In the following question how can we draw the altitude?
Construct a triangle ABC with BC=4.2cm, angle B=45 deg. and altitude through A is 2.5 cm. Draw another triangle similar to this triangle with scale factor ½.
Consider the rough figure of required triangle ABC.
Given : ∠B = 45°
AD is the altitude through A.
∴ AD = 2.5 cm and ∠ADB = 90°
In ΔADB,
∠B + ∠ADB + ∠BAD = 180°
⇒ 45° + 90° + ∠BAD = 180°
∠BAD = 45°
This implies, ΔADB is an isosceles triangle.
∴ AD = BD = 2.5 cm
Step of construction:
1. Draw a line segment BC = 4.2 cm.
2. At B construct an angle of 45° i.e. ∠XBC = 45°.
3. From point B, and a radius of 2.5 cm, draw an arc on BC, intersecting at D on BD.
4. Construct perpendicular at D, such that YD intersecting BX at A.
5. Join A to C. ΔABC in the required triangle.
Steps to draw similar triangle sealing factor .
1. Draw and ray BZ, making and acute angle with BC.
2. Locate 3 (= 1 + 2) points B1 , B2 and B3 on BZ so that BB1 = B1B2 = B2B3.
3. Join CB3
4. Through the point B1, draw a line parallel to CB3 to intersect BC at C'.
5. Draw a line through C', parallel to the AC to intersect AB at A'. ΔA'BC' in the required triangle.