Class Nine Maths NCERT Solutions Chapter 8
Chapter 8: Quadrilaterals (NCERT Solutions)
TEXTBOOK’S EXERCISE – 8.1
1. The angles of a quadrilateral are in the ratio 3: 5: 9:13. Find all the angles of the quadrilateral.
Sol. Suppose the measures of four angles are 3x, 5x, 9x and 13x.
∴ 3x + 5x + 9x + 13x = 360° [Angle sum property of a quadrilateral]
⇒ 30x = 360° ⇒ x = 360/ 30° = 12°
⇒ 3x = 3 × 12° = 36°, 5x = 5 × 12° = 60°,
9x = 9 × 12° = 108°, 13x = 13 × 12° = 156°
∴ the angles of the quadrilateral are 36°, 60°, 108° and 156°.
2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Sol. Given: ABCD is a parallelogram in which AC = BD.
To Prove: ABCD is a rectangle.
Proof : In ΔABC and ΔABD
AB = AB [Common]
BC = AD [Opposite sides of a parallelogram]
AC = BD [Given]
∴ ΔABC ≅ ΔBAD [SSS congruence]
∠ABC = ∠BAD ………………(i) [CPCT]
Since, ABCD is a parallelogram, thus,
∠ABC + ∠BAD = 180° ……………(ii) [Consecutive interior angles]
∠ABC + ∠ABC = 180°
∴ 2∠ABC = 180° [From (i) and (ii)]
⇒ ∠ABC = ∠BAD = 90°
This shows that ABCD is a parallelogram one of whose angle is 90°.
Hence, ABCD is a rectangle. Proved.
3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Sol. Given: A quadrilateral ABCD, in which diagonals AC and BD bisect each other at right angles.
To Prove: ABCD is a rhombus.
Proof: In ΔAOB and ΔBOC
AO = OC
[Diagonals AC and BD bisect each other]
∠AOB = ∠COB [Each = 90°]
BO = BO [Common]
∴ ΔAOB ≅ ΔBOC [SAS congruence]
AB = BC ……….…(i) [CPCT]
Since ABCD is a quadrilateral in which AB = BC [From (i)]
Hence, ABCD is a rhombus.
[Since if the diagonals of a quadrilateral bisect each other, then it is a parallelogram and opposite sides of a parallelogram are equal] Proved.
4. Show that the diagonals of a square are equal and bisect each other at right angles.
Sol. Given: ABCD is a square in which AC and BD are diagonals.
To Prove : AC = BD and AC bisects BD at right angles, i.e. AC ⊥ BD. AO = OC, OB = OD
Proof: In ΔABC and ΔBAD,
AB = AB [Common]
BC = AD [Sides of a square]
∠ABC = ∠BAD = 90° [Angles of a square]
∴ ΔABC ≅ ΔBAD [SAS congruence]
⇒ AC = BD [CPCT]
Now in ΔAOB and ΔCOD,
AB = DC [Sides of a square]
∠AOB = ∠COD [Vertically opposite angles]
∠OAB = ∠OCD [Alternate angles]
∴ ΔAOB ≅ ΔCOD [AAS congruence]
AO = CO [CPCT]
Similarly by taking ΔAOD and ΔBOC, we can show that OB = OD.
In ΔABC, ∠BAC + ∠BCA = 90° [Since ∠B = 90°]
⇒ 2∠BAC = 90° [∠BAC = ∠BCA, as BC = AD]
⇒ ∠BCA = 45° or ∠BCO = 45°
Similarly, ∠CBO = 45°
In ΔBCO, ∠BCO + ∠CBO + ∠BOC = 180°
⇒ 90° + ∠BOC = 180° ⇒ ∠BOC = 90°
⇒ BO ⊥ OC ⇒ BO ⊥ AC
Hence, AC = BD, AC ⊥ BD, AO = OC and OB = OD. Proved.
5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Sol. Given: A quadrilateral ABCD, in which diagonals AC and BD are equal and bisect each other at right angles,
To Prove: ABCD is a square.
Proof : Since ABCD is a quadrilateral whose diagonals bisect each other, so it is a parallelogram. Also, its diagonals bisect each other at right angles, therefore, ABCD is a rhombus.
⇒ AB = BC = CD = DA [Sides of a rhombus]
In ΔABC and ΔBAD, we have
AB = AB [Common]
BC = AD [Sides of a rhombus]
AC = BD [Given]
∴ ΔABC ≅ ΔBAD [SSS congruence]
∴ ∠ABC = ∠BAD [CPCT]
But, ∠ABC + ∠BAD = 180° [Consecutive interior angles]
∠ABC = ∠BAD = 90°.
∠A = ∠B = ∠C = ∠D = 90° [Opposite angles of a | gm]
⇒ ABCD is a rhombus whose angles are of 90° each.
Hence, ABCD is a square. Proved.