# RS Aggarwal Solutions for Class 9 Chapter 10: Quadrilaterals Exercise 10A

## RS Aggarwal Solutions for Class 9 Maths Exercise 10A PDF

RS Aggarwal solutions are an educational aid which helps students to learn easy and difficult problems. It consists of solutions which are prepared in an organized manner by subject experts at BYJUâ€™S. The students can obtain the RS Aggarwal solutions for free in PDF format. This exercise is the introduction which explains types of quadrilaterals. The solutions help students to clear their doubts while solving problems and to prepare more efficiently for the board exams. RS Aggarwal Solutions for Class 9 Maths Chapter 10 Quadrilaterals Exercise 10A are provided here.

## Access RS Aggarwal Solutions for Class 9 Chapter 10: Quadrilaterals Exercise 10A

1. Three angles of a quadrilateral are 75o, 90o and 75o. Find the measure of the fourth angle.

Solution:

Consider the fourth angle as xo

In a quadrilateral we know that the sum of all the angles is 360o

So we can write it as

xo + 75o + 90o + 75o = 360o

On further calculation we get

xo = 360o – 75o – 90o – 75o

By subtraction

xo = 360o – 240o

xo = 120o

Therefore, the measure of the fourth angle is 120o.

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

Solution:

Consider the angles of a quadrilateral as 2x, 4x, 5x and 7x

In a quadrilateral we know that the sum of all the angles is 360o

So we can write it as

2x + 4x + 5x + 7x = 360o

18x = 360o

By division

xo = 20o

Now by substituting the value of xo

2x = 2(20o) = 40o

4x = 4(20o) = 80o

5x = 5(20o) = 100o

7x = 7(20o) = 140o

Therefore, the angles are 40o, 80o, 100o and 140o.

3. In the adjoining figure, ABCD is a trapezium in which AB || DC. If âˆ  A = 55o and âˆ  B = 70o, find âˆ  C and âˆ  D.

Solution:

It is given that AB || DC

From the figure we know that âˆ  A and âˆ  D are consecutive interior angles

So it can written as

âˆ  A + âˆ  D = 180o

To find âˆ  D

âˆ  D = 180o – âˆ  A

By substituting the value in the above equation

âˆ  D = 180o – 55o

By subtraction

âˆ  D = 125o

In a quadrilateral we know that the sum of all the angles is 360o

So we can write it as

âˆ  A + âˆ  B + âˆ  C + âˆ  D = 360o

By substituting the values in the above equation

55o + 70o + âˆ  C + 125o = 360o

On further calculation

âˆ  C = 360o – 55o – 70o – 125o

By subtraction

âˆ  C = 360o – 250o

âˆ  C = 110o

Therefore, âˆ  C = 110o and âˆ  D = 125o.

4. In the adjoining figure, ABCD is a square and â–³ EDC is an equilateral triangle. Prove that

(i) AE = BE,

(ii) âˆ  DAE = 15o

Solution:

(i) From the figure we know that â–³ EDC is an equilateral triangle.

So we get âˆ  EDC = âˆ  ECD = 60o

We know that ABCD is a square

So we get âˆ  CDA = âˆ  DCB = 90o

Consider â–³ EDA

We get

âˆ  EDA = âˆ  EDC + âˆ  CDA

By substituting the values in the above equation

âˆ  EDA = 60o + 90o

So we get

âˆ  EDA = 150o â€¦â€¦. (1)

Consider â–³ ECB

We get

âˆ  ECB = âˆ  ECD + âˆ  DCB

By substituting the values in the above equation

âˆ  ECB = 60o + 90o

So we get

âˆ  ECB = 150o

So we get âˆ  EDA = âˆ  ECB â€¦â€¦ (2)

Consider â–³ EDA and â–³ ECB

From the figure we know that ED and EC are the sides of equilateral triangle

So we get

ED = EC

We also know that the sides of square are equal

DA = CB

By SAS congruence criterion

â–³ EDA â‰… â–³ ECB

AE = BE (c. p. c. t)

(ii) Consider â–³ EDA

We know that

ED = DA

From the figure we know that the base angles are equal

âˆ  DEA = âˆ  DAE

Based on equation (1) we get âˆ  EDA = 150o

By angle sum property

âˆ  EDA + âˆ  DAE + âˆ  DEA = 180o

By substituting the values we get

150o + âˆ  DAE + âˆ  DEA = 180o

We know that âˆ  DEA = âˆ  DAE

So we get

150o + âˆ  DAE + âˆ  DAE = 180o

On further calculation

2 âˆ  DAE = 180o – 150o

By subtraction

2 âˆ  DAE = 30o

By division

âˆ  DAE = 15o

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

Solution:

It is given that BM âŠ¥ AC and DN âŠ¥ AC

From the figure we know that âˆ  DON and âˆ  MOB are vertically opposite angles

âˆ  DON = âˆ  MOB

Since DN and MB are perpendiculars

We know that

âˆ  DNO = âˆ  BMO = 90o

It is given that BM = DN

By AAS congruence criterion

â–³ DNO â‰… â–³ BMO

OD = OB (c. p. c. t)

Therefore, it is proved that AC bisects BD.

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,

(ii) BE = DE,

(iii) âˆ  ABC = âˆ  ADC

Solution:

(i) Consider â–³ ABC and â–³ ADC

It is given that AB = AD and BC = DC

AC is common i.e. AC = AC

By SSS congruence criterion

â–³ ABC â‰… â–³ ADC â€¦â€¦â€¦ (1)

âˆ  BAC = âˆ  DAC (c. p. c. t)

So we get

âˆ  BAE = âˆ  DAE

We know that AC bisects the âˆ  BAD i.e. âˆ  A

So we get

âˆ  BCA = âˆ  DCA (c. p. c. t)

It can be written as

âˆ  BCE = âˆ  DCE

So we know that AC bisects âˆ  BCD i.e. âˆ  C

(ii) Consider â–³ ABE and â–³ ADE

It is given that AB = AD

AE is common i.e. AE = AE

By SAS congruence criterion

BE = DE (c. p. c. t)

(iii) We know that â–³ ABC â‰… â–³ ADC

Therefore, by c. p. c. t âˆ  ABC = âˆ  ADC

7. In the given figure, ABCD is a square and âˆ  PQR = 90o. If PB = QC = DR, prove that

(i) QB = RC,

(ii) PQ = QR,

(iii) âˆ  QPR = 45o

Solution:

(i) We know that the line segment QB can be written as

QB = BC â€“ QC

Since ABCD is a square we know that BC = DC and QC = DR

So we get

QB = CD â€“ DR

From the figure we get

QB = RC

(ii) Consider â–³ PBQ and â–³ QCR

It is given that PB = QC

Since ABCD is a square we get

âˆ  PBQ = âˆ  QCR = 90o

By SAS congruence criterion

â–³ PBQ â‰… â–³ QCR

PQ = QR (c. p. c. t)

(iii) We know that PQ = QR

Consider â–³ PQR

From the figure we know that âˆ  QPR and âˆ  QRP are base angles of isosceles triangle

âˆ  QPR = âˆ  QRP

We know that the sum of all the angles in a triangle is 180o

âˆ  QPR + âˆ  QRP + âˆ  PRQ = 180o

By substituting the values in the above equation

âˆ  QPR + âˆ  QRP + 90o = 180o

On further calculation

âˆ  QPR + âˆ  QRP = 180o – 90o

By subtraction

âˆ  QPR + âˆ  QRP = 90o

We know that âˆ  QPR = âˆ  QRP

So we get

âˆ  QPR + âˆ  QPR = 90o

2 âˆ  QPR = 90o

By division

âˆ  QPR = 45o

Therefore, it is proved that âˆ  QPR = 45o.

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

Solution:

It is given that O is a point within a quadrilateral ABCD

Consider â–³ ACO

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

We get

OA + OC > AC â€¦.. (1)

In the same way

Consider â–³ BOD

We get

OB + OD > BD â€¦.. (2)

By adding both the equations we get

OA + OC + OB + OD > AC + BD

Therefore, it is proved that OA + OC + OB + OD > AC + BD.

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.

Solution:

(i) Consider â–³ ABC

We know that

AB + BC > AC â€¦â€¦ (1)

Consider â–³ ACD

We know that

AD + CD > AC â€¦â€¦. (2)

By adding both the equations we get

AB + BC + AD + CD > AC + AC

So we get

AB + BC + AD + CD > 2AC â€¦â€¦.. (3)

Therefore, it is proved that AB + BC + AD + CD > 2AC.

(ii) Consider â–³ ABC

We know that

AB + BC > AC

Add CD both sides of the equation

AB + BC + CD > AC + CD â€¦â€¦ (4)

Consider â–³ ACD

We know that

AC + CD > DA â€¦â€¦. (5)

By substituting (5) in (4) we get

AB + BC + CD > DA â€¦â€¦.. (6)

(iii) Consider â–³ ABD and â–³ BDC

We know that

AB + DA > BD â€¦â€¦. (7)

So we get

BC + CD > BD â€¦â€¦. (8)

By adding (7) and (8) we get

AB + DA + BC + CD > BD + BD

On further calculation

AB + DA + BC + CD > 2BD â€¦â€¦. (9)

By adding equations (9) and (3) we get

AB + DA + BC + CD + AB + BC + AD + CD > 2BD + 2AC

So we get

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

Dividing by 2

AB + BC + CD + DA > BD + AC

Therefore, it is proved that AB + BC + CD + DA > BD + AC.

10. Prove that the sum of all the angles of a quadrilateral is 360o.

Solution:

Consider â–³ ABC

We can write it as

âˆ  CAB + âˆ  B + âˆ  BCA = 180o â€¦â€¦.. (1)

Consider â–³ ACD

We can write it as

âˆ  DAC + âˆ  ACD + âˆ  D = 180o â€¦â€¦â€¦ (2)

Adding equations (1) and (2) we get

âˆ  CAB + âˆ  B + âˆ  BCA + âˆ  DAC + âˆ  ACD + âˆ  D = 180o + 180o

So we get

âˆ  CAB + âˆ  DAC + âˆ  B + âˆ  BCA + âˆ  ACD + âˆ  D = 360o

So it can be written as

âˆ  A + âˆ  B + âˆ  C + âˆ  D = 360o

### Access other exercise solutions of Class 9 Maths Chapter 10: Quadrilaterals

Exercise 10B Solutions 27 Questions

Exercise 10C Solutions 18 Questions

### RS Aggarwal Solutions Class 9 Maths Chapter 10 – Quadrilaterals Exercise 10A

RS Aggarwal Solutions Class 9 Maths Chapter 10 Quadrilaterals Exercise 10A explains in brief about the types of quadrilaterals and various methods which are used in solving the exercise problems.

### Key features of RS Aggarwal Solutions for Class 9 Maths Chapter 10: Quadrilaterals Exercise 10A

• The solutions are prepared by experts in Mathematics in order to improve logical thinking among students.
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• The main aim is to make students knowledge driven and help them perform better in the exams.
• The faculty explains the concepts in an organized way based on the latest CBSE syllabus.