## RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A.

**Other Exercises**

- RS Aggarwal Solutions Class 9 Chapter 9 Quadrilaterals and Parallelograms Ex 9A
- RS Aggarwal Solutions Class 9 Chapter 9 Quadrilaterals and Parallelograms Ex 9B
- RS Aggarwal Solutions Class 9 Chapter 9 Quadrilaterals and Parallelograms Ex 9C

**Question 1.**

**Three angles of a quadrilateral measure 56°, 115° and 84°. Find the measure of the fourth angle.**

**Solution:**

We know that sum of angles of a quadrilateral is 360°

Now, sum of three angles = 56° + 115° + 84° = 255°

Fourth angle = 360° – 255° = 105° Ans.

**Question 2.**

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

**Solution:**

Sum of angles of a quadrilateral = 360°

Their ratio = 2 : 4 : 5 : 7

Let first angle = 2x

then second angle = 4x

third angle = 5x

and fourth angle = 7x

∴ 2x + 4x + 5x + 7x = 360°

=> 18x = 360°

=> x = = 20°

Hence, first angle = 2x = 2 x 20° = 40°

Second angle = 4x = 4 x 20° = 80°

Third angle = 5x = 5 x 20° = 100°

and fourth angle = 7x = 7 x 20° = 140°Ans.

**Question 3.**

**In the adjoining figure, ABCD is a trapezium is which AB || DC. If ∠ A = 55°, and ∠B = 70°, find ∠C and ∠D.**

**Solution:**

In the trapezium ABCD

DC || AB

∴ ∠ A + ∠ D = 180° (Co-intericr angles)

∴ 55°+ ∠D = 180°

∠D = 180° – 55°

∴ ∠D = 125°

Similarly, ∠B + ∠C = 180°

(Co-interior angles)

=> 70° + ∠C = 180°

=> ∠C = 180° – 70°

∠C = 110°

Hence ∠C = 110° and ∠D = 125° Ans.

**Question 4.**

**In the adjoining figure, ABCD is a square and ∆ EDC is an equilateral triangle. Prove that (i) AE = BE, (ii) ∠ DAE = 15°.**

**Solution:**

**Given :** In the figure, ABCD is a square and ∆ EDC is an equilateral triangles on DC. AE and BE are joined.

**To Prove :** (i) AE = BE

(ii) ∠DAE = 15°

**Question 5.**

**In the adjoining figure, BM ⊥ AC and DN ⊥ AC. It BM = DN, prove that AC bisects BD.**

**Solution:**

**Given :** In the figure,

BM ⊥ AC, DN ⊥ AC.

BM = DN

**Question 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.**

**Solution:**

**Given :** In quadrilateral ABCD,

AB = AD and BC = DC

AC is joined.

**To Prove :** (i) AC bisects ∠ A and ∠ C

(ii) BE = DE

(iii) ∠ABC = ∠ADC

Const. Join BD.

**Proof** **:** (i) In ∆ ABC and ∆ ADB

**Question 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°.**

**Solution:**

**Given :** In square ABCD,

∠ PQR = 90°

PB = QC = DR

**Question 8.**

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

**Solution:**

**Given :** In quadrilateral ABCD, O is any point inside it. OA, OB, OC and OD are joined.

**To Prove :** OA + OB + OC + OD > AC + BD

**Question 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:**

**Given :** In quadrilateral ABCD, AC is its one diagonal.

**Question 10.**

**Prove that the sum of all the angles of a quadrilateral is 360°.**

**Solution:**

**Given :** A quadrilateral ABCD

**To Prove :** ∠A + ∠B + ∠C + ∠D = 360°

Const. Join AC.

Hope given RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9A are helpful to complete your math homework.

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