NCERT Class 9 Solutions: Quadrilaterals (Chapter 8) Exercise 8.1 – Part 2

Vertically opposite angles

Give vertically opposite angle it is opposite each other when two lines cross ,angle a and b is opposite angle

Give Vertically Opposite Angle

Give vertically opposite angle it is opposite each other when two lines cross ,angle a and b is opposite angle

Vertically Opposite Angles are the angles opposite each other when two lines cross. "Vertical" in this case means they share the same Vertex (or corner point), not the usual meaning of up-down.

Q-3 Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus..

Give quadrilateral ABCD,OA=OC,OB=OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90°

Give Quadrilateral ABCD,∠OCD = ∠ODA = 90°

Give quadrilateral ABCD,OA=OC,OB=OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90°

  • Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.Given, OA=OC,OB=ODandAOB=BOC=OCD=ODA=90°

  • To show,

    ABCD is parallelogram and AB=BC=CD=AD

  • Proof,

    In AOB and COB ,

    OA=OC (Given)

    AOB=COB (Each =90° , Opposite sides of a parallelogram are equal)

    OB=OB (Common)

    Therefore, AOBCOB, by Side-Angle-Side congruence condition.

  • Thus , AB=BC by Corresponding Parts of Congruent Triangles

  • Similarly we can prove, AB=BC=CD=AD

  • Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.Thus, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.

Q-4 Show that the diagonals of a square are equal and bisect each other at right angles.

Solution:

Give square ABCD and diagonals AC and BD it is intersect each other at O ,OA=BD,AO=OC and ∠AOB = 90°

Give Square ABCD and Diagonals AC and BD

Give square ABCD and diagonals AC and BD it is intersect each other at O ,OA=BD,AO=OC and ∠AOB = 90°

  • Let ABCD be a square and its diagonals AC and BD intersect each other at O.To show,

    AC=BD , AO=OCandAOB=90°

  • Proof,In ΔABCandΔBAD , BC=BA (Common) ABC=BAD=90° AC=AD (Given)

  • Therefore, ΔABCΔBAD by Side-Angle-Side congruence condition.

  • Thus, AC=BD by Corresponding Parts of Congruent Triangles. Therefore, diagonals are equal.

  • Now,In ΔAOBandΔCOD, BAO=DCO (Alternate interior angles) AOB=COD (Vertically opposite triangle) AB=CD (Given)

  • Therefore ,ΔAOBΔCOD by Angle-Angle-Side congruence condition.

  • Thus, AO=CO by Corresponding Parts of Congruent Triangles (Diagonal bisect each other.)

  • Now, InΔAOBandΔCOB , OB=OB (Given) AO=CO (diagonals are bisected) AB=CB (Sides of the square)

  • Therefore, ΔAOBΔCOB by SSS congruence condition.also, AOB=COB AOB+COB=180° (Linear pair)

  • Thus, AOB=COB=90° (Diagonals bisect each other at right angles)

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