Math, asked by ranitde71, 1 year ago

Construct a triangle ABC, with AB=6cm,BC=7cm and angle ABC=60°. Locate by construction the point P such that (i) P is equidistant from B and C.(ii) P is equidistant from AB and BC.(iii) Measure and record the length of PA.

Answers

Answered by sworna1983
15

Answer:


Step-by-step explanation:

Steps of Construction:

1) Draw a line segment AB = 7 cm.

2) Draw angle ∠ABC = 60° with the help of compass.

3) Cut off BC = 8 cm.

4) Join A and C.

5) The triangle ABC so formed is the required triangle.

i) Draw the perpendicular bisector of BC. The point situated on this line will be equidistant from B and C.

ii) Draw the angle bisector of ∠ABC . Any point situated on this angular bisector is equidistant from lines AB and BC.

The point which fulfills the condition required in (i) and (ii) is the intersection point of bisector of line BC and angular bisector of ∠ABC.

P is the required point which is equidistant from AB and AC as well as from B and C.

On measuring the length of line segment PB, it is equal to 4.5 cm.



ranitde71: But AB is 6cm
ranitde71: and BC is 7cm
sworna1983: oh really sorry
Answered by Anonymous
5

Answer:

Steps of Construction:

1) Draw a line segment AB = 7 cm.

2) Draw angle ∠ABC = 60° with the help of compass.

3) Cut off BC = 8 cm.

4) Join A and C.

5) The triangle ABC so formed is the required triangle.

i) Draw the perpendicular bisector of BC. The point situated on this line will be equidistant from B and C.

ii) Draw the angle bisector of ∠ABC . Any point situated on this angular bisector is equidistant from lines AB and BC.

The point which fulfills the condition required in (i) and (ii) is the intersection point of bisector of line BC and angular bisector of ∠ABC.

P is the required point which is equidistant from AB and AC as well as from B and C.

On measuring the length of line segment PB, it is equal to 4.5 cm.

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