Chapter 6 Class 10 Triangles
Example 8 Important
Question 10 Important Deleted for CBSE Board 2025 Exams
Question 2 Important Deleted for CBSE Board 2025 Exams
Theorem 6.1 - Basic Proportionality Theorem (BPT) Important
Theorem 6.7 Important Deleted for CBSE Board 2025 Exams
Ex 6.2, 4 Important
Ex 6.2, 5 Important
Ex 6.2, 6 Important
Ex 6.2, 9 Important
Ex 6.3, 11 Important
Ex 6.3, 12 Important
Ex 6.3, 13 Important
Ex 6.3, 14 Important
Ex 6.3, 15 Important
Question 1 Important Deleted for CBSE Board 2025 Exams
Question 3 Important Deleted for CBSE Board 2025 Exams
Question 5 Important Deleted for CBSE Board 2025 Exams
Question 2 Important Deleted for CBSE Board 2025 Exams
Question 3 Important Deleted for CBSE Board 2025 Exams
Question 11 Important Deleted for CBSE Board 2025 Exams
Question 8 Important Deleted for CBSE Board 2025 Exams
Question 7 Important Deleted for CBSE Board 2025 Exams
Question 9 Important Deleted for CBSE Board 2025 Exams
Chapter 6 Class 10 Triangles
Last updated at April 16, 2024 by Teachoo
Example 5 Observe figure and then find P. Given: AB = 3.8 , PQ = 6 3 , BC = 6 , QR = 12 , AC = 3 3 , PR = 7.6 To find:- angle P i.e. P Solution:- in Let us find the ratio of sides / =3.8/7.6 = 38/76 = 1/2 / =(3 3)/(6 3)=1/2 / =6/12= 1/2 Thus, / = / = / (Using SSS similarity criterion) So , ~ Now we need to find P We know that corresponding angles of similar triangle are equal . Therefore, C = P In A + B + C = 180 80 +60 + C = 180 140 + C = 180 C = 180 140 C = 40 Thus, C = 40 Hence , P = C = 40