web analytics

Slide27.JPG

Slide28.JPG
Slide29.JPG Slide30.JPG Slide31.JPG

View
  1. Chapter 11 Class 10 Constructions
  2. Serial order wise
Ask Download

Transcript

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 Draw a line segment AB = 4 cm. Draw a ray RA making 90° with it. Taking A as center and 3 cm as radius, draw an arc intersecting the ray RA at C. Join BC. ΔABC is the required triangle. 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. Join A_3B. Draw a line through A_5 parallel to A_3B intersecting extended line segment AB at B'. Through B', draw a line parallel to BC intersecting extended line segment AC at C'. ΔAB'C' is the required triangle. Justification Here,  (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 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^′ )=5/3 This justifies the construction.

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
Jail