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Ex 11.1, 7

Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.

Given sides other than hypotenuse in a right angled triangle.

∴ Both sides will be perpendicular to each other

Steps of construction

  1. Draw a line segment AB = 4 cm. Draw a ray RA making 90° with it.
  2. Taking A as center and 3 cm as radius, draw an arc intersecting the ray RA at C.
  3. Join BC. ΔABC is the required triangle.
  4. Locate 5 points (as 5 is greater in 5 and 3), A_1, A_2, A_3, A_4, A_5, on line segment AX such that 〖AA〗_1 = A_1 A_2 = A_2 A_3 = A_3 A_4= A_4 A_5.
  5. Join A_3B. Draw a line through A_5 parallel to A_3B intersecting extended line segment AB at B'.
  6. Through B', draw a line parallel to BC intersecting extended line segment AC at C'.

ΔAB'C' is the required triangle.



 (AB^′)/AB=(AA_5)/(AA_3 )  = 5/3

Also, A’C’ is parallel to AC

So, the will make the same angle with line BC

∴ ∠ AB’C’ = ∠ ABC


In Δ AB’C’ and Δ ABC

       ∠ A = ∠ A

 ∠ AB’C’ = ∠ 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^′ ) =5/3

This justifies the construction.


  1. Chapter 11 Class 10 Constructions
  2. Concept wise
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CA Maninder Singh is a Chartered Accountant for the past 7 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .