Question

Construct a right-angled triangle whose base is 5 cm and sum of its hypotenuse and other side is 10 cm. Construct another triangle whose sides are 1.4 times the corresponding sides of the previously drawn triangle. Give the justification of the construction.

Answer 1

Let us assume ΔABC that is right-angled at B , with base BC = 5 cm and AC + AB = 10 cm.

A ΔA'BC' whose sides are 1.4 = 7/5 times of ΔABC.

Steps of construction :

1. Draw a line segment BC of length 5 cm.

2. At B, draw ∠XBC = 90º. Taking B as centre and radius as 10 cm, draw an arc that intersects the ray BX at Y.

3. Join CY and draw its perpendicular bisector to intersect BY at A. Join AC.

4. Draw a ray BZ making a acute angle with line segment BC on the opposite side of vertex A.

5. Locate 7 points B₁,B₂,B₃,B₄,B₅,B₆ and B₇ on BZ such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B_7..

6. Join CB₅ and draw a line C'B₇ parallel to CB₅ to intersect extended line segment BC at point C'.

7. Draw a line through C' parallel to AC intersecting the ray BX at A'.

ΔA'BC' is the required triangle.

Justification :

The construction can be justified by proving that:

A'B = 7/5 AB,BC' = 7/5 BC,A'C' = 7/5 AC