A quadrilateral is a polygon which contains four side whereas, a parallelogram is a type of quadrilateral, whose pairs of opposite sides are parallel. One of the most common properties of all quadrilaterals is that sum of all the angles is 360 degrees. Some of the different types of quadrilaterals are:

- Rhombus
- Trapezium
- Square
- Parallelogram
- Rectangle

Parallelograms have some unique properties such as:

- Opposite sides are congruent
- Opposite angles are congruent
- It has adjacent angles which are supplementary
- It has diagonals which bisect each other.

Check out the RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms below:

## RS Aggarwal Class 9 Chapter 9

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

**Sol: **

Let, the measure of the fourth angle be Xâˆ˜

Since the sum of the angle of a quadrilateral is 360âˆ˜, we have:

56 + 115 + 84 + X = 360

â‡’ 255 + X = 360

â‡’ X = 105

Hence, the measure of the fourth angle is 105âˆ˜

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

**Sol:**

Let,Â âˆ A = 2Xâˆ˜

Then âˆ B=(4X)âˆ˜

âˆ C=(5X)âˆ˜ and

âˆ D=(7X)âˆ˜

Since the sum of the angles of a quadrilateral is 360âˆ˜, we have:

2X + 4X + 5X + 7X = 360âˆ˜

â‡’ 18X = 360âˆ˜

â‡’ X = 20âˆ˜

Therefore, Angle A = 40âˆ˜

Angle B = 80âˆ˜

Angle C = 100âˆ˜

Angle D = 140âˆ˜

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

**Sol:**

We have ABâˆ¥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 = 180âˆ˜

â‡’ âˆ D = 180âˆ˜ â€“ âˆ A

Therefore, âˆ D = 180âˆ˜ â€“ 55âˆ˜ = 125âˆ˜

Again, âˆ B + âˆ C = 180âˆ˜

â‡’ âˆ C = 180âˆ˜ â€“ âˆ B

Therefore, âˆ C = 180âˆ˜ â€“ 70âˆ˜ = 110âˆ˜

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

**Sol: **

ABCD is a square in which AB = BC = CD = DA. Î”EDC is an equilateral triangle in which ED = EC = DC

âˆ EDC = âˆ DEC = âˆ DCE = 60âˆ˜.

**To prove:** AE = BE and âˆ DAE = 15âˆ˜

**Proof:** In Î”ADE and Î”BCE, we have:

AD = BC [Sides of a square]

DE = EC [Sides of an equilateral triangle]

âˆ ADE = âˆ BCE = 90âˆ˜ + 60âˆ˜ = 150âˆ˜

Î”ADE â‰… Î”BCE

i.e., AE = BE

Now, âˆ ADE = 150âˆ˜

DA = DC [Sides of a square]

DC = DE [Sides of an equilateral triangle]

So, DA = DE

Î”ADE and Î”BCE are isosceles triangles.

i.e., âˆ DAE = âˆ DEA

= 12(180âˆ˜âˆ’150âˆ˜)

= 30âˆ˜2

= 15âˆ˜

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

**Sol: **

A quadrilateral ABCD, in which BM âŠ¥ AC and DN âŠ¥ AC and BM = DN

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

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

Now, in Î”OND and Î”OMB, we have:

âˆ OND = âˆ OMB (90âˆ˜ each)

âˆ DON = âˆ BOM (Vertically opposite angles)

Also, DN = BM (Given)

i.e., Î”OND â‰… Î”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 âˆ A and âˆ C, Â (ii) BE = DE, (iii) Â âˆ ABC = âˆ ADC.**

**Sol:**

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

**(i)** In Î”ABC and Î”ADC, we have:

AB = AD (Given)

BC = DC (Given)

AC is common

i.e., Î”ABC â‰… Î”ADC (SSS Congruence rule)

Therefore, âˆ BAC = âˆ DAC and

âˆ BCA = âˆ DCA (By CPCT)

Thus, AC bisects âˆ A and âˆ C

**(ii) **Now, in Î”ABE and Î”ADE, we have:

AB = AD (Given)

âˆ BAE = âˆ DAE (Proven above)

AE is common.

Therefore, Î”ABE â‰… Î”ADE (SAS congruence rule)

â‡’ BE = DE (By CPCT)

**(iii)** Î”ABC â‰… Î”ADC (Proven above)

Therefore, Angle ABC = Angle ADC (By CPCT)

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

**Sol:**

Let, ABCD be a parallelogram

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

Let âˆ A be the smallest angle whose measure is xÂ°

âˆ´ âˆ B = (2x â€“ 30)Â°

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

â‡’ x + 2x â€“ 30Â° = 180Â°

â‡’ 3x = 210Â°

â‡’ x = 70Â°

âˆ´ âˆ B = 2 x 70Â°- 30Â° = 110Â°

Hence, âˆ A = âˆ C = 70Â°; âˆ B = âˆ D = 110Â°

**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 Î”AOC, OA + OC > AC

Also, Î”BOD, OB + OD > BD

Adding these inequalities, we get:

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

â‡’ 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 Î”ABC, AB + BC > AC . . . . . . . . (1)

In Î”ACD, CD + DA > AC . . . . . . . . . . (2)

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

AB + BC + CD + DA > 2AC

**(ii)** In Î”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 Î”ACD, we have:

AC > |DAâ€“CD| . . . . . . . (2)

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

AB + BC > |DAâ€“CD|

â‡’ AB + BC + CD > DA

**(iii)** In Î”ABC, AB + BC > AC

In Î”ACD, CD + DA > AC

Î”BCD, BC + CD > BD

Î”ABD, DA + AB > BD

Adding these inequalities, we get:

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

â‡’ (AB + BC + CD + DA) > (AC + BD)

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

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

Join BD which divides ABCD in two triangles, Î”ABD and Î”BCD.

In Î”ABD, we have:

âˆ 1 + âˆ 2 + Â âˆ A = 180Â° . . . . . . . . (i)

In Î”BCD, we have:

âˆ 3 + âˆ 4 + âˆ C = 180Â° . . . . . . . (ii)

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

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

â‡’ âˆ A + âˆ C + âˆ B + âˆ D = 360Â° Â Â ( Since,âˆ 1 + âˆ 3 = âˆ B; âˆ 4 + âˆ 2 = âˆ D)

âˆ´Â âˆ A + âˆ C + âˆ B + âˆ D = 360Â°