In this chapter, we provide RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5 Maths pdf, free RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5 book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 8 |

Subject | Maths |

Chapter | Chapter 18 |

Chapter Name | Practical Geometry |

Exercise | Ex 18.5 |

**RD Sharma Solutions for Class 8 Chapter 18 Practical Geometry Ex 18.5** **Download PDF**

**Question 1.****Construct a quadrilateral ABCD given that AB = 4 cm, BC = 3 cm, ∠A = 75°, ∠B = 80° and ∠C = 120°.****Solution:****Steps of construction :**

(i) Draw a line segment AB = 4 cm.

(ii) At A draw a ray AX making an angle of 75°.

(iii) At B draw another ray BY making an angle of 80° and cut off BC = 3 cm.

(iv) At C, draw another ray CZ making an angle of 120° which intersects AX at D.

Then ABCD is the required quadrilateral.

**Question 2.****Construct a quadrilateral ABCD where AB = 5.5 cm, BC = 3.7 cm, ∠A = 60°, ∠B = 105° and ∠D = 90°.****Solution:**

∠A = 60°, ∠B = 105° and ∠D = 90°

But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral)

⇒ 60° + 105° + ∠C + 90° = 360°

⇒ 255° + ∠C = 360°

⇒ ∠C = 360° – 255° = 105°**Steps of construction :**

(i) Draw a line segment AB = 5.5 cm.

(ii) At A, draw a ray AX making an angle of

(iii) At B, draw another ray BY making an angle of 105° and cut off BC = 3.7 cm.

(iv) At C, draw a ray CZ making an angle of 105° which intersects AX at D.

Then ABCD is the required quadrilateral.

**Question 3.****Construct a quadrilateral PQRS where PQ = 3.5 cm, QR = 6.5 cm, ∠P = ∠R = 105° and ∠S = 75°.****Solution:**

∠P = 105°, ∠R = 105° and ∠S = 75°

But ∠P + ∠Q + ∠R + ∠S = 360° (Sum of angles of a quadrilateral)

⇒ 105° + ∠Q + 105° + 75° = 360°

⇒ 285° + ∠Q = 360°

⇒ ∠Q = 360° – 285° = 75°**Steps of construction :**

(i) Draw a line segment PQ = 3.5 cm.

(ii) At P, draw a ray PX making an angle of 105°.

(iii) At Q, draw another ray QY, making an angle of 75° and cut off QR = 6.5 cm.

(iv) At R, draw a ray RZ making an angle of 105° which intersects PX at S.

Then PQRS is the required quadrilateral.

**Question 4.****Construct a quadrilateral ABCD when BC = 5.5 cm, CD = 4.1 cm, ∠A = 70°, ∠B = 110° and ∠D = 85°.****Solution:**

∠A = 70°, ∠B = 110°, ∠D = 85°

But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral)

⇒ 70° + 110° + ∠C + 85° = 360°

⇒ 265° + ∠C = 360°

⇒ ∠C = 360° – 265° = 95°**Steps of construction:**

(i) Draw a line segment BC = 5.5 cm.

(ii) At B, draw a ray BX making an angle of 110°.

(iii) At C, draw another ray CY making an angle of 95° and cut off CD = 4.1 cm.

(iv) At D, draw a ray DZ making an angle of 85° which intersects BX at A.

Then ABCD is the required quadrilateral.

**Question 5.****Construct a quadrilateral ABCD, where ∠A = 65°, ∠B = 105°, ∠C = 75°, BC = 5.7 cm and CD = 6.8 cm.****Solution:**

∠A = 65°, ∠B = 105°, ∠C = 75°

But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quadrilateral)

⇒ 65° + 105° + 75° + ∠D = 360°

⇒ 245° + ∠D = 360°

⇒ ∠D = 360° – 245° = 115°**Steps of construction:**

(i) Draw a line segment BC = 5.7 cm.

(ii) At B, draw a ray BX making an angle of

(iii) At C draw a another ray CY making an angle of 75° and cut off CD = 6.8 cm.

(iv) At D, draw a ray DZ making an angle of 115° which intersects BX at A.

Then ABCD is the required quadrilateral.

**Question 6.****Construct a quadrilateral PQRS in which PQ = 4 cm, QR = 5 cm, ∠P = 50°, ∠Q = 110° and ∠R = 70°.****Solution:****Steps of construction :**

(i) Draw a line segment PQ = 4 cm.

(ii) At P, draw a ray PX making an angle of 50°.

(iii) At Q, draw another ray QY making an angle of 110° and cut off QR = 5 cm.

(iv) At R, draw a ray RZ making an angle of 70° which intersects PX at S.

Then PQRS is the required quadrilateral.

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