In the given figure, ∠ACB = ∠CDA, AC = 8cm, AD = 3cm, then BD is

(a) 22/3 cm   (b) 26/3 cm   (c) 55/3 cm   (d) 64/3 cm

This question is inspired from Question 17 - Sample Paper for 2020 Boards - Class 10 Sample Paper

Ques 30 (MCQ) - In figure, ∠ACB = ∠CDA, AC = 8cm, AD = 3cm, then BD is - CBSE Class 10 Sample Paper for 2022 Boards - Maths Standard [MCQ]

part 2 - Question 30 - CBSE Class 10 Sample Paper for 2022 Boards - Maths Standard [MCQ] - Solutions of Sample Papers for Class 10 Boards - Class 10
part 3 - Question 30 - CBSE Class 10 Sample Paper for 2022 Boards - Maths Standard [MCQ] - Solutions of Sample Papers for Class 10 Boards - Class 10

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Question 30 In the given figure, ∠ACB = ∠CDA, AC = 8cm, AD = 3cm, then BD is (a) 22/3 cm (b) 26/3 cm (c) 55/3 cm (d) 64/3 cm Given ∠ACB = ∠CDA In Δ ACB and Δ ADC ∠ACB = ∠ADC ∠CAB = ∠DAC ∴ Δ ACB ∼ Δ ADC We know that Sides of similar triangle are in same proportion ∴ 𝐴𝐶/𝐴𝐷 = 𝐴𝐵/𝐴𝐶 Putting values 8/3 = 𝐴𝐵/8 8/3 = 𝐴𝐵/8 8/3 × 8 = AB 64/3 = AB AB = 𝟔𝟒/𝟑 Now, BD = AB − AD = 64/3 − 3 = (64 − 9)/3 = 𝟓𝟓/𝟑 cm So, the correct answer is (c)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo