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Ex 11.1

Ex 11.1, 1
Deleted for CBSE Board 2023 Exams

Ex 11.1, 2 Important Deleted for CBSE Board 2023 Exams

Ex 11.1, 3 Deleted for CBSE Board 2023 Exams

Ex 11.1, 4 Important Deleted for CBSE Board 2023 Exams You are here

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 Aug. 5, 2021 by Teachoo

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Maths Crash Course - Live lectures + all videos + Real time Doubt solving!

Ex 11.1, 4 Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1 1/2 times the corresponding sides of the isosceles triangle. Letβs first draw a rough diagram Let Ξ ABC be the isosceles triangle with AB = AC Base = BC = 8 cm Altitude = AD = 4 cm We first need to draw Ξ ABC, In isosceles triangle, median and altitude are the same So, AD is the median and altitude β΄ AD is the perpendicular bisector of BC So, to draw Ξ ABC, we first draw base BC, then its perpendicular bisector, and then mark 4 cm in perpendicular bisector to mark point A Letβs construct it Steps to draw Ξ ABC Draw base BC of side 8 cm Draw perpendicular bisector of BC Let perpendicular bisector intersect BC at point D Taking point D as center, and 4 cm radius, draw an arc intersecting perpendicular bisector at point A Join AB, and AC β΄ Ξ ABC is the required triangle Now, we need to make a triangle which is 1 1/2 times its size β΄ Scale factor = 1 1/2 = 3/2 > 1 Steps of construction Draw any ray BX making an acute angle with BC on the side opposite to the vertex A. Mark 3 (the greater of 3 and 2 in 3/2 ) points π΅_1, π΅_2, π΅_3 on BX so that γπ΅π΅γ_1=π΅_1 π΅_2=π΅_2 π΅_3 Join π΅_2 πΆ (2nd point as 2 is smaller in 3/2) and draw a line through π΅_3 parallel to π΅_2 πΆ, to intersect BC extended at Cβ². Draw a line through Cβ² parallel to the line AC to intersect AB extended at Aβ². Thus, Ξ Aβ²BCβ² is the required triangle Justification Since scale factor is 3/2, we need to prove (π¨^β² π©)/π¨π©=(π¨^β² πͺ^β²)/π¨πͺ=(π©πͺ^β²)/π©πͺ =π/π By construction, BC^β²/π΅πΆ=(π΅π΅_3)/(π΅π΅_2 )=3/2 Also, AβCβ is parallel to AC So, they will make the same angle with line BC β΄ β AβCβB = β ACB Now, In Ξ AβBCβ and Ξ ABC β B = β B β AβCβB = β ACB Ξ AβBCβ βΌ Ξ ABC Since corresponding sides of similar triangles are in the same ratio (π΄^β² π΅)/π΄π΅=(π΄^β² πΆ^β²)/π΄πΆ=(π΅πΆ^β²)/π΅πΆ So, (π¨^β² π©)/π¨π©=(π¨^β² πͺ^β²)/π¨πͺ=(π©πͺ^β²)/π©πͺ =π/π Thus, our construction is justified