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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.

Given base and altitude of an isosceles triangle.

We make the triangle by drawing its perpendicular bisector, as in isosceles triangle, altitude and perpendicular bisector are the same

  1. Draw a line segment AB of 8 cm. Draw arcs of same radius on both sides of the line segment while taking point A and B as its centre. Let these arcs intersect each other at O and O'. Join OO'. Let OO' intersect AB at D.
  2. Taking D as centre, draw an arc of 4 cm radius which cuts the extended line segment OO' at point C.
     An isosceles ΔABC is formed, having CD  (altitude) 4 cm and AB (base) 8 cm.
  3. Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C.
  4. Locate 3 points (as 3 is greater between 3 and 2) A_1, A_2, and A_3 on AX such that 〖AA〗_1 = A_1 A_2 = A_2 A_3.
  5. Join 〖BA〗_2 and draw a line through A_3 parallel to 〖BA〗_2 to intersect extended line segment AB at point B'.
  6. Draw a line through B' parallel to BC intersecting the extended line segment AC at C'.

ΔAB'C' is the required triangle.

Justification

Here,

 (AB^′)/AB=(AA_3)/(AA_2 )  = 3/2

Also, A’C’ is parallel to AC

So, the will make the same angle with line BC

∴ ∠ AB’C’ = ∠ ABC

Now,

In Δ AB’C’ and Δ ABC

        ∠ A = ∠ A

 ∠ AB’C’ = ∠ ABC

Δ A’BC’ ∼ Δ ABC

Since corresponding sides of
similar triangles are in the same ratio

 (AB^′)/AB=(B^′ C^′)/BC=(AC^′)/AC

So, (AB^′)/AB=(B^′ C^′)/BC=(AC^′)/AC =3/2

This justifies the construction.

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  1. Chapter 11 Class 10 Constructions
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Transcript

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. Given base and altitude of an isosceles triangle. We make the triangle by drawing its perpendicular bisector, as in isosceles triangle, altitude and perpendicular bisector are the same Draw a line segment AB of 8 cm. Draw arcs of same radius on both sides of the line segment while taking point A and B as its centre. Let these arcs intersect each other at O and O'. Join OO'. Let OO' intersect AB at D. Taking D as centre, draw an arc of 4 cm radius which cuts the extended line segment OO' at point C. An isosceles ΔABC is formed, having CD (altitude) as 4 cm and AB (base) as 8 cm. Draw a ray AX making an acute angle with line segment AB on the opposite side of vertex C. Locate 3 points (as 3 is greater between 3 and 2) 𝐴_1, 𝐴_2, and 𝐴_3 on AX such that 〖𝐴𝐴〗_1 = 𝐴_1 𝐴_2 = 𝐴_2 𝐴_3. Join 〖𝐵𝐴〗_2 and draw a line through 𝐴_3 parallel to 〖𝐵𝐴〗_2 to intersect extended line segment AB at point B'. Draw a line through B' parallel to BC intersecting the extended line segment AC at C'. ΔAB'C' is the required triangle. Justification Here, (𝐴𝐵^′)/𝐴𝐵=(𝐴𝐴_3)/(𝐴𝐴_2 ) = 3/2 Also, A’C’ is parallel to AC So, the will make the same angle with line BC ∴ ∠ AB’C’ = ∠ ABC Now, In Δ AB’C’ and Δ ABC ∠ A = ∠ A ∠ AB’C’ = ∠ ABC Δ A’BC’ ∼ Δ ABC Since corresponding sides of similar triangles are in the same ratio 𝐴𝐵/𝐴𝐵′=𝐵𝐶/(𝐵^′ 𝐶^′ )=𝐴𝐶/(𝐴𝐶^′ ) So, 𝑨𝑩/𝑨𝑩′=𝑩𝑪/(𝑩^′ 𝑪^′ )=𝑨𝑪/(𝑨𝑪^′ )=𝟑/𝟐 This justifies the construction.

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CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 7 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
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