBA = \angle ODC\) (i) [D – O – B]
In \( \triangle OBA\) and \( \triangle ODC\)
\( \angle OBA = \angle ODC\) [From (i)]
\( \angle BOA = \angle DOC\) [Vertically opposite angles]
\( \therefore \triangle OBA \sim \triangle ODC\) [AA test of similarity]
\( \therefore \frac{OB}{OD} = \frac{AB}{DC}\) [Corresponding sides of similar triangles]
\( \frac{15}{OD} = \frac{20}{6}\)

\( \implies OD = x = \frac{15 \times 6}{20}\)

\( \implies OD = 4.5\ units\)
In simple words: Since AB is parallel to DC in the trapezium, we can use alternate angles and vertically opposite angles to prove that ∆OBA and ∆ODC are similar. Using the property of corresponding sides, we find OD to be 4.5 units.

🎯 Exam Tip: In problems involving trapeziums and intersecting diagonals, look for parallel lines to establish alternate interior angles, which are key for proving triangle similarity.

Question 7. J ABCD is a parallelogram. Point E is on side BC. Line DE intersects ray AB in point T. Prove that DE × BE = CE × TE.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक समांतर चतुर्भुज ABCD को दिखाता है, जिसमें भुजा BC पर एक बिंदु E स्थित है। एक रेखा DE को आगे बढ़ाया गया है जो भुजा AB को बिंदु T पर काटती है, जिससे एक बड़ा त्रिभुज TBC बनता है।
Answer:
Solution:
Proof:
J ABCD is a parallelogram. [Given]
\( \therefore \) side AB || side CD [Opposite sides of a parallelogram]
\( \therefore \) side AT || side CD [A – B – T]
and seg DT is their transversal.
\( \therefore \angle ATD = \angle CDT\) [Alternate angles]
\( \therefore \angle BTE = \angle CDE\) (i) [A – B – T, T – E – D]
In \( \triangle BTE\) and \( \triangle CDE\),
\( \angle BTE = \angle CDE\) [From (i)]
\( \angle BET = \angle CED\) [Vertically opposite angles]
\( \therefore \triangle BTE \sim \triangle CDE\). [AA test of similarity]
\( \therefore \frac{BE}{CE} = \frac{TE}{DE}\) [Corresponding sides of similar triangles]
\( \therefore DE \times BE = CE \times TE\)
In simple words: By recognizing that AT is parallel to CD in the parallelogram, we use alternate angles and vertically opposite angles to prove that ∆BTE and ∆CDE are similar. The property of corresponding sides in similar triangles then directly leads to the desired product equality.

🎯 Exam Tip: For parallelogram problems, remember properties like opposite sides being parallel. This helps identify alternate angles crucial for similarity proofs.

Question 8. In the adjoining figure, seg AC and seg BD intersect each other in point P and \( \frac{AP}{CP} = \frac{BP}{DP}\). Prove that, \( \triangle ABP \sim \triangle CDP\).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो प्रतिच्छेदी रेखाखंडों AC और BD को दर्शाता है जो बिंदु P पर एक दूसरे को काटते हैं। इस प्रतिच्छेदन से दो त्रिभुज ∆ABP और ∆CDP बनते हैं, जिनके शीर्ष P पर सम्मुख कोण हैं।
Answer:
Solution:
Proof:
In \( \triangle ABP\) and \( \triangle CDP\),
\( \frac{AP}{CP} = \frac{BP}{DP}\) [Given]
\( \angle APB = \angle CPD\) [Vertically opposite angles]
\( \therefore \triangle ABP \sim \triangle CDP\) [SAS test of similarity]
In simple words: Given the ratio of two pairs of corresponding sides and the fact that the included angles (vertically opposite angles) are equal, we can directly prove that ∆ABP and ∆CDP are similar using the SAS (Side-Angle-Side) similarity test.

🎯 Exam Tip: The SAS similarity test requires two pairs of sides to be in proportion and the *included* angle between those sides to be congruent. Ensure the angle is indeed included between the proportional sides.

Question 9. In the adjoining figure, in \( \triangle ABC\), point D is on side BC such that, \( \angle BAC = \angle ADC\). Prove that, \(CA^2 = CB \times CD\).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज ∆ABC को दर्शाता है, जिसमें भुजा BC पर एक बिंदु D स्थित है। बिंदु A को बिंदु D से जोड़ने वाली एक रेखा (AD) खींची गई है, जिससे दो छोटे त्रिभुज ∆ABD और ∆ADC बनते हैं।
Answer:
Solution:
Proof:
In \( \triangle BAC\) and \( \triangle ADC\),
\( \angle BAC = \angle ADC\) [Given]
\( \angle BCA = \angle ACD\) [Common angle]
\( \therefore \triangle BAC \sim \triangle ADC\) [AA test of similarity]
\( \therefore \frac{CA}{CD} = \frac{CB}{CA}\) [Corresponding sides of similar triangles]
\( \therefore CA \times CA = CB \times CD\)
\( \therefore CA^2 = CB \times CD\)
In simple words: By showing that ∆BAC and ∆ADC share a common angle (∠C) and have another pair of equal angles (∠BAC = ∠ADC), we prove their similarity using the AA test. From their similarity, the ratios of corresponding sides allow us to derive the relationship \(CA^2 = CB \times CD\).

🎯 Exam Tip: When proving relations involving squares of sides (like \(CA^2\)), it's a strong indicator to look for similar triangles where that side appears twice in the ratio of corresponding sides.

Question 1. In the adjoining figure, BP ⊥ AC, CQ ⊥ AB, A – P – C, A – Q – B, then prove that ∆APB and ∆AQC are similar.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक त्रिभुज ∆ABC को दर्शाता है, जिसमें शीर्ष B से भुजा AC पर एक लंब BP और शीर्ष C से भुजा AB पर एक लंब CQ खींचा गया है। ये दोनों लंब क्रमशः P और Q बिंदुओं पर 90 डिग्री का कोण बनाते हैं।
Answer:
Solution:
In \( \triangle APB\) and \( \triangle AQC\),
\( \angle APB = 90^\circ\) (i)
\( \angle AQC = 90^\circ\) (ii)
\( \therefore \angle APB \cong \angle AQC\) [From (i) and (ii)]
\( \angle PAB = \angle QAC\) [Common angle]
\( \therefore \triangle APB \sim \triangle AQC\) [AA test of similarity]
In simple words: We prove that ∆APB and ∆AQC are similar by identifying two pairs of congruent angles: the right angles formed by the perpendiculars (∠APB = ∠AQC = 90°) and the common angle ∠A shared by both triangles, satisfying the AA similarity test.

🎯 Exam Tip: The AA similarity test is very efficient. Always look for two pairs of congruent angles (common, right, alternate, corresponding, etc.) to prove similarity.

2. SAS Test For Similarity Of Triangles:

For a given correspondence, if two pairs of corresponding sides are in the same proportion and the angle between them is congruent, then the two triangles are similar.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो त्रिभुजों ∆ABC और ∆PQR को दर्शाता है। इन त्रिभुजों में, भुजा AB का भुजा PQ से अनुपात और भुजा BC का भुजा QR से अनुपात समान है, तथा इन भुजाओं के बीच बनने वाले कोण B और Q सर्वांगसम हैं।

🎯 Exam Tip: When applying SAS, ensure the congruent angle is *specifically* the one included between the two proportional sides. Misidentifying the included angle is a common error.

3. SSS Test For Similarity Of Triangles:

For a given correspondence, if three sides of one triangle are in proportion with the corresponding three sides of the another triangle, then the two triangles are similar.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो त्रिभुजों ∆ABC और ∆PQR को दर्शाता है। इन त्रिभुजों में, भुजा AB का भुजा PQ से अनुपात, भुजा BC का भुजा QR से अनुपात, और भुजा AC का भुजा PR से अनुपात समान है, जो तीनों संगत भुजाओं के अनुपात को दर्शाता है।

🎯 Exam Tip: To use SSS, accurately pair up corresponding sides and calculate all three ratios. All ratios must be exactly equal for the triangles to be similar by SSS.

Properties Of Similar Triangles:

1. Reflexivity: \( \triangle ABC \sim \triangle ABC\) 2. Symmetry: If \( \triangle ABC \sim \triangle DEF\), then \( \triangle DEF \sim \triangle ABC\). 3. Transitivity: If \( \triangle ABC \sim \triangle DEF\) and \( \triangle DEF \sim \triangle GHI\), then \( \triangle ABC \sim \triangle GHI\).

🎯 Exam Tip: These properties (reflexivity, symmetry, transitivity) are fundamental to understanding the nature of similarity as an equivalence relation. They are less about direct problem-solving and more about conceptual understanding.