## RS Aggarwal Class 9 Chapter 9

**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∘

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

56 + 115 + 84 + X = 360

⇒ 255 + X = 360

⇒ X = 105

Hence, the measure of the fourth angle is 105∘

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

**Sol:**

Let, ∠A = 2X∘

Then ∠B=(4X)∘

∠C=(5X)∘ and

∠D=(7X)∘

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

2X + 4X + 5X + 7X = 360∘

⇒ 18X = 360∘

⇒ X = 20∘

Therefore, Angle A = 40∘

Angle B = 80∘

Angle C = 100∘

Angle D = 140∘

**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 AB∥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∘

⇒ ∠D = 180∘ – ∠A

Therefore, ∠D = 180∘ – 55∘ = 125∘

Again, ∠B + ∠C = 180∘

⇒ ∠C = 180∘ – ∠B

Therefore, ∠C = 180∘ – 70∘ = 110∘

**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. ΔEDC is an equilateral triangle in which ED = EC = DC

∠EDC = ∠DEC = ∠DCE = 60∘.

**To prove:** AE = BE and ∠DAE = 15∘

**Proof:** In ΔADE and ΔBCE, we have:

AD = BC [Sides of a square]

DE = EC [Sides of an equilateral triangle]

∠ADE = ∠BCE = 90∘ + 60∘ = 150∘

ΔADE ≅ ΔBCE

i.e., AE = BE

Now, ∠ADE = 150∘

DA = DC [Sides of a square]

DC = DE [Sides of an equilateral triangle]

So, DA = DE

ΔADE and ΔBCE are isosceles triangles.

i.e., ∠DAE = ∠DEA

= 12(180∘−150∘)

= 30∘2

= 15∘

**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 ⊥ AC and DN ⊥ AC and BM = DN

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

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

Now, in ΔOND and ΔOMB, we have:

∠OND = ∠OMB (90∘ each)

∠DON = ∠BOM (Vertically opposite angles)

Also, DN = BM (Given)

i.e., ΔOND ≅ ΔOMB (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 ΔABC and ΔADC, we have:

AB = AD (Given)

BC = DC (Given)

AC is common

i.e., ΔABC ≅ ΔADC (SSS Congruence rule)

Therefore, ∠BAC = ∠DAC and

∠BCA = ∠DCA (By CPCT)

Thus, AC bisects ∠A and ∠C

**(ii) **Now, in ΔABE and ΔADE, we have:

AB = AD (Given)

∠BAE = ∠DAE (Proven above)

AE is common.

Therefore, ΔABE ≅ ΔADE (SAS congruence rule)

⇒ BE = DE (By CPCT)

**(iii)** ΔABC ≅ ΔADC (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)

⇒ x + 2x – 30° = 180°

⇒ 3x = 210°

⇒ 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 ΔAOC, OA + OC > AC

Also, ΔBOD, OB + OD > BD

Adding these inequalities, we get:

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

⇒ 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 ΔABC, AB + BC > AC . . . . . . . . (1)

In ΔACD, CD + DA > AC . . . . . . . . . . (2)

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

AB + BC + CD + DA > 2AC

**(ii)** In ΔABC, 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 ΔACD, we have:

AC > |DA–CD| . . . . . . . (2)

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

AB + BC > |DA–CD|

⇒ AB + BC + CD > DA

**(iii)** In ΔABC, AB + BC > AC

In ΔACD, CD + DA > AC

ΔBCD, BC + CD > BD

ΔABD, DA + AB > BD

Adding these inequalities, we get:

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

⇒ (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, ΔABD and ΔBCD.

In ΔABD, we have:

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

In ΔBCD, we have:

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

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

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

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

∴ ∠ A + ∠ C + ∠ B + ∠ D = 360°