Quadrilaterals and Parallelograms Exercise 9.1 |

Quadrilaterals and Parallelograms Exercise 9.2 |

**Q.1: Three angles of a quadrilateral measure 56°, 115°, 84°. Find the measure of the fourth angle. **

**Sol: **

Let, the measure of the fourth angle be

Since the sum of the angle of a quadrilateral is

56 + 115 + 84 + X = 360

Hence, the measure of the fourth angle is

**Q.2: The angles of a quadrilateral are in the ratio 2 : 4 : 5 : 7. Find the angles.**

**Sol:**

Let,

Then

Since the sum of the angles of a quadrilateral is

2X + 4X + 5X + 7X =

Therefore, Angle A =

Angle B =

Angle C =

Angle D =

**Q.3: In the adjoining figure, ABCD is a trapezium in which AB ∥ DC. If ∠ A = 55°and ∠ B = 70°, find ∠ C and ∠ D.**

**Sol:**

We have

Angle A and Angle D are the interior angles on the same side of the transversal line AD, whereas angle B and Angle C are the same interior angles on the same side of transversal line BC.

Now, Angle A + Angle D =

Therefore,

Again,

Therefore,

**Q.4: In the adjoining figure, ABCD is a square and ΔEDC is an equilateral**

**triangle. Prove that (i) AB = BE, (ii) ∠ DAE = 15°.**

**Sol: **

ABCD is a square in which AB = BC = CD = DA.

**To prove:** AE = BE and

**Proof:** In

AD = BC [Sides of a square]

DE = EC [Sides of an equilateral triangle]

i.e., AE = BE

Now,

DA = DC [Sides of a square]

DC = DE [Sides of an equilateral triangle]

So, DA = DE

i.e.,

=

=

=

**Q.5: In the adjoining figure, BM ⊥ AC and DN ⊥ AC. If BM = DN, prove that AC bisects BD.**

**Sol: **

A quadrilateral ABCD, in which BM

**To prove:** AC bisects BD, or DO = BO

**Proof: **Let, AC and BD intersect at O

Now, in

Also, DN = BM (Given)

i.e.,

Therefore, OD = OB (CPCT)

Hence, AC bisects BD

**Q.6: In the given figure, ABCD is a quadrilateral in which AB = AD and BC = DC. Prove that:**

**(i) AC bisects ∠ A and ∠ C, (ii) BE = DE, (iii) ∠ ABC = ∠ADC.**

**Sol:**

**Given:** ABCD is a quadrilateral in which AB = AD and BC = DC

**(i)** In

AB = AD (Given)

BC = DC (Given)

AC is common

i.e.,

Therefore,

Thus, AC bisects

**(ii) **Now, in

AB = AD (Given)

AE is common.

Therefore,

**(iii)**

Therefore, Angle ABC = Angle ADC (By CPCT)

**Q.7: In the given figure, ABCD is a square and ∠PQR = 90°. If PB = QC = DR, prove that (i) QB = RC, (ii) PQ = QR (iii) ∠ QPR = 45°**

**Sol:**

Let, ABCD be a parallelogram

∠ A = ∠ C and ∠ B = ∠ D (Opposite angles)

Let ∠ A be the smallest angle whose measure is x°

∴ ∠B = (2x – 30)°

Now, ∠ A + ∠ B = 180° (Adjacent angles are supplementary)

∴ ∠B = 2 x 70°- 30° = 110°

Hence, ∠ A = ∠ C = 70°; ∠ B = ∠ D = 110°

**Q.8: If O is a point within a quadrilateral ABCD, show that OA + OB + OC + OD > AC + BD.**

**Sol:**

Let, ABCD be a quadrilateral whose diagonals are AC and BD and O is any point within the quadrilateral

Join O with A, B, C, and D

We know that the sum of any sides of a triangle is greater than the third side

So, in

Also,

Adding these inequalities, we get:

(OA + OC) + (OB + OD) > (AC + BD)

**Q.9: In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals. Prove that: (i) AB + BC + CD + DA > 2AC (ii) AB + BC + CD > DA (iii) AB + BC + CD + DA > AC + BD**

**Sol:**

**Given:** ABCD is a quadrilateral and AC is one of its diagonals

**(i)** In

In

Adding inequalities (1) and (2), we get:

AB + BC + CD + DA > 2AC

**(ii)** In

AB + BC > AC . . . . . . . . (1)

We also know that the length of each side of a triangle is greater than the positive difference of the the other two sides.

In

AC >

**From (1) and (2), we have:**

AB + BC >

**(iii)** In

In

Adding these inequalities, we get:

2 (AB + BC + CD + DA) > 2 (AC + BD)

**Q.10: Prove that the sum of all angles of a quadrilateral is 360°.**

Sol: Let ABCD be a quadrilateral and ∠1, ∠2, ∠3 and ∠4 are its four angles as shown in the figure.

Join BD which divides ABCD in two triangles,

In

∠ 1 + ∠ 2 + ∠ A = 180° . . . . . . . . (i)

In

∠ 3 + ∠ 4 + ∠ C = 180° . . . . . . . (ii)

On adding (i) and (ii), we get:

(∠1 + ∠3) + ∠A + ∠C + (∠ 4 + ∠2) = 360°

∴ ∠ A + ∠ C + ∠ B + ∠ D = 360°

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