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Ex 11.1

Ex 11.1, 1
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Ex 11.1, 2 Important Deleted for CBSE Board 2023 Exams

Ex 11.1, 3 Deleted for CBSE Board 2023 Exams You are here

Ex 11.1, 4 Important Deleted for CBSE Board 2023 Exams

Ex 11.1, 5 Important Deleted for CBSE Board 2023 Exams

Ex 11.1, 6 Deleted for CBSE Board 2023 Exams

Ex 11.1, 7 Important Deleted for CBSE Board 2023 Exams

Last updated at March 23, 2023 by Teachoo

Ex 11.1, 3 Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle. Letβs first construct Ξ ABC with sides 5 cm, 6 cm, 7 cm Steps to draw Ξ ABC Draw base AB of side 5 cm With A as center, and 6 cm as radius, draw an arc With B as center, and 7 cm as radius, draw an arc 3. Let C be the point where the two arcs intersect. Join AC & BC Thus, Ξ ABC is the required triangle Now, letβs make a similar triangle with Scale factor = 7/5 Steps of construction Draw any ray AX making an acute angle with AB on the side opposite to the vertex C. Mark 7 (the greater of 7 and 5 in 7/5 ) points π΄_1, π΄_2, π΄_3, π΄_4, π΄_5, π΄_6, π΄_7 on AX so that γπ΄π΄γ_1=π΄_1 π΄_2=π΄_2 π΄_3 β¦ and so on Join π΄_5 π΅ (5th point as 5 is smaller in 7/5) and draw a line through π΄_7 parallel to π΄_5 π΅, to intersect AB extended at Bβ². Draw a line through Bβ² parallel to the line BC to intersect AC extended at Cβ². Thus, Ξ ABβCβ² is the required triangle Justification Since scale factor is 7/5, we need to prove (π¨π©^β²)/π¨π©=(π¨πͺ^β²)/π¨πͺ=(π©^β² πͺ^β²)/π©πͺ = π/π By construction, (π΄B^β²)/π΄π΅=(π΄π΄_7)/(π΄π΄_5 )= 7/5 Also, BβCβ is parallel to BC So, the will make the same angle with line AB β΄ β ABβCβ = β ABC Now, In Ξ ABβCβ and Ξ ABC β A = β A β ABβCβ = β ABC Ξ ABβCβ βΌ Ξ ABC Since corresponding sides of similar triangles are in the same ratio (π΄π΅^β²)/π΄π΅=(π΄πΆ^β²)/π΄πΆ=(π΅^β² πΆ^β²)/π΅πΆ So,(π¨π©^β²)/π¨π©=(π¨πͺ^β²)/π¨πͺ=(π©^β² πͺ^β²)/π©πͺ =π/π. Thus, our construction is justified