RS Aggarwal Class 9 Quadrilaterals And Parallelograms

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.

https://lh6.googleusercontent.com/xKCBklx1u0ZyV4TFE5HogtljvtCIg3HImWAMXc4bcim4Gbh7dEHms8K-bQWf7DcbVMvoYYx-ZLiJkR3cl8UUBh-xxe3CGkWku10tKwS7p1pShukfbEsUT2E6cuU-XOxiSdGEZvdc2P7wWQfkyg

Sol:

We have ABDC

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 = 180A

Therefore, D = 18055 = 125

Again, B + C = 180

C = 180B

Therefore, C = 18070 = 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°.

https://lh4.googleusercontent.com/dH0ZVEjSO-RES6HYTPMTNWbH3HSF2wxpHg_-fsfWMu49NuYakukmzDySdti9pykyzya2B9Xf3VQnLoOBFzdFS-05ZozgdjV0KYvkSIyVriEyltFevfkqRwjd8AdSdIJNHpEqNedD8j5YAHMM_g

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(180150)

= 302

= 15

 

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

https://lh6.googleusercontent.com/zce4Qj8UrkINR-3JTd2mVxOoPCR-1PFoDYCeJBFg7jkRwVOFf_2scwAd8vdBaD2LQzVb5WFTnIJL5QLqDhmiMGK0TuSiXqWxdG0XJpUuxzWw79FqmpJ62Agfd3V_5LR6MpbYrT2bXL3KtdXD_A

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.

https://lh5.googleusercontent.com/WXdmHJKIISN8ZlG5VaiEcjORs-YLf_qrsjFlRUqxpbXENlSTqIYFn_4HZu5pnaGTV2S7Ki-Tsy_CzdsYHCVA9juKG4By23DyIJkTCN5jbDauIDyqMwUjJInmT3sJlaAJuOiOh0uA1nL5SPvQ5w

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°

https://lh6.googleusercontent.com/i2IHWr0hx95QdGJ2OTrValczBLtU-lHrwv4pAzWsCOvTjBT9xBYWQEvdK6539J49SEE8nMdi-5uunCf9L9NYvnLQhkV5O0R-UY8LMjjnSvs5R8ACl4LmeNrsP5A3KOKzTSJ1iHGmT8Cub8EN6Q

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.

https://lh4.googleusercontent.com/smpzcndTt0bmGnEOJu-45XFqLGbtgDtJScSIqzEd0oGA-Z0wxEBexxNW457PQ7yJLeswrFK1XYFDKrTMHMHCnGDZ5vGYEELoPg0ZRed3w8a9QuZ_TIEK30BlelbrC4mUgR7eKMr5A3bRFtmwhg

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

https://lh3.googleusercontent.com/Jxz2POZVDzjeWrCj2o0g4dB0ZdGRvSTDrKoUDw4NO60zXzIqHBXy2JeKVM_csJcJcJo0AWNhJ5rLSXLF6qgrWSyDjTCoMex3AcMySTAjur3UZ7LIsxs0c1kisjDf6qyoeEa9jyMZIVlmPEkerw

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 > |DACD| . . . . . . . (2)

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

AB + BC > |DACD|

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°.

https://lh3.googleusercontent.com/ZQir6RsxtLYwNMvo21dDPNczDJU1MWWpawvRfsrBLpEqKXbWxwcEOlGMY4xVEmPMQQMrhjiJk2JHSQaKPCqbACu1P30m6o3Jto1VikOLh1NBT2uZOTVOOc6yyNTmvY6Wme-e9gav9Dv3yYofQg

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°

 

 


Practise This Question

A boy weighing 600 N was in a lift. The lift experienced a breakdown and started falling freely under the action of gravity, what would be the weight of the boy in the lift?