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Chapter 6 Class 10 Triangles
Example 8 Important
Example 14 Important Deleted for CBSE Board 2023 Exams
Example 10 Important Deleted for CBSE Board 2023 Exams
Theorem 6.1 - Basic Proportionality Theorem (BPT) Important
Theorem 6.7 Important Deleted for CBSE Board 2023 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
Ex 6.4, 1 Important Deleted for CBSE Board 2023 Exams
Ex 6.4, 3 Important Deleted for CBSE Board 2023 Exams
Ex 6.4, 5 Important Deleted for CBSE Board 2023 Exams
Ex 6.5, 2 Important Deleted for CBSE Board 2023 Exams
Ex 6.5, 3 Important Deleted for CBSE Board 2023 Exams
Ex 6.5, 8 Important Deleted for CBSE Board 2023 Exams
Ex 6.5, 11 Important Deleted for CBSE Board 2023 Exams
Ex 6.5, 12 Important Deleted for CBSE Board 2023 Exams
Ex 6.5, 15 Important Deleted for CBSE Board 2023 Exams
Chapter 6 Class 10 Triangles
Last updated at May 29, 2018 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