ex 11,1class 9

. Construct an angle of 90° at the initial point of a given ray and justify the construction.

Construction Procedure:

To construct an angle 90°, follow the given steps:

1. Draw a ray OA

2. Take O as a centre with any radius, draw an arc DCB is that cuts OA at B.

3. With B as a centre with the same radius, mark a point C on the arc DCB.

4. With C as a centre and the same radius, mark a point D on the arc DCB.

5. Take C and D as centre, draw two arcs which intersect each other with the same radius at P.

6. Finally, the ray OP is joined which makes an angle 90° with OP is formed.

Justification

To prove ∠POA = 90°

In order to prove this, draw a dotted line from the point O to C and O to D and the angles formed are:

From the construction, it is observed that

OB = BC = OC

Therefore, OBC is an equilateral triangle

So that, ∠BOC = 60°.

Similarly,

OD = DC = OC

Therefore, DOC is an equilateral triangle

So that, ∠DOC = 60°.

From SSS triangle congruence rule

△OBC ≅ OCD

So, ∠BOC = ∠DOC [By C.P.C.T]

Therefore, ∠COP = ½ ∠DOC = ½ (60°).

∠COP = 30°

To find the ∠POA = 90°:

∠POA = ∠BOC+∠COP

∠POA = 60°+30°

∠POA = 90°

Hence, justified.

For more click on the given link:

<a href="https://mycbseguide.com/blog/ncert-solutions-for-class-9-maths-exercise-11-1/" ping="/url?sa=t&source=web&rct=j&url=https://mycbseguide.com/blog/ncert-solutions-for-class-9-maths-exercise-11-1/&ved=2ahUKEwjcspLywbHtAhU54zgGHafxBXwQFjALegQICxAC" rel="noopener" target="_blank">NCERT Solutions for Class 9 Maths Exercise 11.1 ...</a>