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 \(X^{\circ}\)

Since the sum of the angle of a quadrilateral is \(360^{\circ}\), we have:

56 + 115 + 84 + X = 360

\(\Rightarrow\) 255 + X = 360

\(\Rightarrow\) X = 105

Hence, the measure of the fourth angle is \(105^{\circ}\)

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

**Sol:**

Let, \(\angle A\) = \(2X^{\circ}\)

Then \(\angle B = (4X)^{\circ}\)

\(\angle C = (5X)^{\circ}\) and

\(\angle D = (7X)^{\circ}\)Since the sum of the angles of a quadrilateral is \(360^{\circ}\), we have:

2X + 4X + 5X + 7X = \(360^{\circ}\)

\(\Rightarrow\) 18X = \(360^{\circ}\)

\(\Rightarrow\) X = \(20^{\circ}\)

Therefore, Angle A = \(40^{\circ}\)

Angle B = \(80^{\circ}\)

Angle C = \(100^{\circ}\)

Angle D = \(140^{\circ}\)

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

**Sol:**

We have \(AB\parallel DC\)

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 = \(180^{\circ}\)

\(\Rightarrow\) \(\angle D\) = \(180^{\circ}\) – \(\angle A\)

Therefore, \(\angle D\) = \(180^{\circ}\) – \(55^{\circ}\) = \(125^{\circ}\)

Again, \(\angle B\) + \(\angle C\) = \(180^{\circ}\)

\(\Rightarrow\) \(\angle C\) = \(180^{\circ}\) – \(\angle B\)

Therefore, \(\angle C\) = \(180^{\circ}\) – \(70^{\circ}\) = \(110^{\circ}\)

**Q.4: In the adjoining figure, ABCD is a square and \(\Delta EDC\) is an equilateral** **triangle. Prove that (i) AB = BE, (ii) ∠ DAE = 15°.**

**Sol: **

ABCD is a square in which AB = BC = CD = DA. \(\Delta EDC\) is an equilateral triangle in which ED = EC = DC

\(\angle EDC\) = \(\angle DEC\) = \(\angle DCE\) = \(60^{\circ}\).

**To prove:** AE = BE and \(\angle DAE\) = \(15^{\circ}\)

**Proof:** In \(\Delta ADE\) and \(\Delta BCE\), we have:

AD = BC [Sides of a square]

DE = EC [Sides of an equilateral triangle]

\(\angle ADE\) = \(\angle BCE\) = \(90^{\circ}\) + \(60^{\circ}\) = \(150^{\circ}\)

\(\Delta ADE\) \(\cong\) \(\Delta BCE\)

i.e., AE = BE

Now, \(\angle ADE\) = \(150^{\circ}\)

DA = DC [Sides of a square]

DC = DE [Sides of an equilateral triangle]

So, DA = DE

\(\Delta ADE\) and \(\Delta BCE\) are isosceles triangles.

i.e., \(\angle DAE\) = \(\angle DEA\)

= \(\frac{1}{2}(180^{\circ}-150^{\circ})\)

= \(\frac{30^{\circ}}{2}\)

= \(15^{\circ}\)

**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 \(\perp\) AC and DN \(\perp\) AC and BM = DN

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

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

Now, in \(\Delta OND\) and \(\Delta OMB\), we have:

\(\angle OND\) = \(\angle OMB\) (\(90^{\circ}\) each)

\(\angle DON\) = \(\angle BOM\) (Vertically opposite angles)

Also, DN = BM (Given)

i.e., \(\Delta OND\) \(\cong\) \(\Delta OMB\) (AAS congruence rule)

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 \(\Delta ABC\) and \(\Delta ADC\), we have:

AB = AD (Given)

BC = DC (Given)

AC is common

i.e., \(\Delta ABC\) \(\cong\) \(\Delta ADC\) (SSS Congruence rule)

Therefore, \(\angle BAC\) = \(\angle DAC\) and

\(\angle BCA\) = \(\angle DCA\) (By CPCT)

Thus, AC bisects \(\angle A\) and \(\angle C\)

**(ii) **Now, in \(\Delta ABE\) and \(\Delta ADE\), we have:

AB = AD (Given)

\(\angle BAE\) = \(\angle DAE\) (Proven above)

AE is common.

Therefore, \(\Delta ABE\) \(\cong\) \(\Delta ADE\) (SAS congruence rule)

\(\Rightarrow\) BE = DE (By CPCT)

**(iii)** \(\Delta ABC\) \(\cong\) \(\Delta ADC\) (Proven above)

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)

\(\Rightarrow\) x + 2x – 30° = 180°

\(\Rightarrow\) 3x = 210°

\(\Rightarrow\) x = 70°

∴ ∠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 \(\Delta AOC\), OA + OC > AC

Also, \(\Delta BOD\), OB + OD > BD

Adding these inequalities, we get:

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

\(\Rightarrow\) OA + OB + OC + 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 \(\Delta ABC\), AB + BC > AC . . . . . . . . (1)

In \(\Delta ACD\), CD + DA > AC . . . . . . . . . . (2)

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

AB + BC + CD + DA > 2AC

**(ii)** In \(\Delta ABC\), we have :

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 \(\Delta ACD\), we have:

AC > \(\left | DA – CD \right |\) . . . . . . . (2)

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

AB + BC > \(\left | DA – CD \right |\)

\(\Rightarrow\) AB + BC + CD > DA

**(iii)** In \(\Delta ABC\), AB + BC > AC

In \(\Delta ACD\), CD + DA > AC

\(\Delta BCD\), BC + CD > BD

\(\Delta ABD\), DA + AB > BD

Adding these inequalities, we get:

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

\(\Rightarrow\) (AB + BC + CD + DA) > (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, \(\Delta ABD\) and \(\Delta BCD\).

In \(\Delta ABD\), we have:

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

In \(\Delta BCD\), we have:

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

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

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

\(\Rightarrow\) ∠ A + ∠ C + ∠ B + ∠ D = 360° ( Since,∠ 1 + ∠ 3 = ∠ B; ∠ 4 + ∠ 2 = ∠ D)

∴ ∠ A + ∠ C + ∠ B + ∠ D = 360°

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