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Ex 11.1, 4 - Construct an isosceles triangle whose base is 8 cm

Ex 11.1, 4 - Chapter 11 Class 10 Constructions - Part 2
Ex 11.1, 4 - Chapter 11 Class 10 Constructions - Part 3
Ex 11.1, 4 - Chapter 11 Class 10 Constructions - Part 4
Ex 11.1, 4 - Chapter 11 Class 10 Constructions - Part 5
Ex 11.1, 4 - Chapter 11 Class 10 Constructions - Part 6
Ex 11.1, 4 - Chapter 11 Class 10 Constructions - Part 7
Ex 11.1, 4 - Chapter 11 Class 10 Constructions - Part 8
Ex 11.1, 4 - Chapter 11 Class 10 Constructions - Part 9

 

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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. Let’s first draw a rough diagram Let Δ ABC be the isosceles triangle with AB = AC Base = BC = 8 cm Altitude = AD = 4 cm We first need to draw Δ ABC, In isosceles triangle, median and altitude are the same So, AD is the median and altitude ∴ AD is the perpendicular bisector of BC So, to draw Δ ABC, we first draw base BC, then its perpendicular bisector, and then mark 4 cm in perpendicular bisector to mark point A Let’s construct it Steps to draw Δ 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 ∴ Δ ABC is the required triangle Now, we need to make a triangle which is 1 1/2 times its size ∴ 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 𝐵_1, 𝐵_2, 𝐵_3 on BX so that 〖𝐵𝐵〗_1=𝐵_1 𝐵_2=𝐵_2 𝐵_3 Join 𝐵_2 𝐶 (2nd point as 2 is smaller in 3/2) and draw a line through 𝐵_3 parallel to 𝐵_2 𝐶, to intersect BC extended at C′. Draw a line through C′ parallel to the line AC to intersect AB extended at A′. Thus, Δ A′BC′ is the required triangle Justification Since scale factor is 3/2, we need to prove (𝑨^′ 𝑩)/𝑨𝑩=(𝑨^′ 𝑪^′)/𝑨𝑪=(𝑩𝑪^′)/𝑩𝑪 =𝟑/𝟐 By construction, BC^′/𝐵𝐶=(𝐵𝐵_3)/(𝐵𝐵_2 )=3/2 Also, A’C’ is parallel to AC So, they will make the same angle with line BC ∴ ∠ A’C’B = ∠ ACB Now, In Δ A’BC’ and Δ ABC ∠ B = ∠ B ∠ A’C’B = ∠ ACB Δ A’BC’ ∼ Δ ABC Since corresponding sides of similar triangles are in the same ratio (𝐴^′ 𝐵)/𝐴𝐵=(𝐴^′ 𝐶^′)/𝐴𝐶=(𝐵𝐶^′)/𝐵𝐶 So, (𝑨^′ 𝑩)/𝑨𝑩=(𝑨^′ 𝑪^′)/𝑨𝑪=(𝑩𝑪^′)/𝑩𝑪 =𝟑/𝟐 Thus, our construction is justified

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.