Maharashtra Board Class 10 Maths Chapter 1 Similarity Set 1.1 Solutions

Question 1. Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of areas of these triangles.
Answer: Solution: Let the base, height and area of the first triangle be b₁, h₁, and A₁ respectively. Let the base, height and area of the second triangle be b₂, h₂ and A₂ respectively. \[ \frac{A_1}{A_2} = \frac{b_1 \times h_1}{b_2 \times h_2} \] \[ \frac{A_1}{A_2} = \frac{9 \times 5}{10 \times 6} \] \[ \frac{A_1}{A_2} = \frac{45}{60} \] \[ \frac{A_1}{A_2} = \frac{3}{4} \] [Since Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights] Therefore, The ratio of areas of the triangles is 3:4.
In simple words: The ratio of the areas of two triangles is found by multiplying their respective bases and heights and then dividing these products. For the given triangles, this ratio simplifies to 3:4.

🎯 Exam Tip: Remember the formula for the ratio of areas of two triangles, which is the product of their bases and corresponding heights. Ensure correct substitution and simplification for full marks.

Question 2. In the adjoining figure, BC ⊥ AB, AD ⊥ AB, BC = 4, AD = 8, then find \( \frac{A(\triangle ABC)}{A(\triangle ADB)} \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख दो समकोण त्रिभुजों, ΔABC और ΔADB को दर्शाता है, जो एक उभयनिष्ठ आधार AB साझा करते हैं। बिंदु C, B के ऊपर है, और D, A के बाईं ओर है। BC और AD लंबवत रेखाएँ क्रमशः C और D से AB तक खींची गई हैं। यह उन दो त्रिभुजों को दर्शाता है जिनके क्षेत्रफल एक उभयनिष्ठ आधार के सापेक्ष उनकी ऊँचाई पर निर्भर करते हैं।
Answer: Solution: ∆ABC and ∆ADB have same base AB. \[ \therefore \frac{A(\triangle ABC)}{A(\triangle ADB)} = \frac{BC}{AD} \] \[ = \frac{4}{8} \] \[ \frac{A(\triangle ABC)}{A(\triangle ADB)} = \frac{1}{2} \] [Since Triangles having equal base]
In simple words: When two triangles share the same base, the ratio of their areas is equal to the ratio of their corresponding heights. Here, with base AB common, the ratio of areas is the ratio of heights BC to AD, which simplifies to 1/2.

🎯 Exam Tip: For triangles with a common base, the ratio of their areas directly corresponds to the ratio of their heights. Accurately identify the common base and the perpendicular heights to score well.

Question 3. In the adjoining figure, seg PS ⊥ seg RQ, seg QT ⊥ seg PR. If RQ = 6, PS = 6 and PR = 12, then find QT.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक त्रिभुज PQR को दर्शाता है, जिसमें P से भुजा RQ पर एक शीर्षलंब PS और Q से भुजा PR पर एक और शीर्षलंब QT खींचा गया है। यह विन्यास दर्शाता है कि त्रिभुज का क्षेत्रफल विभिन्न आधार-ऊँचाई युग्मों का उपयोग करके कैसे ज्ञात किया जा सकता है।
Answer: Solution: In ΔPQR, PR is the base and QT is the corresponding height. Also, RQ is the base and PS is the corresponding height. \[ \frac{A(APQR)}{A(APQR)} = \frac{PR \times QT}{RQ \times PS} \] [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights] \[ \frac{1}{1} = \frac{PR \times QT}{RQ \times PS} \]
\( \implies \) PR × QT = RQ × PS
\( \implies \) 12 x QT = 6×6
\( \implies \) QT = \( \frac{36}{12} \)
\( \implies \) QT = 3 units
In simple words: The area of a triangle can be calculated using any base and its corresponding height. By equating the area of ΔPQR calculated with base RQ and height PS to the area calculated with base PR and height QT, we can find the unknown height QT.

🎯 Exam Tip: Remember that the area of a triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \). When working with the same triangle, equating two different base-height products allows you to find an unknown dimension.

Question 4. In the adjoining figure, AP ⊥ BC, AD || BC, then find A(ΔABC) : A(ABCD).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख दो त्रिभुजों, ΔABC और ΔADC को दर्शाता है, जो दो समांतर रेखाओं AD और BC के बीच स्थित हैं। AP को A से BC तक लंबवत दिखाया गया है, जो ΔABC की ऊँचाई को दर्शाता है। यह विन्यास एक उभयनिष्ठ आधार साझा करने वाले या समांतर रेखाओं के बीच स्थित त्रिभुजों के क्षेत्रों की तुलना करने में मदद करता है।
Answer: Solution: Draw DQ ⊥ BC, B-C-Q.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख प्रश्न 4 के पहले आरेख के समान है, लेकिन इसमें D से BC तक एक अतिरिक्त लंबवत DQ खींचा गया है, जो BC को Q तक बढ़ाता है। यह इस अवधारणा को पुष्ट करता है कि समांतर रेखाओं AD और BC के बीच की लंबवत दूरी स्थिर है, अर्थात् AP = DQ। AD || BC [Given]
\( \therefore \) AP = DQ (i) [Perpendicular distance between two parallel lines is the same] ∆ABC and ∆BCD have same base BC. \[ \therefore \frac{A(\triangle ABC)}{A(\triangle BCD)} = \frac{AP}{DQ} \] \[ = \frac{AP}{AP} \] [From (i)] \[ = 1 \]
\( \therefore \) A(ΔABC) : A(ABCD) = 1:1
In simple words: Triangles lying between the same parallel lines and having the same base will have equal heights. Therefore, their areas will be equal, resulting in a 1:1 ratio.

🎯 Exam Tip: When triangles share a common base and are situated between parallel lines, their heights are equal. This property is crucial for determining the ratio of their areas, which will always be 1:1.

Question 5. In the adjoining figure, PQ ⊥ BC, AD ⊥ BC, then find following ratios.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख रेखा खंड AD और BC के बीच प्रतिच्छेद करने वाली रेखाओं द्वारा निर्मित कई त्रिभुजों को प्रदर्शित करता है, जिसमें PQ, BC पर लंबवत है और AD, BC पर लंबवत है। विशेष रूप से, त्रिभुज APQB, ΔPBC, ΔABC, ΔADC और ΔPQC दिखाई दे रहे हैं, जो उभयनिष्ठ आधारों या उभयनिष्ठ ऊँचाइयों के आधार पर उनके क्षेत्रों के बीच संबंधों को दर्शाते हैं।
(i) A(APQB) / A(APBC)
(ii) A(APBC) / A(AABC)
(iii) A(AABC) / A(AADC)
(iv) A(AADC) / A(APQC)
Answer: Solution: i. APQB and ΔPBC have same height PQ. \[ \frac{A(APQB)}{A(\triangle PBC)} = \frac{BQ}{BC} \] [Triangles having equal height] ii. ΔPBC and ΔABC have same base BC. \[ \frac{A(\triangle PBC)}{A(\triangle ABC)} = \frac{PQ}{AD} \] [Triangles having equal base] iii. ΔABC and ∆ADC have same height AD. \[ \frac{A(\triangle ABC)}{A(\triangle ADC)} = \frac{BC}{DC} \] [Triangles having equal height] iv. \[ \frac{A(\triangle ADC)}{A(APQC)} = \frac{DC \times AD}{QC \times PQ} \] [Ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
In simple words: This question involves finding ratios of areas for various triangles using two fundamental principles: if triangles share the same height, their area ratio is the ratio of their bases; if they share the same base, their area ratio is the ratio of their heights. The final ratio uses the product of bases and heights.

🎯 Exam Tip: Carefully identify common heights or common bases for each pair of triangles. If neither is common, use the general formula for the ratio of areas (product of base and height). Labeling parts of the diagram clearly can prevent errors.

Question 1. Find \( \frac{A(\triangle ABC)}{A(\triangle APQ)} \).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख दो त्रिभुजों, ΔABC और ΔAPQ को दर्शाता है, जो एक उभयनिष्ठ शीर्ष A साझा करते हैं और जिनके आधार BC और PQ एक ही रेखा खंड पर स्थित हैं। AR को A से BC और PQ वाली रेखा तक एक उभयनिष्ठ ऊँचाई के रूप में दिखाया गया है। इस विन्यास का उपयोग उनके क्षेत्रों का अनुपात ज्ञात करने के लिए किया जाता है जब वे एक उभयनिष्ठ ऊँचाई साझा करते हैं।
Answer: Solution: In ∆ABC, BC is the base and AR is the height. In ΔAPQ, PQ is the base and AR is the height. \[ \therefore \frac{A(\triangle ABC)}{A(\triangle APQ)} = \frac{BC \times AR}{PQ \times AR} \] \[ \frac{A(\triangle ABC)}{A(\triangle APQ)} = \frac{BC}{PQ} \] [The ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
In simple words: When two triangles share the same height, the ratio of their areas is simply the ratio of their bases. Here, AR is the common height for ΔABC and ΔAPQ, so the ratio of their areas is BC/PQ.

🎯 Exam Tip: For triangles sharing a common height, the ratio of their areas simplifies to the ratio of their bases. Clearly identify the common height to correctly apply this principle.