#### 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. **