## 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.

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

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

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

**Solution:**

Consider the fourth angle as x^{o}

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

So we can write it as

x^{o} + 75^{o} + 90^{o} + 75^{o} = 360^{o}

On further calculation we get

x^{o} = 360^{o} – 75^{o} – 90^{o} – 75^{o}

By subtraction

x^{o} = 360^{o} – 240^{o}

x^{o} = 120^{o}

Therefore, the measure of the fourth angle is 120^{o}.

**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 360^{o}

So we can write it as

2x + 4x + 5x + 7x = 360^{o}

By addition

18x = 360^{o}

By division

x^{o} = 20^{o}

Now by substituting the value of x^{o}

2x = 2(20^{o}) = 40^{o}

4x = 4(20^{o}) = 80^{o}

5x = 5(20^{o}) = 100^{o}

7x = 7(20^{o}) = 140^{o}

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

**3. In the adjoining figure, ABCD is a trapezium in which AB || DC. If âˆ A = 55 ^{o} and âˆ B = 70^{o}, 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 = 180^{o}

To find âˆ D

âˆ D = 180^{o} – âˆ A

By substituting the value in the above equation

âˆ D = 180^{o} – 55^{o}

By subtraction

âˆ D = 125^{o}

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

So we can write it as

âˆ A + âˆ B + âˆ C + âˆ D = 360^{o}

By substituting the values in the above equation

55^{o} + 70^{o} + âˆ C + 125^{o} = 360^{o}

On further calculation

âˆ C = 360^{o} – 55^{o} – 70^{o} – 125^{o}

By subtraction

âˆ C = 360^{o} – 250^{o}

âˆ C = 110^{o}

Therefore, âˆ C = 110^{o} and âˆ D = 125^{o}.

**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 = 15 ^{o}**

**Solution:**

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

So we get âˆ EDC = âˆ ECD = 60^{o}

We know that ABCD is a square

So we get âˆ CDA = âˆ DCB = 90^{o}

Consider â–³ EDA

We get

âˆ EDA = âˆ EDC + âˆ CDA

By substituting the values in the above equation

âˆ EDA = 60^{o} + 90^{o}

So we get

âˆ EDA = 150^{o} 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 = 60^{o} + 90^{o}

So we get

âˆ ECB = 150^{o}

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 = 150^{o}

By angle sum property

âˆ EDA + âˆ DAE + âˆ DEA = 180^{o}

By substituting the values we get

150^{o} + âˆ DAE + âˆ DEA = 180^{o}

We know that âˆ DEA = âˆ DAE

So we get

150^{o} + âˆ DAE + âˆ DAE = 180^{o}

On further calculation

2 âˆ DAE = 180^{o} – 150^{o}

By subtraction

2 âˆ DAE = 30^{o}

By division

âˆ DAE = 15^{o}

**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 = 90^{o}

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

â–³ ABE â‰… âˆ ADE

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 = 90 ^{o}. If PB = QC = DR, prove that **

**(i) QB = RC,**

**(ii) PQ = QR,**

**(iii) âˆ QPR = 45 ^{o}**

**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 = 90^{o}

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 180^{o}

âˆ QPR + âˆ QRP + âˆ PRQ = 180^{o}

By substituting the values in the above equation

âˆ QPR + âˆ QRP + 90^{o} = 180^{o}

On further calculation

âˆ QPR + âˆ QRP = 180^{o} – 90^{o}

By subtraction

âˆ QPR + âˆ QRP = 90^{o}

We know that âˆ QPR = âˆ QRP

So we get

âˆ QPR + âˆ QPR = 90^{o}

By addition

2 âˆ QPR = 90^{o}

By division

âˆ QPR = 45^{o}

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

**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 360 ^{o}.**

**Solution:**

Construct a quadrilateral ABCD

Consider â–³ ABC

We can write it as

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

Consider â–³ ACD

We can write it as

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

Adding equations (1) and (2) we get

âˆ CAB + âˆ B + âˆ BCA + âˆ DAC + âˆ ACD + âˆ D = 180^{o} + 180^{o}

So we get

âˆ CAB + âˆ DAC + âˆ B + âˆ BCA + âˆ ACD + âˆ D = 360^{o}

So it can be written as

âˆ A + âˆ B + âˆ C + âˆ D = 360^{o}

### 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.