CBSE Class 10 Sample Paper for 2025 Boards - Maths Standard
Question 2
Question 3 Important
Question 4 Important
Question 5 Important
Question 6
Question 7 Important
Question 8 Important
Question 9
Question 10 Important
Question 11 Important
Question 12
Question 13
Question 14
Question 15 Important
Question 16
Question 17
Question 18 Important
Question 19 Important
Question 20 Important
Question 21 (A)
Question 21 (B) Important
Question 22 (A) Important
Question 22 (B) Important
Question 23
Question 24 Important
Question 25
Question 26 (A) Important
Question 26 (B) Important
Question 27
Question 28 Important
Question 29 Important
Question 30 (A)
Question 30 (B) Important
Question 31
Question 32 (A) Important
Question 32 (B) Important
Question 33 - Part 1
Question 33 - Part 2 Important
Question 34 Important
Question 35 (A) Important
Question 35 (B)
Question 36 (i) - Case Based
Question 36 (ii) Important
Question 36 (iii) (A)
Question 36 (iii) (B) Important
Question 37 (i) - Case Based You are here
Question 37 (ii)
Question 37 (iii) (A) Important
Question 37 (iii) (B) Important
Question 38 (i) - Case Based Important
Question 38 (ii)
Question 38 (iii) (A)
Question 38 (iii) (B) Important
CBSE Class 10 Sample Paper for 2025 Boards - Maths Standard
Last updated at Sept. 23, 2024 by Teachoo
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Question 37 [Case Based] Triangle is a very popular shape used in interior designing. The picture given above shows a cabinet designed by a famous interior designer. Here the largest triangle is represented by △ABC and smallest one with shelf is represented by △DEF. PQ is parallel to EF.Question 37 (i) Triangle is a very popular shape used in interior designing. The picture given above shows a cabinet designed by a famous interior designer. Here the largest triangle is represented by △ABC and smallest one with shelf is represented by △DEF. PQ is parallel to EF. (i) Show that △DPQ ∼ △DEF.Given, PQ ∥ EF And, DE as transversal ∴ ∠ DPQ = ∠ DEF Also, PQ ∥ EF And, DE as transversal ∴ ∠ DQP = ∠ DFE (Corresponding Angles) (Corresponding Angles) Now, In Δ DPQ & Δ DEF ∠ DPQ = ∠ DEF ∠ DQP = ∠ DFE ∴ Δ DPQ ~ Δ DEF Hence proved