Construct a ∆ABC in which AB = 6.5 cm, ∠B = 60° and BC = 5.5 cm. Also construct a triangle AB’C’ similar to ∆ABC, whose each side is 3 /2 times the corresponding sides of ∆ABC.
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I take base BC = 5.5 cm. so ∆ formed will be A'BC'.
SOLUTION:
Given , a right angled ∆ ABC , in which BC =5.5 cm , AB= 6.5 cm , ∠B = 60°
STEPS OF CONSTRUCTION:
1. Draw a line segment BC =5.5 cm.
2. At B draw ∠B = 60° and cut off BA =6.5 cm from it.
3. Join AC. Thus ∆ABC is a given right angled Triangle.
4. Now from B draw a ray BX making an acute ∠CBX on the side opposite to the vertex A.
5. Mark 3 points B1,B2,B3, on BY such that BB1 = B1B2 = B2B3.
6.Join B2C and from B3 draw B3C’ || B2C intersecting BC at C.
7. From point C’ draw C’A’ || CA intersecting AB at A’. Thus , ∆A’BC’ is the required triangle whose sides are 3/2 of the corresponding sides of ∆ABC.
HOPE THIS WILL HELP YOU....
SOLUTION:
Given , a right angled ∆ ABC , in which BC =5.5 cm , AB= 6.5 cm , ∠B = 60°
STEPS OF CONSTRUCTION:
1. Draw a line segment BC =5.5 cm.
2. At B draw ∠B = 60° and cut off BA =6.5 cm from it.
3. Join AC. Thus ∆ABC is a given right angled Triangle.
4. Now from B draw a ray BX making an acute ∠CBX on the side opposite to the vertex A.
5. Mark 3 points B1,B2,B3, on BY such that BB1 = B1B2 = B2B3.
6.Join B2C and from B3 draw B3C’ || B2C intersecting BC at C.
7. From point C’ draw C’A’ || CA intersecting AB at A’. Thus , ∆A’BC’ is the required triangle whose sides are 3/2 of the corresponding sides of ∆ABC.
HOPE THIS WILL HELP YOU....
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Hi there!
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