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Construct an angle of 90° at the initial point of a given ray and justify the construction. Q.1

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How i solve the question of class 9th ncert math of Constructions chapter of exercise 11.1 of question no  1. I think it is very important question of class 9th give me the tricky way for solving this question Construct an angle of 90° at the initial point of a given ray and justify the construction

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  1. 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.

    Ncert solutions class 9 chapter 11-1

    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:

    Ncert solutions class 9 chapter 11-2

    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.

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